transport

Principles of Highway Engineering and Traffic Analysis

Principles of Highway Engineering and Traffic Analysis Seventh Edition

Fred L. Mannering University of South Florida

Scott S. Washburn University of Florida

VP AND EDITORIAL DIRECTOR Laurie Rosatone SENIOR DIRECTOR Don Fowley EDITOR Jennifer Brady EDITORIAL MANAGER Judy Howarth CONTENT MANAGEMENT DIRECTOR Lisa Wojcik CONTENT MANAGER Nichole Urban SENIOR CONTENT SPECIALIST Nicole Repasky PRODUCTION EDITOR Mathangi Balasubramanian COVER PHOTO CREDIT © Fred L. Mannering Photos are courtesy of Scott S. Washburn, unless otherwise noted. This book was set in 11/13 Times New Roman MT Std Roman by SPi Global and printed and bound by Quad Graphics. Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company is built on a foundation of principles that include responsibility to the communities we serve and where we live and work. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business. Among the issues we are addressing are carbon impact, paper specifications and procurement, ethical conduct within our business and among our vendors, and community and charitable support. For more information, please visit our website: www.wiley.com/go/citizenship. Copyright © 2020, 2017, 2013, 2009, 2005, 1998, 1990 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923 (Web site: www.copyright.com). Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201) 748-6011, fax (201) 748-6008, or online at: www.wiley.com/go/permissions. Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return shipping label are available at: www.wiley.com/go/returnlabel. If you have chosen to adopt this textbook for use in your course, please accept this book as your complimentary desk copy. Outside of the United States, please contact your local sales representative. ISBN: 978-1-119-49413-3 (PBK) ISBN: 978-1-119-49414-0 (EVAL) Library of Congress Cataloging-in-Publication Data Names: Mannering, Fred L., author. | Washburn, Scott S., author. Title: Principles of highway engineering and traffic analysis / Fred L. Mannering, University of South Florida, Scott S. Washburn, University of Florida. Description: Seventh edition. | Hoboken, NJ : Wiley, 2020. | Includes index. Identifiers: LCCN 2019039405 (print) | LCCN 2019039406 (ebook) | ISBN 9781119494133 (hardback) | ISBN 9781119494102 (adobe pdf) | ISBN 9781119493969 (epub) Subjects: LCSH: Highway engineering. | Traffic engineering. Classification: LCC TE147 .M28 2020 (print) | LCC TE147 (ebook) | DDC 625.7—dc23 LC record available at https://lccn.loc.gov/2019039405 LC ebook record available at https://lccn.loc.gov/2019039406 The inside back cover will contain printing identification and country of origin if omitted from this page. In addition, if the ISBN on the back cover differs from the ISBN on this page, the one on the back cover is correct.

Preface INTRODUCTION

With every edition of Principles of Highway Engineering and Traffic Analysis, we have always sought to set the pedagogical standard for instruction in entry-level transportation engineering courses. When the first edition was published three decades ago, there was a need for an entry-level transportation engineering book that focused exclusively on highway transportation and provided the depth of coverage needed to serve as a basis for future transportation courses as well as the material needed to answer questions likely to appear on the Fundamentals of Engineering (FE) and/or Principles and Practice of Engineering (PE) exams in civil engineering. The subsequent use of the various editions of this book, over the years, at some of the largest and most prestigious schools in the United States and throughout the world suggests that a vision of a concise, highly focused, and well-written entry-level book is shared by many educators.

APPROACH

This seventh edition of Principles of Highway Engineering and Traffic Analysis continues the spirit of the previous six editions by again focusing exclusively on highway transportation and providing the depth of coverage necessary to solve the highway-related problems that are most likely to be encountered in engineering practice. The focus on highway transportation is a natural one given the dominance of highway transportation for people and freight movement in the United States and throughout the world. While the focus on highway transportation is easily accomplished, identifying the highway-related problems most likely to be encountered in practice and providing an appropriate depth of coverage of them is a more challenging task. Using the first six editions as a basis, along with the comments of other instructors and students who have used previous editions of the book, topics that are fundamental to highway engineering and traffic analysis have been carefully selected. The material provided in this book ensures that students learn the fundamentals needed to undertake upper-level transportation courses, enter transportation employment with a basic knowledge of highway engineering and traffic analysis, and have the knowledge necessary to answer transportation-related questions on the civil engineering Fundamentals of Engineering (FE) exam and the Principles and Practice of Engineering (PE) exam.

MATHEMATICAL RIGOR

Within the basic philosophical approach described above, this book addresses the concern of some that traditional highway transportation courses are not as

vi Preface

mathematically challenging or rigorous as other entry-level civil engineering courses, and that this may affect student interest relative to other civil- engineering fields of study. This concern is not easily addressed because there is a dichotomy with regard to mathematical rigor in highway transportation, with relatively simple mathematics used in practice-oriented material and complex mathematics used in research. Thus it is common for instructors to either insult students’ mathematical knowledge or vastly exceed it. This book strives for that elusive middle ground of mathematical rigor that matches junior and senior engineering students’ mathematical abilities.

CHAPTER TOPICS AND ORGANIZATION

The seventh edition of Principles of Highway Engineering and Traffic Analysis has evolved from over three decades of teaching introductory transportation engineering classes at the University of Washington, University of Florida, Purdue University, University of South Florida, and the Pennsylvania State University, feedback from users of the first six editions, and experiences in teaching civil-engineering licensure exam review courses. The book’s material and presentation style (which is characterized by the liberal use of example and practice problems) are largely responsible for transforming much-maligned introductory transportation engineering courses into courses that students consistently rate among the best civil engineering courses.

The book begins with a short introductory chapter that stresses the significance of highway transportation to the social and economic underpinnings of society. Also discussed are environmental impacts including climate change and emerging technologies including connected and automated vehicles. This chapter provides students a basic overview of the problems facing the field of highway engineering and traffic analysis. The chapters that follow are arranged in sequences that focus on highway engineering (Chapters 2, 3, and 4) and traffic analysis (Chapters 5, 6, 7, and 8).

Chapter 2 introduces the basic elements of road vehicle performance. This chapter represents a major departure from the vehicle performance material presented in all other transportation and highway engineering books, in that it is far more involved and detailed. The additional level of detail is justified on two grounds. First, because students own and drive automobiles, they have a basic interest that can be linked to their freshman and sophomore coursework in physics, statics, and dynamics. Traditionally, the absence of such a link has been a common criticism of introductory transportation and highway engineering courses. Second, it is important that engineering students understand the principals involved in vehicle technologies and the effect that continuing advances in vehicle technologies will have on engineering practice.

Chapter 3 presents current design practices for the geometric alignment of highways. This chapter provides details on vertical-curve design and the basic elements of horizontal-curve design. This edition of the book includes the latest design guidelines (Policy on Geometric Design of Highways and Streets, American Association of State Highway and Transportation Officials, Washington, DC, 2018).

Preface vii

Chapter 4 provides a detailed overview of traditional pavement design, covering both flexible and rigid pavements in a thorough and consistent manner. A brief overview of the topics of pavement distresses and mechanistic-empirical approaches to pavement design are also provided. The material in this chapter also links well with the geotechnical and materials courses that are likely to be part of the student’s curriculum.

Chapter 5 presents the fundamentals of traffic flow and queuing theory, which provide the basic tools of traffic analysis. Relationships and models of basic traffic-stream parameters are introduced, as well as queuing analysis models for deterministic and stochastic processes. Considerable effort was expended to make the material in this chapter accessible to junior and senior engineering students.

Chapter 6 presents some of the current methods used to assess highway levels of performance. Fundamentals and concepts are discussed along with the complexities involved in measuring and/or calculating highway level of service. This edition of the book has been updated to the latest analysis standards (Highway Capacity Manual, Transportation Research Board, National Academy of Sciences, Washington, D.C., 2016).

Chapter 7 introduces the basic elements of traffic control at a signalized intersection and applies the traffic analysis tools introduced in Chapter 5 to signalized intersections. The chapter not only focuses on pretimed, isolated signals but also introduces the reader to the fundamentals of actuated and coordinated signal systems. Both theoretical and practical elements associated with traffic signal timing are presented. This edition of the book has been updated to the latest analysis standards (Highway Capacity Manual, Transportation Research Board, National Academy of Sciences, Washington, D.C., 2016).

Chapter 8, the final chapter, provides an overview of travel demand and traffic forecasting. This chapter concentrates on a theoretically and mathematically consistent approach to travel demand and traffic forecasting that closely follows the approach most commonly used in practice, and contains a section on the traditional four-step travel-demand forecasting process. This chapter provides the student with an important understanding of the current state of travel demand and traffic forecasting, and some critical insight into the deficiencies of forecasting methods currently used.

NEW AND REVISED PROBLEMS

This edition, like the sixth edition, includes practice problems as a new pedagogical tool. At the end of chapters that require mathematical solutions to problems (Chapters 2 through 8 inclusive), several partially solved problems are provided for students to practice their problem-solving techniques and more fully understand the material presented in the text. This enables students to follow the thought process involved in solving problems in the chapter, while also engaging them in the solution (as opposed to the traditional example problems, also provided, which present the complete solution). There are also several new and revised end-of-chapter problems relative to the sixth edition.

viii Preface

Users of the book will find the practice and end-of-chapter problems to be extremely useful in supporting the material presented in the book. These problems are precise and challenging, a combination rarely found in transportation/highway engineering books.

NEW TO THIS EDITION

In this edition we have once again made several enhancements to the content and visual presentation, based on suggestions from instructors. Some new features in this edition of the book include: New end-of-chapter problems. Several new end-of- chapter problems have been added to further improve the pedagogical effectiveness of the book. New Enhanced E-Text makes your study time more effective with show/hide solutions and answers for select practice problems.

WEBSITE

The website for this book is www.wiley.com/go/mannering/highwayandtraffic7e and contains the following resources for instructors: Solutions Manual. An example solutions manual (several problems for each chapter). Instructions for obtaining the complete solutions manual—that is, all the problems in the book. In-Class Design Problems. Design problems developed by the authors for in-class use by students in a cooperative-learning context. The problems support the material presented in the chapters and the end-of-chapter problems. In-Class Design Problems with Solutions. A complete solutions manual for all of the In-class design problems. Sample Exams. Sample midterm and final exams are provided to give instructors class-proven ideas relating to successful exam format and problems. A Sample Course Syllabus. Provided to assist instructor’s in structuring their course. Online Metric Conversion Appendix is available to instructors for distribution to students. Image Gallery. All of the text’s figures and tables are provided to instructors to assist them in creating teaching slides and materials. Visit the Instruction Companion Site section of the book website to register for a password to download these resources.

Fred L. Mannering Scott S. Washburn

Contents Content available in eBook Student solution available in interactive e-text

Preface v

Chapter 1 Introduction to Highway Engineering and Traffic Analysis 1 1.1 Introduction 1

1.2 Highways and the Economy 2 1.2.1 The Highway Economy 2 1.2.2 Supply Chains 2 1.2.3 Economic Development 3

1.3 Highways, Energy, the Environment, and Climate Change 3

1.4 Highways as Part of the Transportation System 3

1.5 Highway Transportation and the Human Element 4 1.5.1 Passenger Transportation Modes and Traffic Congestion 4 1.5.2 Highway Safety 5 1.5.3 Demographic Trends 6

1.6 Highways and Evolving Technologies 6 1.6.1 Infrastructure Technologies 6 1.6.2 Traffic Control Technologies 7 1.6.3 Vehicle and Autonomous Vehicle Technologies 8

1.7 Scope of Study 9

Chapter 2 Road Vehicle Performance 11 2.1 Introduction 11

2.2 Tractive Effort and Resistance 11

2.3 Aerodynamic Resistance 12

2.4 Rolling Resistance 15

2.5 Grade Resistance 17

2.6 Available Tractive Effort 18 2.6.1 Maximum Tractive Effort 18 2.6.2 Engine-Generated Tractive Effort 21

2.7 Vehicle Acceleration 25

2.8 Fuel Efficiency 29

2.9 Principles of Braking 30 2.9.1 Braking Forces 30 2.9.2 Braking Force Ratio and Efficiency 32 2.9.3 Antilock Braking Systems 35 2.9.4 Theoretical Stopping Distance 35 2.9.5 Practical Stopping Distance 39 2.9.6 Distance Traveled During Driver Perception/Reaction 42

2.10 Practice Problems 45

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Chapter 3 Geometric Design of Highways 53 3.1 Introduction 53

3.2 Principles of Highway Alignment 54

3.3 Vertical Alignment 55 3.3.1 Vertical Curve Fundamentals 57 3.3.2 Stopping Sight Distance 65 3.3.3 Stopping Sight Distance and Crest Vertical Curve Design 66 3.3.4 Stopping Sight Distance and Sag Vertical Curve Design 70 3.3.5 Passing Sight Distance and Crest Vertical Curve Design 78 3.3.6 Underpass Sight Distance and Sag Vertical Curve Design 81

3.4 Horizontal Alignment 84 3.4.1 Vehicle Cornering 84 3.4.2 Horizontal Curve Fundamentals 86 3.4.3 Stopping Sight Distance and Horizontal Curve Design 90

3.5 Combined Vertical and Horizontal Alignment 92

3.6 Practice Problems 98

Chapter 4 Pavement Design 107 4.1 Introduction 107

4.2 Pavement Types 107 4.2.1 Flexible Pavements 108 4.2.2 Rigid Pavements 109

4.3 Pavement System Design: Principles for Flexible Pavements 109

4.4 Traditional AASHTO Flexible-Pavement Design Procedure 110 4.4.1 Serviceability Concept 111 4.4.2 Flexible-Pavement Design Equation 111 4.4.3 Structural Number 118

4.5 Pavement System Design: Principles for Rigid Pavements 122

4.6 Traditional AASHTO Rigid-Pavement Design Procedure 123

4.7 Design-Lane Loads 132

4.8 Measuring Pavement Quality and Performance 137 4.8.1 International Roughness Index 137 4.8.2 Friction Measurements 138 4.8.3 Rut Depth 139 4.8.4 Cracking 139 4.8.5 Faulting 140 4.8.6 Punchouts 140

4.9 Mechanistic-Empirical Pavement Design 140

4.10 Practice Problems 142

Chapter 5 Fundamentals of Traffic Flow and Queuing Theory 151 5.1 Introduction 151

5.2 Traffic Stream Parameters 151 5.2.1 Traffic Flow, Speed, and Density 152

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5.3 Basic Traffic Stream Models 157 5.3.1 Speed-Density Model 157 5.3.2 Flow-Density Model 159 5.3.3 Speed-Flow Model 160

5.4 Models of Traffic Flow 162 5.4.1 Poisson Model 162 5.4.2 Limitations of the Poisson Model 166

5.5 Queuing Theory and Traffic Flow Analysis 167 5.5.1 Dimensions of Queuing Models 167 5.5.2 D/D/1 Queuing 168 5.5.3 M/D/1 Queuing 175 5.5.4 M/M/1 Queuing 177 5.5.5 M/M/N Queuing 178

5.6 Traffic Analysis at Highway Bottlenecks 181

5.7 Impact of Autonomous Vehicles 184

5.8 Practice Problems 186

Chapter 6 Highway Capacity and Level-of-Service Analysis 191 6.1 Introduction 191

6.2 Level-of-Service Concept 192

6.3 Level-of-Service Determination 195 6.3.1 Base Conditions and Capacity 195 6.3.2 Determine Free-Flow Speed 195 6.3.3 Determine Analysis Flow Rate 196 6.3.4 Calculate Service Measure(s) and Determine LOS 196

6.4 Basic Freeway Segments 196 6.4.1 Speed versus Flow Rate Relationship 196 6.4.2 Base Conditions and Capacity 198 6.4.3 Service Measure 198 6.4.4 Determine Free-Flow Speed 201 6.4.5 Determine Analysis Flow Rate 202 6.4.6 Calculate Density and Determine LOS 208

6.5 Multilane Highway Segments 211 6.5.1 Speed versus Flow Rate Relationship 212 6.5.2 Base Conditions and Capacity 215 6.5.3 Service Measure 215 6.5.4 Determining Free-Flow Speed 215 6.5.5 Determining Analysis Flow Rate 217 6.5.6 Calculate Density and Determine LOS 217

6.6 Two-Lane Highways 221 6.6.1 Analysis Concepts 222

6.7 Design Traffic Volumes 233

6.8 Practice Problems 237

Chapter 7 Traffic Control and Analysis at Signalized Intersections 243 7.1 Introduction 243

7.2 Intersection and Signal Control Characteristics 244

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7.2.1 Actuated Control 247 7.2.2 Signal Controller Operation 250

7.3 Traffic Flow Fundamentals for Signalized Intersections 253

7.4 Development of a Traffic Signal Phasing and Timing Plan 256 7.4.1 Select Signal Phasing 257 7.4.2 Establish Analysis Lane Groups 261 7.4.3 Calculate Analysis Flow Rates and Adjusted Saturation Flow Rates 263 7.4.4 Determine Critical Lane Groups and Total Cycle Lost Time 263 7.4.5 Calculate Cycle Length 266 7.4.6 Allocate Green Time 268 7.4.7 Calculate Change and Clearance Intervals 270 7.4.8 Check Pedestrian Crossing Time 272

7.5 Analysis of Traffic at Signalized Intersections 273 7.5.1 Signalized Intersection Analysis with D/D/1 Queuing 274 7.5.2 Signal Coordination 281 7.5.3 Control Delay Calculation for Level of Service Analysis 289 7.5.4 Level-of-Service Determination 294

7.6 Practice Problems 299

Chapter 8 Travel Demand and Traffic Forecasting 311 8.1 Introduction 311

8.2 Traveler Decisions 313

8.3 Scope of the Travel Demand and Traffic Forecasting Problem 313

8.4 Trip Generation 316 8.4.1 Typical Trip Generation Models 317 8.4.2 Trip Generation with Count Data Models 320

8.5 Mode and Destination Choice 322 8.5.1 Methodological Approach 322 8.5.2 Logit Model Applications 324

8.6 Highway Route Choice 329 8.6.1 Highway Performance Functions 330 8.6.2 User Equilibrium 331 8.6.3 Mathematical Programming Approach to User Equilibrium 336 8.6.4 System Optimization 337

8.7 Autonomous Vehicles, Highway Performance Functions, and System Optimization 341

8.8 Traffic Forecasting in Practice 342

8.9 The Traditional Four-Step Process 346

8.10 The Current State of Travel Demand and Traffic Forecasting 347

8.11 Practice Problems 348

Appendix 8A Least Squares Estimation 352

Appendix 8B Maximum-Likelihood Estimation 354

Problems (Available in e-text for students) P-1

Index I-1

1

Chapter 1

Introduction to Highway Engineering and Traffic Analysis

1.1 INTRODUCTION

In many industrialized nations today, highways present engineers and governments with formidable challenges relating to safety, sustainability, environmental impacts, congestion mitigation, and deteriorating infrastructure. As a result, highways are often viewed from the perspective of the many challenges they present as opposed to the benefits they provide. Historically, highways have always played a key role in the development and sustainability of human civilization. Today, in the United States and throughout the world, highways continue to dominate the transportation system, by providing critical access for the acquisition of natural resources, industrial production, retail marketing, and population mobility. The influence of highway transportation on the economic, social, and political fabric of nations is far-reaching and, as a consequence, highways have been studied for decades as a cultural, political, and economic phenomenon. While industrial needs and economic forces have clearly played an important part in shaping highway networks, societies’ fundamental desire for access to activities and affordable land has generated significant highway demand, which has helped define and shape highway networks.

Without doubt, highways have had a dramatic impact on the environment in terms of the consumption of nonrenewable resources, air pollution, and the generation of greenhouse gases. In addition, vehicle crashes result in well over a million deaths worldwide every year and are the leading cause of death among people 15 to 29 years old [World Health Organization 2015]. As with other critical infrastructures (such as electrical power generation and distributions systems, water distribution systems, and storm-water and sewage systems), highway systems are costly to build, manage, and maintain, and inadequate management and maintenance can result in additional costs with regard to congestion, safety, and a variety of adverse economic impacts.

Given the above, the focus of highway engineering has gone from one of network expansion to one that addresses issues relating to infrastructure maintenance and rehabilitation, improvements in operational efficiency, various traffic-congestion relief measures, energy conservation, improved safety, and environmental mitigation. This shift has forced a new emphasis in highway engineering and traffic analysis and is one that requires a new skill set and a

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deeper understanding of the impact of highway decisions than has historically been the case.

1.2 HIGHWAYS AND THE ECONOMY

It is difficult to overstate the influence that highway transportation has on the world economy. Highway systems have a direct effect on industries that supply vehicles and equipment to support highway transportation and the industries that are involved in highway construction and maintenance. Highway systems are also vital to manufacturing and retail supply chains and distribution systems and serve as regional, national, and international economic engines.

1.2.1 The Highway Economy In the United States, more than 15% of average household income is spent on highway vehicle purchases, maintenance, and other vehicle expenditures. As a consequence, the industries providing vehicles and vehicle services for highway transportation have an enormous economic influence. In the United States alone, in the light-vehicle market (cars, vans, pickup trucks, and so on), as many as 17 million or more new vehicles can be sold annually (depending on economic conditions), which translates to roughly a half a trillion dollars in sales and more than a million jobs in manufacturing and manufacturing-supplier industries. Add to this the additional employment associated with vehicle maintenance and servicing, and more than seven million U.S. jobs can be tied directly to highway vehicles. The influence of the highway economy extends further to the heavy- vehicle sector as well, with more than 1.3 million jobs and trucking industry revenue of roughly three-quarters of a trillion dollars annually in the United States.

The direct influence that highways have also includes the construction and maintenance of highways, with over 100 billion dollars in annual expenditures in the United States alone. This too has an enormous impact on employment and other aspects of the economy.

1.2.2 Supply Chains The survival of modern economies is predicated on efficient, reliable, and resilient supply chains. Industries have become increasingly dependent on their supply chains to reduce costs and remain competitive. As an example, most manufacturing industries today rely on just-in-time delivery to reduce inventory- related costs, which can be a substantial percentage of total costs in many industries. The idea of just-in-time delivery is that the materials required for production are supplied just before they are needed. While such a strategy significantly reduces inventory costs, it requires a very high degree of certainty that the required materials will be delivered on time. If not, the entire production process could be adversely affected and costs could rise dramatically.

In retail applications, effective supply chains can significantly reduce consumer costs and ensure that a sufficient quantity of goods is available to satisfy consumer demand. The ability of highways to provide reliable service for just-in-time inventory control and other supply chain–related industrial and retail applications has made highways critical to the function of modern economies.

1.4 Highways as Part of the Transportation System 3

1.2.3 Economic Development It has long been recognized that highway construction and improvements to the highway network can positively influence economic development. Such improve- ments can increase accessibility and thus attract new industries and spur local economies. To be sure, measuring the economic development impacts of specific highway projects is not an easy task because such measurements must be made in the context of regional and national economic trends. Still, the effect that highways can have on economic development is yet another example of the far- reaching economic influences of highway transportation.

1.3 HIGHWAYS, ENERGY, THE ENVIRONMENT, AND CLIMATE CHANGE

As energy demands fluctuate and supplies vary, and nations become increasingly concerned about environmental impacts, the role that highway transportation plays has come under close scrutiny. As a primary consumer of fossil fuels and a major contributor to air-borne pollution and greenhouse gas emissions, highway transportation is an obvious target for energy conservation and environmental impact mitigation efforts.

In the United States, highway transportation is responsible for roughly 60% of all petroleum consumption. This translates into about 12 million barrels of oil a day. In light of the limitations of oil reserves, this is an astonishing rate of consumption. Highway transportation’s contribution to other pollutants is also substantial. Highway travel is responsible for about 35% of all nitrous oxide emissions and 25% of volatile organic compound emissions, both major contributors to the formation of ozone. Highway travel also contributes more than 50% of all carbon monoxide emissions in the United States and is a major source of fine particulate matter (2.5 microns or smaller), which is a known carcinogen.

The effect of highways on climate change is also formidable. Highway transportation is responsible for roughly 25% of U.S. greenhouse gas emissions (including over 30% of carbon dioxide emissions). While highways affect climate change, the effect that climate change will ultimately have on highways is an issue that has just begun to be addressed [National Cooperative Research Program 2014]. The effect of climate change on extreme weather events, geographic shifts in temperature and moisture, and rising sea levels present highway engineers with extraordinary challenges with regard to highway design, infrastructure maintenance, highway operations, and highway safety.

1.4 HIGHWAYS AS PART OF THE TRANSPORTATION SYSTEM

It is important to keep in mind that highway transportation is part of a larger transportation system that includes air, rail, water, and pipeline transportation. In this system, highways are the dominant mode of most passenger and freight movements. For passenger travel, highways account for about 90% of all passenger-miles. On the freight side, commercial trucks account for about 37% of the freight ton-miles and, because commercial trucks transport higher-valued goods than other modes of transportation (with the exception of air

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transportation), nearly 80% of the dollar value of all goods is transported by commercial trucks.

While highways play a dominant role in both passenger and freight movement, in many applications there are critical interfaces among the various transportation modes. For example, many air, rail, water, and pipeline freight movements involve highway transportation at some point for their initial collection and final distribution. Interfaces between modes, such as those at water ports, airports, and rail terminals, create interesting transportation problems but, if handled correctly, can greatly improve the efficiency of the overall transportation system. However, inter-modal coordination can be problematic because of institutional, regulatory, and other barriers.

1.5 HIGHWAY TRANSPORTATION AND THE HUMAN ELEMENT

Within the highway transportation system, traveler options include single- occupant private vehicles, multi-occupant private vehicles, and public trans- portation modes (such as bus). It is critical to develop a basic understanding of the effect that highway-related projects and policies may have on the individual highway modes of travel (single-occupant private vehicles, bus, and so on) because the distribution of travel among modes will strongly influence overall highway-system performance. In addition, highway safety and the changing demographics of highway users are important considerations.

1.5.1 Passenger Transportation Modes and Traffic Congestion Of the available urban transportation modes (bus, commuter train, subway, private vehicle, and others), private vehicles, and single-occupant private vehicles in particular, offer an unequaled level of mobility. The single-occupant private vehicle has been such a dominant choice that travelers have been willing to pay substantial capital and operating costs, confront high levels of congestion, and struggle with parking-related problems just to have the flexibility in travel departure time and destination choices that is uniquely provided by private vehicles. In the last 50 years, the percentage of trips taken in private vehicles has risen from slightly less than 70% to over 90% (public transit and other modes make up the balance). Over this same period, the average private-vehicle occupancy has dropped from 1.22 to 1.09 persons per vehicle, reflecting the fact that the single-occupant vehicle has become an increasingly dominant mode of travel.

Traffic congestion that has arisen as a result of extensive private-vehicle use and low-vehicle occupancy presents a perplexing problem. The high cost of new highway construction (including monetary, environmental, and social costs) often makes building new highways or adding additional highway capacity an unattractive option. Trying to manage the demand for highways also has its problems. For example, programs aimed at reducing congestion by encouraging travelers to take alternate modes of transportation (bus-fare incentives, increases in private-vehicle parking fees, tolls and traffic-congestion pricing, rail- and bus- transit incentives) or increasing vehicle occupancy (high-occupancy vehicle lanes and employer-based ridesharing programs) can be considered as viable options.

1.5 Highway Transportation and the Human Element 5

However, such programs have the adverse effect of directing people toward travel modes that inherently provide lower levels of mobility because no other mode offers the departure-time and destination-choice flexibility provided by private, single-occupant vehicles. Managing traffic congestion is an extremely complex problem with significant economic, social, environmental, and political implications.

1.5.2 Highway Safety The mobility and opportunities that highway infrastructure provides also have a human cost. Although safety has always been a primary consideration in highway design and operation, highways continue to exact a terrible toll in loss of life, injuries, property damage, and reduced productivity as a result of vehicle accidents. Highway safety involves technical and behavioral components and the complexities of the human/machine interface. Because of the high costs of highway accidents, efforts to improve highway safety have been intensified dramatically in recent decades. This has resulted in the implementation of new highway design guidelines and countermeasures (some technical and some behavioral) aimed at reducing the frequency and severity of highway accidents. Fortunately, efforts to improve highway design (such as more stringent design guidelines, breakaway signs, and so on), vehicle occupant protection (safety belts, padded dashboards, collapsible steering columns, driver- and passenger- side airbags, improved bumper design), as well as advances in vehicle technologies (antilock braking, traction control systems, electronic stability control) and new accident countermeasures (campaigns to reduce drunk driving), have managed to gradually reduce the fatality rate (the number of fatalities per mile driven). However, in spite of continuing efforts and unprecedented advancements in vehicle safety technologies, the total number of fatalities per year has remained unacceptably high worldwide (in the United States, the fatality rate has remained at more than 30,000 per year).

To understand why highway fatality numbers have not dramatically decreased or why fatality rates (fatalities per distance driven) have not dropped more than they have as a result of all the safety efforts, a number of possible explanations arise including: an increase in the overall level of aggressive driving; increasing levels of disrespect for traffic control devices (red light and stop-sign running being two of the more notable examples); in-vehicle driving distractions (such as cell phones); and poor driving skills in the younger and older driving populations. Two other phenomena are being observed that may be contributing to the persistently stable number of fatalities. One is that some people drive more aggressively (speeding, following too closely, frequent lane changing) in vehicles with advanced safety features, thus offsetting some or all of the benefits of new safety technologies [Winston et al., 2006]. Another possibility is that many people are more influenced by style and function than safety features when making vehicle purchase decisions. This is evidenced by the growing popularity of vehicles such as sport utility vehicles, mini-vans, and pickup trucks, despite their consistently overall lower rankings in certain safety categories, such as roll-over probability, relative to traditional passenger cars. These issues underscore the

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overall complexity of the highway safety problem and the trade-offs that must be made with regard to cost, safety, and mobility (speed).

1.5.3 Demographic Trends Travelers’ commuting patterns (which lead to traffic congestion) are inextricably intertwined with socioeconomic characteristics such as age, income, household size, education, and job type, as well as the distribution of residential, commercial, and industrial developments within the region. Many American metropolitan areas have experienced population declines in central cities accompanied by a growth in suburban areas. One could argue that the population shift from the central cities to the suburbs has been made possible by the increased mobility provided by the major highway projects undertaken during the 1960s and 1970s. This mobility enabled people to improve their quality of life by gaining access to affordable housing and land, while still being able to get to jobs in the central city with acceptable travel times. Conventional wisdom suggested that as overall metropolitan traffic congestion grew (making the suburb-to-city commuting pattern much less attractive), commuters would seek to avoid traffic congestion by reverting back to public transport modes and/or once again choosing to reside in the central city. This has certainly happened to some extent, but a different trend has also emerged. Employment centers have developed in the suburbs and now provide a viable alternative to the suburb-to-city commute (the suburb-to-suburb commute). The result is a continuing tendency toward low-density, private vehicle–based development as people seek to retain the high quality of life associated with such development.

Ongoing demographic trends also present engineers with an ever-moving target that further complicates the problem of providing mobility and safety. An example is the rising average age of the U.S. population that has resulted from population cohorts and advances in medical technology that prolong life. Because older people tend to have slower reaction times, taking longer to respond to driving situations that require action, engineers must confront the possibility of changing highway design guidelines and practices to accommodate slower reaction times and the potentially higher variance of reaction times among highway users.

1.6 HIGHWAYS AND EVOLVING TECHNOLOGIES

As in all fields, technological advances at least offer the promise of solving complex problems. For highways, technologies can be classified into those impacting infrastructure, vehicles, and traffic control.

1.6.1 Infrastructure Technologies Investments in highway infrastructure have been made continuously throughout the twentieth and twenty-first centuries. Such investments have understandably varied over the years in response to need and political and national priorities. For example, in the United States, an extraordinary capital investment in highways during the 1960s and 1970s was undertaken by constructing the interstate highway system and upgrading and constructing many other

1.6 Highways and Evolving Technologies 7

highways. The economic and political climate that permitted such an ambitious construction program has not been replicated before or since. It is difficult to imagine, in today’s economic and political environment, that a project of the magnitude of the interstate highway system would ever be seriously considered. This is because of the prohibitive costs associated with land acquisition and construction and the community and environmental impacts that would result.

It is also important to realize that highways are long-lasting investments that require maintenance and rehabilitation at regular intervals. The legacy of a major capital investment in highway infrastructure is the proportionate maintenance and rehabilitation schedules that will follow. Although there are sometimes compelling reasons to defer maintenance and rehabilitation (including the associated construction costs and the impact of the reconstruction on traffic), such deferral can result in unacceptable losses in mobility and safety as well as more costly rehabilitation later.

As a consequence of past capital investments in highway infrastructure and the current high cost of highway construction and rehabilitation, there is a strong emphasis on developing and applying new technologies to more economically construct and extend the life of new facilities and to effectively combat an aging highway infrastructure. Included in this effort are the extensive development and application of new sensing technologies in the emerging field of structural health monitoring. There are also opportunities to extend the life expectancy of new infrastructure with the ongoing advances in material science. Such technological advances are essential elements in the future of highway infrastructure.

1.6.2 Traffic Control Technologies Traffic signals at highway intersections are a familiar traffic control technology. At signalized intersections, the trade-off between mobility and safety is brought into sharp focus. Procedures for developing traffic-signal control plans (allocating green time to conflicting traffic movements) have made significant advances over the years. Today, signals at critical intersections can be designed to respond quickly to prevailing traffic flows, groups of signals can be coordinated to provide a smooth through-flow of traffic, and, in some cases, computers control entire networks of signals. Still, at some level, the effectiveness of improvements in signal control is fundamentally limited by the reaction times and the driving behavior of the motoring public as well as the braking and acceleration performance of the vehicles they drive. This presents highway engineers with a formidable barrier.

In addition to traffic signal controls, numerous safety, navigational, and congestion-mitigation technologies are reaching the market under the broad heading of Intelligent Transportation Systems (ITS). Such technological efforts offer the potential to reduce traffic congestion and improve safety on highways by providing an unprecedented level of traffic control. There are, however, many obstacles associated with ITS implementation, including system reliability, human response, and the human/machine interface. Numerous traffic control technologies offer the potential for considerable improvement in the efficient use of the highway infrastructure, but there remain limits defined by vehicle performance characteristics and their human operators.

8 Chapter 1 Introduction to Highway Engineering and Traffic Analysis

1.6.3 Vehicle and Autonomous Vehicle Technologies Until the 1970s, vehicle technologies evolved slowly and often in response to mild trends in the vehicle market as opposed to an underlying trend toward technological development. Beginning in the 1970s, however, three factors began a cycle of unparalleled advances in vehicle technology that continues to this day: (1) government regulations on air quality, fuel efficiency, and vehicle-occupant safety; (2) energy shortages and fuel-price increases; and (3) intense competition among vehicle manufacturers (foreign and domestic). The aggregate effect of these factors has resulted in vehicle consumers that demand new technology at highly competitive prices. Vehicle manufacturers have found it necessary to reallocate resources and to restructure manufacturing and inventory control processes to meet this demand. In recent years, consumer demand and competition among vehicle manufacturers has resulted in the widespread implementation of new technologies including supplemental restraint systems, antilock brake systems, traction control systems, electronic stability control, and a host of other applications of new technologies to improve the safety and comfort in highway vehicles. There is little doubt that the combination of consumer demand and intense competition in the vehicle industry will continue to spur vehicle technological innovations.

As vehicle technologies have progressed, the highway engineering profession is approaching a potential breakthrough in how traffic can be managed and how future highways may be designed. Specifically, the prospect of connected vehicles (being able to receive and send information to other vehicles and central controls and potentially from infrastructure sensors that may transmit pavement conditions and other factors relevant to vehicle operation) and fully automated vehicles has the potential to reshape the landscape of highway engineering. With connected and fully automated vehicle operation, one can imagine immense changes in highway design because the need to account for variation in human abilities (reaction times, sight distances, detection of changing road conditions with regard to weather, ability to predict the actions of other drivers, etc.) and human behavior (choice of car-following distances, speed choice, route choice, lane choice on multilane highways, choices at traffic signals, etc.) would now be replaced simply by the need to assess and account for variation in vehicle characteristics (vehicle dimensions, acceleration, deceleration, etc.), system operating software, and possible component and system failures, all of which are much more “predictable” than the human-related factors that currently dominate highway design. The eventual result is that, with connected and autonomous vehicles, there could be substantial increases in highway capacity for some situations, vast improvements in safety, and essentially a complete change in how highway transportation affects the economy, energy consumption, the environment, the spatial structure of cities, the impact on human activities, and so on.

As autonomous vehicles begin to enter the highway traffic mix, presumably in the coming decade, there will be enormous challenges. While operating a completely connected and autonomous vehicle fleet would be relatively easy because of the complete control available, having autonomous vehicles mixed with human drivers creates a serious challenge because sensor systems must be

1.7 Scope of Study 9

able to detect, and software systems must be able to quickly respond to the incredible range of human-driver behaviors.

To better understand the range of vehicle technologies that are likely to be present in highway traffic of the future, the U.S. Department of Transportation’s National Highway Traffic Safety Administration has chosen to define levels of vehicle automation as follows (NHTSA, 2018):

No-Automation (Level 0): The driver is in complete and sole control of the primary vehicle controls (brake, steering, throttle, and motive power) at all times. Driver Assistance (Level 1): Automation at this level involves the driver with driver assistance features in the design of the vehicle. Partial Automation (Level 2): This level involves engagement from the driver, but the vehicle has combined automated functions controlling both the steering and acceleration/deceleration capabilities. Conditional Automation (Level 3): Vehicles at this level of automation enable the driver to cede full control of all safety-critical functions under certain traffic or environmental conditions and in those conditions to rely heavily on the vehicle to monitor for changes in those conditions requiring transition back to driver control. The driver is expected to be available for occasional control, but with sufficiently comfortable transition time. High Automation (Level 4): The vehicle is designed to perform all safety-critical driving functions and monitor roadway conditions for an entire trip. Such a design anticipates that the driver will provide destination or navigation input, but is not expected to be available for control at any time during the trip. The driver has the option to control the vehicle. Full Automation (Level 5): The vehicle functions under all conditions without a driver and does not require a steering wheel, throttle, or brake pedals. Level 5 vehicles have the most advanced environment detection system.

In today’s vehicle fleet, the vast majority of vehicles are at Levels 0 and 1. Many newer cars can be purchased at automation Level 2, and it is quite likely that within the coming decade, that substantial numbers of Level 3 and Level 4 vehicles will be on highways worldwide. While the presence of these vehicles will inevitably change highway design, the fundamental principles covered in this text will still apply with appropriate modification.

1.7 SCOPE OF STUDY

Highway engineering and traffic analysis involves an extremely complex interaction of economic, behavioral, social, political, environmental, and technological factors. This complexity makes highway engineering and traffic analysis far more challenging than many typical engineering disciplines that tend to have an overriding focus on only the technical aspects of the problem. To be sure, the technical challenges encountered in highway engineering and traffic analysis easily rival the most complex technical problems encountered in any other engineering discipline (and this will be even more pronounced with the

10 Chapter 1 Introduction to Highway Engineering and Traffic Analysis

potential introduction of autonomous vehicles). However, it is the economic, behavioral, social, political, and environmental elements that introduce a level of complexity unequalled in any other engineering discipline.

The remaining chapters in this book do not intend to provide a comprehensive assessment of the many factors that influence highway engi- neering and traffic analysis. Instead, Chapters 2 through 8 seek to provide readers with the fundamental elements and methodological approaches that are used to design and maintain highways and assess their operating performance. This material constitutes the fundamental principles of highway engineering and traffic analysis that are needed to begin to grasp the many complex elements and considerations that come into play (now and in the future) during the design, construction, maintenance, and operation of highways.

REFERENCES

National Cooperative Research Program. Climate Change, Extreme Weather Events, and the Highway System: Practitioner’s Guide and Research Report. Strategic Issues Facing Transportation, Volume 2, NCHRP Report 750, Transportation Research Board, Washington, D.C., 2014.

National Highway Traffic Safety Administration (NHTSA). https://www.nhtsa.gov/technology- innovation/automated-vehicles-safety

Winston, C., V. Maheshri, and F. Mannering. An exploration of the offset hypothesis using disaggregate data: The case of airbags and antilock brakes. Journal of Risk and Uncertainty, vol. 32, no. 2, 2006.

World Health Organization. Global Status Report on Road Safety 2015. Geneva, Switzerland, 2015.

P-1

CHAPTER 2 PROBLEMS

Students solution available in interactive e-text.

Resistance, Tractive Effort, and Acceleration (Sections 2.2-2.7)

2.1 A new sports car has a drag coefficient of 0.30 and a frontal area of 21 ft2, and is traveling at 110 mi/h. How much power is required to overcome aerodynamic drag if = 0.002378 slugs/ft3?

2.2 For Example 2.3, how far back from the front axle would the center of gravity have to be to ensure that the maximum tractive effort developed for front- and rear-wheel–drive options is equal (assume that all other variables are unchanged)?

2.3 A vehicle manufacturer is considering an engine for a new sedan (CD = 0.34, Af = 22 ft2). The car is being designed to achieve a top speed of 100 mi/h on a paved surface at sea level ( = 0.002378 slugs/ft3). The car currently weighs 2500 lb, but the designers initially selected an underpowered engine because they did not account for aerodynamic and rolling resistances. If 2 lb of additional vehicle weight is added for each unit of horsepower needed to overcome the neglected resistance, what will be the final weight of the car if it is to achieve the 100-mi/h top speed?

2.4 A 2650-lb car is traveling at sea level at a constant speed. Its engine is running at 4500 rev/min and is producing 175 ft-lb of torque. It has a drivetrain efficiency of 90%, a drive axle slippage of 2%, 15-inch–radius wheels, and an overall gear reduction ratio of 3 to 1. If the car’s frontal area is 21.5 ft2, what is its drag coefficient?

2.5 A 3000-lb car has a maximum speed (at sea level and on a level, paved surface) of 140 mi/h with 16- inch–radius wheels, a gear reduction of 3.5 to 1, and a drivetrain efficiency of 92%. It is known that at the car’s top speed the engine is producing 220 ft-lb of torque. If the car’s frontal area is 25 ft2, what is its drag coefficient?

2.6 A 3200-lb car (CD = 0.35, Af = 25 ft2, and ρ = 0.002378 slugs/ft3) has 14-inch–radius wheels, a drivetrain efficiency of 93%, an overall gear reduction ratio of 3.2 to 1, and drive axle slippage of 3.5%. The engine develops a maximum torque of 210 ft-lb at 3600 rev/min. What is the maximum grade this vehicle could ascend, on a paved surface, while the engine is developing maximum torque?

(Assume that the available tractive effort is the engine-generated tractive effort.)

2.7 A 3400-lb car is traveling in third gear (overall gear reduction ratio of 2.5 to 1) on a level road at its top speed of 130 mi/h. The air density is 0.00206 slugs/ft3. The car has a frontal area of 19.8 ft2, a drag coefficient of 0.28, a wheel radius of 12.6 inches, a drive axle slippage of 3%, and a drivetrain efficiency of 88%. At this vehicle speed, what torque is the engine producing and what is the engine speed (in revolutions per minute)?

2.8 A rear-wheel–drive car weighs 2600 lb and has an 84-inch wheelbase, a center of gravity 20 inches above the roadway surface and 30 inches behind the front axle, a drivetrain efficiency of 85%, 14-inch– radius wheels, and an overall gear reduction of 7 to 1. The car’s torque/engine speed curve is given by the equation 26 0.045e e eM n n= − . If the car is on a paved, level roadway surface with a coefficient of adhesion of 0.75, determine its maximum acceleration from rest.

2.9 Consider the car in Problem 2.8. If it is known that the car achieves maximum speed at an overall gear reduction ratio of 2.7 to 1 with a drive axle slippage of 3.5%, how fast would the car be going if it could achieve its maximum speed when its engine is producing maximum power?

2.10 An engineer designs a rear-wheel–drive car (without an engine) that weighs 2000 lb and has a 100-inch wheelbase, drivetrain efficiency of 80%, 14-inch–radius wheels, an overall gear reduction ratio of 10 to 1, and a center of gravity (without engine) that is 22 inches above the roadway surface and 55 inches behind the front axle. An engine that weighs 3 lb for each ft-lb of developed torque is to be placed in the front portion of the car. Calculations show that for every 20 lb of engine weight added, the car’s center of gravity moves 1 inch closer to the front axle (but stays at the same height above the roadway surface). If the car is starting from rest on a level paved roadway with a coefficient of adhesion of 0.8, select an engine size (weight and associated torque) that will result in the highest possible available tractive effort.

2.11 A 3000-lb car is traveling on a paved road with CD = 0.35, Af = 21 ft2, and ρ = 0.002378 slugs/ft3. Its

P-2 Chapter 2 Problems engine is running at 3000 rev/min and is producing 250 ft-lb of torque. The car’s gear reduction ratio is 4.5 to 1, drivetrain efficiency is 90%, drive axle slippage is 3.5%, and the wheel radius is 16 inches. What will the car’s maximum acceleration rate be under these conditions on a level road? (Assume that the available tractive effort is the engine- generated tractive effort.)

2.12 A rear-wheel–drive car weighs 3600 lb, has 15- inch–radius wheels, a drivetrain efficiency of 95%, and an engine that develops 520 ft-lb of torque. Its wheelbase is 8.2 ft, and the center of gravity is 18 inches above the road surface and 3.3 ft behind the front axle. What is the lowest gear reduction ratio that would allow this car to achieve the highest possible acceleration from rest on good, dry pavement?

2.13 A newly designed car has a 9.0-ft wheelbase, is rear-wheel drive, and has a center of gravity 18 inches above the road and 4.3 ft behind the front axle. The car weighs 2450 lb, the mechanical efficiency of the drivetrain is 90%, and the wheel radius is 14 inches. The base engine develops 200 ft- lb of torque, and a modified version of the engine develops 240 ft-lb of torque. If the overall gear reduction ratio is 8 to 1, what is the maximum acceleration from rest for the car with the base engine and for the car with the modified engine? (It is on good, dry, and level pavement.)

2.14 A rear-wheel–drive 3000-lb drag race car has a 200-inch wheelbase and a center of gravity 20 inches above the pavement and 140 inches behind the front axle. The owners wish to achieve an initial acceleration from rest of 22 ft/s2 on a level paved surface. What is the minimum coefficient of road adhesion needed to achieve this acceleration? (Assume γm = 1.00.)

2.15 If the race car in Problem 2.14 has a center of gravity 32 inches above the roadway and is run on a pavement with a coefficient of adhesion of 1.0, how far back from the front axle would the center of gravity have to be to develop a maximum acceleration from rest of 1.0 g (32.2 ft/s2)? (Assume γm = 1.00.)

2.16 Consider the situation described in Example 2.5. If the vehicle is redesigned with wheels that have a 13-inch radius (assume that the mass factor is unchanged) and a center of gravity located at the same height but at the midpoint of the wheelbase, determine the acceleration for front- and rear- wheel–drive options.

Braking and Stopping Distance (Section 2.9)

2.17 If the car in Example 2.8 had CD = 0.45 and Af = 25 ft2, what is the difference in minimum theoretical stopping distances with and without aerodynamic resistance considered (all other factors the same as in Example 2.8)?

2.18 A 3500-lb vehicle (CD = 0.38, Af = 26 ft2, ρ = 0.002378 slugs/ft3) is driven on a surface with a coefficient of adhesion of 0.5, and the coefficient of rolling friction is approximated as 0.015 for all speeds. Assuming minimum theoretical stopping distances, if the vehicle comes to a stop 260 ft after brake application on a level surface and has a braking efficiency of 0.82, what was its initial speed (a) if aerodynamic resistance is considered and (b) if aerodynamic resistance is ignored?

2.19 A level test track has a coefficient of road adhesion of 0.80, and a car being tested has a coefficient of rolling friction that is approximated as 0.018 for all speeds. The vehicle is tested unloaded and achieves the theoretical minimum stop in 180 ft (from brake application). The initial speed was 60 mi/h. Ignoring aerodynamic resistance, what is the unloaded braking efficiency?

2.20 A driver is traveling at 90 mi/h down a 3% grade on good, wet pavement. An accident investigation team noted that braking skid marks started 410 ft before a parked car was hit at an estimated 45 mi/h. Ignoring air resistance, and using theoretical stopping distance, what was the braking efficiency of the car?

2.21 A small truck is to be driven down a 4% grade at 70 mi/h. The coefficient of road adhesion is 0.95, and it is known that the braking efficiency is 80% when the truck is empty and decreases by one percentage point for every 100 lb of cargo added. Ignoring aerodynamic resistance, if the driver wants the truck to be able to achieve a minimum theoretical stopping distance of 275 ft from the point of brake application, what is the maximum amount of cargo (in pounds) that can be carried?

2.22 Consider the conditions in Example 2.9. The car has W = 3500 lb, CD = 0.5, Af = 25 ft2, ρ = 0.002378 slugs/ft3, and a coefficient of rolling friction approximated as 0.018 for all speed conditions. If aerodynamic resistance is considered in stopping, estimate how fast the car will be going when it strikes the object on a level and a +5% grade [all other conditions (speed, etc.) as described in Example 2.11].

Chapter 2 Problems P-3 2.23 A race car with a 106-inch wheelbase has its weight evenly distributed between front and rear axles. At 150 mi/h, on a race track with μ = 1.0, the optimal brake force has 67.32% of the braking force on the front brakes. A new racing tire generates μ = 1.2. At 150 mi/h, what percentage of the braking force should now be allocated to the front to achieve optimal braking?

2.24 A car is traveling up a 2% grade at 70 mi/h on good, wet pavement. The driver brakes to try to avoid hitting stopped traffic that is 250 ft ahead. The driver’s reaction time is 0.5 s. At first, when the driver applies the brakes, a software flaw causes the antilock braking system to fail (brakes work in non- antilock mode with 80% efficiency), leaving 80 ft skid marks. After the 80 ft skid, the antilock brakes work with 100% efficiency. How fast will the driver be going when the stopped traffic is hit if the coefficient of rolling resistance is constant at 0.013? (Assume minimum theoretical stopping distance and ignore aerodynamic resistance.)

2.25 A car is traveling at 76 mi/h down a 3% grade on poor, wet pavement. The car’s braking efficiency is 90%. The brakes were applied 320 ft before impacting an object. The car had an antilock braking system, but the system failed 200 ft after the brakes had been applied (wheels locked). What speed was the car traveling at just before it impacted the object? (Assume theoretical stopping distance, ignore air resistance, and let frl = 0.015.)

2.26 A driver traveling down a 4% grade collides with a roadside object in rainy conditions, and is issued a ticket for driving too fast for conditions. The posted speed limit is 65 mi/h. The accident investigation team determined the following: The vehicle was traveling 40 mi/h when it struck the object, braking skid marks started 205 ft before the struck object, the pavement is in good condition, and the braking efficiency of the vehicle was 93%. Using theoretical stopping distance, assuming aerodynamic resistance is negligible, and with the coefficient rolling resistance approximated as 0.015, should the driver appeal the ticket? Why or why not?

2.27 A driver is traveling 68 mi/h on a road with a −3% grade. There is a stalled car on the road 1000 ft ahead of the driver. The driver’s vehicle has a braking efficiency of 90%, and it has antilock brakes. The road is in good condition and is initially dry, but it becomes wet 160 ft before the stalled car (and stays wet until the car is reached). What is the

minimum distance from the stalled car at which the driver could apply the brakes and still stop before hitting it? (Assume theoretical stopping distance, ignore air resistance, and let frl = 0.013.)

2.28 A car is traveling at 70 mi/h on a level section of road with good, wet pavement. Its antilock braking system (ABS) only starts to work after the brakes have been locked for 100 ft. If the driver holds the brake pedal down completely, immediately locking the wheels, and keeps the pedal down during the entire process, how many feet will it take the car to stop from the point of initial brake application? (The braking efficiency is 80% with the ABS not working and 100% with the ABS working. Use theoretical stopping distance and ignore air resistance. Let frl = 0.02 when the brakes are locked, but compute the frl once the ABS becomes active.)

2.29 Two cars are traveling on level terrain at 60 mi/h on a road with a coefficient of adhesion of 0.8. The driver of car 1 has a 2.5-s perception/reaction time and the driver of car 2 has a 2.1-s perception/reaction time. Both cars are traveling side by side and the drivers are able to stop their respective cars in the same distance after first seeing a roadway obstacle (perception and reaction plus vehicle stopping distance). If the braking efficiency of car 2 is 0.78, determine the braking efficiency of car 1. (Assume minimum theoretical stopping distance and ignore aerodynamic resistance.)

2.30 An engineering student is driving on a level roadway and sees a construction sign 500 ft ahead in the middle of the roadway. The student strikes the sign at a speed of 25 mi/h. If the student was traveling at 55 mi/h when the sign was first spotted, what was the student’s associated perception/reaction time (use practical stopping distance)?

2.31 An engineering student claims that a country road can be safely negotiated at 65 mi/h in rainy weather. Because of the winding nature of the road, one stretch of level pavement has a sight distance of only 510 ft. Assuming practical stopping distance, comment on the student’s claim.

2.32 A driver is traveling at 52 mi/h on a wet road. An object is spotted on the road 415 ft ahead and the driver is able to come to a stop just before hitting the object. Assuming standard perception/reaction time and practical stopping distance, determine the grade of the road.

2.33 A test of a driver’s perception/reaction time is being conducted on a special testing track with level,

P-4 Chapter 2 Problems wet pavement and a driving speed of 50 mi/h. When the driver is sober, a stop can be made just in time to avoid hitting an object that is first visible 385 ft ahead. After a few drinks under exactly the same conditions, the driver fails to stop in time and strikes the object at a speed of 30 mi/h. Determine the driver’s perception/reaction time before and after drinking. (Assume practical stopping distance.)

Acceleration and Braking (Sections 2.7 and 2.9)

2.34 On a level test track, a car with antilock brakes and 90% braking efficiency is determined to have a theoretical stopping distance (ignoring aerodynamic resistance) of 408 ft (after the brakes are applied) from 100 mi/h. The car is rear-wheel drive with a 110-inch wheelbase, weighs 3200 lb, and has a 50/50 weight distribution (front to back), a center of gravity that is 22 inches above the road surface, an engine that generates 300 ft-lb of torque, an overall gear reduction of 8.5 to 1 (in first gear), a wheel radius of 15 inches and a driveline efficiency of 95%. What is the maximum acceleration from rest of this car on this test track?

Multiple Choice Problems (Multiple Sections)

2.35 A 2500-lb vehicle has a drag coefficient of 0.35 and a frontal area of 20 ft2. What is the minimum tractive effort required for this vehicle to maintain a 70 mi/h speed on a 5% upgrade through an air density of 0.002045-slugs/ft3?

a) 242.9 lb b) 197.0 lb c) 161.9 lb d) 160.1 lb

2.36 A car is traveling at 20 mi/h on good, dry pavement at 5000 ft elevation. The front-wheel– drive car has a drag coefficient of 0.30, a frontal area of 20 ft2 and a weight of 2500 lb. The wheelbase is 110 inches and the center of gravity is 20 inches from the ground, 50 inches behind the front axle. The engine is producing 95 ft-lb of torque and is in a gear that gives an overall gear reduction ratio of 4.5. The radius of the drive wheels is 14 inches and the mechanical efficiency of the drivetrain is 90%. What would the acceleration of the car be if the driver was accelerating quickly to avoid a collision?

a) 3.65 ft/s2 b) 13.26 ft/s2 c) 15.90 ft/s2 d) 3.48 ft/s2

2.37 A car is traveling at 60 mi/h on good, wet pavement. It has a wheelbase of 110 inches with the center of gravity 50 inches behind the front axle and at a height of 24 inches above the pavement surface. Determine the percentage of braking force that the braking system should allocate to the rear axle.

a) 74.5% b) 65.4% c) 25.5% d) 34.6%

2.38 A truck traveling at 75 mi/h has a braking efficiency of 70%. The coefficient of road adhesion is 0.80. Ignoring aerodynamic resistance, determine the theoretical stopping distance on a level grade.

a) 340.9 ft b) 180.6 ft c) 425.6 ft d) 338.6 ft

2.39 A child accidentally runs into the street in front of an approaching vehicle. The vehicle is traveling at 40 mi/h. Assuming the road is level, at what distance must the driver first see the child to stop just in time?

a) 153.7 ft b) 300.3 ft c) 318.8 ft d) 146.7 ft

2.40 A car is traveling at sea level at 78 mi/h on a 4% upgrade before the driver sees a fallen tree in the roadway 150 ft away. The coefficient of road adhesion is 0.8. The car weighs 2700 lb, has a drag coefficient of 0.35, a frontal area of 18 ft2, and a coefficient of rolling friction approximated as 0.017 for all speed conditions. The car has an antilock braking system that gives it a braking efficiency of 100%. If the driver first applies the brakes 150 ft from the tree, how fast will the car be traveling when it reaches the tree? Include the effect of aerodynamic resistance.

a) 49.5 mi/h b) 48.8 mi/h c) 50.5 mi/h d) 47.7 mi/h

11

Chapter 2

Road Vehicle Performance 2.1 INTRODUCTION

Road vehicle performance forms the basis for highway design guidelines and traffic analysis. As an example, in highway design, the determination of the length required for freeway acceleration and deceleration lanes, maximum highway grades, stopping-sight distances, passing-sight distances, and numerous crash mitigation highway features all rely on a basic understanding of vehicle performance. Vehicle performance is also a major consideration in the selection and design of traffic control devices, the determination of speed limits, and the timing and control of traffic signal systems.

Studying vehicle performance serves two important functions. First, it provides insight into highway design and traffic operations and the compromises that are necessary to accommodate the wide variety of vehicles that use highways (from high-powered sports cars to heavily laden trucks). Second, vehicle performance is critical to the assessment of the impact that new vehicle technologies will have on existing highway design guidelines. This second function is particularly important in light of the ongoing unprecedented advances in vehicle technologies, including autonomous vehicles. Such advances will necessitate more frequent updating of highway design guidelines as well as engineers who have a better understanding of the fundamental principles underlying vehicle performance.

The objective of this chapter is to introduce the basic principles of road vehicle performance. Primary attention will be given to the straight-line performance of vehicles (acceleration, deceleration, top speed, and the ability to ascend grades). Cornering performance of vehicles is overviewed in Chapter 3, but detailed presentations of this material are better suited to more specialized sources [Wong, 2008].

2.2 TRACTIVE EFFORT AND RESISTANCE

Tractive effort (also referred to as thrust) and resistance are the two primary opposing forces that determine the straight-line performance of road vehicles. Tractive effort is simply the force available, at the roadway surface, to perform work and is expressed in lb. Resistance, also expressed in lb, is defined as the force impeding vehicle motion. The three major sources of vehicle resistance are (1) aerodynamic resistance, (2) rolling resistance (which originates from the roadway surface and tire interface), and (3) grade or gravitational resistance. To illustrate these forces, consider the vehicle force diagram shown in Fig. 2.1.

12 Chapter 2 Road Vehicle Performance

Figure 2.1 Forces acting on a road vehicle.

Ra = aerodynamic resistance in lb, W = total vehicle weight in lb,

Rrlf = rolling resistance of the front tires in lb, θg = angle of the grade in degrees,

Rrlr = rolling resistance of the rear tires in lb, m = vehicle mass in slugs, and

Ff = available tractive effort of the front tires in lb, a = acceleration in ft/s2.

Fr = available tractive effort of the rear tires in lb,

Summing the forces along the vehicle’s longitudinal axis provides the basic

equation of vehicle motion:

f r a rlf rlr gF F ma R R R R+ = + + + + (2.1)

where Rg is the grade resistance and is equal to W sin θg. For exposition purposes, it is convenient to let F be the sum of available tractive effort delivered by the front and rear tires (Ff + Fr) and similarly to let Rrl be the sum of rolling resistance (Rrlf + Rrlr). This notation allows Eq. 2.1 to be written as

a rl gF ma R R R= + + + (2.2)

Sections 2.3 to 2.8 present a thorough discussion of the components and implications of Eq. 2.2.

2.3 AERODYNAMIC RESISTANCE

Aerodynamic resistance is a resistive force that can have significant impacts on vehicle performance. At high speeds, where this component of resistance can become overwhelming, proper vehicle aerodynamic design is essential. Aerodynamic efficiency has long been given primary consideration in the design of racing and sports cars. More recently, concerns over fuel efficiency and overall vehicle performance have resulted in more efficient aerodynamic designs in common passenger cars, although not necessarily in pickup trucks or sport utility vehicles where the need for cargo space often limits what can be done with aerodynamic efficiency.

2.3 Aerodynamic Resistance 13

Aerodynamic resistance originates from a number of sources. The primary source (typically accounting for over 85% of total aerodynamic resistance) is the turbulent flow of air around the vehicle body. This turbulence is a function of the shape of the vehicle, particularly the rear portion, which has been shown to be a major source of air turbulence. To a much lesser extent (on the order of 12% of total aerodynamic resistance), the friction of the air passing over the body of the vehicle contributes to resistance. Finally, approximately 3% of the total aerodynamic resistance can be attributed to air flow through vehicle components such as radiators and air vents.

Based on these sources, the equation for determining aerodynamic resistance is

ρ= 2

2a D fR C A V (2.3)

where

Ra = aerodynamic resistance in lb,

ρ = air density in slugs/ft3,

CD = coefficient of drag (unitless), Af = frontal area of the vehicle (projected area of the vehicle in the direction of

travel) in ft2, and V = speed of the vehicle in ft/s.

To be truly accurate, for aerodynamic resistance computations, V is actually the speed of the vehicle relative to the prevailing wind speed. To simplify the exposition of concepts soon to be presented, the wind speed is assumed to be equal to zero for all problems and derivations in this book.

Air density is a function of both elevation and temperature, as indicated in Table 2.1. Equation 2.3 indicates that as the air becomes denser, total aerodynamic resistance increases. The drag coefficient (CD) is a term that implicitly accounts for all three of the aerodynamic resistance sources discussed above. Drag coefficients are measured from empirical data either from wind tunnel experiments or actual field tests in which the vehicle is allowed to decelerate from a known speed with other sources of resistance (rolling and grade) taken into account. Table 2.2 provides a range of drag coefficients for different types of road vehicles, and Table 2.3 presents drag coefficients for specific automobiles starting with those in the mid-twentieth century.

Table 2.3 shows a wide range of drag coefficients over time. Vehicles with an emphasis on fuel efficiency tend to have lower drag coefficients. High-performance sports cars also seek low drag coefficients, but these cars can have bodies designed to generate downforce for performance braking and cornering, and this can adversely affect the drag coefficient. Table 2.3 also includes some larger personal vehicles, such as pickup trucks and sport utility vehicles, which generally represent the upper range of drag coefficients. Also, operating conditions can have a significant effect on drag coefficients. For example, even a small operational change, such as the opening of windows, can increase drag coefficients by 5% or more. More significant operational changes, such as having the top down on a convertible automobile, can increase the drag coefficient by more than 25%.

14 Chapter 2 Road Vehicle Performance

Finally, projected frontal area (approximated as the height of the vehicle multiplied by its width) typically ranges from 10 ft2 to 30 ft2 for passenger cars and frontal area is also a major factor in determining aerodynamic resistance.

Table 2.1 Typical Values of Air Density under Specified Atmospheric Conditions

Altitude (ft) Temperature

(°F) Pressure (lb/in2) Air density

(slugs/ft3)

0 59.0 14.7 0.002378 5,000 41.2 12.2 0.002045

10,000 23.4 10.1 0.001755

Table 2.2 Ranges of Drag Coefficients for Typical Road Vehicles

Vehicle type Drag coefficient (CD)

Automobile 0.25–0.55 Bus 0.5–0.7 Tractor-Trailer 0.6–1.3 Motorcycle 0.27–1.8

Table 2.3 Drag Coefficients of Selected Automobiles

Vehicle Drag coefficient (CD)

1955 Mercedes Benz 300SL Gullwing 0.38 1967 Volkswagen Beetle 0.46 1977 Jaguar XJS 0.36 1987 Acura Integra 0.34 1987 Ford Taurus 0.32 1992 Mazda MX5 Miata 0.38 1993 Ford Ranger (truck) 0.45 1996 Dodge Viper RT/10 0.45 1997 Lexus LS400 0.29 2002 Acura NSX 0.30 2003 Dodge Caravan (minivan) 0.35 2003 Dodge Ram (truck) 0.53 2003 Hummer H2 0.57 2005 Chevrolet Corvette C5 0.29 2009 Nissan Cube 0.35 2012 Tesla Model S 0.24 2014 BMW I8 0.26 2016 Audi A5 0.31 2016 Dodge Charger 0.35 2016 Nissan Leaf 0.28 2016 Mercedes Benz GT AMG 0.36

2.4 Rolling Resistance 15

Because aerodynamic resistance is proportional to the square of the vehicle’s speed, it is clear that such resistance will increase rapidly at higher speeds. The magnitude of this increase can be underscored by considering an expression for the power (hp

aR ) required to overcome aerodynamic resistance. With power being the product of force and speed, the multiplication of Eq. 2.3 by speed gives

3

hp 1100a

D f R

C A V

ρ = (2.4)

where

hp aR = horsepower required to overcome aerodynamic resistance

(1 horsepower = 550 ft-lb/s),

other terms are as defined previously. Thus, the amount of power required to overcome aerodynamic resistance increases with the cube of speed, indicating, for example, that eight times as much power is required to overcome aerodynamic resistance if the vehicle speed is doubled.

2.4 ROLLING RESISTANCE

Rolling resistance refers to the resistance generated from a vehicle’s internal mechanical friction and from pneumatic tires and their interaction with the roadway surface. The primary source of this resistance is the deformation of the tire as it passes over the roadway surface. The force needed to overcome this deformation accounts for approximately 90% of the total rolling resistance. Depending on the vehicle’s weight and the material composition of the roadway surface, the penetration of the tire into the surface and the corresponding surface compression can also be a significant source of rolling resistance. However, for typical vehicle weights and pavement types, penetration and compression constitute only around 4% of the total rolling resistance. Finally, frictional motion due to the slippage of the tire on the roadway surface and, to a lesser extent, air circulation around the tire and wheel (the fanning effect) are sources accounting for roughly 6% of the total rolling resistance [Taborek, 1957].

In considering the sources of rolling resistance, three factors are worthy of note. First, the rigidity of the tire and the roadway surface influence the degree of tire penetration, surface compression, and tire deformation. Hard, smooth, and dry roadway surfaces provide the lowest rolling resistance. Second, tire conditions, including inflation pressure and temperature, can have a substantial impact on rolling resistance. High tire inflation decreases rolling resistance on hard paved surfaces as a result of reduced friction but increases rolling resistance on soft unpaved surfaces due to additional surface penetration. Also, higher tire temperatures make the tire body more flexible, and thus less resistance is encountered during tire deformation. The third and final factor is the vehicle’s operating speed, which affects tire deformation. Increasing speed results in additional tire flexing and vibration and thus a higher rolling resistance.

16 Chapter 2 Road Vehicle Performance

Due to the wide range of factors that determine rolling resistance, a simplifying approximation is used. Studies have shown that overall rolling resistance can be approximated as the product of a friction term (coefficient of rolling resistance) and the weight of the vehicle acting normal to the roadway surface. The coefficient of rolling resistance for road vehicles operating on paved surfaces is approximated as

 = +   

0.01 1 147rl V

f (2.5)

where

frl = coefficient of rolling resistance (unitless), and V = vehicle speed in ft/s.

By inspection of Fig. 2.1, the rolling resistance, in lb, will simply be the coefficient of rolling resistance multiplied by W cos θg, the vehicle weight acting normal to the roadway surface. For most highway applications, θg is quite small, so it can be assumed that cos θg = 1, giving the equation for rolling resistance (Rrl) as

rl rlR f W= (2.6)

From this, the amount of power required to overcome rolling resistance is

hp 550rl

rl R

f WV= (2.7)

where

hp rlR = horsepower required to overcome rolling resistance

(1 horsepower = 550 ft-lb/s), W = total vehicle weight in lb.

EXAMPLE 2.1 AERODYNAMIC AND ROLLING RESISTANCE

A 2500-lb car is driven at sea level (ρ = 0.002378 slugs/ft3) on a level paved surface. The car has CD = 0.38 and 20 ft2 of frontal area. It is known that at maximum speed, 50 hp is being expended to overcome rolling and aerodynamic resistance. Determine the car’s maximum speed.

SOLUTION

It is known that at maximum speed (Vm),

available horsepower a m rl mR V R V= +

2.5 Grade Resistance 17

or

3

2available hp 550

D f m rl mC A V f WV ρ +

=

Substituting, we have

 + +   =

30.002378 (0.38)(20) 0.01 1 (2500)

2 147 50

550

m m m

V V V

or

3 227,500 0.00904 0.17 25m m m V V V= + +

Solving for Vm gives

Vm = 133 ft/s or 90 mi/h

2.5 GRADE RESISTANCE

Grade resistance is simply the gravitational force (the component parallel to the roadway) acting on the vehicle. As suggested in Fig. 2.1, the expression for grade resistance (Rg) is

sing gR W θ= (2.8)

As in the development of the rolling resistance formula (Eq. 2.6), highway grades are usually very small, so sin θg ≅ tan θg. Rewriting Eq. 2.8, we get

tang gR W WGθ≅ = (2.9)

where

G = grade, defined as the vertical rise per some specified horizontal distance (opposite side of the force triangle, Fig. 2.1, divided by the adjacent side) in ft/ft.

Grades are generally specified as percentages for ease of understanding. Thus a roadway that rises 5 ft vertically per 100 ft horizontally (G = 0.05 and θg = 2.86°) is said to have a 5% grade.

EXAMPLE 2.2 GRADE RESISTANCE

A 2000-lb car has CD = 0.40, Af = 20 ft2, and an available tractive effort of 255 lb. If the car is traveling at an elevation of 5000 ft (ρ = 0.002045 slugs/ft3) on a paved surface at a speed of 70 mi/h, what is the maximum grade that this car could ascend and still maintain the 70-mi/h speed?

18 Chapter 2 Road Vehicle Performance

SOLUTION

To maintain the speed, the available tractive effort will be exactly equal to the summation of resistances. Thus no tractive effort will remain for vehicle acceleration (ma = 0). Therefore, Eq. 2.2 can be written as

a rl gF R R R= + +

For grade resistance (using Eq. 2.9),

2000gR WG G= =

for aerodynamic resistance (using Eq. 2.3),

( )( )( )

2

2

2 0.002045

0.4 20 70 5280 / 3600 2

86.22 lb

a D fR C A V ρ=

= ×

=

and for rolling resistance (using Eq. 2.6),

70 5280 3600 0.01 1 2000

147

33.97 lb

rl rlR f W= × = + × 

  =

Therefore,

255 86.22 33.97 2000

0.0674 or a 6.74% grade

F G

G

= = + + =

2.6 AVAILABLE TRACTIVE EFFORT

With the resistance terms in the basic equation of vehicle motion (Eq. 2.2) discussed, attention can now be directed toward available tractive effort (F ) as used in Example 2.2. The tractive effort available to overcome resistance and/or to accelerate the vehicle is determined either by the force generated by the vehicle’s engine or by some maximum value that will be a function of the vehicle’s weight distribution and the characteristics of the roadway surface–tire interface. The basic concepts underlying these two determinants of available tractive effort are presented here.

2.6.1 Maximum Tractive Effort No matter how much force a vehicle’s engine makes available at the roadway surface, there is a point beyond which additional force merely results in the spinning of tires and does not overcome resistance or accelerate the vehicle. To explain what determines this point of maximum tractive effort (the limiting value beyond which tire spinning begins), a force and moment-generating diagram is provided in Fig. 2.2.

2.6 Available Tractive Effort 19

Figure 2.2 Vehicle forces and moment-generating distances.

Ra = aerodynamic resistance in lb, θ g = angle of the grade in degrees,

Rrlf = rolling resistance of the front tires in lb, m = vehicle mass in slugs,

Rrlr = rolling resistance of the rear tires in lb, a = acceleration in ft/s2, Ff = available tractive effort of the front tires in lb, L = length of wheelbase,

Fr = available tractive effort of the rear tires in lb, h = height of the center of gravity above the roadway surface, W = total vehicle weight in lb,

Wf = weight of the vehicle on the front axle in lb, lf = distance from the front axle to the center of gravity, and Wr = weight of the vehicle on the rear axle in lb,

lr = distance from the rear axle to the center of

gravity. To determine the maximum tractive effort that the roadway surface-tire contact can support, it is necessary to examine the normal loads on the axles. The normal load on the rear axle (Wr) is given by summing the moments about point A (see Fig. 2.2):

cos sina f g g

r

R h Wl mah Wh W

L

θ θ+ + ± = (2.10)

In this equation, the grade moment (Wh sin θg) is positive for an upward slope and negative for a downward slope. Rearranging terms (assuming cos θg = 1 for the small grades encountered in highway applications) and substituting into Eq. 2.2 gives

( )f r rl

l h W W F R

L L = + − (2.11)

From basic physics, the maximum tractive effort as determined by the roadway surface–tire interaction will be the normal force multiplied by the coefficient of road adhesion (μ), so for a rear-wheel–drive car

rF Wμ= (2.12)

20 Chapter 2 Road Vehicle Performance

and substituting Eq. 2.11 into Eq. 2.12,

μ  

= + −   

max( )f rl

l h F W F R

L L (2.13)

( ) 1

f rlW l f h L F

h L

μ μ −

= −

(2.14)

Similarly, by summing moments about point B (see Fig. 2.2), it can be shown that for a front-wheel–drive vehicle

( ) 1

r rlW l f h L F

h L

μ μ +

= +

(2.15)

Note that in Eqs. 2.14 and 2.15, because of canceling of units, h, lf, lr, and L can be in any unit of length (feet, inches, etc.). However, all of these terms must be in the same chosen unit of measure. The units of Fmax will be the same as the units for W (lb). Typical values of the coefficient of road adhesion (μ) are provided in Table 2.4.

Table 2.4 Typical Coefficients of Road Adhesion (μ)

Pavement Coefficient of road adhesion

*In some instances, the coefficient of road adhesion values can exceed 1.0. See discussion near the end of Section 2.6.1.

Source: S. G. Shadle, L. H. Emery, and H. K. Brewer, “Vehicle Braking, Stability, and Control,” SAE Transactions, vol. 92, paper 830562, 1983.

Good, dry 1.00* Good, wet 0.90 Poor, dry 0.80 Poor, wet 0.60 Packed snow or ice 0.25

EXAMPLE 2.3 MAXIMUM TRACTIVE EFFORT A 2500-lb car is designed with a 120-inch wheelbase. The center of gravity is located 22 inches above the pavement and 40 inches behind the front axle. If the car is on a road with poor, wet pavement, what is the maximum tractive effort that can be developed if the car is (a) front-wheel drive and (b) rear-wheel drive?

SOLUTION For the front-wheel–drive case Eq. 2.15 is used:

( ) /

1 / r rlW l f h L

F h L

μ μ +

= +

With poor, wet pavement, the coefficient of road adhesion μ = 0.60 from Table 2.4, and from Eq. 2.5, frl = 0.01 because V = 0 ft/s, so

2.6 Available Tractive Effort 21

( ) ( )

 × × + = + ×

=

0.60 2500 80 0.01(22) 120

1 0.60 22 120

903.38 lb

maxF

For the rear-wheel–drive case, Eq. 2.14 is used:

( ) ( )

 × × − = − ×

=

0.60 2500 40 0.01(22) 120

1 0.60 22 120

558.71 lb

maxF

2.6.2 Engine-Generated Tractive Effort The amount of tractive effort generated by the vehicle’s engine is a function of a variety of engine and drivetrain design factors. For engine design, critical factors in determining output include the shape of the combustion chamber, the quantity of air drawn into the combustion chamber during the induction phase, the type of fuel used, fuel intake design, and so on. Although a complete description of engine design is beyond the scope of this book, an understanding of how engine output is measured and used is important in the study of vehicle performance. The two most commonly used measures of engine output are torque and power. Torque is the work generated by the engine (the twisting moment) and is expressed in foot-pounds (ft-lb). Power is the rate of engine work, expressed in horsepower (hp), and is related to the engine’s torque by the following equation:

2

hp 550

e e e

M n = (2.16)

where

hpe = engine-generated horsepower (1 horsepower equals 550 ft-lb/s), Me = engine torque in ft-lb, and ne = engine speed in crankshaft revolutions per second.

Figure 2.3 presents a torque-power diagram for a typical gasoline-powered engine.

22 Chapter 2 Road Vehicle Performance

Figure 2.3 Example torque-power curves for a gasoline-powered automobile engine.

EXAMPLE 2.4 ENGINE TORQUE AND POWER

It is known that an experimental engine has a torque curve of the form 2 e e eM an bn= −

where Me is engine torque in ft-lb, ne is engine speed in revolutions per second, and a and b are unknown parameters. If the engine develops a maximum torque of 92 ft-lb at 3200 rev/min (revolutions per minute), what is the engine’s maximum power?

SOLUTION

At maximum torque, ne = 53.33 rev/s (3200/60) and

( )

0 2

2 53.33 106.67

e e

e

dM a bn

dn

a b b

= = −

= =

Also, at maximum torque,

( ) ( )

2

2 92 53.33 53.33

e e eM an bn

a b

= −

= −

Using these two equations to solve for the two unknowns (a and b), we find that b = 0.032 and a = 3.450. Using Eq. 2.16 and 2

e e eM an bn= − ,

π

π

− =

− =

2

2

2 ( ) hp

550

2 (3.450 0.032 )

550

e e e e

e e e

an bn n

n n n

The first derivative of the power equation is used to solve for the engine speed at maximum power:

Medium-performance vehicle High-performance vehicle

250

200

150

Torque

Power

T or

qu e

(f t-

lb )

P ow

er (

hp )

100

50

0

250

200

150

100

50

0 0 1000 2000 3000

Engine speed (rev/min)

4000 5000 6000 7000

Torque

Power

P ow

er (

hp )

T or

qu e

(f t-

lb )

700

600

500

400

300

200

100

0 0 1000 2000 3000 4000 5000 6000 7000

700

600

500

400

300

200

100

0

Engine speed (rev/min)

2.6 Available Tractive Effort 23

( )2hp 0 (0.01142) 6.90 0.096

71.88 rev/s

e e e

e

e

d n n

dn

n

= = −

=

so the engine’s maximum power is

( )

π

π

− =

− =

=

2

2

2 (3.450 0.032 ) hp

550

2 3.450(71.88) 0.032(71.88) 71.88

550 67.87 hp

e e e e

n n n

Given the output measures of output of a vehicle’s engine, focus can be directed toward the relationship between engine-generated torque and the tractive effort ultimately delivered to the driving wheels. The tractive effort needed for acceptable vehicle performance (to provide adequate acceleration characteristics) is greater at lower vehicle speeds, and because maximum engine torque is developed at fairly high engine speeds (crankshaft revolutions) for common engine types, such as gasoline- and diesel-powered engines, some form of gear reduction is necessary, as illustrated in Fig. 2.4 (in contrast, electric vehicles provide the same “maximum” torque at all engine speeds, so a gear reduction is not necessary). For gasoline and diesel engines, this gear reduction provides the mechanical advantage necessary for acceptable vehicle acceleration.

Figure 2.4 Tractive effort requirements and tractive effort generated by a typical gasoline-powered vehicle.

24 Chapter 2 Road Vehicle Performance

With gear reductions there are two factors that determine the amount of tractive effort reaching the drive wheels. First, the mechanical efficiency of the drivetrain (the engine and gear reduction devices, including the transmission and differential) must be considered. Typically 5% to 25% of the tractive effort generated by the engine is lost in gear reduction devices, which corresponds to a mechanical efficiency of the drivetrain (η d) of 0.75 to 0.95. Second, the overall gear reduction ratio (ε0), which includes the gear reductions of the transmission and differential, plays a key role in the determination of tractive effort.

By definition, the overall gear reduction ratio refers to the relationship between the revolutions of the engine’s crankshaft and the revolutions of the drive wheels. For example, an overall gear reduction ratio of 4 to 1 (ε0 = 4) means that the engine’s crankshaft turns four revolutions for every one revolution of the drive wheels.

With these terms defined, the engine-generated tractive effort reaching the drive wheels is given as

ε η

= 0e d e

M F

r (2.17)

where

Fe = engine-generated tractive effort reaching the drive wheels in lb, Me = engine torque in ft-lb,

ε0 = overall gear reduction ratio,

dη = mechanical efficiency of the drivetrain, and

r = radius of the drive wheels in ft. It follows that the relationship between vehicle speed and engine speed is

( ) 0

2 1ern i V

π ε

− = (2.18)

where

V = vehicle speed in ft/s, ne = engine speed in crankshaft revolutions per second, i = slippage of the drive axle, generally taken as 2 to 5% (i = 0.02 to 0.05) for

passenger vehicles, and other terms are as defined previously.

To summarize this section, the available tractive effort (F in Eq. 2.2) at any

given speed is the lesser of the maximum tractive effort (Fmax) and the engine- generated tractive effort (Fe).

2.7 Vehicle Acceleration 25

2.7 VEHICLE ACCELERATION

As defined in the previous section, available tractive effort (F) can be used to determine a number of vehicle performance characteristics including vehicle acceleration and top speed. For determining vehicle acceleration, Eq. 2.2 can be applied with an additional term to account for the inertia of the vehicle’s rotating parts that must be overcome during acceleration. This term is referred to as the mass factor (γm) and is introduced in Eq. 2.2 as

γ− = mF R ma (2.19)

where the mass factor is approximated as

2 01.04 0.0025m γ ε= + (2.20)

Two measures of vehicle acceleration are worthy of note: the time to accelerate and the distance to accelerate. For both, the force available to accelerate is Fnet = F − ΣR. The basic relationship between the force available to accelerate, Fnet, the available tractive effort, F (which again is the lesser of Fmax and Fe), and the summation of resistances is illustrated in Fig. 2.5. In this figure, Fnet is the vertical distance between the lesser of the Fmax and the Fe curves and the total resistance curve. So, referring to Fig. 2.5, at speed V′, Fnet will be Fmax − ΣR, and at V″, Fnet will be Fe − ΣR. It follows that when Fnet = 0, the vehicle cannot accelerate and is at its maximum speed for specified conditions (grade, air density, engine torque, and so on). Such was the case for the vehicle described in Example 2.2. When Fnet is greater than zero (the vehicle is traveling at a speed less than its maximum speed), Eq. 2.19 can be written in differential form as

or m net m

net

mdVdV F m dt

dt F

γγ= =

and because Fnet is itself a function of vehicle speed [Fnet = f (V)], integration gives the time to accelerate as

( ) 2

1

V

m

V

dV t m

f V γ=  (2.21)

where V1 is the initial vehicle speed and V2 is the final vehicle speed.

Similarly, it can be shown that the distance to accelerate is

( ) 2

1

V

a m

V

VdV d m

f V γ=  (2.22)

To solve Eqs. 2.21 and 2.22, numerical integration is necessary because the functional forms of these equations do not lend themselves to closed-form solutions. Such numerical integration is straightforward but requires a computer. Consequently, we do not provide an example of solving these equations.

26 Chapter 2 Road Vehicle Performance

Figure 2.5 Relationship among the forces available to accelerate, available tractive effort, and total vehicle resistance.

EXAMPLE 2.5 VEHICLE ACCELERATION

A car is traveling at 10 mi/h on a roadway covered with hard-packed snow. The car has CD = 0.30, Af = 20 ft2, and W = 3000 lb. The wheelbase is 120 inches, and the center of gravity is 20 inches above the roadway surface and 50 inches behind the front axle. The air density is 0.002045 slugs/ft3. The car’s engine is producing 95 ft-lb of torque and is in a gear that gives an overall gear reduction ratio of 4.5 to 1, the wheel radius is 14 inches, and the mechanical efficiency of the drivetrain is 80%. If the driver needs to accelerate quickly to avoid an accident, what would the acceleration be if the car is (a) front-wheel drive and (b) rear-wheel drive?

SOLUTION We begin by computing the resistances, tractive effort generated by the engine, and mass factor because all of these factors will be the same for both front- and rear-wheel drive. The aerodynamic resistance is (from Eq. 2.3)

( ) ( ) ( )

2

2

2 0.002045

0.3 20 10 5280 / 3600 2

1.32 lb

a D fR C A V

ρ =

= ×

=

The rolling resistance is (from Eq. 2.6)

=

× = + ×   

=

10 5280 3600 0.01 1 3000

147

32.99 lb

rl rlR f W

2.7 Vehicle Acceleration 27

The engine-generated tractive effort is (from Eq. 2.17)

( ) ( )

0

95 4.5 0.8

14 12

293.14 lb

e d e

M F

r

ε η =

=

=

The mass factor is (from Eq. 2.20)

( )

2 0

2

1.04 0.0025

1.04 0.0025 4.5

1.091

m

γ ε= +

= +

=

Recall that, to determine acceleration, we need the resistances (already computed) and the available tractive effort, F, which is the lesser of Fe or Fmax. For the case of the front-wheel–drive car, Eq. 2.15 can be applied to determine Fmax (with the coefficient of road adhesion μ = 0.25 for hard-packed snow from Table 2.4):

( )

μ μ +

= +

 × × + = + ×

=

max ( ) /

1 /

0.25 3000 70 0.011(20) 120

1 (0.25 20) 120

421.32 lb

r rlW l f h L F

h L

Thus for a front-wheel–drive car F = 293.14 lb (the lesser of Fe = 293.14 and Fmax = 421.32) and the acceleration is (from Eq. 2.19)

( ) 2293.14 34.31

2.546 ft/s 1.091 3000 32.2

m

m

F R ma

F R a

m

γ

γ

− =

− − = = =





For the case of the rear-wheel–drive car, Eq. 2.14 can be applied to determine Fmax :

( ) × × − = = − ×

0.2 3000 50 0.011(20) 120 257.48 lb

1 (0.2 20) 120 F

Thus for a rear-wheel–drive car F = 257.48 lb (the lesser of 293.14 and 257.48) and the acceleration is (from Eq. 2.19)

( ) 2257.48 34.31

2.196 ft/s 1.091 3000 32.2m

F R a

mγ − −

= = =

EXAMPLE 2.6 ENGINE-TORQUE AND VEHICLE ACCELERATION

A front-wheel–drive car is in gear with a gear-reduction ratio 8 to 1 and is traveling at 25 mi/h. The engine torque of the car is given by the equation Me = 10ne – 0.06ne

2. The car has a frontal area of 20 ft2, CD of 0.30 and is traveling at sea level (59 degrees F) on

28 Chapter 2 Road Vehicle Performance

a level road. The wheelbase is 120 inches and the center of gravity is 40 inches behind the front axle and 30 inches above the road surface and the car weighs 2800 pounds. If the car is on a road that is wet with poor pavement, what is the maximum acceleration from 25 mi/h (driveline efficiency is 90%, slippage of the drive axle is 2%, wheel radius is 15 inches)?

SOLUTION To begin, the engine speed must be computed so that the torque equation provided in the problem statement can be applied to determine the engine-generated tractive effort. This is done by applying Eq. 2.18 with i = 0.02, 0 = 8, r = 15/12 ft, and V = 20 mi/h:

( )

( ) 0

2 1 2(3.141)(15 /12) (1 0.02) 25 5280 / 3600

8

25 5280 / 3600 8 38.22 revolutions per second

(2)(3.141)(15 /12)(1 0.02)

e e

e

rn i n V

n

π ε

− − = = × =

× = =

−

Substituting this engine speed into the torque equation provided in the problem statement gives

( ) ( )2210 – 0.06 10 38.22 – 0.06 38.22 294.55 ft-lbe e eM n n= = =

The engine-generated tractive effort is (from Eq. 2.17) with dη = 0.9

( ) ( )0 294.55 8 0.9 1696.63 lb

15 12 e d

e M

F r

ε η = = =

As in Example 2.5, the available tractive effort for maximum acceleration, F, is the lesser of Fe or Fmax. Fmax for a front-wheel–drive car is computed using Eq. 2.15. With given values of lr = 80 inches, h = 30 inches, L = 120 inches, W = 2800 lb, μ = 0.6 (from Table 2.4 with poor, wet pavement), and frl from Eq. 2.5, we find

25 5280 3600 0.01 1 0.01 1 0.0125

147 147rl V

f ×   = + = + =   

   

Application of Eq. 2.15 yields (with the coefficient of road adhesion μ = 0.60 for poor, wet pavement from Table 2.4)

( )

( ) 1 /

0.60 2800 80 0.0125(30) 120

1 (0.60 30) 120

978.48 lb

r rl max

W l f h L F

h L

μ μ +

= +

 × × + = + ×

=

Thus F = 978.48 lb (the lesser of 978.48 and 1696.63). With this, acceleration is determined by applying Eq. 2.19. For input into Eq. 2.19, the rolling resistance (from Eq. 2.6) is

2.8 Fuel Efficiency 29

= = × =

0.0125 2800

35 lb

rl rlR f W

The aerodynamic resistance with = 0.002378 slugs/ft3 (from Table 2.1), CD = 0.30, Af = 20 ft2 and V = 25 mi/h (from Eq. 2.3) is

( )( )( )

2

2

2 0.002378

0.3 20 25 5280 / 3600 2

9.63 lb

a D fR C A V

ρ=

= ×

=

The mass factor (from Eq. 2.20) is

( )

2 0

2

1.04 0.0025

1.04 0.0025 8

1.2

m

γ ε= +

= +

=

Thus, application of Eq. 2.19 gives the maximum acceleration as

( ) 2978.48 35 9.63

8.95 ft/s 1.2 2800 32.2

m

m

F R ma

F R a

m

γ

γ

− =

− − − = = =





The values given for the coefficient of road adhesion (μ) in the previous

examples are drawn from Table 2.4. However, in determining acceleration (as well as braking and cornering, as will be shown later in this book) two points are worthy of note. First, it is possible for the coefficient of road adhesion to exceed 1.0. This is because a micro-interaction at the tire–pavement interface results in a “cog-type” effect that, for some high-performance tires that use softer compounds, can increase μ to above 1.0. This explains why many race cars, particularly drag-racing cars, have initial acceleration rates well in excess of 1 g (32.2 ft/s2). Second, at high speed, vehicle aerodynamics can create downward forces that effectively increase W in the preceding equations, and this facilitates greater acceleration. Drag-racing cars and some open-wheel race cars (such as Formula One–style cars) are examples of aerodynamic designs that use air deflectors designed to generate significant downward forces to enhance acceleration, braking, and cornering.

2.8 FUEL EFFICIENCY

Given the factors discussed in the preceding sections of this chapter, the elements that determine a vehicle’s fuel efficiency are clear. One of the most critical

30 Chapter 2 Road Vehicle Performance

determinants relates to engine design (how the engine-generated tractive effort is produced). Engine designs that increase the quantity of air entering the combustion chamber, improve fuel delivery to the combustion chamber, and decrease internal engine friction lead to improved fuel efficiency. Improvements in other mechanical components, such as decreasing slippage and improving the mechanical efficiency of the transmission and driveshaft, also increase the overall fuel efficiency.

In terms of resistance-reducing options, decreasing overall vehicle weight (W) will lower grade and rolling resistances, thus reducing fuel consumption (all other factors held constant). Similarly, aerodynamic improvements such as a lower drag coefficient (CD) and a reduced frontal area (Af) can produce significant fuel savings. Finally, improved tire designs with lower rolling resistance can improve overall fuel efficiency. However, there can sometimes be a trade-off between improved efficiency and safety. For example, low rolling resistance tires may not be able to the same high coefficient of road adhesion values that high-performance race tires can. Possible safety-efficiency trade-offs must be given careful consideration.

2.9 PRINCIPLES OF BRAKING

In highway design and traffic analysis, the braking characteristics of road vehicles are arguably the single most important aspect of vehicle performance. The braking behavior of road vehicles is critical in the determination of stopping sight distance, roadway surface design, and accident avoidance and mitigation systems. Moreover, ongoing advances in braking technology make it essential that transportation engineers have a basic comprehension of the underlying principles involved.

2.9.1 Braking Forces To begin the discussion of braking principles, consider the force and moment- generating diagram in Fig. 2.6. During vehicle braking there is a load transfer from the rear to the front axle. To illustrate this, expressions for the normal loads on the front and rear axles can be written by summing the moments about roadway surface–tire contact points A and B (as was done in deriving Eqs. 2.14 and 2.15, with cos θg assumed equal to 1 because of the small grades encountered in highway applications):

( )θ = + − ±  1

sinf r a gW Wl h ma R W L

(2.23)

and

2.9 Principles of Braking 31

Figure 2.6 Forces acting on a vehicle during braking, with drivetrain resistance ignored.

Ra = aerodynamic resistance in lb, gθ = angle of the grade in degrees,

Rrlf = rolling resistance of the front tires in lb, m = vehicle mass in slugs,

Rrlr = rolling resistance of the rear tires in lb, a = acceleration in ft/s2, Fbf = braking force on the front tires in lb, L = length of wheelbase,

Fbr = braking force on the rear tires in lb, h = height of the center of gravity above the roadway surface, W = total vehicle weight in lb,

Wf = weight of the vehicle on the front axle in lb, lf = distance from the front axle to the center of gravity, and Wr = weight of the vehicle on the rear axle in lb,

lr = distance from the rear axle to the center

of gravity.

θ = − − ±  1

( sin )r f a gW Wl h ma R W L

(2.24)

where, in this case, the contribution of grade resistance (W sin θg) is negative for uphill grades and positive for downhill grades.

From the summation of forces along the vehicle’s longitudinal axis,

sinb rl a gF + f W = ma R W− ± (2.25)

with the rolling resistance equal to the coefficient of rolling resistance multiplied by the vehicle weight (from Eq. 2.6, Rrl = frlW) and Fb = Fbf + Fbr. Substituting Eq. 2.25 into Eqs. 2.23 and 2.24 gives

( ) = + +  1

f r b rlW Wl h F f W L

(2.26)

and

( ) = − +  1

r f b rlW Wl h F f W L

(2.27)

32 Chapter 2 Road Vehicle Performance

Because the maximum vehicle braking force (Fb max) is equal to the coefficient of road adhesion (μ), multiplied by the vehicle weights normal to the roadway surface,

[ ]

μ μ μ

=

= + +( )

bf f

r rl

F W

W l h f

L

(2.28)

and

( )

μ μ μ

=

 = − + 

br r

f rl

F W

W l h f

L

(2.29)

To develop maximum braking forces, the tires should be at the point of an impending slide. If the tires begin to slide (the brakes lock), a significant reduction in road adhesion results. An indication of the extent of the reduction in the coefficient of road adhesion as the result of tire slide, under various pavement and weather conditions, is presented in Table 2.5. It is clear from this table that the braking forces decline dramatically when the wheels are locked (resulting in tire slide). Avoiding this locked condition is the function of antilock braking systems in cars. Such systems are discussed later in this chapter.

2.9.2 Braking Force Ratio and Efficiency On a given roadway surface, the maximum attainable vehicle deceleration (using the vehicle’s braking system) is equal to μg, where μ is the coefficient of road adhesion and g is the gravitational constant (32.2 ft/s2). To achieve this maximum vehicle deceleration, vehicle braking systems must correctly distribute braking forces between the vehicle’s front and rear brakes. This is typically done by allocation of hydraulic pressures within the braking system. This front-rear proportioning of braking forces (within the vehicle’s braking system) will be optimal (achieving a deceleration rate equal to μg) when it is in exactly the same proportion as the ratio of the maximum braking forces on the front and rear axles (Fbf max / Fbr max). Thus maximum braking forces (with the tires at the point of impending slide) will be developed when the brake force ratio (front force over rear force) is

Table 2.5 Typical Values of Coefficients of Road Adhesion (Maximum and Slide)

Coefficient of road adhesion

*In some instances, the coefficient of road adhesion values can exceed 1.0. See discussion near the end of Section 2.6.1. Source: S. G. Shadle, L. H. Emery, and H. K. Brewer, “Vehicle Braking, Stability, and Control,” SAE Transactions, vol. 92, paper 830562, 1983.

Pavement Maximum Slide

Good, dry 1.00* 0.80 Good, wet 0.90 0.60 Poor, dry 0.80 0.55 Poor, wet 0.60 0.30 Packed snow or ice 0.25 0.10

2.9 Principles of Braking 33

( ) ( )/

r rl f r

f rl

l h f BFR

l h f

μ μ

+ + =

− + (2.30)

where

BFRf/r max = the brake force ratio, allocated by the vehicle’s braking system, that results in maximum (optimal) braking forces, and

other terms are as defined previously.

It follows that the percentage of braking force that the braking system should allocate to the front axle (PBFf) for maximum braking is

= − + /

100 100

1f f r max

PBF BFR

(2.31)

and the percentage of braking force that the braking system should allocate to the rear axle (PBFr) for maximum braking is

= + /

100 1r

f r max

PBF BFR

(2.32)

EXAMPLE 2.7 BRAKE-FORCE PROPORTIONING

A car has a wheelbase of 100 inches and a center of gravity that is 40 inches behind the front axle at a height of 24 inches. If the car is traveling at 80 mi/h on a road with poor pavement that is wet, determine the percentages of braking force that should be allocated to the front and rear brakes (by the vehicle’s braking system) to ensure that maximum braking forces are developed.

SOLUTION The coefficient of rolling resistance is

× = + =   

80 5280 / 3600 0.01 1 0.018

147rlf

and μ = 0.6 from Table 2.5 (maximum because we want the tires to be at the point of impending slide). Applying Eq. 2.30 gives

( ) ( )

( ) ( )

μ μ

+ + =

− +

+ + =

− +

=

/

60 24 0.6 0.018

40 24 0.6 0.018

2.973

r rl f r max

f rl

l h f BFR

l h f

Using Eq. 2.31, the percentage of the force allocated to the front brakes should be

34 Chapter 2 Road Vehicle Performance

= − +

= − +

=

/

100 100

1

100 100

1 2.973 74.83 %

f f r max

PBF BFR

and using Eq. 2.32 (or simply 100 − PBFf), the percentage of the force allocated to the rear brakes should be

= +

= +

=

/

100 1

100 1 2.973 25.17 %

r f r max

PBF BFR

It is clear from Eq. 2.30 that the design of a vehicle’s braking system is not an

easy task because the optimal brake force proportioning changes with both vehicle and road conditions. For example, the addition of vehicle cargo and/or passengers will change not only the weight of the vehicle (which affects frl in Eq. 2.30) but also the distribution of the weight, shifting the height of the center of gravity and its location along the vehicle’s longitudinal axis, and this will change the optimal brake force proportioning (BFRf/r max). This is particularly problematic for trucks because of the large weight and center of gravity differences between loaded and unloaded conditions. Similarly, changes in road conditions produce different coefficients of adhesion, again changing optimal brake force proportioning. As a result of the uncertainties in weight and road conditions, vehicle designers often choose a compromise value of brake force proportioning that, on average, provides good braking but is rarely, if ever, optimal.

It is important to note that studies have indicated that if wheel lockup is to occur, it is preferable to have the front wheels lock first because having the rear wheels lock first can result in uncontrollable vehicle spin. Front-wheel lockup results in loss of steering control, but the vehicle will at least continue to brake in a straight line. Technological advancements in braking systems since the late 1970s have resulted in vehicles that are increasingly capable of proportioning brake forces in a manner that is closer to optimal and avoids the dangerous rear- wheel–first lockup due to front-wheel under-braking.

Because true optimal brake force proportioning is seldom achieved in standard non-antilock braking systems, it is useful to define a braking-efficiency term that reflects the degree to which the braking system is operating below optimal. Simply stated, braking efficiency is defined as the ratio of the maximum rate of deceleration, expressed in g’s (gmax), achievable prior to any wheel lockup to the coefficient of road adhesion:

max b

g = η

μ (2.33)

2.9 Principles of Braking 35

where

ηb = braking efficiency,

gmax = maximum deceleration in g units (with the absolute maximum = μ), and

μ = coefficient of road adhesion.

2.9.3 Antilock Braking Systems Many modern cars have braking systems designed to prevent the wheels from locking during braking applications (antilock braking systems). In theory, antilock braking systems serve two purposes. First, they prevent the coefficient of road adhesion from dropping to slide values (see Table 2.5). Second, they have the potential to raise the braking efficiency to 100%. In practice, designing an antilock braking system that avoids slide coefficients of adhesion and achieves 100% braking efficiency (ηb = 1.0) is a difficult task. This is because most antilock braking system technologies detect which wheels have locked and release them momentarily before reapplying the brake on the locking wheel. The wheel lock detection speed, speed of brake force reallocation, and braking system design (the amount of braking forces that can be accommodated by the vehicle’s front and rear brake discs and calipers) all impact the overall effectiveness of the antilock braking system. Early antilock braking systems often fell short of achieving 100% braking efficiency, and in many cases, an expert driver operating a non-antilock braking car could modulate the brakes to achieve shorter stopping distances than cars equipped with antilock brakes. However, advances in antilock braking system technology continue to bring us closer to 100% braking efficiency.

2.9.4 Theoretical Stopping Distance With a basic understanding of brake force proportioning and the resulting braking efficiency, attention can now be directed toward developing expressions for minimum stopping distances. By inspection of Fig. 2.6, it can be seen that the relationship among stopping distance, braking force, vehicle mass, and vehicle speed is

γ  +

=     

=



b

b

F R a ds ds

m

V dV

(2.34)

where

γb = mass factor accounting for moments of inertia during braking, which is given the value of 1.04 for automobiles [Wong, 2008], and

other terms are as defined in Fig. 2.6.

36 Chapter 2 Road Vehicle Performance

Integrating to determine stopping distance (S) gives

1

2

V

b bV

V dV S m

F R γ=

+  (2.35)

Substituting in the resistances (see Fig. 2.6), we obtain

1

2

sin

V

b b a rl gV

V dV S m

F R f W W γ

θ =

+ + ± (2.36)

where

V1 = initial vehicle speed in ft/s,

V2 = final vehicle speed in ft/s,

frlW = rolling resistance,

W sin θg = grade resistance (positive for uphill slopes and negative for downhill slopes), and

other terms are as defined previously. To simplify notation, let

2a D fK C A ρ= (2.37)

so that Eq. 2.3 is

2 a aR K V= (2.38)

Continuing, assume that the effect of speed on the coefficient of rolling resistance, frl, is constant and can be approximated by using the average of initial (V1) and final (V2) speeds in Eq. 2.5 [V = (V1 + V2) / 2]. With this assumption (which introduces only a very small amount of error), and letting m = W/g and Fb = μW, integration of Eq. 2.36 gives

2

1 2

2

sin ln

2 sin a rl gb

a a rl g

W K V f W WW S

gK W K V f W W

μ θγ μ θ

 + + ± =  

+ + ±   (2.39)

If the vehicle is assumed to stop (V2 = 0),

2

1ln 1 2 sin

b a

a rl g

W K V S

gK W f W W

γ μ θ

  = + 

+ ±   (2.40)

With braking efficiency considered, the actual braking force is

b bF Wη μ= (2.41)

2.9 Principles of Braking 37

Therefore, by substitution into Eq. 2.40, the theoretical stopping distance is

2

1ln 1 2 sin

b a

a b rl g

W K V S

gK W f W W

γ η μ θ

  = + 

+ ±   (2.42)

Similarly, Eq. 2.39 can be written to include braking efficiency. Finally, if aerodynamic resistance is ignored (due to its comparatively small contribution to braking), integration of Eq. 2.35 gives the theoretical stopping distance as

( )

( ) 2 2

1 2

2 sin

b

b rl g

V V S

g f

γ

η μ θ

− =

+ ± (2.43)

EXAMPLE 2.8 THEORETICAL MINIMUM STOPPING DISTANCE

A new experimental 2500-lb car, with CD = 0.25 and Af = 18 ft2, is traveling at 90 mi/h down a 10% grade. The coefficient of road adhesion is 0.7 and the air density is 0.0024 slugs/ft3. The car has an advanced antilock braking system that gives it a braking efficiency of 100%. Determine the theoretical minimum stopping distance for the case where aerodynamic resistance is considered and the case where aerodynamic resistance is ignored.

SOLUTION

With aerodynamic resistance considered, Eq. 2.42 can be applied with γb = 1.04, gθ = 5.71°, and

( )( )

90 5280 3600 0 2

0.01 1 0.0145 147

0.0024 0.25 18 0.0054

2

rl

a

f

K

 × +    

  = + =      

= =

Then

21.04(2500) 0.0054(90 5280 3600) ln 1

2(32.2)(0.0054) (1.0)(0.7)(2500) (0.0145)(2500) 2500sin(5.71 )

444.07 ft

S  ×= + + − °  

=

With aerodynamic resistance excluded, Eq. 2.43 is used:

( ) ( ) ( )( )

2 1.04 90 5280 / 3600

457.53 ft 2 32.2 0.7 0.0145 sin 5.71

S

× = =

+ − °

38 Chapter 2 Road Vehicle Performance

EXAMPLE 2.9 EFFECTS OF GRADE ON THEORETICAL MINIMUM STOPPING DISTANCE A car is traveling at 80 mi/h and has a braking efficiency of 80%. The brakes are applied to miss an object that is 150 ft from the point of brake application, and the coefficient of road adhesion is 0.85. Ignoring aerodynamic resistance and assuming the theoretical minimum stopping distance, estimate how fast the car will be going when it strikes the object if (a) the surface is level and (b) the surface is on a 5% upgrade.

SOLUTION In both cases, rolling resistance is approximated as

 × +    

  = + = +      

2

2

80 5280 / 3600 2

0.01 1 0.014 0.000034 147rl

V

f V

Applying Eq. 2.43 for the level grade with γb = 1.04, θg = 0°,

( ) ( )

2 2 1 2

2 sin

b

b rl g

V V S

g f

γ

η μ θ

− =

+ ±

( ) [ ]

2 2 2

2

1.04 (80 5280 3600) 150

2(32.2) 0.8(0.85) (0.014 0.000034 ) 0

V

V

× − =

+ + ±

2 85.40 ft/s or 58.23 mi/hV =

On a 5% grade with θg = 2.86°,

( ) [ ]

2 2 2

2

1.04 (80 5280 3600) 150

2(32.2) 0.8(0.85) (0.014 0.000034 ) 0.05

V

V

× − =

+ + +

2 82.64 ft/s or 56.35 mi/hV =

EXAMPLE 2.10 THEORETICAL MINIMUM STOPPING DISTANCE WITH AND WITHOUT ANTILOCK BRAKES A car is traveling up a 3% grade on a road that has good, wet pavement. The engine is running at 2500 revolutions per minute. The radius of the wheels is 15 inches, the driveline slippage is 3%, and the overall gear reduction ratio is 2.5 to 1. A deer jumps out onto the road and the driver applies the brakes 291 ft from it. The driver hits the deer at a speed of 20mi/h. If the driver did not have antilock brakes, and the wheels were locked the entire distance, would a deer-impact speed of 20 mi/h be possible?

SOLUTION The speed of the car must first be determined by applying Eq. 2.18 with i = 0.03,

0 = 2.5, r = 15/12 ft, and ne = 2500/60 revolutions per second:

2.9 Principles of Braking 39

( ) 0

2 1

2(3.141)(15 /12)(2500 / 60)(1 0.03)

2.5 126.92 ft/s (86.52 mi/h)

ern i V

π ε

− =

−=

=

Next, using Eq. 2.5 the coefficient of rolling resistance is computed using the average speed, (V1 + V2)/2 as an approximation of V (with a V2 of 20 mi/h):

+    = + = + =  

      

126.92 20(5280 3600) 20.01 1 0.01 1 0.01532

147 147rl V

f

If the car did not have an antilock braking system and the brakes were locked, the slide value on good, wet pavement is a coefficient of road adhesion of 0.6 (μ = 0.6) from Table 2.5. Applying Eq. 2.43 with μ = 0.6, a 3% grade (so sinθg ≈ 0.03), and γb = 1.04,

( )

( )( ) ( )

2 2 1 2

22

2 ( sin )

1.04 126.92 20(5280 3600) 291

2 32.2 (0.6) 0.01532 0.03

or 1.33

b

b rl g

b

b

V V S

g f

γ

η μ θ

η η

− =

+ ±

− =

+ +   =

Because a braking efficiency of 1.33 is not possible, the driver would hit the deer at a higher speed if the wheels were locked the entire distance. Note that to achieve a deer- impact speed of 20 mi/h or less, bη μ in Eq. 2.43 must be 0.8 (1.33× 0.6) or greater. The maximum coefficient of road adhesion for good, wet pavements is 0.9 (from Table 2.5). So, to achieve a deer-impact speed of 20 mi/h or less (with μ = 0.9),

0.8 so,

0.8

0.8 0.9

0.89

b

b

η μ η μ

= = = =

Thus a braking efficiency of at least 89% is needed (if the vehicle has a functional antilock braking system) in order to achieve a deer-impact speed of 20 mi/h or less.

2.9.5 Practical Stopping Distance As mentioned earlier, one of the most critical concerns in the design of a highway is the provision of adequate driver sight distance to permit a safe stop. The theoretical assessment of vehicle stopping distance presented in the previous section provided the principles of braking for an individual vehicle under specified roadway surface conditions. However, highway engineers face a more complex problem because they must design for a variety of driver skill levels (which can affect whether or not the brakes lock and reduce the coefficient of road adhesion to slide values), vehicle types (with varying aerodynamics, weight

40 Chapter 2 Road Vehicle Performance

distributions, and brake efficiencies), and weather conditions (which change the roadway’s coefficient of adhesion). As a result of the wide variability inherent in the determination of braking distance, an equation is required that provides an estimate of typical observed braking distances and is more simplistic and usable than Eq. 2.42.

The basic physics equation on rectilinear motion, assuming constant deceleration, is chosen as the basis of a practical equation for stopping distance:

2 2 2 1 2V V ad= + (2.44)

where

V2 = final vehicle speed in ft/s, V1 = initial vehicle speed in ft/s,

a = acceleration (negative for deceleration) in ft/s2, and d = deceleration distance (practical stopping distance) in ft.

Rearranging Eq. 2.44 and assuming a is negative for deceleration gives

2 2

1 2

2 V V

d a −

= (2.45)

If V2 = 0 (the vehicle comes to a complete stop), the practical stopping distance equation is

2

1

2 V

d a

= (2.46)

To make this equation generally applicable for design purposes, a deceleration rate, a, must be chosen that is representative of appropriately conservative braking behavior. AASHTO [2018] recommends a deceleration rate of 11.2 ft/s2. Empirical studies [Fambro et al., 1997] have shown that approximately 90% of drivers decelerate at rates greater than this, and that this deceleration rate is well within a driver’s capability to maintain steering control during a braking maneuver on wet surfaces. Additionally, empirical studies [Fambro et al., 1997] have confirmed that most vehicle braking systems and tire-pavement friction levels are capable of supporting this deceleration rate, even under wet conditions. To account for the effect of grade, Eq. 2.46 is modified as follows:

2

1

2

V d

a g G

g

=    ±     

(2.47)

2.9 Principles of Braking 41

where

g = gravitational constant, 32.2 ft/s2, G = roadway grade (+ for uphill, − for downhill) in percent/100, and

other terms are as defined previously. It is important to note that Eq. 2.47 is consistent with Eq. 2.43 (the theoretical stopping distance ignoring aerodynamic resistance). Rewriting Eq. 2.43 with the assumption that the vehicle comes to a stop (V2 = 0), that sin θg = tan θg = G (for small grades), and that γb and frl can be ignored due to their small and essentially offsetting effects, we have

( ) 2

1

2 b

V S

g Gη μ =

± (2.48)

Recall that ηbμ = gmax (Eq. 2.33). However, rather than determining the maximum deceleration rate (in g’s) for a specific vehicle braking efficiency and specific coefficient of road adhesion, the AASHTO-recommended maximum deceleration rate (again, an appropriately conservative value for the overall driver and vehicle population) is used. Thus, a maximum deceleration of 0.35 g’s (11.2/32.2) is used for Eq. 2.47.

The recommended deceleration rate as determined empirically already accounts for the effects of aerodynamic resistance, braking efficiency, coefficient of road adhesion, and inertia during braking (the braking mass factor). This value reflects current vehicle technologies and driving behavior. It is important to recognize that as vehicle braking technology and other vehicle characteristics change, as well as possibly driver behavior, the recommended value of a should be reviewed to determine if it is still applicable for highway design purposes. The relationship between changing vehicle characteristics and changing highway design guidelines is one that must always be kept in the design engineer’s mind.

EXAMPLE 2.11 BRAKING EFFICIENCY AND STOPPING DISTANCE

A car [W = 2200 lb, CD = 0.25, Af = 21.5 ft2] has an antilock braking system that gives it a braking efficiency of 100%. The car’s stopping distance is tested on a level roadway with poor, wet pavement (with tires at the point of impending skid), and ρ = 0.00238 slugs/ft3. How inaccurate will the stopping distance predicted by the practical- stopping-distance equation be compared with the theoretical stopping distance, assuming the car is initially traveling at 60 mi/h? How inaccurate will the practical- stopping-distance equation be if the same car has a braking efficiency of 85%?

SOLUTION First, to calculate the theoretical minimum stopping distance, Eq. 2.42 is applied with γb = 1.04, θg = 0°, μ = 0.60 (maximum for poor, wet pavement, from Table 2.5), and

42 Chapter 2 Road Vehicle Performance

( )( )

60 5280 3600 0 2

0.01 1 0.013 147

0.00238 0.25 21.5 0.0064

2

rl

a

f

K

 × +    

  = + =      

= =

Thus, from Eq. 2.42,

21.04(2200) 0.0064(60 5280 3600) ln 1 200.35 ft

2(32.2)(0.0064) (1.0)(0.60)(2200) (0.013)(2200) 0 S

 ×= + = + ±  

For the same conditions but with a vehicle braking efficiency of 85%, Eq. 2.42 gives

 ×= + + ±   =

21.04(2200) 0.0064(60 5280 3600) ln 1

2(32.2)(0.0064) (0.85)(0.60)(2200) (0.013)(2200) 0

234.11ft

S

Now applying Eq. 2.46 (since G = 0) for the practical stopping distance, we find

( ) ( )

× = =

2 60 5280 / 3600

345.71 ft 2 11.2

d

In the first case, the error is 145.36 ft. In the case of 85% braking efficiency, the error is 111.60 ft. Rearranging Eq. 2.46 to solve for a, we find that stopping distances of 200.35 ft and 234.11 ft correspond to deceleration rates of 19.33 ft/s2 and 16.54 ft/s2, respectively. Studies [Fambro et al., 1997] have shown that most drivers decelerate at rates of 18.4 ft/s2 or greater in emergency stopping situations. Thus, this range of theoretical values is consistent with observed distances for situations in which minimum stopping distances are being attempted. Comparing these theoretical values to the AASHTO-recommended deceleration rate of 11.2 ft/s2, it is readily apparent that a considerable level of conservatism is built into the deceleration rate for practical stopping distance.

2.9.6 Distance Traveled during Driver Perception/Reaction Until now the focus has been directed toward the distance required to stop the vehicle from the point of brake application. However, in providing sufficient sight distance for a driver to stop safely, it is also necessary to consider the distance traveled during the time the driver is perceiving and reacting to the need to stop. The distance traveled during perception/reaction (dr) is given by

1r rd V t= × (2.49)

where V1 = initial vehicle speed in ft/s, and tr = time required to perceive and react to the need to stop, in s.

2.9 Principles of Braking 43

The perception/reaction time of a driver is a function of a number of factors, including the driver’s age, physical condition, and emotional state, as well as the complexity of the situation and the strength of the stimuli requiring a stopping action. For highway design, a conservative perception/reaction time has been determined to be 2.5 seconds [AASHTO, 2018]. For comparison, average drivers have perception/reaction times of approximately 1.0 to 1.5 seconds.

Thus, the total required stopping distance is a combination of the braking distance, either theoretical (Eq. 2.42 or 2.43) or practical (Eq. 2.47), and the distance traveled during perception/reaction (Eq. 2.49), as shown in Eq. 2.50:

s rd d d= + (2.50)

where

ds = total stopping distance (including perception/reaction) in ft, d = distance traveled during braking in ft, and dr = distance traveled during perception/reaction in ft.

The combination of practical stopping distance and the distance traveled

during perception/reaction is a primary consideration in highway design, as will be discussed in detail in Chapter 3.

EXAMPLE 2.12 PRACTICAL STOPPING DISTANCE AND PERCEPTION/REACTION TIMES Two drivers each have a reaction time of 2.5 seconds. One is obeying a 55-mi/h speed limit and the other is traveling illegally at 70 mi/h. How much distance will each of the drivers cover while perceiving/reacting to the need to stop, and what will the total stopping distance be for each driver (using practical stopping distance and assuming G = −2.5%)?

SOLUTION The distances traveled by each driver during perception/reaction will be calculated

first, using Eq. 2.49. For the driver traveling at 55 mi/h,

( )( )1 55 5280 / 3600 2.5 201.67 ftr rd V t= × = × =

For the driver traveling at 70 mi/h,

( )( )1 70 5280 / 3600 2.5 256.67 ftrd V tr = × = × =

Therefore, driving at 70 mi/h increases the distance traveled during perception/reaction by 55.0 ft.

Next, the distance traveled during braking will be calculated for each driver, using the equation for practical stopping distance (Eq. 2.47). For the driver traveling at 55 mi/h,

2(55 5280 3600) 6507.11 312.99 ft

20.7911.2 2(32.2) 0.025

32.2

d ×= = =    −     

44 Chapter 2 Road Vehicle Performance

For the driver traveling at 70 mi/h,

2(70 5280 3600) 10540.44 507.00 ft

20.7911.2 2(32.2) 0.025

32.2

d ×= = =    −     

The total stopping distance for each driver is now calculated with Eq. 2.50. For the driver traveling at 55 mi/h,

312.99 201.67 514.66 fts rd d d= + = + =

For the driver traveling at 70 mi/h,

507.00 256.67 763.67 fts rd d d= + = + =

Therefore, driving at 70 mi/h increases the total stopping distance by a very substantial 249.01 ft.

EXAMPLE 2.13 PRACTICAL STOPPING DISTANCES FOR HUMAN-DRIVEN VEHICLE AND AUTONOMOUS VEHICLE Two vehicles, one operated by a human and the other computer controlled, are traveling on a level highway. If the autonomous vehicle is traveling at 70 mi/h, and its perception/reaction time is considered negligible, what is the maximum speed the human-controlled vehicle can travel and have a stopping distance that does not exceed that of the autonomous vehicle?

SOLUTION While autonomous vehicle perception/reaction time will not be zero in practice, it will

be very small. So for the purposes of this example, we will assume tr to be negligible and thus set it to zero. Correspondingly, dr for the autonomous vehicle will be zero and the total stopping distance will just be equal to the distance traveled during braking. Using the practical stopping distance equation (Eq. 2.47) to calculate the braking distance gives,

2(70 5280 3600) 470.56 ft

2(11.2) d

×= =

Substituting Eqs. 2.47 and 2.49 into Eq. 2.50 yields (for level grade)

2

= + 2s r V

d V t a

×

Solving the resulting quadratic for V, with coefficient a = 1/(2×11.2), coefficient b = 2.5, and constant c = −470.56 gives two solutions. One solution is negative, thus disregarded. The other solution is 78.4 ft/s (53.5 mi/h), which is reasonable. Thus, the human-controlled vehicle must travel 16.5 mi/h (70 – 53.5) slower than the autonomous vehicle to attain the same stopping distance.

This problem assumed the same currently recommended deceleration rate of 11.2 ft/s2 (0.35g) for the autonomous vehicle. This value is rooted in empirical studies of human behavior, as well as roadway surface condition considerations (such as wet pavement), but corresponds to a conservative value to account for the variability in driver, vehicle, and roadway characteristics. By eliminating the human driver

2.10 Practice Problems 45

variability component with autonomous vehicles, it can be argued that higher deceleration rates could be applied, thus reducing the stopping distances used for roadway design purposes (which will be discussed in Chapter 3). While deceleration rates under good pavement and tire conditions for a modern automobile can approach 1g, the design value adopted in practice will still have to consider poor pavement conditions and human, tolerances for rapid deceleration. Vehicle types with lower braking efficiencies, such as commercial trucks, may impose additional practical constraints to the design deceleration rate used for roadway design.

2.10 PRACTICE PROBLEMS

PRACTICE PROBLEM 2.1

ENGINE-GENERATED TRACTIVE EFFORT, AERODYNAMIC AND ROLLING RESISTANCE, ACCELERATION A car weighs 2200 lb and is traveling 100 mi/h on a race track that is on a 3% upgrade. The car is preparing to pass a slower car and its torque/engine speed curve is given by the equation,

28 0.05e e eM n n= −

with Me in ft-lb and ne in revolutions per second. Drivetrain efficiency is 90%, drive axle slippage is 2%, wheel radius is 15 inches, frontal area is 22 ft2, drag coefficient is 0.35, and air density is 0.0022 slugs/ft3. If the car is in a gear that produces maximum torque, what would the car’s maximum acceleration be?

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

This problem combines a number of concepts covered in this chapter. We want to determine the acceleration, so eventually we want to apply Eq. 2.19,

mF R maγ− = ,

and solve for a. In this equation, all necessary terms for determining a can be readily calculated with the information provided except for the available tractive effort F. It is known that the available tractive effort is the lesser of the engine-generated tractive effort Fe, and the maximum tractive effort Fmax. Because the car is traveling at 100 mi/h, the engine-generated tractive effort will almost certainly be less than the maximum tractive effort since at this speed the amount of engine-generated force required to have Fe exceed Fmax would require an unrealistically powerful engine. This can be seen in Fig. 2.5 where the Fmax curve is well above the Fe curves at higher speeds. Given this, we know from Eq. 2.17 that,

0e d e

M F F

r

ε η = =

The problem provides all of the terms in this equation except the torque Me and the gear reduction ratio 0. To determine the values of these terms, the following steps are taken: 1. Note that in the last sentence there is the wording “the car is in a gear that produces maximum torque”. First consider the latter portion of this wording and find the maximum torque. As in Example 2.4, this is done by taking the first derivative

46 Chapter 2 Road Vehicle Performance

of 28 0.05e e eM n n= − with respect to engine speed ne and setting it equal to zero (to get the maximum):

8 0

8 80 revolutions per second

e e

e

e

dM n

dn

n

= − × =

= =

Note that this gives an equation with engine speed ne at maximum torque being the only unknown. Solving, it is found that ne = 80 revolutions per second which is the engine speed at maximum torque. Next, substitute this value of ne into the given torque equation:

2 8 0.05 320 ft-lbeM = × − × = .

2. The problem states that the car is in “a gear that produces maximum torque.” We need to find the gear reduction ratio 0, and Eq. 2.18 [ ( ) 02 1eV rn iπ ε = −  ] is the equation that relates gear reduction ratio to engine speed. This equation is rearranged to get the gear reduction ratio as,

0

2 / 12 1 4.22

100

π ε

   × × × −   = = ×

With the above, and all elements of Eq. 2.17 known, so

0 972.29 lb

12 e d

e M

F F r

ε η × ×= = = =

Next, the resistances are now computed as follows: for grade resistance (using Eq. 2.9),

66 lbgR WG = = × =

for aerodynamic resistance (using Eq. 2.3),

ρ=

 = × × × × 

=

2

2

2

1.467

2 182.3 lb

a D fR C A V

and for rolling resistance (using Eqs. 2.5 and 2.6),

1.467 0.01 1

147

44 lb

rl rlR f W=

  ×= + ×       =

2.10 Practice Problems 47

The last variable needed to compute the maximum acceleration is the mass factor. This is determined from Eq. 2.20 as,

2 0

2

1.04 0.0025

1.04 0.0025

1.085

m

γ ε= +

= + × =

Rearranging Eq. 2.19 to solve for acceleration gives,

γ

γ

− =

− − − −= = =  ×  

  2

9.17 ft/s . 32.2

m

m

F R ma

F R a

m

PRACTICE PROBLEM 2.2

VEHICLE DESIGN, ENGINE-GENERATED TRACTIVE EFFORT, ROLLING RESISTANCE, ACCELERATION A rear-wheel–drive car has an engine running at 3296 revolutions per minute. It is known that at this engine speed the engine produces 80 horsepower. The car has an overall gear reduction ratio of 10, a wheel radius of 16 inches, and a 95% drivetrain mechanical efficiency. The weight of the car is 2600 lb, the wheelbase is 95 inches, and the center of gravity is 22 inches above the roadway surface. What is the closest distance the center of gravity can be behind the front axle to have the vehicle achieve its maximum acceleration from rest on good, wet pavement?

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

This problem illustrates the relationship between engine-generated tractive effort Fe, and the maximum tractive effort Fmax. To achieve maximum acceleration, we can see from Eq. 2.19 ( mF R maγ− = ) that we want the available tractive effort F (which is

the lesser of the engine-generated tractive effort Fe, and the maximum tractive effort Fmax) to be as high as possible. The problem provides information that will allow us to calculate the engine-generated tractive effort Fe at the conditions specified. So the point of interest in this problem is to determine an Fmax that will enable the vehicle to use all of the available engine-generated tractive effort under the conditions given. It is known from Eq. 2.14 that Fmax is a function of the location of the center of gravity from the front axle. In this problem, if the center of gravity is too close to the front axle, Fmax will be less than Fe and the vehicle will not be able to achieve it maximum acceleration based on it engine-generated tractive effort. This condition is illustrated in the lowest speed region of Fig. 2.5 where Fe curve exceeds the Fmax line. If Fmax < Fe the vehicle will not be able to achieve maximum acceleration based on current engine operating conditions because Fmax will limit the available tractive effort F in Eq. 2.19. For maximum acceleration to be achieved Fmax ≥ Fe. As shown in Eq. 2.14, as the center of gravity gets farther behind the front axle, Fmax will become larger (see Eq. 2.14). Since we want the center of gravity as close to the front axle as possible, we need to solve for the limiting case where Fmax = Fe. To show this equality using Eq. 2.17 we will have,

0e d max e

M F F

r

ε η = =

48 Chapter 2 Road Vehicle Performance

All values in the right-side of the above equation are given in the problem except Me. However, we are given horsepower and engine speed so Eq. 2.16 can be rearranged to solve for Me as,

550hp 550 127.54 ft-lb

2 6.28 ( ) e

e e

M = n

×= = ×

Using this value of Me, Eq. 2.17 can now be applied to give,

0 908.27 lb

e d

max e M

F F r

ε η × ×= = = =

With Fmax known, Eq. 2.14 can be applied to solve for lf which will be the closest distance the center of gravity can be behind the front axle (with the coefficient of road adhesion μ = 0.9 for good, wet pavement from Table 2.4),

( ) 1

f rlW l f h L F

h L

μ μ −

= −

The coefficient of rolling resistance frl, which can be determined from Eq. 2.5 (with V = since acceleration is from rest) as,

0.01 1 0.01 1

147 147rl V

f   = + = + =       

Substituting given and calculated values into Eq. 2.14 gives,

( ) 95 908.72

1 22

fl

× × − × =

− ×

Which gives lf = 29.42 inches.

PRACTICE PROBLEM 2.3

ENGINE-GENERATED TRACTIVE EFFORT AND THEORETICAL MINIMUM STOPPING DISTANCE A car has 11-inch radius wheels and is traveling up a 3% grade with a gear reduction ratio of 2 to 1 and a driveline slippage of 2%. The car has a torque/engine speed curve that is given by the equation,

27 0.08e e eM n n= −

with Me in ft-lb and ne in revolutions per second. It is also known that the car is traveling at a speed where its engine is generating maximum torque. If the car is on good, wet pavement with antilock brakes and a 95% braking efficiency, ignoring air resistance, what would be the theoretical stopping distance (distance traveled until a stop after the brakes are applied)?

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

To solve this problem we must first determine the speed at which the vehicle was traveling before the brakes were applied. The problem states that the car was traveling is traveling at a speed that has its engine producing maximum torque. So we need to find the engine speed at this maximum torque, and then use this engine speed along

2.10 Practice Problems 49

with the given gear reduction ratio and driveline slippage to arrive at vehicle speed (by applying Eq. 2.18). As in Practice Problem 2.1, finding the engine speed ne at maximum torque is achieved by taking the first derivative of 27 0.08e e eM n n= − with respect to engine speed ne and setting it equal to zero (to get the maximum):

7 0

7 43.75 revolutions per second

e e

e

e

dM n

dn

n

= − × =

= =

With the engine speed, Eq. 2.18 is applied as follows,

( ) 0

2 1ern i V

π ε

− =

( ) 1 123.45 ft/s

V

× × − = =

Eq. 2.43 can now be applied to determine the theoretical stopping distance,

( ) ( )

2 2 1 2

2 sin

b

b rl g

V V S

g f

γ

η μ θ

− =

+ ±

In the above equation, b and sin g are given in the problem, b is a known constant (see text), μ can be obtained from Table 2.5, and frl can be determined from Eq. 2.5 as,

0.01 1 0.0142

147rlf

  −      = + =       

So Eq. 2.43 is solved as,

( ) ( )

2 2

273.69 ft. 2

S

× − = =

× × + ±

50 Chapter 2 Road Vehicle Performance

PRACTICE PROBLEM 2.4

THEORETICAL MINIMUM STOPPING DISTANCE WITH INTERMITENT ANTILOCK BRAKE FAILURE A car is traveling 50 mi/h when the driver applies the brakes. The car is traveling on good wet pavement and is going down a 3% grade. The antilock braking system works intermittently during the stop. When the antilock brakes system is functioning, the braking efficiency is 100%. When the antilock system is not functioning the wheels lock and the braking efficiency is 70%. The antilock braking system turns off/on every 60 ft while the brakes are being applied (it is on for the first 60 ft of the stop, off for the second 60 ft, on for the third 60 ft of the stop, and so on). Under these conditions, what is the total distance needed for the car to stop after the brakes are applied? (Assume theoretical stopping distance, ignore aerodynamic resistance, and let frl = 0.011.)

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

Eq. 2.43 can be applied to solve this problem,

( ) ( )

2 2 1 2

2 sin

b

b rl g

V V S

g f

γ

η μ θ

− =

+ ±

but it will have to be applied in 60-ft distance increments to account for the intermittent antilock brake failure. For the first 60 ft with the antilock brakes functioning, in Eq. 2.43 S will be 60 ft, frl and sin g are given in the problem, b is a known constant (see text), μ can be obtained from Table 2.5 and will alternate between maximum and slide values depending on whether or not the antilock braking system is functioning, and b will alternate between 1.0 and 0.7 depending on whether the antilock braking system is functioning or not. We wish to determine the vehicle speed at the end of the first 60-ft interval (an interval where the antilock brakes are working), so rearranging Eq. 2.43 we have,

( ) ( )× × × + − = × − =

2

2

60 2 0.9 1.467 45.91 ft/s

V

The final vehicle speed of 45.91 ft/s after the first 60-ft interval will become the initial speed of the second 60-ft interval, an interval in which the antilock braking system does not function. For the second 60-ft interval, μ and b will change to their non-antilock brake values and Eq. 2.43 is applied again to give,

( ) ( )× × × + − = − =2

2

60 2 46.14 24.86 ft/s

V

The final vehicle speed of 24.86 ft/s after the second 60-ft interval will become the initial speed of the third 60-ft interval, an interval in which the antilock braking system functions again. For the third 60-ft interval μ and b will change to their antilock brake values. The reader can show that the car will come to a stop at some time in this third 60-ft interval. So in the third distance interval Eq. 2.43 is applied to give,

( ) ( )

× − = =

× × × + −

2 0

11.3 ft 2

S

Thus, the total stopping distance is,

= + + = 131.3 ftS

References 51

NOMENCLATURE FOR CHAPTER 2

a acceleration (deceleration if negative) Af frontal area of vehicle BFRf/r max brake force ratio (front over rear) for

maximum braking force CD coefficient of aerodynamic drag d practical stopping distance da distance to accelerate dr distance traveled during driver

perception/reaction ds total stopping distance (vehicle braking

distance plus perception/reaction distance)

F total available tractive effort Fb total braking force Fbf front-axle braking force Fbf max maximum front-axle braking force Fbr rear-axle braking force Fbr max maximum rear-axle braking force Fe engine-generated tractive effort Ff available tractive effort at the front axle Fmax maximum tractive effort Fr available tractive effort at the rear axle frl coefficient of rolling resistance G roadway grade in ft/ft (percent grade

divided by 100; G = 0.05 is a 5% grade) g gravitational constant (32.2 ft/s2) gmax maximum deceleration achieved before

wheel lockup h height of vehicle’s center of gravity

above the roadway surface hpe engine-generated power, measured in

horsepower i drive axle slippage Ka elements of aerodynamic resistance that

are not a function of speed

L vehicle wheelbase lf distance from vehicle’s center of gravity

to front axle lr distance from vehicle’s center of gravity

to rear axle Me engine torque m mass ne engine speed in crankshaft revolutions

per second PBFf optimal percent of braking force on the

front axle PBFr optimal percent of braking force on the

rear axle Ra aerodynamic resistance Rg grade resistance Rrl rolling resistance r radius of vehicle drive wheels S minimum theoretical stopping distance tr driver perception/reaction time V vehicle speed W total vehicle weight Wf vehicle weight acting normal to the

roadway surface on the front axle Wr vehicle weight acting normal to the

roadway surface on the rear axle γb braking mass factor γm acceleration mass factor ε0 gear reduction ratio ηb braking efficiency ηd drivetrain efficiency θg angle of grade μ coefficient of road adhesion ρ air density

REFERENCES

AASHTO (American Association of State Highway and Transportation Officials). A Policy on Geometric Design of Highways and Streets, 7th ed. Washington, DC: AASHTO, 2018.

Fambro, D. B., K. Fitzpatrick, and R. J. Koppa. Determination of Stopping Sight Distances. NCHRP Report 400. Transportation Research Board, Washington, DC, 1997.

S. G. Shadle, L. H. Emery, and H. K. Brewer, “Vehicle Braking, Stability, and Control,” SAE Transactions, vol. 92, paper 830562, 1983.

Taborek, J. J. “Mechanics of Vehicles.” Machine Design, 1957.

Wong, J. Y. Theory of Ground Vehicles. Fourth Edition. New York: John Wiley & Sons, 2008.

P-5

CHAPTER 3 PROBLEMS

Crest Vertical Curves (Section 3.3)

3.1 A 520-ft long equal-tangent crest vertical curve connects tangents that intersect at station 340 + 00 and elevation 1325 ft. The initial grade is +4.0% and the final grade is −2.5%. Determine the elevation and stationing of the high point, PVC, and PVT.

3.2 Consider Example 3.4. Solve this problem with the parabolic equation (Eq. 3.1) rather than by using offsets.

3.3 Again consider Example 3.4. Does this curve provide sufficient stopping sight distance for a speed of 60 mi/h?

3.4 An equal-tangent crest vertical curve is designed for 70 mi/h. The high point is at elevation 1011.4 ft. The initial grade is +2% and the final grade is −1%. What is the elevation of the PVT?

3.5 An equal-tangent crest curve has been designed for 70 mi/h to connect a +2% initial grade and a −1% final grade for a new vehicle that has a 3 ft driver’s eye height; the curve was designed to avoid an object that is 1 ft high. Standard practical stopping distance design was used but, unlike current design standards, the vehicle was assumed to make a 0.5g stop, although driver reactions are assumed to be the same as in current highway design standards. If the PVC of the curve is at elevation 848 ft and station 43 + 48, what is the station and elevation of the high point of the curve?

3.6 A vertical curve is designed for 55 mi/h and has an initial grade of +2.5% and a final grade of −1.0%. The PVT is at station 114 + 50. It is known that a point on the curve at station 112 + 35 is at elevation 245 ft. What is the stationing and elevation of the PVC? What is the stationing and elevation of the high point on the curve?

3.7 An equal-tangent crest vertical curve is designed for 65 mi/h. The initial grade is +3.4% and the final grade is negative. What is the elevation difference between the PVC and the high point of the curve?

3.8 An equal-tangent crest vertical curve has a 50- mi/h design speed. The initial grade is +3%. The high point is at station 33 + 40.76 and the PVT is at station 37 + 24.66. What is the elevation difference between the high point and the PVT?

3.9 An equal-tangent crest curve connects a +2% initial grade with a −1% final grade and is designed for 55 mi/h. The station of the PVI is 233 + 40 with an elevation of 1203 ft. What is the elevation of the curve at station 234 + 00?

3.10 An equal-tangent vertical curve was designed in 2012 (to 2018 AASHTO guidelines) for a design speed of 70 mi/h to connect grades G1 = +1.2% and G2 = −2.1%. The curve is to be redesigned for a 70-mi/h design speed in the year 2025. Vehicle braking technology has advanced so that the recommended design deceleration rate is 25% greater than the 2011 value used to develop Table 3.1, but due to the higher percentage of older persons in the driving population, design reaction times have increased by 20%. Also, vehicles have become smaller so that the driver’s eye height is assumed to be 3.0 ft above the pavement and roadway objects are assumed to be 1.0 ft above the pavement surface. Compute the difference in design curve lengths for the 2012 and 2025 designs.

3.11 An equal-tangent crest vertical curve is designed with a PVI at station 110 + 00 (elevation 927.2 ft) and a PVC at station 107 + 43.3 (elevation 921.55 ft). If the high point is at station 110 + 75.5, what is the design speed of the curve?

3.12 An equal-tangent crest vertical curve connects a +3.2% and a −1.1% grade. The PVI is at station 98 + 20. Due to drainage considerations, the highest point of the curve is at station 100 + 79.35. Determine the station of the PVC and PVT and the design speed of the curve.

3.13 A 1200-ft equal-tangent crest vertical curve is currently designed for 50 mi/h. A civil engineering student contends that 60 mi/h is safe in a van because of the higher driver’s eye height. If all other design inputs are standard, what must the driver’s eye height (in the van) be for the student’s claim to be valid?

3.14 A highway reconstruction project is being undertaken to reduce crash rates. The reconstruction involves a major realignment of the highway such that a 60-mi/h design speed is attained. At one point on the highway, a 720-ft equal-tangent crest vertical curve exists. Measurements show that at 3 + 40 stations from the PVC, the vertical curve offset is 3.5 ft. Assess the adequacy of this existing curve in light of the reconstruction design speed of 60 mi/h and, if the existing curve is inadequate, compute a satisfactory curve length.

3.15 An equal-tangent crest curve connects a +1.0% and a −0.5% grade. The PVC is at station 54 + 24 and the PVI is at station 56 + 92. Is this curve long enough to provide passing sight distance for a 60- mi/h design speed?

P-6 Chapter 3 Problems

3.16 An equal-tangent crest vertical curve connects an initial grade of +2.5% and a final grade of –0.5%. The curve is designed for 70 mi/h and the station of the PVT is 132+62 and is at elevation 833 ft. What is the station and elevation of the curve’s high point?

3.17 An equal-tangent vertical curve connects a +2.25% initial grade and a −2.75% final grade. It is known that the highpoint of the curve is at elevation 1497.4 ft. If the PVI is at elevation 1500 ft, what is the design speed of the curve?

3.18 An equal-tangent vertical curve connects a +2% initial grade and a –1% final grade. It is known that the offset at 400 ft from the PVC is 5.298 ft and that the elevation on the curve at this point is 562 ft. What is the design speed of the curve, the elevation of the PVT, and the elevation of the PVI?

3.19 An equal-tangent vertical curve connects a +1.5% initial grade and a –3% final grade. The PVT is at station 120+52 and elevation 197.665 ft. If the design speed is 50 mi/h, what is the elevation of the curve at station 118+00?

Sag Vertical Curves (Section 3.3)

3.20 A 1400-ft–long sag vertical curve (equal tangent) has a PVC at station 115 + 00 and elevation 750 ft. The initial grade is −3.5% and the final grade is +6.5%. Determine the elevation and stationing of the low point, PVI, and PVT.

3.21 An equal-tangent sag vertical curve is designed with the PVC at station 109 + 00 and elevation 950 ft, the PVI at station 110 + 77 and elevation 947.34 ft, and the low point at station 110 + 50. Determine the design speed of the curve.

3.22 An equal-tangent vertical curve connects a −2% and a +3% grade. The low point of the curve is at elevation 297.88 ft. If the PVI is at elevation 295 ft, what is the design speed of the curve?

3.23 An equal-tangent sag vertical curve is designed for 45 mi/h. The low point is 237 ft from the PVC at station 112 + 37 and the final offset at the PVT is 19.355 ft. If the PVC is at station 110 + 00, what is the elevation difference between the PVT and a point on the curve at station 111 + 00?

3.24 An equal-tangent vertical curve connects an initial grade of 3% and a final grade of +1% and is designed for 60 mi/h. The PVI is at station 250+50 and elevation 732 ft. What is the station and elevation of the lowest point on the curve?

3.25 An overpass is being built over the PVI of an existing equal-tangent sag curve. The sag curve has a 70-mi/h design speed and G1 = −5%, G2 = +3%. Determine the minimum necessary clearance height of the overpass and the resultant elevation of the bottom of the overpass over the PVI. (Ignore the cross-sectional width of the overpass.)

3.26 An existing highway-railway at-grade crossing is being redesigned as grade separated to improve traffic operations. The railway must remain at the same elevation. The highway is being reconstructed to travel under the railway. The underpass will be a sag curve that connects to 2.25% tangent sections on both ends, and the PVI will be centered under the railway (a symmetrical alignment). The sag curve design speed is 45 mi/h. How many feet below the railway should the curve PVI be located?

3.27 An existing equal-tangent sag vertical curve is designed for 60 mi/h. The initial grade is −3% and the elevation of the PVT is 754 ft. The PVC of the curve is at station 134 + 16 and the PVI is at 137 + 32. An overpass is being constructed directly above the PVI. The highway is for cars only (AASHTO minimum and recommended structure clearances do not apply) and the overpass design assumes the driver’s eye height is set conservatively to 5 ft. What is the lowest possible elevation of the bottom of the overpass structure to ensure sufficient stopping sight distance at 60 mi/h?

3.28 An equal-tangent sag curve has its PVI at station 10 + 00 and elevation at 138 ft. Directly above the PVI, the bottom of an overpass structure is at elevation 162 ft. The PVC is at station 4 + 00. If the initial grade is −4%, what is the highest possible value of the final grade given that a 70-mi/h design speed is to be provided in daytime conditions? What is the highest possible final grade in nighttime conditions? (Note: Be careful of units of A and ignore the cross-sectional width of the overpass.)

3.29 An equal-tangent sag vertical curve connects a +1% and +3% initial and final grades, respectively, and is designed for 70 mi/h. The high point on the curve is at elevation 822 ft. If the PVC is at station 110+00, what is the elevation of the curve at station 112+12?

3.30 An equal-tangent sag vertical curve connects a –1.5% initial grade and a +1.25% final grade. The design speed is 70 mi/h and the station and elevation of the PVT are 138+70 and 851.5 ft, respectively. A storm drain is to be installed at the low point of the curve. What is the station and elevation of the low point?

Chapter 3 Problems P-7

Combined Crest and Sag Vertical Curves (Section 3.3)

3.31 Consider the bridge-tunnel problem in Example 3.9. Suppose that a 70 mi/h interstate design speed is needed. If so, what would be the minimum bridge- tunnel separation distance (something higher than the current 1200 ft separation) needed to connect the elevations of the bridge and tunnel with 70 mi/h design-speed curves?

3.32 Two level sections of an east-west highway (G = 0) are to be connected. Currently, the two sections of highway are separated by a 4000-ft (horizontal distance), 2% grade. The westernmost section of highway is the higher of the two and is at elevation 100 ft. If the highway has a 60-mi/h design speed, determine, for the crest and sag vertical curves required, the stationing and elevation of the PVCs and PVTs given that the PVC of the crest curve (on the westernmost level highway section) is at station 0 + 00 and elevation 100 ft. In solving this problem, assume that the curve PVIs are at the intersection of G = 0 and the 2% grade, that is, A = 2.

3.33 Consider Problem 3.32. Suppose it is necessary to keep the entire alignment within the 4000 ft that currently separate the two level sections. It is determined that the crest and sag curves should be connected (the PVT of the crest and PVC of the sag) with a constant-grade section that has the lowest grade possible. Again using a 60-mi/h design speed, determine, for the crest and sag vertical curves, the stationing and elevation of the PVCs and PVTs given that the westernmost level section ends at station 0 + 00 and elevation 100 ft. (Note that A must now be determined and will not be equal to 2.)

3.34 Due to crashes at a railroad crossing, an overpass (with a roadway surface 26 ft above the existing road) is to be constructed on an existing level highway. The existing highway has a design speed of 50 mi/h. The overpass structure is to be level, centered above the railroad, and 180 ft long. What length of the existing level highway must be reconstructed to provide an appropriate vertical alignment?

3.35 A section of a freeway ramp has a +4.0% grade and ends at station 127 + 00 and elevation 138 ft. It must be connected to another section of the ramp (which has a 0.0% grade) that is at station 162 + 00 and elevation 97 ft. It is determined that the crest and sag curves required to connect the ramp should be connected (the PVT of the crest and PVC of the sag)

with a constant-grade section that has the lowest grade possible. Design a vertical alignment to connect between these two stations using a 50-mi/h design speed. Provide the lengths of the curves and constant-grade section.

3.36 A tangent section of highway has a −1.0% grade and ends at station 4 + 75 and elevation 82 ft. It must be connected to another section of highway that has a −1.0% grade and that begins at station 44 + 12 and elevation 131.2 ft. The connecting alignment should consist of a sag curve, constant- grade section, and crest curve, and be designed for a speed of 50 mi/h. What is the lowest grade possible for the constant-grade section that will complete this alignment?

3.37 A roadway has a design speed of 50 mi/h, and at station 105 + 00 a +3.0% grade roadway section ends and at station 125 + 00 a +2.0% grade roadway section begins. The +3.0% grade section of highway (at station 105 + 00) is at a higher elevation than the +2.0% grade section of highway (at station 125 + 00). If a −4% constant-grade section is used to connect the crest and sag vertical curves that are needed to link the +3.0 and +2.0% grade sections, what is the elevation difference between stations 105 + 00 and 125 + 00? (The entire alignment, crest and sag curves, and constant-grade section must fit between stations 105 + 00 and 125 + 00.)

3.38 A sag curve and crest curve connect a −3.5% tangent section of highway (to the west) with a +2.5% tangent section of highway (to the east). The +2.5% tangent section is at a higher elevation than the −3.5% tangent section. The two tangent sections are separated by 1150 ft of horizontal distance. If the design speed of the curves is 50 mi/h, what is the common grade between the sag and crest curves (G2 of sag and G1 of crest, from west to east), and what is the elevation difference between the PVCs and PVTc?

3.39 A level section of highway is to be connected to a section of highway with a −5% grade. The level highway section ends at station 108 + 40 (elevation 865 ft) and is to connect with the −5% section of highway at station 139 + 20 (elevation 758 ft). Using a design speed of 50 mi/h, determine the stations and elevations of the PVCs, PVIs, and PVTs of the two vertical curves required to connect the highway segments, as well as the length of the constant-grade section (connecting grade is to be as small as possible).

P-8 Chapter 3 Problems

3.40 A crest and sag curve connect a 0% west highway segment (left) with a +2% east highway segment (right). The 0% west highway segment is at a higher elevation than the start of the +2% east highway segment. The two vertical curves connect with each other (PVTc = PVCs) and share a 3% common grade. If the design speed of the curves is 45 mi/h, what is the elevation difference between the two road segments?

Horizontal Curves (Section 3.4)

3.41 You are asked to design a horizontal curve for a two-lane road. The road has 12-ft lanes. Due to expensive excavation, it is determined that a maximum of 34 ft can be cleared from the road’s centerline toward the inside lane to provide for stopping sight distance. Also, local guidelines dictate a maximum superelevation of 0.08 ft/ft. What is the highest possible design speed for this curve?

3.42 A horizontal curve on a two-lane highway (10- ft lanes) is designed for 50 mi/h with a 6% superelevation. The central angle of the curve is 35 degrees and the PI is at station 482 + 72. What is the station of the PT and how many feet have to be cleared from the lane's shoulder edge to provide adequate stopping sight distance?

3.43 A horizontal curve on a single-lane highway has its PC at station 123 + 70 and its PI at station 130 + 90. The curve has a superelevation of 0.06 ft/ft and is designed for 70 mi/h. What is the station of the PT?

3.44 A horizontal curve is being designed through mountainous terrain for a four-lane road with lanes that are 10 ft wide. The central angle ( ) is known to be 40 degrees, the tangent distance is 520 ft, and the stationing of the tangent intersection (PI) is 2600 + 00. Under specified conditions and vehicle speed, the roadway surface is determined to have a coefficient of side friction of 0.08, and the curve’s superelevation is 0.09 ft/ft. What is the stationing of the PC and PT and what is the safe vehicle speed?

3.45 A new interstate highway is being built with a design speed of 70 mi/h. For one of the horizontal curves, the radius (measured to the innermost vehicle path) is tentatively planned as 2500 ft. What rate of superelevation is required for this curve?

3.46 On a roadway with two 12-ft lanes, a horizontal curve is designed for 35 mi/h with a 4% superelevation. It is known that = 2 s. The PI of the curve is at station 30 + 00. What is the station of the PT of the curve?

3.47 A developer is having a single-lane raceway constructed with a 200-mi/h design speed. A curve on

the raceway has a radius of 4500 ft, a central angle of 30 degrees, and PI stationing at 1125 + 10. If the design coefficient of side friction is 0.20, determine the superelevation required at the design speed (do not ignore the normal component of the centripetal force). Also, compute the degree of curve, length of curve, and stationing of the PC and PT.

3.48 A horizontal curve is being designed for a new two-lane highway (12-ft lanes). The PI is at station 250 + 50, the design speed is 65 mi/h, and a maximum superelevation of 0.07 ft/ft is to be used. If the central angle of the curve is 38 degrees, design a curve for the highway by computing the radius and stationing of the PC and PT.

3.49 You are asked to design a horizontal curve with a 40-degree central angle (Δ = 40) for a two-lane road with 11-ft lanes. The design speed is 70 mi/h and superelevation is limited to 0.06 ft/ft. Give the radius, degree of curvature, and length of curve that you would recommend.

3.50 For the horizontal curve in Problem 3.49, what distance must be cleared from the inside edge of the inside lane to provide adequate stopping sight distance?

3.51 A horizontal curve on a single-lane freeway ramp is 400 ft long, and the design speed of the ramp is 45 mi/h. If the superelevation is 10% and the station of the PC is 18 + 25, what is the station of the PI and how much distance must be cleared from the center of the lane to provide adequate stopping sight distance?

3.52 A freeway exit ramp has a single lane and consists entirely of a horizontal curve with a central angle of 90 degrees and a length of 628 ft. If the distance cleared from the centerline for sight distance is 19.4 ft, what design speed was used?

3.53 A horizontal curve on a two-lane highway (12- ft lanes) has PC at station 123 + 80 and PT at station 129 + 60. The central angle is 35 degrees, the superelevation is 0.08, and 20.6 ft is cleared (for sight distance) from the inside edge of the innermost lane. Determine a maximum safe speed (assuming current design standards) to the nearest 5 mi/h.

3.54 A horizontal curve was designed for a four-lane highway for adequate SSD. Lane widths are 12 ft, and the superelevation is 0.06 and was set assuming maximum fs. If the necessary sight distance required 52 ft of lateral clearance from the roadway centerline, what design speed was used for the curve?

3.55 On a two-lane road with 12-ft lanes, a horizontal curve is designed for 50 mi/h with a superelevation of

Chapter 3 Problems P-9

10%. The PI of the curve is at 220+48 and the PC is at station 216+74. Determine the station of the PT and the middle ordinate for stopping sight distance.

3.56 A horizontal curve is designed for 35 mi/h with a central angle of 85 degrees. The curve has four 11- ft lanes (two in each direction) and a superelevation of 8%. If the PI is at station 148+40 what is the station of the PT?

Combined Vertical and Horizontal Curves (Section 3.5)

3.57 A section of highway has vertical and horizontal curves with the same design speed. A vertical curve on this highway connects a +1% and a +3% grade and is 420 ft long. If a horizontal curve on this highway is on a two-lane section with 12-ft lanes and has a central angle of 37 degrees and a superelevation of 6%, what is the length of the horizontal curve?

3.58 A section of a two-lane highway (12-ft lanes) is designed for 75 mi/h. At one point a vertical curve connects a −2.5% and +1.5% grade. The PVT of this curve is at station 36 + 50. It is known that a horizontal curve starts (has PC) 294 ft before the vertical curve’s PVC. If the superelevation of the horizontal curve is 0.08 and the central angle is 38 degrees, what is the station of the PT?

3.59 Two straight sections of freeway cross at a right angle. At the point of crossing, the east-west highway is at elevation 150 ft and has a constant +5.0% grade (upgrade in the east direction), and the north-south highway is at elevation 125 ft and has a constant −3.0% grade (downgrade in the north direction). Design a 90-degree ramp that connects the northbound direction of travel to the eastbound direction of travel. Design the ramp for the highest design speed (to nearest 5 mi/h) with the constraint that the minimum allowable value of D is 8.0. (Assume that the PC of the horizontal curve is at station 15 + 00, and the vertical curve PVIs are at the PC and PT.) Give the stationing and elevations of the PC, PT, PVCs, and PVTs.

3.60 A crest vertical curve and a horizontal curve on the same highway have the same design speed. The equal-tangent vertical curve connects a +3% initial grade with a +1% final grade and has a PVC at 101 + 78 and a PVT at 106 + 72, The horizontal curve has a PI at 150 + 10 and a central angle of 75 degrees. If the superelevation of the horizontal curve is 8% and the road has two 12-ft lanes, what is the stationing of the PT?

3.61 West and east highway segments are separated by 600 ft horizontally. The west segment has a +2% grade and the east segment has a +1% grade. The west-grade segment ends at a higher elevation than the east segment and the two segments are connected by a joining sag and crest curve combination (so PVTs = PVCc). If the road is designed for 45 mi/h, what is the elevation difference between the west and east highway segments?

Multiple Choice Problems (Multiple Sections)

3.62 A 400-ft equal-tangent sag vertical curve has its PVC at station 100 + 00 and elevation 500 ft. The initial grade is −4.0% and the final grade is +2.5%. Determine the elevation of the lowest point of the curve.

a) 495.077 ft b) 495.250 ft c) 485.231 ft d) 492.043 ft

3.63 A horizontal curve is being designed around a pond with a tangent length of 1200 ft and central angle of 0.5211 radians. If the PI is at station 145 + 00, determine the station of PT.

a) 168 + 45.44 b) 156 + 45.44 c) 173 + 94.00 d) 156 + 72.72

3.64 A car is traveling over a 1400-ft vertical curve. One of the passengers decides to calculate the current offset from the PVC. By looking at the onboard navigation device, the passenger knows that the car is 750 feet from the PVC. The initial grade is +5.5% while the final roadway grade is +3.0%. What is the current offset?

a) 4.38 ft b) 17.50 ft c) 17.08 ft d) 5.02 ft

3.65 You are designing a highway to AASHTO guidelines on rolling terrain where the design speed will be 65 mi/h. At one section, a +1.25% grade and a −2.25% grade must be connected with an equal-tangent vertical curve. Determine the minimum length of curve that can be designed while meeting SSD requirements.

a) 864.30 ft b) 645.00 ft c) 674.74 ft d) 673.43 ft

P-10 Chapter 3 Problems

3.66 A car is traveling downhill on a suburban road with a grade of 4% at a speed of 35 mi/h. Determine the required stopping sight distance.

a) 149.29 ft b) 245.97 ft c) 233.84 ft d) 261.26 ft

3.67 A tow truck is searching a city street at 40 mi/h for illegally parked vehicles. It travels over an equal-tangent vertical curve with an initial grade of +4.0% and final grade of −2.0%. If the height of the driver's eye is 6.0 ft and the driver spots a car 450 ft away with a height of 4.0 ft, what is the minimum length of the vertical curve for this situation?

a) 562.94 ft b) 1304.15 ft c) 240.07 ft d) 306.85 ft

53

Chapter 3

Geometric Design of Highways 3.1 INTRODUCTION

The information on vehicle performance provided in Chapter 2 provides the fundamentals needed for highway design. Highway design encompasses a variety of design elements including the alignment required to provide adequate stopping sight distances (SSD), adequate passing sight distances on two-lane roads, length of acceleration and deceleration lanes for on- and off-ramps, number of lanes required to provide adequate mobility, and identification of the need for truck climbing lanes on steep grades. The relationship with these design elements and vehicle performance is obvious. For example, vehicle acceleration and deceleration characteristics have a direct impact on the design of acceleration and deceleration lanes (the length needed to provide a safe and orderly flow of traffic) and the highway alignment needed to provide adequate passing and stopping sight distances. Furthermore, vehicle performance characteristics determine the need for truck climbing lanes on steep grades (where the poor performance of large trucks necessitates a separate lane) as well as the number of lanes required because the observed spacing between vehicles in traffic is directly related to vehicle performance characteristics (this will be discussed further in Chapter 5). In addition, the physical dimensions of vehicles affect a number of design elements, such as the curve radii required for low- speed turning, height of highway overpasses, and lane widths.

Considering the range of vehicle performance characteristics likely to be encountered in highway traffic (from high-performance sports cars to heavily- loaded trucks), as well as the variance in the physical dimensions of highway vehicles, and the interaction of these characteristics with the many elements constituting highway design, it is clear that highway geometric design is a complex procedure that requires numerous compromises. Moreover, it is important that highway design guidelines evolve over time in response to changes in vehicle performance and dimensions, and in response to evidence collected on the effectiveness of existing highway design practices, such as the relationship between crash rates and various roadway design characteristics. Current guidelines of highway design are presented in detail in A Policy on Geometric Design of Highways and Streets, 7th Edition, published by the American Association of State Highway and Transportation Officials [AASHTO, 2018].

54 Chapter 3 Geometric Design of Highways

Because of the sheer number of geometric elements involved in highway design, a detailed discussion of each design element is beyond the scope of this book (the reader is referred to [AASHTO, 2018] for a complete discussion of current design practices). Instead, this book focuses exclusively on the key elements of highway alignment, which are arguably the most important components of geometric design. As will be shown, the alignment topic is particularly well suited for demonstrating the effect of vehicle performance (specifically braking performance) and vehicle dimensions (which determine critical factors such as the driver’s eye height, vehicle headlight height, and vehicle taillight height) on the design of highways. By concentrating on the specifics of the highway alignment problem, the reader will develop an understanding of the procedures and compromises inherent in the design of all highway-related geometric elements.

3.2 PRINCIPLES OF HIGHWAY ALIGNMENT

The alignment of a highway is a three-dimensional problem measured in x, y, and z coordinates. This is illustrated, from a driver’s perspective, in Fig. 3.1. However, in highway design practice, three-dimensional design computations are cumbersome, and, what is perhaps more important, the actual implementation and construction of a design based on three-dimensional coordinates has historically been prohibitively difficult. As a consequence, the three-dimensional highway alignment problem is reduced to two two- dimensional alignment problems, as illustrated in Fig. 3.2. One of the alignment problems in this figure corresponds roughly to x and z coordinates and is referred to as horizontal alignment. The other corresponds to highway length (measured along some constant elevation) and y coordinates (elevation) and is referred to as vertical alignment. Referring to Fig. 3.2, note that the horizontal alignment of a highway is referred to as the plan view, which is roughly equivalent to the perspective of an aerial photo of the highway. The vertical alignment is represented in a profile view, which gives the elevation of all points measured along the length of the highway (again, with length measured along a constant elevation reference).

Aside from considering the alignment problem as two two-dimensional problems, one further simplification is made: instead of using x and z coordinates, highway positioning and length are defined as the distance along the highway (usually measured along the centerline of the highway, on a horizontal, constant-elevation plane) from a specified point. This distance is measured in terms of stations, with each station consisting of 100 ft of highway alignment distance.

The notation for stationing distance is such that a point on a highway 4250 ft from a specified point of origin is said to be at station 42 + 50 ft, that is, 42 stations and 50 ft, with the point of origin being at station 0 + 00. This stationing

3.3 Vertical Alignment 55

Figure 3.1 Highway alignment in three dimensions.

Figure 3.2 Highway alignment in two-dimensional views. concept, combined with the highway’s alignment direction given in the plan view (horizontal alignment) and the elevation corresponding to stations given in the profile view (vertical alignment), gives a unique identification of all highway points in a manner that is virtually equivalent to using true x, y, and z coordinates.

3.3 VERTICAL ALIGNMENT

Vertical alignment specifies the elevation of points along a roadway. The elevation of these roadway points is usually determined by the need to provide an acceptable level of driver safety, driver comfort, and proper drainage (from rainfall runoff). A primary concern in vertical alignment is establishing the transition of roadway elevations between two grades. This transition is achieved by means of a vertical curve.

56 Chapter 3 Geometric Design of Highways

Vertical curves can be broadly classified into crest vertical curves and sag vertical curves, as illustrated in Fig. 3.3. Note that in Fig. 3.3, the distance from the PVC to the PVI is L/2. This is used in this figure because in practice the vast majority of vertical curves are arranged such that half of the curve length is positioned before the PVI and half after. Curves that satisfy this criterion are called equal-tangent vertical curves.

For referencing points on a vertical curve, it is important to note that the profile views presented in Fig. 3.3 correspond to all highway points even if a horizontal curve occurs concurrently with a vertical curve (as in Figs. 3.1 and 3.2). Thus, each roadway point is uniquely defined by stationing (which is measured along a horizontal plane) and elevation. This will be made clearer through forthcoming examples.

Figure 3.3 Types of vertical curves.

Used by permission from American Association of State Highway and Transportation Officials, A Policy on Geometric Design of Highways and Streets, 7th Edition, Washington, DC, 2018.

G1 = initial roadway grade in percent or ft/ft (this grade is also referred to as the initial tangent grade, viewing Fig. 3.3 from left to right),

PVC = point of the vertical curve (the initial point of the curve),

PVI = point of vertical intersection (intersection of initial and final grades), and

G2 = final roadway (tangent) grade in percent or ft/ft,

PVT = point of vertical tangent, which is the final point of the vertical curve (the point where the curve returns to the final grade or, equivalently, the final tangent).

A = absolute value of the difference in grades (initial minus final, usually expressed in percent),

L = length of the curve in stations or ft measured in a constant-elevation horizontal plane.

3.3 Vertical Alignment 57

3.3.1 Vertical Curve Fundamentals In connecting roadway grades (tangents) with an appropriate vertical curve, a mathematical relationship defining elevations at all points (or equivalently, stations) along the vertical curve is needed. A parabolic function has been found suitable in this regard because, among other things, it provides a constant rate of change of slope and implies equal curve tangents. The general form of the parabolic equation, as applied to vertical curves, is

= + +2y ax bx c (3.1)

where

y = roadway elevation at distance x from the beginning of the vertical curve (the PVC) in stations or ft,

x = distance from the beginning of the vertical curve in stations or ft, a, b = coefficients defined below, and

c = elevation of the PVC (because x = 0 corresponds to the PVC) in ft. In defining a and b, note that the first derivative of Eq. 3.1 gives the slope

and is

= +2

dy ax b

dx (3.2)

At the PVC, x = 0, so, using Eq. 3.2,

= = 1

dy b G

dx (3.3)

where G1 is the initial slope in ft/ft, as defined in Fig. 3.3. Also note that the second derivative of Eq. 3.1 is the rate of change of slope and is

=

2

2 2

d y a

dx (3.4)

However, the average rate of change of slope, by observation of Fig. 3.3, can also be written as

−= 2

2 1 2

G Gd y Ldx

(3.5)

Equating Eqs. 3.4 and 3.5 gives

−= 2 1

2 G G

a L

(3.6)

with all terms as defined previously (see Fig. 3.3). Please note that the units for coefficients a and b in Eqs. 3.3 and 3.6 must be such that they provide ft when multiplied by x2 and x, respectively. The preceding equations define all the terms in the parabolic vertical curve equation (Eq. 3.1). The following example gives a typical application of this equation.

58 Chapter 3 Geometric Design of Highways

EXAMPLE 3.1 VERTICAL CURVE STATIONS AND ELEVATIONS

A 600-ft equal-tangent sag vertical curve has the PVC at station 170 + 00 and elevation 1000 ft. The initial grade is −3.5% and the final grade is +0.5%. Determine the stationing and elevation of the PVI, the PVT, and the lowest point on the curve.

SOLUTION

Since the curve is equal tangent, the PVI will be 300 ft or three stations (measured in a horizontal plane) from the PVC, and the PVT will be 600 ft or six stations from the PVC. Therefore, the stationing of the PVI and PVT is 173 00+ and 176 00+ ,

respectively. For the elevations of the PVI and PVT, it is known that a −3.5% grade can be equivalently written as −3.5 ft/station (a 3.5 ft drop per 100 ft of horizontal distance). Since the PVI is three stations from the PVC, which is known to be at elevation 1000 ft, the elevation of the PVI is

1000 − 3.5 ft/station × (3 stations) = 989.5 ft

Similarly, with the PVI at elevation 989.5 ft, the elevation of the PVT is

989.5 + 0.5 ft/station × (3 stations) = 991.0 ft

It is clear from the values of the initial and final grades that the lowest point on the vertical curve will occur when the first derivative of the parabolic function (Eq. 3.1) is zero because the initial and final grades are opposite in sign. When initial and final grades are not opposite in sign, the low (or high) point on the curve will not be where the first derivative is zero because the slope along the curve will never be zero. For example, a sag curve with an initial grade of −2.0% and a final grade of −1.0% will have its lowest elevation at the PVT, and the first derivative of Eq. 3.1 will not be zero at any point along the curve. However, in our example problem the derivative will be equal to zero at some point, so the low point will occur when

= + =2 0 dy

ax b dx

From Eq. 3.3, we have

= = −1 3.5b G with G1 in percent. From Eq. 3.6 (with L in stations and G1 and G2 in percent),

− −= =0.5 ( 3.5) 0.33333

2(6) a

Substituting for a and b gives

= + − =

=

2(0.33333) ( 3.5) 0

5.25 stations

dy x

dx x

3.3 Vertical Alignment 59

This gives the stationing of the low point at 175 25+ (5 + 25 stations from the PVC).

For the elevation of the lowest point on the vertical curve, the values of a, b, c (elevation of the PVC), and x are substituted into Eq. 3.1, giving

= + − + =

20.33333(5.25) ( 3.5)(5.25) 1000

990.81 ft

y

Note that the preceding equations can also be solved with grades expressed as the decimal equivalent of percent (for example, 0.02 ft/ft for 2%) if x is expressed in feet instead of stations. Care must be taken not to mix units. A dimensional analysis of Eq. 3.1 must ensure that each right-side element of the equation has resulting units of feet.

Another interesting vertical curve problem that is sometimes encountered is one in which the curve must be designed so that the elevation of a specific location is met. An example might be to have the roadway connect with another (at the same elevation) or to have the roadway at some specified elevation so as to pass under another roadway. This type of problem is referred to as a curve-through- a-point problem and is demonstrated by the following example.

EXAMPLE 3.2 ELEMENTS OF VERTICAL CURVE DESIGN

An equal-tangent vertical curve is to be constructed between grades of −2.0% (initial) and +1.0% (final). The PVI is at station 110 + 00 and at elevation 420 ft. Due to a street crossing the roadway, the elevation of the roadway at station 112 + 00 must be at 424.5 ft. Design the curve.

SOLUTION

The design problem is one of determining the length of the curve required to ensure that station 112 + 00 is at elevation 424.5 ft. To begin, we use Eq. 3.1:

= + +2y ax bx c From Eq. 3.3,

= = −1 2.0b G and from Eq. 3.6,

− = 2 1

2

G G a

L

Substituting G1 = −2.0 and G2 = 1.0, we have

( )− −− = = =2 1 1.0 2.0 1.5

2 2

G G a

L L L

Now note that c (the elevation of the PVC) in Eq. 3.1 will be equal to the elevation of the PVI plus G1 × 0.5L (this is simply using the slope of the initial grade to determine

60 Chapter 3 Geometric Design of Highways

the elevation difference between the PVI and PVC). With G1 in percent (which is ft/station) and the curve length L in stations, we have

( )= + = +420 2.0 0.5 420c L L

Finally, the value of x to be used in Eq. 3.1 will be 0.5L + 2 because the point of interest (station 112 + 00) is two stations from the PVI (which is at station 110 + 00). Substituting b = −2.0, the expressions for a, c, and x, and y = 424.5 ft (the given elevation) into Eq. 3.1 give

( )( ) ( )( ) ( )= + + − + + + = + + −

= − + −

2

2

424.5 1.5 0.5 2 2.0 0.5 2 420

4.5 0.375 3 6 4

0 0.375 5.5 6

L L L L

L L

L L Solving this quadratic equation gives L = 1.187 stations (which is not feasible because we know that the point of interest is 2.00 stations beyond the PVI, so the curve must be longer than 1.187 stations) or L = 13.466 stations (which is the only feasible solution). This means that the curve must be 1346.6 ft long. Using this value of L,

( ) = = + = + = = + − + = +

elevation of 420 420 13.466 433.47 ft

station of 110 00 13 46.6 / 2 103 26.7

PVC c L

PVC

( ) ( ) ( ) ( )

 = + = + =  = + + + = +

2elevation of elevation of 0.5 420 0.5 13.466 1.0 426.73 ft

station of 110 00 13 46.6 / 2 116 73.3

PVT PVI L G

PVT

and

x = 0.5L + 2.0 = 6.733 + 2.0 = 8.733 stations from the PVC To check the elevation of the curve at station 112 + 00, we apply Eq. 3.1 with x = 8.733:

( ) ( ) ( )( )

= + +

  = + − +    

=

2

2

3 8.733 2.0 8.733 433.47

2 13.466

424.5 ft

y ax bx c

Therefore, all calculations are correct.

Some additional properties of vertical curves can now be formalized. For example, offsets, which are vertical distances from the initial tangent to the curve, as illustrated in Fig. 3.4, are extremely important in vertical curve design and construction.

3.3 Vertical Alignment 61

Figure 3.4 Offsets for equal-tangent vertical curves.

G1 = initial roadway grade in percent or ft/ft (this grade is also referred to as the initial tangent grade, viewing Fig. 3.4 from left to right),

x = distance from the PVC in ft,

L = length of the curve in stations or ft measured in a constant-elevation horizontal plane,

PVC = point of the vertical curve (the initial point of

G2 = final roadway (tangent) grade in percent or ft/ft,

the curve),

PVI = point of vertical intersection (intersection of

Y = offset at any distance x from the PVC in ft,

initial and final grades), and PVT = point of vertical tangent, which is the final

Ym = midcurve offset in ft, point of the vertical curve (the point where the curve returns to the final grade or, equivalently, the final tangent).

Yf = offset at the end of the vertical curve in ft,

Referring to the elements shown in Fig. 3.4, the properties of an equal-tangent parabola can be used to give

= 2

200 A

Y x L

(3.7)

where

A = absolute value of the difference in grades (|G1 − G2|) expressed in percent, and other terms are as defined in Fig. 3.4. Note that in this equation, 200 is used in the denominator instead of 2 because A is expressed in percent instead of ft/ft (this division by 100 also applies to Eqs. 3.8 and 3.9 below). It follows from Fig. 3.4 that

=

800m AL

Y (3.8)

62 Chapter 3 Geometric Design of Highways

and

=

200f AL

Y (3.9)

Another useful vertical curve property is one that gives the length of curve required to effect a 1% change in slope. Because the parabolic equation used for roadway elevations (Eq. 3.1) gives a constant rate of change of slope, it can be shown that the horizontal distance required to change the slope by 1% is

= L

K A

(3.10)

where

K = value that is the horizontal distance, in ft, required to affect a 1% change in the slope of the vertical curve,

L = length of curve in ft, and A = absolute value of the difference in grades (|G1 − G2|) expressed as a percentage.

This K-value can also be used to compute the high and low point locations of

crest and sag vertical curves, respectively (provided the high or low point does not occur at the PVC or PVT). As shown in Example 3.1, setting dy/dx = 0 in Eq. 3.2 and solving for x gives the distance from the PVC to the high/low point. If Eq. 3.6 is used to substitute for a in Eq. 3.2 (with L = KA), it can be shown that setting dy/dx = 0 in Eq. 3.2 gives

= × 1hlx K G (3.11)

where

xhl = distance from the PVC to the high/low point in ft, and other terms are as defined previously.

In addition to high/low point computations, K-values have an important

application in the design of vertical curves, as wil1 be demonstrated in Sections 3.3.3 and 3.3.4.

EXAMPLE 3.3 VERTICAL CURVE DESIGN WITH K-VALUES

A curve has initial and final grades of +3% and −4%, respectively, and is 700 ft long. The PVC is at elevation of 100 ft. Graph the vertical curve elevations and the slope of the curve against the length of curve. Compute the K-value and use it to locate the high point of the curve (distance from the PVC).

SOLUTION

Recall that to find the slope at any point on the curve, we take the derivative of Eq. 3.1, which gives Eq. 3.2. To apply this equation, a and b need to be determined. From Eq. 3.6,

3.3 Vertical Alignment 63

( ) − −= = −4.0 3.0

0.5 2 7

a

and from Eq. 3.3,

b = G1 = 3 The results of applying Eq. 3.2 and solving for the slope at all points along the curve, as well as a profile view of the curve itself (by application of Eq. 3.1), are shown graphically in Fig. 3.5 (exaggerating the vertical scale). Figure 3.5 shows the constant rate of change of the slope along the length of the curve. The circular points on the slope-of-curve line correspond to changes in grade of 1%, and these points occur at equal intervals of 100 ft.

To show that this is consistent with the K-value, Eq. 3.10 gives

= = = − − 700

100 ft 3 ( 4)

L K

A

Figure 3.5 Profile view of vertical curve for Example 3.3 with the graph of the slope at all points along the curve overlaid.

This indicates that there should be a change in grade of 1% for every 100 ft of curve length (measured in the horizontal plane), and this is consistent with Fig. 3.5. Applying Eq. 3.11 with the K-value of 100 ft gives the high point at 300 ft from the beginning of

the curve (xhl = 100 × 3 = 300 ft). This is shown in Fig. 3.5, where the slope of the curve at 300 ft is zero (the same result obtained by setting the derivative of Eq. 3.2 equal to zero and solving for x). This result can also be explained conceptually based on the definition of the K-value. The K-value gives the horizontal distance required to effect a 1% change in the slope of the curve, and for this curve that value is 100 ft. Thus, to go from an initial grade (G1) of 3% to a grade of 0% (the high point) requires a horizontal distance equal to K × 3, or 300 ft.

64 Chapter 3 Geometric Design of Highways

EXAMPLE 3.4 VERTICAL CURVE DESIGN USING OFFSETS

A vertical curve crosses a 4-ft diameter pipe at right angles. The pipe is located at station 110 + 85 and its centerline is at elevation 1091.60 ft. The PVI of the vertical curve is at station 110 + 00 and elevation 1098.4 ft. The vertical curve is equal tangent, 600 ft long, and connects an initial grade of +1.20% and a final grade of −1.08%. Using offsets, determine the depth, below the surface of the curve, of the top of the pipe and determine the station of the highest point on the curve.

SOLUTION

The PVC is at station 107 + 00 (110 + 00 minus 3 + 00, which is half of the curve length), so the pipe is 385 ft (110 + 85 minus 107 + 00) from the beginning of the curve (PVC). The elevation of the PVC will be the elevation of the PVI minus the drop in grade over one-half the curve length,

1098.4 − (3 stations × 1.2 ft/station) = 1094.8 ft Using this, the elevation of the initial tangent above the pipe is

1094.8 + (3.85 stations × 1.2 ft/station) = 1099.42 ft Using Eq. 3.7 to determine the offset above the pipe at x = 385 ft (the distance of the pipe from the PVC), we have

= 2

200 A

Y x L

( ) ( ) ( )

− − = =21.2 1.08

385 2.82 ft 200 600

Y

Thus, the elevation of the curve above the pipe is 1096.6 ft (1099.42 − 2.82). The elevation of the top of the pipe is 1093.60 ft (elevation of the centerline plus one-half of the pipe’s diameter), so the pipe is 3.0 ft below the surface of the curve (1096.6 − 1093.6).

To determine the location of the highest point on the curve, we find K from Eq. 3.10 as

( ) = =

− − 600

263.16 1.2 1.08

K

and the distance from the PVC to the highest point is (from Eq. 3.11)

= × = × =1 263.16 1.2 315.79 fthlx K G

This gives the station of the highest point at 110 15.79+ (107 + 00 plus 3 + 15.79).

Note that this example could also be solved by applying Eq. 3.1, setting Eq. 3.2 equal to zero (for determining the location of the highest point on the curve), and following the procedure used in Example 3.1.

3.3 Vertical Alignment 65

3.3.2 Stopping Sight Distance Construction of a vertical curve is generally a costly operation requiring the movement of significant amounts of earthen material. Thus, one of the primary challenges facing highway designers is to minimize construction costs (usually by making the vertical curve as short as possible) while still providing an adequate level of safety. An appropriate level of safety is usually defined as that level of safety that gives drivers sufficient sight distance to allow them to safely stop their vehicles to avoid collisions with objects obstructing their forward motion. The provision of adequate roadway drainage is sometimes an important concern as well, but is not discussed in terms of vertical curves in this book (see [AASHTO, 2018]). Referring back to the vehicle braking performance concepts discussed in Chapter 2, we can compute this necessary SSD as the summation of vehicle practical stopping distance (Eq. 2.47) and the distance traveled during driver perception/reaction time (Eq. 2.49). That is,

×    ±     

2 1

1SSD = +

2 r

V V t

a g G

g

(3.12)

where

SSD = stopping sight distance in ft, V1 = initial vehicle speed in ft/s,

g = gravitational constant, 32.2 ft/s2, a = deceleration rate in ft/s2, G = roadway grade (+ for uphill and − for downhill) in percent/100, and tr = perception/reaction time in s.

Recall from Sections 2.9.5 and 2.9.6 that a value of 11.2 ft/s2 for a and a value

of 2.5 s for tr were recommended for roadway design purposes. The design speed of the highway is defined as the maximum safe speed at which a highway can be negotiated assuming near-worst-case conditions (wet-weather conditions). The application of Eq. 3.12 (assuming G = 0) produces the SSDs presented in Table 3.1.

66 Chapter 3 Geometric Design of Highways

Table 3.1 Stopping Sight Distance

Design speed (mi/h)

Brake reaction distance

(ft)

Braking distance on level

(ft)

Stopping sight distance

Note: Brake reaction distance is based on a time of 2.5 s; a deceleration rate of 11.2 ft/s2 is used to determine calculated stopping sight distance. Source: American Association of State Highway and Transportation Officials, A Policy on Geometric Design of Highways and Streets, 7th Edition, Washington, DC, 2018. Used by permission.

Calculated (ft)

Design (ft)

15 55.1 21.6 76.7 80 20 73.5 38.4 111.9 115 25 91.9 60.0 151.9 155 30 110.3 86.4 196.7 200 35 128.6 117.6 246.2 250 40 147.0 153.6 300.6 305 45 165.4 194.4 359.8 360 50 183.8 240.0 423.8 425 55 202.1 290.3 492.4 495 60 220.5 345.5 566.0 570 65 238.9 405.5 644.4 645 70 257.3 470.3 727.6 730 75 275.6 539.9 815.5 820 80 294.0 614.3 908.3 910

3.3.3 Stopping Sight Distance and Crest Vertical Curve Design The length of curve (L in Fig. 3.3) is the critical element in providing sufficient SSD on a vertical curve. Longer curve lengths provide more SSD, all else being equal, but are more costly to construct. Shorter curve lengths are less expensive to construct but may not provide adequate SSD due to more rapid changes in slope. What is needed, then, is an expression for minimum curve length given a required SSD. In developing such an expression, crest and sag vertical curves are considered separately.

The case of designing a crest vertical curve for adequate SSD is illustrated in Fig. 3.6. To determine the minimum length of curve for a required sight distance, the properties of a parabola for an equal-tangent curve can be used to show that For S < L

( )

= +

2

2

1 2200 m

AS L

H H (3.13)

For S > L

( )+ = −

2

1 2200 2m

H H L S

A (3.14)

3.3 Vertical Alignment 67

where

Lm = minimum length of vertical curve in ft, A = absolute value of the difference in grades (|G1−G2|), expressed as a percentage,

and other terms are as defined in Fig. 3.6.

For the sight distance required to provide adequate SSD, current AASHTO design guidelines [2018] use a driver eye height, H1, of 3.5 ft and a roadway object height, H2, of 2.0 ft (the height of an object to be avoided by stopping before a collision). In applying Eqs. 3.13 and 3.14 to determine the minimum length of curve required to provide adequate SSD, we set the sight distance, S, equal to the stopping sight distance, SSD (note that the relatively small distance from the driver’s eye position to the front of the vehicle is ignored). Substituting AASHTO guidelines for H1 and H2 and letting S = SSD in Eqs. 3.13 and 3.14 gives For SSD < L ×=

2SSD 2158m

A L (3.15)

For SSD > L

= × − 2158 2 SSDmL

A (3.16)

Figure 3.6 Stopping sight distance considerations for crest vertical curves.

S = sight distance in ft, PVC = point of vertical curve (the initial point of the curve), H1 = height of driver’s eye above roadway

surface in ft, PVI = point of vertical intersection (intersection of initial

and final grades), and H2 = height of object above roadway

surface in ft, PVT = point of vertical tangent, which is the final point of

the vertical curve (the point where the curve returns to the final grade or, equivalently, the final tangent).

L = length of the curve in ft,

68 Chapter 3 Geometric Design of Highways

EXAMPLE 3.5 DESIGN SPEED AND CREST VERTICAL CURVE DESIGN

A highway is being designed to AASHTO guidelines with a 70-mi/h design speed, and at one section, an equal-tangent vertical curve must be designed to connect grades of +1.0% and −2.0%. Determine the minimum length of curve necessary to meet SSD requirements.

SOLUTION

If we ignore the effect of grades (Gs = 0), the SSD can be read directly from Table 3.1. In this case, the SSD corresponding to a speed of 70 mi/h is 730 ft. If we assume that L > SSD (an assumption that is typically made), Eq. 3.15 gives

× ×= = = 2 23 730

740.82 ft 2158 2158m

A SSD L

Since 740.82 > 730, the assumption that L > SSD was correct.

The assumption that G = 0, made at the beginning of Example 3.5, is not

really correct. If G ≠ 0, we cannot use the SSD values in Table 3.1 and instead must apply Eq. 3.12 with the appropriate G value. In this problem, if we use the initial grade in Eq. 3.12 (+1.0%), we will underestimate the SSD because the vertical curve has a slope as steeply positive as this only at the PVC. If we use the final grade in Eq. 3.12 (−2.0%), we will overestimate the SSD because the vertical curve has a slope as steeply negative as this only at the PVT. If we knew where the vehicle began to brake, we could use the first derivative of the parabolic curve function (from Eq. 3.2) to give G in Eq. 3.12 and set up the equation to solve for SSD exactly. In practice, policies vary as to how this grade issue is handled. Fortunately, because sight distance tends to be greater on downgrades (which require longer stopping distances) than on upgrades, a self- correction for the effect of grades is generally provided. As a consequence, some design agencies ignore the effect of grades completely, while others assume G is equal to zero for grades less than 3% and use simple adjustments to the SSD, depending on the initial and final grades, for grades of 3% or more. For the remainder of this chapter, we will ignore the effect of grades (G = 0 will be used in Eq. 3.12). However, it must be pointed out that the use of SSD grade corrections is very easy and straightforward, and all of the equations presented herein still apply.

The use of Eqs. 3.15 and 3.16 can be simplified if the initial assumption that L > SSD is made, in which case Eq. 3.15 is always used. The advantage of this assumption is that the relationship between A and Lm is linear, and Eq. 3.10 can be used to give

=mL KA (3.17)

where K = horizontal distance, in ft, required to effect a 1% change in the slope (as in Eq. 3.10), defined as

3.3 Vertical Alignment 69

= 2SSD

2158 K (3.18)

With known SSD for a given design speed (assuming G = 0), K-values can be computed for crest vertical curves as shown in Table 3.2. Thus, the minimum curve length can be obtained (as shown in Eq. 3.17) simply by multiplying A by the K-value read from Table 3.2.

Some discussion about the assumption that L > SSD is warranted. This assumption is made because there are two complications that could arise when SSD > L. First, if SSD > L, the relationship between A and Lm is not linear, so K-values cannot be used in the L = KA formula (Eq. 3.10). Second, at low values of A, it is possible to get negative minimum curve lengths (see Eq. 3.16). As a result of these complications, the assumption that L > SSD is almost always made in practice, and Eqs. 3.17 and 3.18 and the K-values presented in Table 3.2 are used. It is important to note that the assumption that L > SSD (upon which Eqs. 3.17 and 3.18 are based) is a good one because in many cases, L is greater than SSD, and when it is not (SSD > L), use of the L > SSD formula (Eq. 3.15 instead of Eq. 3.16) gives longer curve lengths and thus the error is on the conservative, safe side.

A final point relates to the smallest allowable length of curve. Very short vertical curves can be difficult to construct and may not be warranted for safety purposes. As a result, it is common practice to set minimum curve length limits that range from 100 to 325 ft depending on individual jurisdictional guidelines. A common alternative to these limits is to set the minimum curve length limit at three times the design speed (with speed in mi/h and length in ft) [AASHTO, 2018].

Table 3.2 Design Controls for Crest Vertical Curves Based on Stopping Sight Distance

Design speed (mi/h)

Stopping sight distance

(ft)

Rate of vertical curvature, K*

*Rate of vertical curvature, K, is the length of curve per percent algebraic difference in intersecting grades (A): K = L/A. Source: American Association of State Highway and Transportation Officials, A Policy on Geometric Design of Highways and Streets, 7th Edition, Washington, DC, 2018. Used by permission.

Calculated Design

15 80 3.0 3 20 115 6.1 7

25 155 11.1 12

30 200 18.5 19 35 250 29.0 29

40 305 43.1 44

45 360 60.1 61 50 425 83.7 84

55 495 113.5 114

60 570 150.6 151 65 645 192.8 193

70 730 246.9 247

75 820 311.6 312 80 910 383.7 384

70 Chapter 3 Geometric Design of Highways

EXAMPLE 3.6 DESIGN SPEED AND CREST VERTICAL CURVE DESIGN WITH K-VALUES

Solve Example 3.5 using the K-values in Table 3.2.

SOLUTION

From Example 3.5, A = 3. For a 70-mi/h design speed, K = 247 (from Table 3.2). Therefore, application of Eq. 3.17 gives

( )= = =247 3 741.00 ftmL KA

which is almost identical to the 740.82 ft obtained in Example 3.5. This difference is due to rounding. In this example, the rounded K of 247 was used as opposed to the calculated K of 246.9. The rounded values are typically used in design for computational convenience. Note, however, that fractional calculated values are always rounded up to the nearest integer value, to be conservative.

EXAMPLE 3.7 STOPPING SIGHT DISTANCE AND CREST VERTICAL CURVE DESIGN

If the grades in Example 3.5 intersect at station 100 + 00, determine the stationing of the PVC, PVT, and curve high point for the minimum curve length based on SSD requirements.

SOLUTION

Using the curve length from Example 3.6, L = 741 ft. Since the curve is equal tangent (as are virtually all curves used in practice), one-half of the curve will occur before the PVI and one-half after, so that

PVC is at 100 + 00 − L/2 = 100 + 00 minus 3 + 70.5 = 96 29.5+

PVT is at 100 + 00 + L/2 = 100 + 00 plus 3 + 70.5 = 103 70.5+

For the stationing of the high point, Eq. 3.11 is used:

( )× 1 = 247 1 = 247 fthlx = K G or

station 96 + 29.5 plus 2 + 47 = 98 76.5+

3.3.4 Stopping Sight Distance and Sag Vertical Curve Design Sag vertical curve design differs from crest vertical curve design in the sense that sight distance is governed by nighttime conditions because in daylight, sight distance on a sag vertical curve is unrestricted. Thus, the critical concern for sag vertical curve design is the length of roadway illuminated by the vehicle headlights, which is a function of the height of the headlight above the roadway and the

3.3 Vertical Alignment 71

inclined angle of the headlight beam, relative to the horizontal plane of the car. The sag vertical curve sight distance design problem is illustrated in Fig. 3.7.

Figure 3.7 Stopping sight distance considerations for sag vertical curves.

S = sight distance in ft, PVC = point of the vertical curve (the initial point of the curve), H = height of headlight in ft,

β = inclined angle of headlight beam in degrees,

PVI = point of vertical intersection (intersection of initial and final grades), and

L = length of the curve in ft, PVT = point of vertical tangent, which is the final point of the vertical curve (the point where the curve returns to the final grade or, equivalently, the final tangent).

To determine the minimum length of curve for a required sight distance, the properties of a parabola for an equal-tangent curve can be used to show that For S < L

( )β =

+

2

200 tanm AS

L H S (3.19)

For S > L

( )β+ = −

200 tan 2m

H S L S

A (3.20)

where

Lm = minimum length of vertical curve in ft, A = absolute value of the difference in grades (|G1−G2|), expressed as a percentage,

and other terms are as defined in Fig. 3.7.

72 Chapter 3 Geometric Design of Highways

For the sight distance required to provide adequate SSD, current AASHTO design guidelines [2018] use a headlight height of 2.0 ft and an upward angle of one degree. Substituting these design guidelines and S = SSD (as was done in the crest vertical curve case, and again ignoring the relatively small distance from the driver’s eye position to the front of the vehicle) into Eqs. 3.19 and 3.20 gives For SSD < L

×= + ×

2SSD 400 3 5 SSDm

A L

. (3.21)

For SSD > L + ×= × − 400 3 5 SSD

2 SSDm .

L A

(3.22)

where

SSD = stopping sight distance in ft, and other terms are as defined previously.

As was the case for crest vertical curves, K-values can be computed by assuming L > SSD, which gives us the linear relationship between Lm and A as shown in Eq. 3.21. Thus, for sag vertical curves (with Lm = KA),

= +

2SSD 400 3.5 SSD

K (3.23)

where

K = horizontal distance, in ft, required to effect a 1% change in the slope (as in Eq. 3.10), and

other terms are as defined previously.

The K-values corresponding to design-speed–based SSDs are presented in Table 3.3. As was the case for crest vertical curves, some caution should be exercised in using this table because the assumption that G = 0 (for determining SSD) is used. Also, assume that L > SSD is a safe, conservative assumption (as was the case for crest vertical curves), and the smallest allowable curve lengths for sag curves are the same as those for crest curves (see discussion in Section 3.3.3).

3.3 Vertical Alignment 73

Table 3.3 Design Controls for Sag Vertical Curves Based on Stopping Sight Distance

Design speed (mi/h)

Stopping

sight distance

(ft)

Rate of vertical curvature, K*

*Rate of vertical curvature, K, is the length of curve per percent algebraic difference in intersecting grades (A): K = L/A.

Source: American Association of State Highway and Transportation Officials, A Policy on Geometric Design of Highways and Streets, 7th Edition, Washington, DC, 2018. Used by permission.

Calculated Design

15 80 9.4 10 20 115 16.5 17 25 155 25.5 26 30 200 36.4 37 35 250 49.0 49 40 305 63.4 64 45 360 78.1 79 50 425 95.7 96 55 495 114.9 115 60 570 135.7 136 65 645 156.5 157 70 730 180.3 181 75 820 205.6 206 80 910 231.0 231

EXAMPLE 3.8 SAG VERTICAL CURVE FUNDAMENTALS WITH DESIGN SPEED

An equal-tangent sag vertical curve has an initial grade of –2.5%. It is known that the final grade is positive and that the low point is at elevation 270 ft and station 141 + 00. The PVT of the curve is at elevation 274 ft and the design speed of the curve is 35 mi/h. Determine the station and elevation of the PVC and PVI.

SOLUTION

From Table 3.3, it can be seen the K = 49 for a design speed of 35 mi/h. With this, Eq. 3.11 is used to find the distance of the low point from the PVC:

( )× 1 = 49 2.5 = 122.5 fthlx = K G

Knowing the elevation of the low point (270 ft) and the distance of the low point from the PVC (122.5 ft), Eq. 3.1 can be applied to determine the elevation of the PVC (c in the Eq. 3.1):

= + +2y ax bx c

From Eq. 3.3,

= = −1 0.025b G

and from Eq. 3.6,

− = 2 1

2

G G a

L

74 Chapter 3 Geometric Design of Highways

Because the final grade is known to be positive (and with G2 being negative),

G2 − G1 = |G1 − G2| = A,

Using L = KA from Eq. 3.17, Eq. 3.6 becomes

( ) −

= = = = =2 1 /100 0.01 0.01 0.000102

2 2 2 2 49

G G A a

L KA K Note that A is divided by 100 to make certain that the units are consistent with the denominator since L is in feet from the L = KA equation. At the low point, y = 270 ft so solving for c in Eq. 3.1 with x = 122.5, a = 0.000102 and b = −0.025 gives

= + − + = =

2270 0.000102(122.5) ( 0.025)(122.5)

271.53 elevation of the

c

c PVC

Knowing that the station of the low point is 141 + 00 and the distance from the PVC to the low point is 122.5 ft,

= + + = +station of 141 00 minus 1 22.5 139 77.5PVC

Next, the length of the curve is determined by applying Eq. 3.1. Because it is known that the elevation of the PVT is 274 ft, using y = 274 means that x = L in Eq. 3.1, so (with c = 271.53)

( ) = + +

= + − +

=

2

2

274 0.000102 0.025 271.53

320.624 ft

y aL bL c

L L

L

The station of the PVI is

= = = +

station of station of + /2

139 + 77.5 plus (3 + 20.624/2) 141 37.812

PVI PVC L

Finally, the elevation of the PVI is determined as

( ) ( )

= +

= − = 1elevation of elevation of /2

271.53 0.025 320.624 / 2 267.52

PVI PVC G L

EXAMPLE 3.9 COMBINED SAG AND CREST VERTICAL CURVES WITHOUT A CONSTANT-GRADE CONNECTION

An existing tunnel needs to be connected to a newly constructed bridge with sag and crest vertical curves. The profile view of the tunnel and bridge is shown in Fig. 3.8. Develop a vertical alignment to connect the tunnel and bridge by determining the highest possible common design speed for the sag and crest (equal-tangent) vertical curves needed. Compute the stationing and elevations of PVC, PVI, and PVT curve points.

3.3 Vertical Alignment 75

Figure 3.8 Profile view (vertical alignment diagram) for Example 3.9.

SOLUTION

From left to right (see Fig. 3.8), a sag vertical curve (with subscript s) and a crest vertical curve (with subscript c) are needed to connect the tunnel and bridge. From the given information, it is known that G1s = 0% (the initial slope of the sag vertical curve) and G2c = 0% (the final slope of the crest vertical curve). To obtain the highest possible design speed, we want to use all of the horizontal distance available. This means we want to connect the curve so that the PVT of the sag curve (PVTs) will be the PVC of the crest curve (PVCc). If this is the case, G2s = G1c and since G1s = G2c = 0, As = Ac = A, the common algebraic difference in the grades.

Since 1200 ft separates the tunnel and bridge,

+ = 1200s cL L Also, the summation of the end-of-curve offset for the sag curve and the beginning-of- curve offset (relative to the final grade) for the crest curve must equal 40 ft. Using the equation for the final offset, Eq. 3.9, we have

+ = 40 200 200

s cAL AL

Rearranging,

( )+ = 40 200 s c A

L L

and since Ls + Lc = 1200,

( ) =1200 40 200 A

Solving for A gives A = 6.667%. The problem now becomes one of finding K-values that allow Ls + Lc = 1200. Since L = KA (Eq. 3.17), we can write

+ = 1200s cK A K A Substituting A = 6.667,

+ = 180s cK K To find the highest possible design speed, Tables 3.2 and 3.3 are used to arrive at K- values to solve Ks + Kc = 180. From Tables 3.2 and 3.3, it is apparent that the highest possible design speed is 50 mi/h, at which speed Kc = 84 and Ks = 96 (the summation of K’s is 180).

76 Chapter 3 Geometric Design of Highways

To arrive at the stationing of curve points, we first determine curve lengths as

( )= = =96 6.667 640.0 fts sL K A

( )= = =84 6.667 560.0 ftc cL K A

Since the station of the PVCs is 0 00+ (given), it is clear that the PVIs = +3 20.0,

PVTs = PVCc = +6 40.0, PVIc = +9 20.0, and PVTc = +12 00.0. For elevations,

PVCs = PVIs = 100 ft and PVIc = PVTc = 140 ft. Finally, the elevation of PVTs and

PVCc can be computed as

( ) + = + =

6.667 640.0 100 100 121.33 ft

200 200 sAL

EXAMPLE 3.10 COMBINED SAG AND CREST VERTICAL CURVES WITH A CONSTANT-GRADE CONNECTION

Consider the conditions described in Example 3.9. Suppose that a design speed of only 35 mi/h is needed. Determine the lengths of curves required to connect the bridge and tunnel while keeping the connecting grade as small as possible.

SOLUTION

It is known that the 1200 ft separating the tunnel and bridge are more than enough to connect a 35-mi/h alignment because Example 3.9 showed that 50 mi/h is possible. Therefore, to connect the tunnel and bridge and keep the connecting grade as small as possible, we will place a constant-grade section between the sag and crest curves (as shown in Fig. 3.9).

The elevation change will be the final offsets of the sag and crest curves plus the change in elevation resulting from the constant-grade section connecting the two curves. Let Gcon be the grade of the constant-grade section. This means that G2s = G1c = Gcon, and since G1s = G2c = 0 (as in Example 3.9), Gcon = As = Ac = A. The equation that will solve the vertical alignment for this problem is

( )− − + + =

1200 40

200 200 100 s cs c A L LAL AL

where the third term accounts for the elevation difference attributable to the constant-grade section connecting the sag and crest curves (the 100 in the denominator of this term converts A from percent to ft/ft). Using L = KA, we have

( )− − + + =

2 2 1200 40

200 200 100 s cs c A K A K AA K A K

From Table 3.2, Kc = 29, and from Table 3.3, Ks = 49. Putting these values in the above equation gives

+ − =

− + − =

2 2

2

0.39 12 0.78 40

0.39 12 40 0

A A A

A A

3.3 Vertical Alignment 77

Solving this gives A = 3.803 and A = 26.966; A = 3.803% is chosen because we want to minimize the grade. For this value of A, the curve lengths are

( )= = =49 3.803 186.35 fts sL K A

( )= = =29 3.803 110.29 ftc cL K A

and the length of the constant-grade section will be 903.36 ft. This means that about 34.35 ft of the elevation difference will occur in the constant-grade section, with the remainder of the elevation difference attributable to the final curve offsets.

Figure 3.9 Profile view (vertical alignment diagram) for Example 3.10.

Another variation of this type of problem is the case when the initial and

final grades are not equal to zero, as in the following example.

EXAMPLE 3.11 COMBINED SAG AND CREST VERTICAL CONNECTING HIGHWAY SEGMENTS WITH NONZERO GRADES

Two sections of highway are separated by 1800 ft, as shown in Fig. 3.10. Determine the curve lengths required for a 60-mi/h vertical alignment to connect these two highway segments while keeping the connecting grade as small as possible.

Figure 3.10 Profile view (vertical alignment diagram) for Example 3.11.

SOLUTION

Let Yfc and Yfs be the final offsets of the crest and sag curves, respectively. Let Gcon be the slope of a constant-grade section connecting the crest and sag curves (we will assume that the horizontal distance is sufficient to connect the highway with a 60-mi/h alignment; if this assumption is incorrect, the following equations will

78 Chapter 3 Geometric Design of Highways

produce an obviously erroneous answer and a lower design speed will have to be chosen). Finally, let ∆ycon be the change in elevation over the constant-grade section, and let ∆yc and ∆ys be the changes in elevation due to the extended curve tangents. The elevation equation is then (see Fig. 3.10)

+ + Δ + Δ = + Δ30 fc fs con s cY Y y y y

Substituting offset equations and equations for elevation changes (with subscripts c for crest and s for sag),

( )− − + + + = +

1800 1.0 3.0 30

200 200 100 100 100 con c sc c s s s cG L LA L A L L L

Using L = KA, this equation becomes

( )− − + + + = +

2 2 1800 1.0 3.0 30

200 200 100 100 100 con c c s sc c s s s s c cG K A K AA K A K K A K A

From Tables 3.2 and 3.3, Kc = 151 and Ks = 136. Substituting and defining A’s (and arranging the equation so that Gcon will be positive, and assuming Gcon will be greater than 1%) gives

( ) ( ) ( ) ( )( )

( )( ) ( )( )

− + − −+ − + +

− + + = +

2 2 1800 151 3 136 13 151 1 136

200 200 100

1.0 136 1 3.0 151 3 30

100 100

con con concon con

con con

G G GG G

G G

or

− + − =21.435 14.83 37.475 0con conG G which gives Gcon = 4.40 (the other possible solution is 5.93, which is rejected because we want to minimize the grade). Using L = KA gives Lc = 1117.40 ft (151 × 7.40) and

Ls = 462.40 ft (136 × 3.40). Accordingly, the length of the constant-grade section is

220.20 ft (1800 − Lc − Ls). Elevations and the locations of curve points can be readily computed with this information.

3.3.5 Passing Sight Distance and Crest Vertical Curve Design In addition to SSD, in some instances, it may be desirable to provide adequate passing sight distance, which can be an important issue in two-lane highway design (one lane in each direction). Passing sight distance is a factor only in crest vertical curve design because, for sag curves, the sight distance is unobstructed looking up or down the grade, and at night, the headlights of oncoming or opposing vehicles will be seen. In determining the sight distance required to pass on a crest vertical curve, Eqs. 3.13 and 3.14 will apply; however, whereas the driver’s eye height, H1, will remain at 3.5 ft, H2 will now also be set to 3.5 ft. This value for H2 is the assumed value for the portion of a vehicle’s height necessary to be visible such that it can be recognized as an opposing vehicle to a driver performing a passing maneuver. Using the same height for both H1 and H2 provides a reciprocal design relationship; that is, if the driver of the passing vehicle can see the opposing vehicle, then the opposing

3.3 Vertical Alignment 79

vehicle driver can see the passing vehicle. Substituting these H values into Eqs. 3.13 and 3.14 and letting the sight distance S equal the passing sight distance, PSD, gives

For PSD < L

×= 2PSD

2800m A

L (3.24)

For PSD > L

= × − 2800 2 PSDmL

A (3.25)

where

Lm = minimum length of vertical curve in ft,

A = absolute value of the difference in grades (|G1−G2|), expressed as a percentage, and

PSD = passing sight distance in ft.

As was the case for SSD, it is typically assumed that the length of curve is greater than the required sight distance (in this case, L > PSD), so

= 2PSD

2800 K (3.26)

where

K = horizontal distance, in ft, required to effect a 1% change in the slope (as in Eq. 3.10), and

PSD = passing sight distance in ft.

The passing sight distance (PSD) used for design is assumed to consist of four distances: (1) the initial maneuver distance (which includes the driver’s perception/reaction time and the time it takes to bring the vehicle from its trailing speed to the point of encroachment on the left lane), (2) the distance that the passing vehicle traverses while occupying the left lane, (3) the clearance length between the passing and opposing vehicles at the end of the passing maneuver, and (4) the distance traversed by the opposing vehicle during two-thirds of the time the passing vehicle occupies the left lane. The determination of these distances is undertaken using assumptions regarding the time of the initial maneuver, average vehicle acceleration, and the speeds of passing, passed, and opposing vehicles. The sum of these four distances gives the required passing sight distance. The reader is referred to [AASHTO, 2018] for a complete description of the assumptions made in determining required passing sight distances.

80 Chapter 3 Geometric Design of Highways

The minimum distances needed to pass (PSDs) at various design speeds, along with the corresponding K-values as computed from Eq. 3.26, are presented in Table 3.4. Notice that the K-values in this table are much higher than those required for SSD (as given in Table 3.2). As a result, designing a crest curve to provide adequate passing sight distance is often an expensive proposition (due to the length of curve required).

Table 3.4 Design Controls for Crest Vertical Curves Based on Passing Sight Distance.

Design speed (mi/h)

Passing sight distance (ft)

Rate of vertical curvature, K*

*Rate of vertical curvature, K, is the length of curve per percent algebraic difference in intersecting grades (A): K = L/A.

Source: American Association of State Highway and Transportation Officials, A Policy on Geometric Design of Highways and Streets, 7th Edition, Washington, DC, 2018. Used by permission.

20 400 57 25 450 72 30 500 89 35 550 108 40 600 129 45 700 175 50 800 229 55 900 289 60 1000 357 65 1100 432 70 1200 514 75 1300 604 80 1400 700

EXAMPLE 3.12 VERTICAL CURVE DESIGN WITH PASSING SIGHT DISTANCE

An equal-tangent crest vertical curve is 1000 ft long and connects a +2.5% and a −1.5% grade. If the design speed of the roadway is 55 mi/h, does this curve have adequate passing sight distance?

SOLUTION

To determine the length of curve required to provide adequate passing sight distance at a design speed of 55 mi/h, we use L = KA with K = 289 (as read from Table 3.4). This gives

L = 289(4.0) = 1156 ft Since the curve is only 1000 ft long, it is not long enough to provide adequate passing sight distance. Alternatively, the K-value for the existing design can be compared with that required for a PSD-based design. The K-value for the existing design is

1000 250

4 K = =

Since the K-value of 250 for the existing curve design is less than 289, this curve does not provide adequate PSD for a 55-mi/h design speed.

3.3 Vertical Alignment 81

3.3.6 Underpass Sight Distance and Sag Vertical Curve Design As mentioned in Section 3.3.4, design for sag curves is based on nighttime conditions because during daytime conditions a driver can see the entire sag curve. However, in the case of a sag curve being built under an overhead structure (such as roadway or railroad crossing), a driver’s line of sight may be restricted so that the entire curve length is not visible. An example of this situation is shown in Fig. 3.11.

In designing the sag curve, it is essential that the curve be long enough to provide a suitably gradual rate of curvature such that the overhead structure does not block the line of sight and allows the required SSD for the specified design speed to be maintained.

Figure 3.11 Stopping sight distance considerations for underpass sag curves. Used by permission from American Association of State Highway and Transportation Officials, A Policy on Geometric Design of Highways and Streets, 7th Edition, Washington, DC, 2018.

S = sight distance in ft, G1 = initial roadway grade in percent or ft/ft,

H1 = height of driver’s eye in ft, G2 = final roadway grade in percent or ft/ft,

H2 = height of object in ft, PVC = point of the vertical curve (the initial point of the curve), and Hc = clearance height of overpass structure

above roadway in ft, PVT = point of vertical tangent, which is the final point of the vertical curve (the point where the curve returns to the final grade or, equivalently, the final tangent).

L = length of the curve in ft

Again, by using the properties of a parabola for an equal-tangent vertical curve, it can be shown that the minimum length of sag curve for a required sight distance and clearance height is

82 Chapter 3 Geometric Design of Highways

For S < L

=  + −  

  

2

1 2800 2

m

c

AS L

H H H

(3.27)

For S > L

 + −     = −

1 2800 2

2 c

m

H H H

L S A

(3.28)

where

Lm = minimum length of vertical curve in ft, A = absolute value of the difference in grades (|G1−G2|), expressed as a percentage,

and other terms are as defined in Fig. 3.11.

Current AASHTO design guidelines [2018] use a driver eye height, H1, of 8 ft for a truck driver, and an object height, H2, of 2 ft for the taillights of a vehicle. Substituting these values and S = SSD into Eqs. 3.27 and 3.28 gives For SSD < L

( ) ×=

−

2SSD 800 5c

A L

H (3.29)

For SSD > L

( )− = × −

800 5 2 SSD cH

L A

(3.30)

where

SSD = stopping sight distance in ft, and other terms are as defined previously.

In the case where there is an existing sag curve alignment and a new

overpass structure is to be built over it, the above equations can be rearranged to solve for the necessary clearance height, Hc, of the overpass structure to provide for the required SSD. When the clearance height is determined in this manner, it is necessary to check this value against the minimum clearance heights based on maximum vehicle height regulations and AASHTO recommendations. Maximum vehicle heights as regulated by state laws range from 13.5 to 14.5 ft. AASHTO [2018] recommends a minimum structure clearance height of 14.5 ft and a desirable clearance height of 16.5 ft. AASHTO [2018] also recommends that clearance heights be no less than 1 ft greater than

3.3 Vertical Alignment 83

the maximum allowable vehicle height. This provides a margin for snow or ice accumulation, some over-height vehicles, and future roadway resurfacings. Thus, in building a new overpass structure over an existing sag curve alignment, the clearance height must be determined for both required SSD and maximum allowable vehicle height for that roadway, and the greater of the two values should be used.

EXAMPLE 3.13 UNDERPASS VERTICAL CURVE CLEARANCE

An equal-tangent sag curve has an initial grade of −4.0%, a final grade of +3.0%, and a length of 1270 ft. An overpass is being placed directly over the PVI of this curve. At what height above the roadway should the bottom of this sign be placed?

SOLUTION

For this situation, Eq. 3.29 or 3.30 must be used to solve for the necessary clearance height based on SSD. Thus, the required SSD must be determined for the given sag curve specifications, based on the design speed. The design speed for the curve can be determined from the K-value by applying Eq. 3.10 as follows:

= = = − − 1270

181.4 4 3

L K

A From Table 3.3, this K-value corresponds approximately to a design speed of 70 mi/h (K = 181). For a 70-mi/h design speed, the required SSD is 730 ft. Since the curve length is greater than the required SSD (1270 > 730), Eq. 3.29 applies:

( ) ×=

−

2SSD 800 5c

A L

H Rearranging this equation to solve for the clearance height, Hc, and substituting A = 7%, SSD = 730 ft, and L = 1270 ft gives

( )

×= +

×= +

=

2

2

SSD 5

800

7 730 5

800 1270

8.67 ft

c A

H L

Although only 8.67 ft is needed for SSD requirements, AASHTO [2018] recommends a minimum clearance height of 14.5 ft to take maximum vehicle height into account. Thus, the bottom of the overpass should be placed at least 14.5 ft

above the roadway surface (at the PVI), but desirably at a height of 16.5 ft according to AASHTO [2018].

84 Chapter 3 Geometric Design of Highways

3.4 HORIZONTAL ALIGNMENT

The critical aspect of horizontal alignment is the horizontal curve, with the focus on design of the directional transition of the roadway in a horizontal plane. Stated differently, a horizontal curve provides a transition between two straight (or tangent) sections of roadway. A key concern in this directional transition is the ability of the vehicle to negotiate a horizontal curve. (Provision of adequate drainage is also important, but is not discussed in this book; see [AASHTO, 2018].) As was the case with the straight-line vehicle performance characteristics discussed at length in Chapter 2, the highway engineer must design a horizontal alignment to accommodate the cornering capabilities of a variety of vehicles, ranging from nimble sports cars to ponderous trucks. A theoretical assessment of vehicle cornering at the level of detail given to straight-line performance in Chapter 2 is beyond the scope of this book (see [Campbell, 1978] and [Wong, 2008]). Instead, vehicle cornering performance is viewed only at the practical design-oriented level, with equations simplified in a manner similar to that used for the stopping-distance equation discussed in Section 2.9.5.

3.4.1 Vehicle Cornering Figure 3.12 illustrates the forces acting on a vehicle during cornering.

Figure 3.12 Vehicle cornering forces.

Rv = radius defined to the vehicle’s traveled path in ft,

Wp = vehicle weight parallel to the roadway surface in lb,

α = angle of incline in degrees, Ff = side frictional force (centripetal, in lb),

e = number of vertical ft of rise per 100 ft of horizontal distance,

Fc = centripetal force (lateral acceleration × mass, in lb),

W = weight of the vehicle in lb, Fcp = centripetal force acting parallel to the roadway surface in lb, and

Wn = vehicle weight normal to the roadway surface in lb,

Fcn = centripetal force acting normal to the roadway surface in lb.

Some basic horizontal curve relationships can be derived by noting that

+ =p f cpW F F (3.31)

3.4 Horizontal Alignment 85

From basic physics this equation can be written as [with Ff = fs(Wn + Fcn)]

α α α α

  + + = 

 

2 2

sin cos sin cos s v v

WV WV W f W

gR gR (3.32)

where

fs = coefficient of side friction (unitless), V = vehicle speed in ft/s, g = gravitational constant, 32.2 ft/s2, and

other terms are as defined in Fig. 3.12. Dividing both sides of Eq. 3.32 by W cos gives

( ) 2

tan 1 tans s v

V f f

gR α α+ = − (3.33)

The term tanα indicates the superelevation of the curve (banking) and can be expressed in percent; it is denoted e (e = 100 tan α). In words, the superelevation is the number of vertical feet of rise per 100 feet of horizontal distance (see Fig. 3.12). The term fs tanα in Eq. 3.33 is conservatively set equal to zero for practical applications due to the small values that fs and α typically assume (this is equivalent to ignoring the normal component of centripetal force). With e = 100 tan α, Eq. 3.33 can be arranged as

=  +   

2

100

v

s

V R

e g f

(3.34)

EXAMPLE 3.14 SUPERELEVATION ON HORIZONTAL CURVES

A roadway is being designed for a speed of 70 mi/h. At one horizontal curve, it is known that the superelevation is 8.0% and the coefficient of side friction is 0.10. Determine the minimum radius of curve (measured to the traveled path) that will provide for safe vehicle operation.

SOLUTION

The application of Eq. 3.34 gives [with 1.467 (5280/3600) converting mi/h to ft/s]

( ) ( )

× = = =

+ +   

22 70 1.467 1819.40 ft

32.2 0.10 0.08 100

v

s

V R

e g f

This value is the minimum radius because radii smaller than 1819.40 ft will generate centripetal forces higher than those that can be safely supported by the superelevation and the side frictional force.

In the actual design of a horizontal curve, the engineer must select appropriate values of e and fs. The value selected for superelevation, e, is critical because high

86 Chapter 3 Geometric Design of Highways

rates of superelevation can cause vehicle steering problems on the horizontal curve, and in cold climates, ice on the roadway can reduce fs such that vehicles traveling at less than the design speed on an excessively superelevated curve could slide inward off the curve due to gravitational forces. AASHTO provides general guidelines for the selection of e and fs for horizontal curve design, as shown in Table 3.5. The values presented in this table are grouped by five values of maximum e. The selection of any one of these five maximum e values is dependent on the type of road (for example, higher maximum e’s are permitted on freeways than on arterials and local roads) and local design practice. Limiting values of fs are simply a function of design speed. Table 3.5 also presents calculated radii (given V, e, and fs) by applying Eq. 3.34.

3.4.2 Horizontal Curve Fundamentals In connecting straight (tangent) sections of roadway with a horizontal curve, several options are available. The most obvious of these is the simple circular curve, which is just a curve with a single, constant radius. Other options include reverse curves, compound curves, and spiral curves. Reverse curves generally consist of two consecutive curves that turn in opposite directions. They are used to shift the alignment of a highway laterally. The curves used are usually circular and have equal radii. Reverse curves, however, are not recommended because drivers may find it difficult to stay within their lane as a result of sudden changes to the alignment. Compound curves consist of two or more curves, usually circular, in succession. Compound curves are used to fit horizontal curves to very specific alignment needs, such as interchange ramps, intersection curves, or difficult topography. In designing compound curves, care must be taken not to have successive curves with widely different radii, as this will make it difficult for drivers to maintain their lane position as they transition from one curve to the next. Spiral curves are curves with a continuously changing radius. Spiral curves are sometimes used to transition a tangent section of roadway to a circular curve. In such a case, the radius of the spiral curve is equal to infinity where it connects to the tangent section and ends with the radius value of the connecting circular curve at the other end. Because motorists usually create their own transition paths between tangent sections and circular curves by utilizing the full lane width available, spiral curves are not often used. However, there are exceptions. Spiral curves are sometimes used on high-speed roadways with sharp horizontal curves and also to gradually introduce the superelevation of an upcoming horizontal curve. To illustrate the basic principles involved in horizontal curve design, this book will focus only on the single simple circular curve. For detailed information regarding these additional horizontal curve types, refer to standard route-surveying texts, such as Hickerson [1964]. Figure 3.13 shows the basic elements of a simple horizontal curve.

3.4 Horizontal Alignment 87

Table 3.5 Minimum Radius Using Limiting Values of e and fs

Design speed (mi/h)

Maximum e (%)

Limiting values of fs

Total (e/100 +

fs)

Calculated radius, Rv (ft)

Rounded radius, Rv (ft)

Design speed (mi/h)

Maximum e (%)

Limiting values of fs

Total (e/100 +

fs)

Calculated radius, Rv (ft)

Rounded radius, Rv (ft)

10 4.0 0.38 0.42 15.9 16 10 10.0 0.38 0.48 13.9 14 15 4.0 0.32 0.36 41.7 42 15 10.0 0.32 0.42 35.7 36 20 4.0 0.27 0.32 86.0 86 20 10.0 0.27 0.37 72.1 72 25 4.0 0.23 0.27 154.3 154 25 10.0 0.23 0.33 126.3 126 30 4.0 0.20 0.24 250.0 250 30 10.0 0.20 0.30 200.0 200 35 4.0 0.18 0.22 371.2 371 35 10.0 0.18 0.28 291.7 292 40 4.0 0.16 0.20 533.3 533 40 10.0 0.16 0.26 410.3 410 45 4.0 0.15 0.19 710.5 711 45 10.0 0.15 0.25 540.0 540 50 4.0 0.14 0.18 925.9 926 50 10.0 0.14 0.24 694.4 694 55 4.0 0.13 0.17 1186.3 1190 55 10.0 0.13 0.23 876.8 877 60 4.0 0.12 0.16 1500.0 1500 60 10.0 0.12 0.22 1090.9 1090

65 10.0 0.11 0.21 1341.3 1340 10 6.0 0.38 0.44 15.2 15 70 10.0 0.10 0.20 1633.3 1630 15 6.0 0.32 0.38 39.5 39 75 10.0 0.09 0.19 1973.7 1970 20 6.0 0.27 0.33 80.8 81 80 10.0 0.08 0.18 2370.4 2370 25 6.0 0.23 0.29 143.7 144 30 6.0 0.20 0.26 230.8 231 10 12.0 0.38 0.50 13.3 13 35 6.0 0.18 0.24 340.3 340 15 12.0 0.32 0.44 34.1 34 40 6.0 0.16 0.22 484.8 485 20 12.0 0.27 0.39 68.4 68 45 6.0 0.15 0.21 642.9 643 25 12.0 0.23 0.35 119.0 119 50 6.0 0.14 0.20 833.3 833 30 12.0 0.20 0.32 187.5 188 55 6.0 0.13 0.19 1061.4 1060 35 12.0 0.18 0.30 272.2 272 60 6.0 0.12 0.18 1333.3 1330 40 12.0 0.16 0.28 381.0 381 65 6.0 0.11 0.17 1656.9 1660 45 12.0 0.15 0.27 500.0 500 70 6.0 0.10 0.16 2041.7 2040 50 12.0 0.14 0.26 641.0 641 75 6.0 0.09 0.15 2500.0 2500 55 12.0 0.13 0.25 806.7 807 80 6.0 0.08 0.14 3047.6 3050 60 12.0 0.12 0.24 1000.0 1000

65 12.0 0.11 0.23 1224.6 1220 10 8.0 0.38 0.46 14.5 14 70 12.0 0.10 0.22 1484.8 1480 15 8.0 0.32 0.40 37.5 38 75 12.0 0.09 0.21 1785.7 1790 20 8.0 0.27 0.35 76.2 76 80 12.0 0.08 0.20 2133.3 2130 25 8.0 0.23 0.31 134.4 134 30 8.0 0.20 0.28 214.3 214 35 8.0 0.18 0.26 314.1 314 40 8.0 0.16 0.24 444.4 444 45 8.0 0.15 0.23 587.0 587 50 8.0 0.14 0.22 757.6 758 55 8.0 0.13 0.21 960.3 960 60 8.0 0.12 0.20 1200.0 1200 65 8.0 0.11 0.19 1482.5 1480 70 8.0 0.10 0.18 1814.8 1810 75 8.0 0.09 0.17 2205.9 2210 80 8.0 0.08 0.16 2666.7 2670

Note: In recognition of safety considerations, use of emax = 4.0% should be limited to urban conditions.

Source: American Association of State Highway and Transportation Officials, A Policy on Geometric Design of Highways and Streets, 7th Edition, Washington, DC, 2018. Used by permission.

88 Chapter 3 Geometric Design of Highways

Figure 3.13 Elements of a simple circular horizontal curve.

R = radius, usually measured to the centerline of the road in ft,

PC = point of curve (the beginning point of the horizontal curve),

∆ = central angle of the curve in degrees, PI = point of tangent intersection, T E

= =

tangent length in ft, external distance in ft,

PT = point of tangent (the ending point of the horizontal curve), and

M = middle ordinate in ft, L = length of curve in ft.

Another important term is the degree of curve, which is defined as the angle subtended by a 100-ft arc along the horizontal curve. It is a measure of the sharpness of the curve and is frequently used instead of the radius in the construction of the curve. The degree of curve is directly related to the radius of the horizontal curve by

π π

     = =

100 100

18000 D

R R (3.35)

where

D = degree of curve [angle subtended by a 100-ft arc along the horizontal curve], and

other terms are as defined in Fig. 3.13. Note that the quantity 180/π converts from radians to degrees.

Geometric and trigonometric analyses of Fig. 3.13 reveal the following relationships:

Δ= tan 2

T R (3.36)

3.4 Horizontal Alignment 89

( )

  = −  Δ 

1 1

cos 2 E R (3.37)

Δ = −    1 cos

2 M R (3.38)

π= Δ 180

L R (3.39)

where all terms are as defined in Fig. 3.13. It is important to note that horizontal curve stationing, curve length, and

curve radius (R) are usually measured to the centerline of the road. In contrast, the radius determined on the basis of vehicle forces (Rv in Eq. 3.34) is measured from the innermost vehicle path, which is assumed to be the midpoint of the innermost vehicle lane. Thus, a slight correction for lane width is required in equating the Rv of Eq. 3.34 with the R in Eqs. 3.35 through 3.39.

EXAMPLE 3.15 STATIONING ON HORIZONTAL CURVES

A horizontal curve is designed with a 2000-ft radius. The curve has a tangent length of 400 ft and the PI is at station 103 + 00. Determine the stationing of the PT.

SOLUTION

Equation 3.36 is applied to determine the central angle, ∆:

Δ=

Δ=

Δ = °

tan 2

400 2000 tan 2

22.62

T R

So, from Eq. 3.39, the length of the curve is

π= Δ

= =

180 3.1416

2000(22.62) 789.58 ft 180

L R

L

Given that the tangent length is 400 ft,

station of PC = 103 + 00 minus 4 + 00 = 99 + 00 Since horizontal curve stationing is measured along the alignment of the road,

= + = + + = +

station of station of

99 00 plus 7 89.58 106 89.58

PT PC L

90 Chapter 3 Geometric Design of Highways

3.4.3 Stopping Sight Distance and Horizontal Curve Design As is the case for vertical curve design, adequate SSD must be provided in the design of horizontal curves. Sight distance restrictions on horizontal curves occur when obstructions are present, as shown in Fig. 3.14. Such obstructions are frequently encountered in highway design due to the cost of right-of-way acquisition or the cost of moving earthen materials, such as rock outcroppings. When such an obstruction exists, the SSD is measured along the horizontal curve from the center of the traveled lane (the assumed location of the driver’s eyes). As shown in Fig. 3.14, for a specified stopping distance, some distance Ms (the middle ordinate of a curve that has an arc length equal to the SSD) must be visually cleared so that the line of sight is such that sufficient SSD is available.

Figure 3.14 Stopping sight distance considerations for horizontal curves.

R = radius measured to the centerline of the road in ft,

Ms = middle ordinate necessary to provide adequate stopping sight distance (SSD) in ft.

Rv = radius to the vehicle’s traveled path (usually measured to the center of the innermost lane of the road) in ft,

SSD = stopping sight distance in ft,

PC = point of curve (the beginning point of the

Δ = central angle of the curve in degrees, horizontal curve), and

Δs = angle (in degrees) subtended by an arc equal in length to the required stopping sight distance (SSD),

PT = point of tangent (the ending point of the horizontal curve).

L = length of curve in ft,

Equations for computing SSD relationships for horizontal curves can be derived by first determining the angle, ∆s, for an arc length equal to the required SSD (see Fig. 3.14 and note that this is not the central angle, ∆ of the horizontal curve whose arc length is equal to L). Assuming that the length of the horizontal curve exceeds the required SSD (as shown in Fig. 3.14), we have (as with Eq. 3.39)

3.4 Horizontal Alignment 91

π= ΔSSD 180 v sR (3.40)

Rearranging terms,

π Δ = 180 SSD

s vR

(3.41)

Substituting this into the general equation for the middle ordinate of a simple horizontal curve (Eq. 3.38) to get an expression for Ms gives

π  

= −   

90SSD 1 coss v

v

M R R

(3.42)

Solving Eq. 3.42 for SSD gives

π − −  =   

   

1SSD cos 90

v sv

v

R MR

R (3.43)

Note that Eqs. 3.40 to 3.43 can also be applied directly to determine sight distance requirements for passing. If these equations are to be used for passing, distance values given in Table 3.4 would apply and SSD in the equations would be replaced by PSD.

EXAMPLE 3.16 HORIZONTAL CURVE STATIONING AND DESIGN SPEED

A horizontal curve on a four-lane highway (two lanes each direction with no median) has a superelevation of 6% and a central angle of 40 degrees. The PT of the curve is at station 322 + 50 and the PI is at 320 + 08. The road has 10-ft lanes and 8-ft shoulders on both sides with high retaining walls going up immediately next to the shoulders. What is the highest safe speed of this curve (highest in 5 mi/h increments) and what is the station of the PC?

SOLUTION

The tangent will be equal to the PI – PC so T = 320 + 08 – PC. The length of the curve will be equal to the PT – PC so L = 322 + 50 – PC. With the equations for the tangent and length of curve both put in terms of the PC, Eqs. 3.36 and 3.39 can be rearranged, respectively, as,

π π= = = Δ ΔΔ Δ

and so that tan tan

2 180 2 180

T L T L R R

92 Chapter 3 Geometric Design of Highways

Substituting the previous tangent and length-of-curve equations (T = 32008 – PC and L = 32250 – PC),

π − −=32008 32250 40

tan 40 2 180

PC PC

Which gives = +317 44.25.PC Using this value of PC, the tangent can be computed as T = 32008 – 31744.25 = 263.75 ft. This value of T can then be used to determine R (see equations above),

= = Δ

263.75 = 724.59 ft

40 tan tan

2 2

T R

Because the curve radius is usually taken to the centerline of the roadway and there are two 10-ft lanes before the centerline (working from the inside of the curve to the outside), Rv = R − 10 − 10/2 = 724.59 − 15 = 709.59 ft. From Table 3.5 with a superelevation of 6%, at 45 mi/h a radius of 643 ft is needed; and at 50 mi/h a radius of 833 ft is needed. Therefore, the highest deign speed for centripetal force is 45 mi/h (since 709.59 > 643, the design is acceptable for 45 mi/h because more than the needed radius is available, but with 833 > 709.59 the design is not acceptable for 50 mi/h since insufficient radius is available). To check for adequate sight distance, Ms is going to be the shoulder width plus half of the inside lane width or 8 + 10/2 = 13 ft. Consider the SSD required at 40 mi/h. At 40 mi/h, the required SSD is 305 ft (from Table 3.1). Applying Eq. 3.42 gives,

( ) ( )π π

   = − = − =       

90 30590SSD 1 cos 709.59 1 cos 16.34 ft

709.59s v v

M R R

Because 16.34 ft is greater than the 13 ft of available Ms, 40 mi/h is too fast. Consider a speed of 35 mi/h which gives SSD = 250 ft (from Table 3.1). The application of Eq. 3.42 then gives,

( ) ( )π π

   = − = − =       

90 20590SSD 1 cos 709.59 1 cos 10.99 ft

709.59s v v

M R R

Because 10.99 ft is less than 13 ft, the highway is safe at 35 mi/h. Considering both the maximum safe speeds for centripetal force (45 mi/h) and sight distance (35 mi/h), the lower of the two speeds will govern. Thus, 35 mi/h (the highest safe speed for sight distance) is the lower of the two speeds and is the highest safe speed for this curve.

3.5 COMBINED VERTICAL AND HORIZONTAL ALIGNMENT

Thus far the discussion on highway alignment has treated vertical and horizontal curves independently. The combination of vertical and horizontal curves, however, is quite common in geometric design, and often necessary. Obvious examples are highways through mountainous terrain and freeway interchange

3.5 Combined Vertical and Horizontal Alignment 93

ramp roadways, which typically have to make significant changes in direction and elevation over a relatively short distance.

As previously mentioned, the design of an alignment that consists of a vertical and horizontal curve in combination usually consists of 2 two- dimensional alignment problems. The following examples illustrate this process.

EXAMPLE 3.17 COMBINED HORIZONTAL/VERTICAL ALIGNMENT DESIGN ADEQUACY

A two-lane highway (two 12-ft lanes) has a posted speed limit of 50 mi/h and, on one section, has both horizontal and vertical curves, as shown in Fig. 3.15. A recent daytime crash (driver traveling eastbound and striking a stationary roadway object) resulted in a fatality and a lawsuit alleging that the 50-mi/h posted speed limit is an unsafe speed for the curves in question and was a major cause of the crash. Evaluate and comment on the roadway design.

Figure 3.15 Horizontal and vertical alignment for Example 3.17.

SOLUTION

Begin with an assessment of the horizontal alignment. Two concerns must be considered: the adequacy of the curve radius and superelevation and the adequacy of the sight distance on the eastbound (inside) lane. For the curve radius, note from Fig. 3.15 that

= − = + + =

station of station of

32 75 minus 16 00 1675 ft

L PT PC

94 Chapter 3 Geometric Design of Highways

Rearranging Eq. 3.39, we get

( ) ( ) π π

= = = Δ

180 180 1675 1199.63 ft

80 R L

Using the posted speed limit of 50 mi/h with e = 8.0%, we find that Eq. 3.34 can be rearranged to give (with the vehicle traveling in the middle of the inside lane, Rv = R − half the lane width, or Rv = 1199.63 − 6 = 1193.63 ft)

( ) ( )

22 50 1.467 0.08 0.060

32.2 1193.63s v

V f e

gR

× = − = − =

From Table 3.5, the maximum fs for 50 mi/h is 0.14. Since 0.060 does not exceed 0.14, the radius and superelevation are sufficient for the 50-mi/h design speed. For sight distance, the available Ms is 18 ft plus the 6-ft distance to the center of the eastbound (inside) lane, or 24 ft. Application of Eq. 3.43 gives

( )

π

π

−

−

  − =   

     − =   

   =

1

1

SSD cos 90

1193.63 1193.63 24 cos

90 1193.63

479.5 ft

v v s

v

R R M

R

From Table 3.1, the required SSD at 50 mi/h is 425 ft, so the 479.5 ft of SSD provided is sufficient. Turning to the sag vertical curve, the length of curve is

= − = + + =

station of station of

18 80 minus 14 00 480 ft

L PVT PVC

Using A = 6% (from Fig. 3.15) and applying Eq. 3.10, we obtain

= = =480 80

6 L

K A

For the 50-mi/h design speed, Table 3.3 indicates a necessary K-value of 96. Thus, the K-value of 80 reveals that the curve is inadequate for the 50-mi/h speed. However, because the crash occurred in daylight and sight distances on sag vertical curves are governed by nighttime conditions, this design did not contribute to the crash.

EXAMPLE 3.18 DESIGN OF A COMBINED HORIZONTAL/VERTICAL ALIGNMENT

A new highway is to be constructed over an existing highway. The two highways will intersect at right angles and are to be grade-separated. Both highways are level grade (constant elevation). The new highway will run east-west and the existing highway runs north-south at elevation of 565.5 ft. The proposed bridge structure for the new highway is such that the bridge girder thickness is 6 ft (measured from the road surface to the bottom of the girder). A single-lane ramp is to be constructed to allow eastbound traffic to go southbound. A single horizontal curve, with a central angle of 90 degrees, is to be used. With a design speed of 40 mi/h and a required superelevation of 4%, determine the following: the stationing of the PC, PI, and PT, assuming that the curve begins at station 40 + 00; the stationing and elevation of all key points along the vertical alignment; the distance that must be cleared from the

3.5 Combined Vertical and Horizontal Alignment 95

inside of the horizontal curve so that the line of sight is sufficient to provide sufficient SSD. Fig. 3.16 displays the horizontal and vertical alignments.

Figure 3.16 Horizontal and vertical alignment for Example 3.18.

SOLUTION

To begin, the required radius to the vehicle path (Rv) is determined to be 533 ft from Table 3.5 with a 40-mi/h design speed and 4% superelevation. Because the ramp is a single lane, the horizontal curve radius will be equal to the radius to the vehicle path (R = Rv). Applying Eq. 3.39 gives the length of the horizontal curve as (with R = 533 ft and Δ = 90 degrees):

( ) 180 3.1416

533 90 837.24 ft 180

L R π

= Δ

= =

Also, by inspection of Fig. 3.16 (or application of Eq. 3.36), the tangent length T = R = 533 ft. The stationing for the horizontal curve is as follows:

96 Chapter 3 Geometric Design of Highways

= +

= + = + + = +

= + = + + = +

station of   40 00

station of   station of

40 00 plus 5 33 45 33

station of   station of

40 00 plus 8 37.24 48 37.24

PC

PI PC T

PT PC L

For the vertical alignment, both a sag and crest curve will be needed, and adequate clearance must be provided over the existing highway. As shown in Section 3.3.6, AASHTO [2018] specifies a desirable clearance height of 16.5 ft. The bridge girder thickness is given as 6 ft so the total elevation difference between the two highways is 22.5 ft (16.5 + 6).

For the vertical alignment, the elevation change will be the final offsets of the sag and crest curves plus the change in elevation resulting from the constant-grade section connecting the two curves. The constant-grade section is included because the available distance of 837.24 ft (known from the length of the horizontal curve) is likely to be more than sufficient to affect a 22.5 ft elevation difference at a 40-mi/h design speed. Because both the sag and crest curves connect to a level grade at one end and have the constant grade in common at the other end, the A value will be the same for both curves and will also be the grade for the constant-grade section. That is,

= =s cA A A With this information, the equation that will solve the vertical alignment for this problem is

( )− − + + =

837.24 22.5

200 200 100 s cs c A L LAL AL

where the third term accounts for the elevation difference attributable to the constant-grade section connecting the sag and crest curves (see also Example 3.9 for comparison). Using L = KA, we have

( )− − + + =

2 2 837.24 22.5

200 200 100 s cs c A K A K AA K A K

From Table 3.2 for 40 mi/h, Kc = 44, and from Table 3.3 for 40 mi/h, Ks = 64. Putting these values in the above equation gives

+ − =

− + − =

2 2

2

0.54 8.374 1.08 22.5

0.54 8.374 22.5 0

A A A

A A Solving this gives A = 3.458 and A = 12.049; A = 3.458% is chosen because we want to minimize the grade. For this value of A, the curve lengths are

( )= = =64 3.458 221.31 fts sL K A

( )= = =44 3.458 152.15 ftc cL K A

and the length of the constant-grade section (Lcon) will be 463.78 ft (837.24 – 221.31 – 152.15). This means that about 16.04 ft of the elevation difference will occur in the constant-grade section, with the remainder of the elevation difference attributable to the final curve offsets.

The stationing and elevation of the key points along the vertical alignment can now be calculated:

3.5 Combined Vertical and Horizontal Alignment 97

( )

= = +

= + = + + = +

= +

= + + = +

= + =

station of station of

40 00

station of 40 00 plus / 2 40 00 plus (1 52.15) / 2 40 76.075

station of station of

40 00 plus 1 52.15 41 52.15

station of station of

41

c

c c

c c c

s c con

PVC PC

PVI L

PVT PVC L

PVC PVT L

( )

( )

+ + = +

= +

= + + = +

= = +

= + + = +

52.15 plus 4 63.78 46 15.93

station of station of / 2

46 15.93 plus 1 10.66 / 2 47 26.59

station of station of

station of

46 15.93 plus 2 21.31 48 37.24

s s s

s

s s

PVI PVC L

PVT PT

PVC L

The elevation of the new east-west road will be 588 ft which is determined by adding 22.5 ft (the 16.5 ft clearance plus the 6 ft girder depth) above the north–south road elevation of 565.5 ft. The PC and PVCc are both at station 40 + 00 and elevation 588 ft (by inspection, the PVIc will also be at this elevation)

( )

= −

= − =

elevation elev 200

3.458 152.15 588 585.37 ft

200

c c c

AL PVT PVI

( )

= −

= − =

elevation elev 100

3.458 463.78 585.37 569.33 ft

100

con s c

AL PVC PVT

( )

= −

= − =

elevation elev 200

3.458 221.31 569.33 565.00 ft

200

s s s

AL PVT PVC

The PT and PVTs are both at station 48 + 37.24 and will thus both be at 565 ft (by inspection, the PVIs will also be at this elevation).

Finally, the distance that must be cleared from the inside of the horizontal curve to provide sufficient SSD is determined by applying Equation 3.42:

98 Chapter 3 Geometric Design of Highways

π

  = − 

 

90SSD 1 coss v

v

M R R

With Rv = 533 and SSD = 305 ft (from Table 3.1 at 40 mi/h),

( ) ( )

  = −  

  =

90 305 533 1 cos

3.1416 533

21.67 ft

sM

Thus, a distance of at least 21.67 ft must be cleared from the center of the ramp’s lane to the nearest sight obstruction on the inside of the curve.

3.6 PRACTICE PROBLEMS

PRACTICE PROBLEM 3.1

DESIGN SPEED AND VERTICAL CURVE STATIONS AND ELEVATIONS An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at station 240 + 00 and elevation 122 ft. The PVT is at station 242 + 30 and elevation 127.75 ft. What is the station and elevation of the lowest point on the curve?

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

This problem shows the basic principles of vertical curve design as well as the determination of the high/low point on a curve. To begin, the length of the curve is determined as,

( ) −

−

1 2 = station of station of

= 2 = .

L PVT PVI

L

The elevations of both PVT and PVI are given, so G2 can be determined as,

− −= = =2 elevation elevation 127.75

0.025 or 2.5%. 1 2 1 2

PVT PVI G

L

For a sag curve with a 55-mi/h design speed, Table 3.3 shows that the horizontal distance required to affect a 1% change in slope (the K-value) is equal to 115. With this, Eq. 3.10 can be rearranged to solve for A (the absolute value of the algebraic difference in the grades, in percent) such that,

= = =

115 L

A K

,

3.6 Practice Problems 99

which means, because A = |G1 − G2| and the problem states that the initial grade is negative and the final grade is positive, that

− = − = − = −1 2 1 2.5, so 1.5.G A G G

With this initial grade, the elevation of the PVC is,

( ) = =1elevation = elevation + 1 2 + ( ) 125.45 ft.PVC PVI G L

Also, the station of the PVC is determined as,

−station of = station of 2 = minus = 237 + 70.PVC PVI L

With the station and elevation of the PVC known, we can now move forward to determine the station and elevation of the lowest point on the curve. For the station of the lowest point on the curve, we can first apply Eq. 3.11 to determine the distance of the low point from the PVC,

= × = × =1 115 172.5 ft.hlx K G

With this,

+station of low point = station of distance from the to the low point

station of low point = 237 + 70 plus 1 + = 239 + 42.50.

PVC PVC

Solving for the elevation of the low point, Eq. 3.1 can be applied

( ) ( )

= + +

− −= + +

=

2

2 ( 1.5) 1.725 1.725 125.45

2( )

124.16 ft.

y ax bx c

PRACTICE PROBLEM 3.2

COMBINED SAG AND CREST VERTICAL CURVES

West and east highway segments are separated by 1000 ft horizontally. The west segment has a 0% constant grade and the east segment has a –1% grade. The east segment has a higher elevation than the west segment, and the two segments are connected by a joining sag and crest curve combination (so PVTs = PVCc). If the road is designed for 60 mi/h, what is the elevation difference between the west and east highway segments?

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

The intent of this problem is to reinforce the concepts covered in Example 3.11, and the set up for this problem is similar to that shown in Example 3.11. In looking at Example 3.11, the current problem will be similar to what is shown in Fig. 3.10 except it will be

100 Chapter 3 Geometric Design of Highways

reversed with the lower highway segment on the left of that figure (instead of on the right) and the higher roadway segment on the right instead of the left. Also, in comparison to Example 3.11, in the current problem the ycon = 0 (since the problem states that sag and crest curves are joined) and ys = 0 since the initial grade of the sag curve (the grade of the west segment) is 0%.

To begin this problem, it is important to first get the K-values for both the sag and crest curves. With the 60-mi/h design speed given in the problem, from Table 3.2 we find the K-value for the crest curve is Kc =151, and from Table 3.3 we find the sag curve’s K-value is Ks = 136. As in Example 3.11, let Gcon be the constant grade connecting the two curves so that G2s = G1c = Gcon. To determine the absolute value of the algebraic difference in the grades (in percent) for the sag and crest curves, we have As = Gcon (since G1s = 0 as given in the problem) and Ac = Gcon + 1 (since G1c = Gcon and G2c = −1, note that the sign of G2c becomes +1 since A’s are based on the absolute value in the differences of the grades which means you add the absolute values of two grades if they are opposite in sign, as they are here, and you subtract the lower absolute value grade from the higher absolute value grade if the two grades are the same sign).

With these values of Ac and As, and with 1000 ft separating the two highway segments, we have,

( )

= + = +

× + × +

1000

= 136 1

s c

s s c c

con

L L

K A K A

G

which gives Gcon = 2.958%. To develop an equation for the elevation difference between the two highway

segments, let Yfs and Yfc be the final offsets of the sag and vertical curves, respectively, and let ∆yc be the change in elevation due to the extended curve tangent for the crest curve, recalling that the extended curve tangent for the sag curve, ∆ys is now zero since the initial grade for the sag curve, G1s = 0 with the west highway section a 0% grade (see these terms in Example 3.11 and Fig. 3.10 for comparison with this practice problem). With these terms and the elevation differences between the two highway segments ( elevation), the elevation equality is (compare to Example 3.11),

Δ + Δ = +elevation .c fs fcy Y Y

Rearranging terms and using Eq. 3.9 for final offsets and the 1% slope of the east highway segment (which produces a positive ∆yc), we have,

Δ = + − 1elevation . 200 200

s s c c c

A L A L G L

With Ls = KsAs and Lc = KcAc, substitution gives,

( ) ( ) ( )× × Δ = + − ×

elevation

200 200 ,

and a final answer for the elevation difference between the west and east highway segments is elevation = 11.803 ft.

3.6 Practice Problems 101

PRACTICE PROBLEM 3.3

HORIZONTAL CURVE STATIONING, DESIGN SPEED, AND SIGHT DISTANCE A horizontal curve on a two-lane highway (with 11-ft lanes) has a central angle of 33 degrees and is designed for 60 mi/h with a 8% superelevation. First, if the PI is at station 300 + 00, what is the station of the PT and how many feet have to be cleared from the inside lane's lane-shoulder edge to provide adequate stopping sight distance? Second, if the normal component of centripetal force was considered, how much shorter would the radius of the curve be?

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

This problem illustrates a number of principles in horizontal curve design, including the magnitude of imprecision introduced by ignoring the normal component of centripetal force (which is standard practice in highway design). To begin, the radius to the vehicle (Rv) for a 60-mi/h design speed and a superelevation of 8% can be determined from Table 3.5. With this, to determine R (the radius to the center of the highway), half of the lane width is added to Rv obtained from Table 3.5 to get to the centerline of the roadway so,

= + = × =1 lane-width .

2 vR R

With R known and given as 33 degrees, Eq. 3.39 can be applied to get length of the curve,

π π= Δ = × = 694.19 ft. 180 180

L R

Next, the tangent of the curve is determined by applying Eq. 3.36,

Δ= = × = tan tan .

2 2 T R

To determine the station of the PT, we must first determine the station of the PC and then add the curve length L. The station of the PC is,

= −

= − = +

station of station of

296 42.92

PC PI T

and the station PT is therefore,

102 Chapter 3 Geometric Design of Highways

= +

= + = +

station of station of

303 37.01.

PT PC L

To determine how many feet have to be cleared from the inside lane’s lane-shoulder edge to provide adequate stopping sight distance, Eq. 3.42 can be used to arrive at Ms,

π  

= −   

90SSD 1 coss v

v

M R R

with SSD determined to be from Table 3.1 (with a design speed of 60 mi/h), we have,

π  ×= − =  × 

90 1 cos 33.70 ft.

sM

Because the problem asks for the distance from the inside lane’s lane-shoulder edge and MS is the distance to the path of the vehicle (which is assumed to be in the center of the lane), we must subtract half of the lane with from MS to get the answer which is,

= − − =1 1 required clearance lane width= 28.20 ft.

2 2sM

Turning to the second part of the problem, to determine how much shorter the radius of the curve would be if the normal component of the centripetal force were considered, recall that highway design ignores the normal component of the centripetal force which gives a conservative answer (larger radius than actually needed). Thus, in highway design, Eq. 3.34 is used to arrive at the RV values listed in Table 3.5. However, if the normal component of the centripetal force is considered, Eq. 3.34 becomes, with superscript N denoting the normal component of centripetal force (see also Eq. 3.33 and discussion below it),

 = −    +   

2

1 tan . 100

100

N v s

s

V e R f

e g f

With fs = 0.12 (from Table 3.5 with the given design speed of 60 mi/h), solving for the radius in the equation above gives,

  = − =     +  

 

2

1 tan . 100

32.2 100

N vR

3.6 Practice Problems 103

With this, Rv minus N vR will tell us how much shorter the radius would be, so the

answer is,

− = − = 9.06 ft.N v vR R

This rather modest difference shows that, with the small superelevations used in typical highway design, ignoring the normal component of centripetal force does not introduce great error.

PRACTICE PROBLEM 3.4

COMBINED VERTICAL AND HORIZONTAL ALIGNMENT

A highway segment has an equal-tangent sag curve and a horizontal curve with the same design speed. It is known that the sag curve has an initial grade of –1.8% and a positive final grade that is not known. The station of the PVC is 232 + 70, and the station of the low point of the sag vertical curve is 235 + 14.8. The horizontal curve has a superelevation of 8%, central angle of 30 degrees, and a PT at Station 302 + 20. The road has two 11-ft lanes (one in each direction). What is the station of the PI?

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

This problem shows how common design elements from horizontal and vertical alignments can be used to solve for unknown curve characteristics. The approach to this problem is to use the information from the equal-tangent vertical curve to determine the design speed, and then use this design speed to solve for the station of the horizontal curve’s PI. By framing the problem in this way, the reader will gain additional insight into the various elements involved in the vertical and horizontal alignment of highways.

To begin, the design speed of the highway is determined from the vertical curve information provided. First, the distance from the PVC to the highpoint is determined as

= − − =station of low point = 244.80 ft.hlx PVC

With this, Eq. 3.11 can be rearranged to solve for the K-value such that,

= = = 1

136.

hlx

K G

104 Chapter 3 Geometric Design of Highways

Table 3.3 for a sag vertical shows that a K-value of 136 corresponds to a design speed of 60 mi/h. With this design speed, the radius to the curve of the vehicle’s traveled path Rv can be obtained from Table 3.5. With an 8% superelevation given and a 60-mi/h design speed, Table 3.5 gives,

= vR

and, with 11-ft lanes (on a two-lane road), the radius of the curve measured to the roadway centerline is,

= + = + =1 1 lane width .

2 2vR R

Eq. 3.39 can now be applied to arrive at the horizontal curve length L as,

π π= Δ = × = 631.08 ft. 180 180

L R

With this length of curve,

= −

= − = +

station of station of

295 88.92.

PC PT L

To arrive at the station of the PI, we must add the tangent length to the station of the PC (note that this is not the same value as subtracting the tangent length from the PT since the tangents are not on the highway alignment). The tangent length is determined from Eq. 3.36 to be,

Δ= = × = tan tan

2 2 T R

So, = −

= − = +

station of station of

299 11.93.

PI PC T

References 105

NOMENCLATURE FOR CHAPTER 3

A absolute value of the algebraic difference in grades (in percent)

a coefficient in the parabolic curve equation or the deceleration in the stopping distance equation

b coefficient in the parabolic curve equation

c elevation of the PVC D degree of curvature e rate of superelevation Ff frictional side force Fc centripetal force Fcn centripetal force normal to the roadway

surface Fcp centripetal force parallel to the roadway

surface fs coefficient of side friction G grade G1 initial roadway grade G2 final roadway grade g gravitational constant H height of vehicle headlights Hc clearance height of structure above sag

curve H1 height of driver’s eye H2 height of roadway object for stopping,

height of oncoming car for passing K horizontal distance required to effect a

1% change in slope L length of curve Lm minimum length of curve M middle ordinate Ms middle ordinate for stopping sight

distance PC initial point of horizontal curve

PI point of tangent intersection (horizontal curve)

PSD passing sight distance PT final point of horizontal curve PVC initial point of vertical curve PVI point of tangent intersection (vertical

curve) PVT final point of vertical curve R radius of curve measured to the roadway

centerline Rv radius of curve to the vehicle’s traveled

path S sight distance SSD stopping sight distance T tangent length V vehicle speed V1 initial vehicle speed W vehicle weight Wn vehicle weight normal to the roadway

surface Wp vehicle weight parallel to the roadway

surface x distance from the beginning of the

vertical curve to specified point xhl distance from the beginning of the

vertical curve to high or low point Y vertical curve offset Yf end-of-curve offset (vertical curve) Ym midcurve offset (vertical curve) α angle of superelevation β upward angle of headlight beam Δ central angle Δs angle subtended by the stopping sight

distance (SSD) arc

REFERENCES

AASHTO (American Association of State Highway and Transportation Officials). A Policy on Geometric Design of Highways and Streets, 7th ed. Washington, DC, 2018.

Campbell, C. The Sports Car: Its Design and Performance. Cambridge, MA: Robert Bently, 1978.

Hickerson, T. F. Route Location and Design, 5th ed. New York: McGraw-Hill, 1964.

Wong, J. Y. Theory of Ground Vehicles. New York: John Wiley & Sons, 2008.

P-11

CHAPTER 4 PROBLEMS

Flexible-Pavement Design (Sections 4.3–4.4)

4.1 Truck A has two single axles. One axle weighs 12,000 lb and the other weighs 23,000 lb. Truck B has an 8000-lb single axle and a 43,000-lb tandem axle. On a flexible pavement with a 3-inch HMA wearing surface, a 6-inch soil-cement base, and an 8- inch crushed stone subbase, which truck will cause more pavement damage? (Assume that drainage coefficients are 1.0.)

4.2 A flexible pavement has a 4-inch HMA wearing surface, a 7-inch dense-graded crushed stone base, and a 10-inch crushed stone subbase. The pavement is on a soil with a resilient modulus of 5000 lb/in2. The pavement was designed with 90% reliability, an overall standard deviation of 0.4, and a ΔPSI of 2.0 (a TSI of 2.5). The drainage coefficients are 0.9 and 0.8 for the base and subbase, respectively. How many 25-kip single-axle loads can be carried before the pavement reaches its TSI (with given reliability)?

4.3 A highway has the following pavement design daily traffic: 300 single axles at 10,000 lb each, 120 single axles at 18,000 lb each, 100 single axles at 23,000 lb each, 100 tandem axles at 32,000 lb each, 30 single axles at 32,000 lb each, and 100 triple axles at 40,000 lb each. A flexible pavement is designed to have 4 inches of sand-mix asphalt wearing surface, 6 inches of soil-cement base, and 7 inches of crushed stone subbase. The pavement has a 10-year design life, a reliability of 85%, an overall standard deviation of 0.30, drainage coefficients of 1.0, an initial PSI of 4.7, and a TSI of 2.5. What is the minimum acceptable soil resilient modulus?

4.4 Consider the conditions in Problem 4.3. Suppose the state has relaxed its truck weight limits and the impact has been to reduce the number of 18,000-lb single-axle loads from 120 to 20 and increase the number of 32,000-lb single-axle loads from 30 to 90 (all other traffic is unaffected). Under these revised daily counts, what is the minimum acceptable soil resilient modulus?

4.5 A flexible pavement was designed for the following daily traffic with a 12-year design life: 1300 single axles at 8000 lb each, 900 tandem axles at 15,000 lb each, 20 single axles at 40,000 lb each, and 200 tandem axles at 40,000 lb each. The highway was designed with 4 inches of HMA wearing surface,

4 inches of hot-mix asphaltic base, and 8 inches of crushed stone subbase. The reliability was 70%, overall standard deviation was 0.5, ΔPSI was 2.0 (with a TSI of 2.5), and all drainage coefficients were 1.0. What was the soil resilient modulus of the subgrade used in design?

4.6 A flexible pavement has a SN of 3.8 (all drainage coefficients are equal to 1.0). The initial PSI is 4.7 and the terminal serviceability is 2.5. The soil has a CBR of 9. The overall standard deviation is 0.40 and the reliability is 95%. The pavement is currently designed for 1800 equivalent 18-kip single-axle loads per day. If the number of 18-kip single-axle loads were to increase by 30%, by how many years would the pavement’s design life be reduced?

4.7 An engineer plans to replace the rigid pavement in Example 4.3 with a flexible pavement. The chosen design has 6 inches of sand-mix asphalt wearing surface, 9 inches of soil-cement base, and 10 inches of crushed stone subbase. All drainage coefficients are 1.0 and the soil resilient modulus is 5000 lb/in2. If the highway’s traffic is the same (same axle loadings per vehicle as in Example 4.3), for how many years could you be 95% sure that this pavement will last? (Assume that any parameters not given in this problem are the same as those given in Example 4.3.)

4.8 A flexible pavement is designed with 5 inches of HMA wearing surface, 6 inches of hot-mix asphaltic base, and 10 inches of crushed stone subbase. All drainage coefficients are 1.0. Daily traffic is 200 passes of a 20-kip single axle, 200 passes of a 40-kip tandem axle, and 80 passes of a 22-kip single axle. If the initial minus the terminal PSI is 2.0 (the TSI is 2.5), the soil resilient modulus is 3000 lb/in2, and the overall standard deviation is 0.6, what is the probability (reliability) that this pavement will last 20 years before reaching its terminal serviceability?

4.9 A flexible pavement is designed with 4 inches of sand-mix asphalt wearing surface, 6 inches of dense- graded crushed stone base, and 8 inches of crushed stone subbase. All drainage coefficients are 1.0. The pavement is designed for 18-kip single-axle loads (1290 per day). The initial PSI is 4.5 and the TSI is 2.5. The soil has a resilient modulus of 12,000 lb/in2. If the overall standard deviation is 0.40, what is the probability that this pavement will have a PSI greater than 2.5 after 20 years?

P-12 Chapter 4 Problems

4.10 A flexible pavement has a 4-inch sand-mix asphalt wearing surface, 10-inch soil-cement base, and a 10-inch crushed stone subbase. It is designed to withstand 400 20-kip single-axle loads and 900 35- kip tandem-axle loads per day. The subgrade CBR is 8, the overall standard deviation is 0.45, the initial PSI is 4.2, and the final PSI is 2.5. What is the probability that this pavement will have a PSI above 2.5 after 25 years? (Drainage coefficients are 1.0.)

Rigid-Pavement Design (Sections 4.5–4.6)

4.11 Consider the two trucks in Problem 4.1. Which truck will cause more pavement damage on a rigid pavement with a 10-inch slab?

4.12 You have been asked to design the pavement for an access highway to a major truck terminal. The design daily truck traffic consists of the following: 80 single axles at 22,500 lb each, 570 tandem axles at 25,000 lb each, 50 tandem axles at 39,000 lb each, and 80 triple axles at 48,000 lb each. The highway is to be designed with rigid pavement having a modulus of rupture of 600 lb/in2 and a modulus of elasticity of 5 million lb/in2. The reliability is to be 95%, the overall standard deviation is 0.4, the drainage coefficient is 0.9, ΔPSI is 1.7 (with a TSI of 2.5), and the load transfer coefficient is 3.2. The modulus of subgrade reaction is 200 lb/in3. If a 20-year design life is to be used, determine the required slab thickness.

4.13 A rigid pavement is being designed with the same parameters as used in Problem 4.5. The modulus of subgrade reaction is 300 lb/in3 and the slab thickness is determined to be 8.5 inches. The load transfer coefficient is 3.0, the drainage coefficient is 1.0, and the modulus of elasticity is 4 million lb/in2. What is the design modulus of rupture? (Assume that any parameters not given in this problem are the same as those given in Problem 4.5.)

4.14 A rigid pavement is designed with a 10-inch slab, an Ec of 6 million lb/in2, a concrete modulus of rupture of 432 lb/in2, a load transfer coefficient of 3.0, an initial PSI of 4.7, and a TSI of 2.5. The overall standard deviation is 0.35, the modulus of subgrade reaction is 190 lb/in3, and a reliability of 90% is used along with a drainage coefficient of 0.8. The pavement is designed assuming that traffic is composed entirely of trucks (100 per day). Each truck has one 20-kip single axle and one 42-kip tandem axle (the effect of all other vehicles is ignored). A section of this road is to be replaced (due to different subgrade characteristics) with a flexible pavement having a SN of 4 and is expected to last the

same number of years as the rigid pavement. What is the assumed soil resilient modulus? (Assume that all other factors are the same as for the rigid pavement.)

4.15 Consider the loading conditions in Problem 4.3. A rigid pavement is used with a modulus of subgrade reaction of 200 lb/in3, a slab thickness of 8 inches, a load transfer coefficient of 3.2, a modulus of elasticity of 5 million lb/in2, a modulus of rupture of 600 lb/in2, and a drainage coefficient of 1.0. How many years is the pavement expected to last using the same reliability as in Problem 4.3? (Assume that all other factors are as in Problem 4.3.)

4.16 Consider Problem 4.15. How long would the rigid pavement be expected to last if you wanted to be 95% sure that the pavement would stay above the 2.5 TSI?

4.17 Consider the traffic conditions in Example 4.3. Suppose a 10-inch slab was used and all other parameters are as described in Example 4.3. What would the design life be if the drainage coefficient was 0.8, and what would it be if it was 0.6?

4.18 Consider the conditions in Example 4.4. Suppose all of the parameters are the same, but further soil tests found that the modulus of subgrade reaction was only 150 lb/in3. In light of this new soil finding, how would the design life of the pavement change?

4.19 Consider the conditions in Example 4.4. Suppose all of the parameters are the same, but a quality control problem resulted in a modulus of rupture of 600 lb/in2 instead of 800 lb/in2. How would the design life of the pavement change?

Pavement Design with Design-Lane Traffic (Sections 4.3–4.7)

4.20 You have been asked to design a flexible pavement, and the following daily traffic is expected for design: 5000 single axles at 10,000 lb each, 400 single axles at 24,000 lb each, 1000 tandem axles at 30,000 lb each, and 100 tandem axles at 50,000 lb each. There are three lanes in the design direction (conservative design is to be used). Reliability is 90%, overall standard deviation is 0.40, ΔPSI is 1.8, and the design life is 15 years. The soil has a resilient modulus of 13,750 lb/in2. If the TSI is 2.5, what is the required SN?

4.21 A three-lane northbound section of interstate (with the design lane conservatively designed) has rigid pavement (PCC) and was designed with a 10-inch slab, 90% reliability, 700 lb/in2 concrete

Chapter 4 Problems P-13

modulus of rupture, 4.5 million lb/in2 modulus of elasticity, 3.0 load transfer coefficient, and an overall standard deviation of 0.35. The initial PSI is 4.6 and the TSI is 2.5. The CBR is 2 with a drainage coefficient of 1.0. The road was designed exclusively for trucks that have one 24-kip tandem axle and one 12-kip single axle. It is known from weigh-in-motion scales that there have been 13 million 18-kip– equivalent single-axle loads in the entire northbound direction of this freeway so far. If a section of flexible pavement is used to replace a section of the PCC that was removed for utility work, what SN should be used so that the PCC and flexible pavements have the same life expectancy (the new life of the flexible pavement and the remaining life of the PCC)?

4.22 A rigid pavement is designed with an 11-inch slab thickness, 90% reliability, Ec = 4 million lb/in2, modulus of rupture of 600 lb/in2, modulus of subgrade reaction of 150 lb/in3, a 2.8 load transfer coefficient, initial PSI of 4.8, final PSI of 2.5, overall standard deviation of 0.35, and a drainage coefficient of 0.8. The pavement has a 20-year design life. The pavement has three lanes and is conservatively designed for trucks that have one 20,000-lb single axle, one 26,000-lb tandem axle, and one 34,000-lb triple axle. What is the daily estimated truck traffic on the three lanes?

4.23 A rigid pavement is on a highway with two lanes in one direction, and the pavement is conservatively designed. The pavement has an 11-inch slab with a modulus of elasticity of 5,000,000 lb/in2 and a concrete modulus of rupture of 700 lb/in2, and it is on a soil with a CBR of 25. The design drainage coefficient is 1.0, the overall standard deviation is 0.3, and the load transfer coefficient is 3.0. The pavement was designed to last 20 years (initial PSI of 4.7 and a final PSI of 2.5) with 95% reliability carrying trucks with one 18-kip single axle and one 28-kip tandem axle. However, after the pavement was designed, one more lane was added in the design direction (conservative design still used), and the weight limits on the trucks were increased to a 20-kip single and a 34-kip tandem axle (the slab thickness was unchanged from the original two-lane design with lighter trucks). If climate change has caused the drainage coefficient to drop to 0.8, how long will the pavement last with the new loading and the additional lane (same volume of truck traffic)?

4.24 A four-lane northbound section of interstate has rigid pavement and was designed with an 8-inch

slab, 90% reliability, a 700 lb/in2 concrete modulus of rupture, a 5 million lb/in2 modulus of elasticity, a 3.0 load transfer coefficient, and an overall standard deviation of 0.3. The initial PSI is 4.6 and the TSI is 2.5. The pavement was conservatively designed (assuming the upper limit of the W18 design lane load) to last 20 years, and the CBR is 25 with a drainage coefficient of 1.0. A design mistake was made that ignored 1000 total northbound (daily) passes of trucks with 22-kip single and 30-kip tandem axles. What slab thickness should have been used?

4.25 A flexible pavement is designed with 4 inches of an experimental asphalt wearing surface that has a structural layer coefficient of 0.575, with 4 inches of a hot-mix asphaltic base, and 10 inches of a crushed stone subbase (all drainage coefficients are 1.0). The roadway is designed to accommodate 200 delivery trucks per day and each truck has one 12-kip single front axle and one 20-kip single rear axle. The subgrade CBR is 8, the initial present serviceability index is 4.5, the TSI is 2.5, and the overall standard deviation is 0.5. How confident would you be that this pavement would last 20 years?

4.26 For the conditions in Practice Problem 4.4 (again conservatively designed), suppose that a rigid pavement was designed with a 12-inch slab (modulus of rupture of 900 lb/in2, modulus of elasticity of 6.0 × 106 lb/in2, load transfer coefficient of 3.2, and drainage coefficient of 1.0) for the four-lane northbound direction described in Practice Problem 4.4 (all other traffic and PSI information is the same as in that problem). How confident would you be that this rigid pavement would have a PSI above 2.5 after 20 years?

4.27 A rigid pavement on a multilane highway (two lanes each direction) has been conservatively designed (for its design-lane traffic). The pavement engineer uses an 11-inch slab, Ec of 4 × 106 lb/in2, a concrete modulus of rupture of 695 lb/in2, a load transfer coefficient of 2.8, an initial present serviceability index of 4.2, and a TSI of 2.5. The overall standard deviation is 0.6, the subgrade CBR is 20, and the drainage coefficient is 1.0. The pavement was designed for four hundred 30,000 lb triple axles, nine hundred 20,000 lb tandem axles per day, and twelve hundred 10,000 lb single axle loads per day. If the pavement was designed to last 40 years, what reliability was used?

P-14 Chapter 4 Problems

Multiple Choice Problems (Multiple Sections)

4.28 A flexible pavement is constructed with 5 inches of sand-mix asphaltic wearing surface, 9 inches of dense-graded crushed stone base, and 10 inches of crushed stone subbase. The base has a drainage coefficient of 0.90, while the subbase drainage coefficient is 1.0. Determine the SN of the pavement.

a) 4.47 b) 4.31 c) 4.76 d) 3.98

4.29 A flexible pavement is designed to last 10 years to withstand truck traffic that consists only of trucks with two 18-kip single axles. The pavement is designed for a soil CBR of 2, an initial PSI of 5.0, a TSI of 2.5, an overall standard deviation of 0.40, and a reliability of 90%, and the SN was determined to be 6. On one section of this roadway, beneath an underpass, an engineer uses a 10-inch rigid pavement in an attempt to have it last longer before resurfacing. How many years will this rigid- pavement section last? (Given the same traffic conditions, modulus of rupture = 800 lb/in2, modulus of elasticity = 5,000,000 lb/in2, load transfer coefficient of 3.0, and drainage coefficient of 1.0.)

a) 11.33 b) 13.22 c) 17.14 d) 25.65

4.30 A flexible pavement at an access road to a sports stadium parking lot is designed with a 4-inch sand-mix asphaltic concrete surface, 5-inch aggregate bituminous emulsion base, and a 10-inch crushed stone subbase. There are 95 scheduled baseball and football games at the stadium per year. The access road to the parking lot is three lanes in each direction (conservatively designed). The pavement was designed for recreational vehicles with one 20-kip single axle and one 20-kip tandem axle. There are 9000 recreational vehicles estimated at each event. Given that drainage coefficients are 1.0, the overall

standard deviation of traffic is 0.45, reliability is 90%, and the soil’s resilient modulus is 15,000 lb/in2, how many years will the access road last if the initial PSI is 4.0 and the TSI is 2.5?

a) 7 b) 9 c) 12 d) 14

4.31 A rigid pavement on a new interstate (3 lanes each direction) has been conservatively designed with a 12-inch slab, an Ec of 5.5 × 106 lb/in2, a concrete modulus of rupture of 700 lb/in2, a load transfer coefficient of 3.0, an initial present serviceability index of 4.5, and a TSI of 2.5. The overall standard deviation is 0.35, the subgrade CBR is 25, and the drainage coefficient is 0.9. The pavement was designed for 600 30-kip tandem axles per day and 1400 20-kip single-axle loads per day. If the desired reliability was 90%, how long was this pavement designed to last?

a) 18 b) 32 c) 42 d) 46

4.32 A flexible pavement was designed to have a 6- inch sand-mix asphaltic surface, 8-inch soil-cement base, and a 21-inch crushed stone subbase (all drainage coefficients are 1.0). The pavement was designed for 800 12-kip single axles and 1600 34-kip tandem axles per day in the design direction. The reliability used was 90%, the overall standard deviation was 0.35, initial PSI was 4.7, the TSI was 2.5, and the soil resilient modulus was 2582 lb/in2. If the road has three lanes in the design direction (and was conservatively designed), for how many years was the pavement designed to last?

a) 12 b) 15 c) 20 d) 24

107

Chapter 4

Pavement Design 4.1 INTRODUCTION

Pavements are among the most financially important highway-system assets due to their high construction and maintenance costs. Their costs are largely responsible for making the U.S. highway system the most expensive public works project undertaken by any society. Because pavements are so costly to construct and maintain, it is important for highway engineers to have an understanding of basic pavement design principles.

In the United States, there are over 4 million miles of highways. Of these, roughly 45% are lower-volume roads that are not paved (these roads generally have a gravel surface, or are composed of a stabilized material consisting of an aggregate bound together with a cementing agent such as Portland cement, lime fly ash, or asphaltic cement). However, environmental conditions such as moisture can seriously compromise the structural integrity of unpaved highway surfaces. As a result, highways that carry higher volumes of traffic, typically with heavy axle loads, require surfaces with asphalt concrete or Portland cement concrete (PCC) to provide for all-weather operations. Paved highways can cost several million dollars per mile to construct (based on the number of lanes, shoulders, soil conditions, etc.) and many thousands of dollars annually (per mile) to maintain. In the United States, roughly $40 billion dollars are spent annually on the construction and maintenance of pavements. Given the magnitude of this pavement-asset investment, it is easy to understand why the construction, maintenance, and rehabilitation of the pavement infrastructure must be done in a cost-effective manner.

While paved highway surfaces (and their associated lane markings) provide visual guidance to help vehicle operators traverse the highway alignment, the primary functions of pavement are to provide adequate friction, particularly to ensure safe stopping distances and acceptable vehicle cornering (the emphasis of Chapters 2 and 3), and to support vehicle loads while maintaining a safe and comfortable ride surface. This latter function (supporting vehicle loads and the quality of the riding surface) is the emphasis of the current chapter.

4.2 PAVEMENT TYPES

Pavement structures can be broadly categorized into two types: flexible pavements and rigid pavements. There are, however, many variations of these pavement types

108 Chapter 4 Pavement Design

including composite pavements (which are made of both rigid and flexible pavement layers), continuously reinforced pavements, and post-tensioned pavements (these variations require specialized design procedures that not covered in this chapter).

As with most conventional civil-engineering structures, the underlying soil must ultimately carry the design-load forces. The intent of a pavement structure is to distribute traffic loads to the soil (subgrade) at a magnitude that will not shear or distort the soil. Typical soil-bearing capacities can be less than 50 lb/in2, and in some cases can be as low as 2 to 3 lb/in2. When soil is saturated with water, the load-bearing capacity can be very low, and in these cases it is very important for pavement to distribute tire loads to the supporting soil in such a way as to prevent permanent deformation of the underlying soil, which would classify as a failure of the pavement structure.

A typical automobile weighs approximately 4000 lb, with tire pressures around 35 lb/in2. These loads are small compared with a typical tractor–semi- trailer truck, which can weigh up to 80,000 lb (the legal limit, in many states), on five axles with tire pressures of 100 lb/in2 or higher. Truck loads such as these represent the standard loading used in pavement design. In this chapter, attention is directed toward an accepted procedure that can be used to design pavement structures for high–traffic-volume highways subjected to heavy truck traffic. The design of lower-volume facilities, which may have stabilized-soil and gravel-surfaced structures, is discussed elsewhere [Yoder and Witczak, 1991].

4.2.1 Flexible Pavements A flexible pavement is constructed with asphaltic cement and aggregates and usually consists of several layers, as shown in Fig. 4.1. The lowest layer is referred to as the subgrade (the soil itself). The upper 6 to 8 inches of the subgrade is usually scarified and blended to provide a uniform material before it is compacted to a high density to assist with the overall stability and uniformity of the pavement structure. The next layer up in the structure, the subbase, usually consists of crushed aggregate (rock). This material has better engineering properties (higher modulus values) and more uniformity than the subgrade material in terms of its bearing capacity. The next layer up in the structure is the base layer and it is also often made of crushed aggregates (of a higher strength than those used in the subbase), which can in some cases be stabilized with a cementing material such as Portland cement, lime fly ash, or asphaltic cement.

The top layer of the flexible pavement structure is referred to as the wearing surface and usually consists of asphaltic concrete, which is a mixture of asphalt cement and aggregates. The purpose of the wearing layer is to protect the base layer from wheel abrasion and to waterproof the entire pavement structure. It also provides a skid-resistant surface that is important for safe vehicle stops and cornering. Typical thicknesses of the individual layers are shown in Fig. 4.1. These thicknesses vary with the types and magnitudes of axle loadings, available materials, and expected pavement design life (which is defined as the number of years the pavement is expected to provide adequate service before it must undergo major rehabilitation).

4.3 Pavement System Design: Principles for Flexible Pavements 109

Figure 4.1 Typical flexible- pavement cross section.

4.2.2 Rigid Pavements Rigid pavements are constructed with PCC and aggregates, as shown in Fig. 4.2. As with flexible pavements, the subgrade (the lower layer) is often scarified, blended, and compacted to high density before additional layers are added. In rigid pavements, the base layer (see Fig. 4.2) is optional, depending on the engineering properties of the subgrade. If the subgrade soil is poor and erodable, then it is advisable to use a base layer. However, if the soil has good engineering properties and drains well, a base layer need not be used. The top layer (wearing surface) is the PCC slab. Slab length varies from a spacing of 10 to 13 ft to a spacing of 40 ft or more.

Transverse contraction joints are built into the pavement to control cracking due to shrinkage of the concrete during the curing process. Load-transfer devices, such as dowel bars, are placed in the joints to minimize deflections between adjacent slabs and reduce stresses near the edges of the slabs. Slab thicknesses for PCC highway pavements usually vary from 8 to 12 inches, as shown in Fig. 4.2.

Figure 4.2 Typical rigid- pavement cross section.

4.3 PAVEMENT SYSTEM DESIGN: PRINCIPLES FOR FLEXIBLE PAVEMENTS

In supporting vehicle loads, the primary function of the pavement structure is to distribute the surface stresses (contact tire pressure) to an acceptable magnitude at the subgrade (to a magnitude that prevents permanent deformation). A flexible pavement reduces the stresses by distributing the traffic wheel loads over greater and greater areas, down through the individual layers, until the stress at the subgrade is at an acceptably low level. The traffic loads are transmitted to the subgrade by aggregate-to-aggregate particle contact. Confining pressures

110 Chapter 4 Pavement Design

(lateral forces due to material weight) in the base and subbase layers increase the bearing strength of the materials in these layers. A cone of distributed loads spreads the stresses to the subgrade, as shown in Fig. 4.3.

Figure 4.3 Distribution of load on a flexible pavement.

4.4 TRADITIONAL AASHTO FLEXIBLE-PAVEMENT DESIGN PROCEDURE

There are several accepted flexible-pavement design procedures available, including the Asphalt Institute method, the National Stone Association procedure, the Shell procedure, and the Mechanistic-Empirical Pavement Design Guide (which is briefly discussed later in this chapter). Most of the procedures have been field verified and used by highway agencies for many years and the selection of one procedure over another is usually based on highway-agency experience and satisfaction with previous design results.

A traditional and widely accepted flexible-pavement design procedure is presented in the AASHTO Guide for Design of Pavement Structures, which is published by the American Association of State Highway and Transportation Officials (AASHTO). The procedure was first published in 1972, with the latest revisions in 1993. Test data, used for the development of the design procedure, were collected at the AASHO Road Test in Illinois from 1958 to 1960 (AASHO, which stands for American Association of State Highway Officials, was the prior name of AASHTO).

Pavement designs must account for various failure possibilities as well as uncertainties in environmental conditions. For example, pavements can be subjected to a number of detrimental effects from traffic loads, including fatigue failures (cracking), which are the result of the accumulating effects of traffic passing over the pavement surface. The pavement is also placed in an uncontrolled environment that produces temperature extremes and moisture variations. In addition, pavement structures can have considerable variability in material quality (materials are often drawn from local sources due to the expense of transportation) and construction quality. The combination of the environment, traffic loads, material variations, and construction variations requires a robust design procedure. The AASHTO pavement design procedure accounts for most of the critical elements needed for an effective pavement design. It considers the environment, traffic loads, and material strengths in a methodology that is relatively easy to use. The AASHTO pavement design procedure has been widely accepted throughout the United States and around the world. Details of this procedure are presented in the following sections.

4.4 Traditional AASHTO Flexible-Pavement Design Procedure 111

4.4.1 Serviceability Concept Prior to the AASHO Road Test, there was no real consensus on the definition of pavement failure. In the eyes of an engineer, pavement failure occurred whenever cracking, rutting, or other surface distresses became visible. In contrast, the motoring public usually associated pavement failure with poor ride quality. Pavement engineers conducting the AASHO Road Tests of the late 1950s were faced with the task of accounting for these two failure definitions (engineer and public) in a single design procedure. The Pavement Serviceability-Performance Concept was developed by Carey and Irick [1962] to address the definition of pavement failure. Carey and Irick considered extensive pavement performance histories and noted that pavements usually begin their service life in excellent condition and deteriorate as traffic loads are applied in conjunction with prevailing environmental conditions. This gives rise to a pavement performance curve, which charts typical pavement conditions over time. Pavement performance, at any point in time, is known as the present serviceability index, or PSI. Examples of pavement performance trends (or PSI trends) are shown in Fig. 4.4.

The present serviceability index (PSI) of a pavement can be measured at any point in time. The PSI was originally based on the present serviceability rating (PSR) which was measured by a panel of raters who drove over the pavement section and rated the pavement performance on a scale of 0 to 5, with 5 being the smoothest ride and 0 being virtually impassible. Because having a panel of raters evaluate all pavements is not practical, the PSI was developed by correlating panel-rated PSRs with actual pavement measurements (roughness, cracking, etc.). Thus PSI values are based on a variety of pavement measurements and does not require a panel of raters. As shown in Fig. 4.4, the accumulation of traffic loads causes the pavement to deteriorate, and, as expected, the serviceability rating (PSI) drops. At some point, a terminal serviceability index (TSI) is reached and the pavement is in need of rehabilitation or replacement.

It has been found that new pavements usually have an initial PSI rating of approximately 4.2 to 4.5. The point at which pavements are considered to have failed (the TSI) varies by type of highway. Highway facilities such as interstate highways or principal arterials usually have TSIs of 2.5 or 3.0, whereas local roads can have TSIs as low as 2.0.

4.4.2 Flexible-Pavement Design Equation The basic equation for flexible-pavement design given in the 1993 AASHTO design guide enables engineers to determine a structural number (SN) necessary to carry a designated traffic loading. The AASHTO equation is

( ) [ ]

( )

10 18 o 10

10

5.19

10

log = +9.36 log SN 1 0.20

log PSI / 2.7 +

0.40 1094 SN 1

+2.32 log 8.07

R

R

W Z S

M

 + −  Δ

 + +  −

(4.1)

where

112 Chapter 4 Pavement Design

Figure 4.4 Pavement performance trends. Redrawn from AASHTO Guide for Design of Pavement Structures, Washington, DC, The American Association of State Highway and Transportation Officials, 1993. Used by permission.

W18 = 18-kip–equivalent single-axle load (1 kip = 1000 lb),

ZR = reliability (z-statistic from the standard normal curve),

So = overall standard deviation of traffic,

SN = structural number,

ΔPSI = loss in serviceability from the time the pavement is new until it reaches its TSI, and

MR = soil resilient modulus of the subgrade in lb/in2.

A graphical solution to Eq. 4.1 is shown in Fig. 4.5. Details on the variables that serve as inputs to Eq. 4.1 and Fig. 4.5 are as follows:

W18 Automobiles and truck traffic provide a wide range of vehicle axle types and axle loads. If one were to attempt to account for individual-vehicle traffic loads on a pavement, this input variable would require an incomprehensible amount of data collection and design evaluation. Instead, the problem of handling mixed traffic loadings is solved by axle classifications (by weight and axle type) and the adoption of a standard 18-thousand-pound- equivalent single-axle load or (with 1 kip = 1000 lb) an 18-kip–equivalent single-axle load (ESAL).

4.4 Traditional AASHTO Flexible-Pavement Design Procedure 113

The idea is to determine the impact of any individual axle load on the pavement in terms of the equivalent amount of pavement impact that an 18-kip single-axle load would have. For example, if a 44-kip tandem-axle (double-axle) load has 2.88 times the impact on pavement structure as an 18-kip single-axle load, 2.88 would be the W18 value assigned to this tandem-axle load. The AASHO Road Test also found that the 18-kip– equivalent axle load is a function of the TSI of the pavement structure. The axle-load equivalency factors for flexible-pavement design, with a TSI of 2.5, are presented in Tables 4.1 (for single axles), 4.2 (for tandem axles), and 4.3 (for triple axles).

ZR Represents the probability that serviceability will be maintained at adequate levels from a user’s point of view throughout the design life of the facility. This factor estimates the likelihood that the pavement will perform at or above the TSI level during the design period and takes into account the inherent uncertainty in design. Equation 4.1 uses the z- statistic, which is obtained from the cumulative probabilities of the standard normal distribution (a normal distribution with mean equal to 0 and variance equal to 1). The z-statistics corresponding to various probability levels are given in Table 4.4. In the flexible-pavement–design nomograph (Fig. 4.5), the probabilities (in percent) are used directly (instead of ZR as in the case of Eq. 4.1), and these percent probabilities are denoted R, the reliability (see Table 4.4).

Highways such as interstates and major arterials, which are costly to reconstruct (have their pavements rehabilitated) because of resulting traffic delay and disruption, require a high reliability level, whereas local roads, which will have lower impacts on users in the event of pavement rehabilitation, do not. Typical reliability values for interstate highways are 90% or higher, whereas local roads can have a reliability as low as 50%.

So The overall standard deviation, So, takes into account the designers’ inability to accurately estimate the variation in future 18-kip–equivalent axle loads, and the statistical error in the equations resulting from variability in materials and construction practices. Typical values of So are on the order of 0.30 to 0.50.

SN The structural number, SN, represents the overall structural requirement needed to sustain the design’s traffic loadings. The SN is discussed further in Section 4.4.3.

ΔPSI The amount of serviceability loss over the life of the pavement, ΔPSI, is determined during the pavement design process. The engineer must decide on the final PSI level for a particular pavement. Loss of serviceability is caused by pavement roughness, cracking, patching, and rutting. As pavement distress increases, serviceability decreases. If the design is for a pavement with heavy traffic loads, such as an interstate highway, then the serviceability loss may only be 1.2 (an initial PSI of 4.2 and a TSI of 3.0), whereas a low-volume road can be allowed to deteriorate further, with a possible total serviceability loss of 2.7 or more.

114

Figure 4.5 Design chart for flexible pavements based on the use of mean values for each input. Redrawn from AASHTO Guide for Design of Pavement Structures, Washington, DC, The American Association of State Highway and Transportation Officials, 1993. Used by permission.

4.4 Traditional AASHTO Flexible-Pavement Design Procedure 115

Table 4.1 Axle-Load Equivalency Factors for Flexible Pavements, Single Axles, and TSI = 2.5

Axle load

(kips)

Pavement structural number (SN)

1 2 3 4 5 6

2 0.0004 0.0004 0.0003 0.0002 0.0002 0.0002

4 0.003 0.004 0.004 0.003 0.002 0.002

6 0.011 0.017 0.017 0.013 0.010 0.009

8 0.032 0.047 0.051 0.041 0.034 0.031

10 0.078 0.102 0.118 0.102 0.088 0.080

12 0.168 0.198 0.229 0.213 0.189 0.176

14 0.328 0.358 0.399 0.388 0.360 0.342

16 0.591 0.613 0.646 0.645 0.623 0.606

18 1.00 1.00 1.00 1.00 1.00 1.00

20 1.61 1.57 1.49 1.47 1.51 1.55

22 2.48 2.38 2.17 2.09 2.18 2.30

24 3.69 3.49 3.09 2.89 3.03 3.27

26 5.33 4.99 4.31 3.91 4.09 4.48

28 7.49 6.98 5.90 5.21 5.39 5.98

30 10.3 9.5 7.9 6.8 7.0 7.8

32 13.9 12.8 10.5 8.8 8.9 10.0

34 18.4 16.9 13.7 11.3 11.2 12.5

36 24.0 22.0 17.7 14.4 13.9 15.5

38 30.9 28.3 22.6 18.1 17.2 19.0

40 39.3 35.9 28.5 22.5 21.1 23.0

42 49.3 45.0 35.6 27.8 25.6 27.7

44 61.3 55.9 44.0 34.0 31.0 33.1

46 75.5 68.8 54.0 41.4 37.2 39.3

48 92.2 83.9 65.7 50.1 44.5 46.5

50 112.0 102.0 79.0 60.0 53.0 55.0

Source: AASHTO Guide for Design of Pavement Structures, The American Association of State Highway and Transportation Officials, Washington, DC, 1993. Used by permission.

116 Chapter 4 Pavement Design

Table 4.2 Axle-Load Equivalency Factors for Flexible Pavements, Tandem Axles, and TSI = 2.5

Axle load

(kips)

Pavement structural number (SN)

1 2 3 4 5 6

2 0.0001 0.0001 0.0001 0.0000 0.0000 0.0000 4 0.0005 0.0005 0.0004 0.0003 0.0003 0.0002 6 0.002 0.002 0.002 0.001 0.001 0.001 8 0.004 0.006 0.005 0.004 0.003 0.003

10 0.008 0.013 0.011 0.009 0.007 0.006 12 0.015 0.024 0.023 0.018 0.014 0.013 14 0.026 0.041 0.042 0.033 0.027 0.024 16 0.044 0.065 0.070 0.057 0.047 0.043 18 0.070 0.097 0.109 0.092 0.077 0.070 20 0.107 0.141 0.162 0.141 0.121 0.110 22 0.160 0.198 0.229 0.207 0.180 0.166 24 0.231 0.273 0.315 0.292 0.260 0.242 26 0.327 0.370 0.420 0.401 0.364 0.342 28 0.451 0.493 0.548 0.534 0.495 0.470 30 0.611 0.648 0.703 0.695 0.658 0.633 32 0.813 0.843 0.889 0.887 0.857 0.834 34 1.06 1.08 1.11 1.11 1.09 1.08 36 1.38 1.38 1.38 1.38 1.38 1.38 38 1.75 1.73 1.69 1.68 1.70 1.73 40 2.21 2.16 2.06 2.03 2.08 2.14 42 2.76 2.67 2.49 2.43 2.51 2.61 44 3.41 3.27 2.99 2.88 3.00 3.16 46 4.18 3.98 3.58 3.40 3.55 3.79 48 5.08 4.80 4.25 3.98 4.17 4.49 50 6.12 5.76 5.03 4.64 4.86 5.28 52 7.33 6.87 5.93 5.38 5.63 6.17 54 8.72 8.14 6.95 6.22 6.47 7.15 56 10.3 9.6 8.1 7.2 7.4 8.2 58 12.1 11.3 9.4 8.2 8.4 9.4 60 14.2 13.1 10.9 9.4 9.6 10.7 62 16.5 15.3 12.6 10.7 10.8 12.1 64 19.1 17.6 14.5 12.2 12.2 13.7 66 22.1 20.3 16.6 13.8 13.7 15.4 68 25.3 23.3 18.9 15.6 15.4 17.2 70 29.0 26.6 21.5 17.6 17.2 19.2 72 33.0 30.3 24.4 19.8 19.2 21.3 74 37.5 34.4 27.6 22.2 21.3 23.6 76 42.5 38.9 31.1 24.8 23.7 26.1 78 48.0 43.9 35.0 27.8 26.2 28.8 80 54.0 49.4 39.2 30.9 29.0 31.7 82 60.6 55.4 43.9 34.4 32.0 34.8 84 67.8 61.9 49.0 38.2 35.3 38.1 86 75.7 69.1 54.5 42.3 38.8 41.7 88 84.3 76.9 60.6 46.8 42.6 45.6 90 93.7 85.4 67.1 51.7 46.8 49.7

Source: AASHTO Guide for Design of Pavement Structures, The American Association of State Highway and Transportation Officials, Washington, DC, 1993. Used by permission.

4.4 Traditional AASHTO Flexible-Pavement Design Procedure 117

Table 4.3 Axle-Load Equivalency Factors for Flexible Pavements, Triple Axles, and TSI = 2.5

Axle load

(kips)

Pavement structural number (SN)

1 2 3 4 5 6

2 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 4 0.0002 0.0002 0.0002 0.0001 0.0001 0.0001 6 0.0006 0.0007 0.0005 0.0004 0.0003 0.0003 8 0.001 0.002 0.001 0.001 0.001 0.001

10 0.003 0.004 0.003 0.002 0.002 0.002 12 0.005 0.007 0.006 0.004 0.003 0.003 14 0.008 0.012 0.010 0.008 0.006 0.006 16 0.012 0.019 0.018 0.013 0.011 0.010 18 0.018 0.029 0.028 0.021 0.017 0.016 20 0.027 0.042 0.042 0.032 0.027 0.024 22 0.038 0.058 0.060 0.048 0.040 0.036 24 0.053 0.078 0.084 0.068 0.057 0.051 26 0.072 0.103 0.114 0.095 0.080 0.072 28 0.098 0.133 0.151 0.128 0.109 0.099 30 0.129 0.169 0.195 0.170 0.145 0.133 32 0.169 0.213 0.247 0.220 0.191 0.175 34 0.219 0.266 0.308 0.281 0.246 0.228 36 0.279 0.329 0.379 0.352 0.313 0.292 38 0.352 0.403 0.461 0.436 0.393 0.368 40 0.439 0.491 0.554 0.533 0.487 0.459 42 0.543 0.594 0.661 0.644 0.597 0.567 44 0.666 0.714 0.781 0.769 0.723 0.692 46 0.811 0.854 0.918 0.911 0.868 0.838 48 0.979 1.015 1.072 1.069 1.033 1.005 50 1.17 1.20 1.24 1.25 1.22 1.20 52 1.40 1.41 1.44 1.44 1.43 1.41 54 1.66 1.66 1.66 1.66 1.66 1.66 56 1.95 1.93 1.90 1.90 1.91 1.93 58 2.29 2.25 2.17 2.16 2.20 2.24 60 2.67 2.60 2.48 2.44 2.51 2.58 62 3.09 3.00 2.82 2.76 2.85 2.95 64 3.57 3.44 3.19 3.10 3.22 3.36 66 4.11 3.94 3.61 3.47 3.62 3.81 68 4.71 4.49 4.06 3.88 4.05 4.30 70 5.38 5.11 4.57 4.32 4.52 4.84 72 6.12 5.79 5.13 4.80 5.03 5.41 74 6.93 6.54 5.74 5.32 5.57 6.04 76 7.84 7.37 6.41 5.88 6.15 6.71 78 8.83 8.28 7.14 6.49 6.78 7.43 80 9.92 9.28 7.95 7.15 7.45 8.21 82 11.1 10.4 8.8 7.9 8.2 9.0 84 12.4 11.6 9.8 8.6 8.9 9.9 86 13.8 12.9 10.8 9.5 9.8 10.9 88 15.4 14.3 11.9 10.4 10.6 11.9 90 17.1 15.8 13.2 11.3 11.6 12.9

Source: AASHTO Guide for Design of Pavement Structures, The American Association of State Highway and Transportation Officials, Washington, DC, 1993. Used by permission.

118 Chapter 4 Pavement Design

Table 4.4 Cumulative Percent Probabilities of Reliability, R, of the Standard Normal Distribution, and Corresponding ZR

R 0 1 2 3 4 5 6 7 8 9 9.5 9.9

90 –1.282 –1.341 –1.405 –1.476 –1.555 –1.645 –1.751 –1.881 –2.054 –2.326 –2.576 –3.080

80 –0.842 –0.878 –0.915 –0.954 –0.994 –1.036 –1.080 –1.126 –1.175 –1.227 –1.253 –1.272

70 –0.524 –0.553 –0.583 –0.613 –0.643 –0.675 –0.706 –0.739 –0.772 –0.806 –0.824 –0.838 60 –0.253 –0.279 –0.305 –0.332 –0.358 –0.385 –0.412 –0.440 –0.468 –0.496 –0.510 –0.522

50 0 –0.025 –0.050 –0.075 –0.100 –0.125 –0.151 –0.176 –0.202 –0.228 –0.241 –0.251

Example: To be 95% confident that the pavement will remain at or above its TSI (R = 95 for use in Fig. 4.7), a ZR value of –1.645 would be used in Eq. 4.1 (and in Eq. 4.4).

MR The soil resilient modulus, MR, is used to reflect the engineering properties

of the subgrade (the soil). Each time a vehicle passes over pavement, stresses are developed in the subgrade. After the load passes, the subgrade soil relaxes and the stress is relieved. The resilient modulus test is used to determine the properties of the soil under this repeated load. The resilient modulus can be determined by AASHTO test method T274. Measurement of the resilient modulus is not performed by all transportation agencies; therefore, a relationship between MR and the California bearing ratio (CBR) has been determined. The CBR has been widely used to determine the supporting characteristics of soils since the mid-1930s, and a significant amount of historical information is available. The CBR is the ratio of the load-bearing capacity of the soil to the load-bearing capacity of a high- quality aggregate, multiplied by 100. The relationship, used to provide a very basic approximation of MR (in lb/in2) from a known CBR, is

MR = 1500 × CBR (4.2)

The coefficient of 1500 in Eq. 4.2 is used for CBR values less than 10. Caution must be exercised in applying this equation to higher CBRs because the coefficient (the value 1500 shown in Eq. 4.2) has a range of 750 to 3000.

4.4.3 Structural Number The objective of Eq. 4.1 and the nomograph in Fig. 4.5 is to determine a required SN for given axle loadings, reliability, overall standard deviation, change in PSI, and soil resilient modulus. As previously mentioned, there are many pavement material combinations and thicknesses that will provide satisfactory pavement service life. The following equation can be used to relate individual material types and thicknesses to the SN:

SN = a1D1 + a2D2M2 + a3D3M3 (4.3)

where

4.4 Traditional AASHTO Flexible-Pavement Design Procedure 119

a1, a2, a3 = structural-layer coefficients of the wearing surface, base, and subbase layers, respectively,

D1, D2, D3 = thickness of the wearing surface, base, and subbase layers in inches, respectively, and

M2, M3 = drainage coefficients for the base and subbase, respectively.

Values for the structural-layer coefficients for various types of material are presented in Table 4.5. Drainage coefficients are used to modify the thickness of the lower pavement layers (base and subbase) to take into account a material’s drainage characteristics. A value of 1.0 for a drainage coefficient represents a material with good drainage characteristics (a sandy material). A soil such as clay does not drain very well and, consequently, will have a lower drainage coefficient (less than 1.0) than a sandy material. The reader is referred to [AASHTO, 1993] for further information on drainage coefficients.

Because there are many combinations of structural-layer coefficients and thicknesses that solve Eq. 4.3, some guidelines are used to narrow the number of solutions. Experience has shown that wearing layers are typically 2 to 4 inches thick, whereas subbases and bases range from 4 to 10 inches thick. Knowing which of the materials is the most costly per inch of depth will assist in the determination of an initial layer thickness.

Table 4.5 Structural-Layer Coefficients

Pavement component Coefficient

Wearing surface

Sand-mix asphaltic concrete 0.35 Hot-mix asphaltic (HMA) concrete 0.44

Base

Crushed stone 0.14 Dense-graded crushed stone 0.18

Soil cement 0.20

Emulsion/aggregate-bituminous 0.30 Portland cement/aggregate 0.40

Lime-pozzolan/aggregate 0.40

Hot-mix asphaltic (HMA) concrete 0.40 Subbase

Crushed stone 0.11

120 Chapter 4 Pavement Design

EXAMPLE 4.1 FLEXIBLE-PAVEMENT DESIGN—STRUCTURAL NUMBER DETERMINATION

A pavement is to be designed to last 10 years. The initial PSI is 4.2, and the TSI (the final PSI) is determined to be 2.5. The subgrade has a soil resilient modulus of 15,000 lb/in2. Reliability is 95% with an overall standard deviation of 0.4. For design, the daily car, pickup truck, and light van traffic is 30,000, and the daily truck traffic consists of 1000 passes of single-unit trucks with two single axles and 350 passes of tractor semi- trailer trucks with single, tandem, and triple axles. The axle weights are

cars, pickups, light vans = two 2000-lb single axles single-unit truck = 8000-lb steering, single axle

= 22,000-lb drive, single axle tractor semi-trailer truck = 10,000-lb steering, single axle

= 16,000-lb drive, tandem axle = 44,000-lb trailer, triple axle

M2 and M3 are equal to 1.0 for the materials in the pavement structure. Four

inches of hot-mix asphalt (HMA) is to be used as the wearing surface and 10 inches of crushed stone as the subbase. Determine the thickness required for the base if soil cement is the material to be used.

SOLUTION

Because the axle-load equivalency factors presented in Tables 4.1, 4.2, and 4.3 are a function of the SN, we have to assume an SN to start the problem (later we will arrive at a SN and check to make sure that it is consistent with our assumed value). A typical assumption is to let SN = 4. Given this, the 18-kip–equivalent single-axle load for cars, pickups, and light vans is

2-kip single-axle equivalent = 0.0002 (Table 4.1)

This gives an 18-kip ESAL total of 0.0004 for each vehicle. For single-unit trucks,

8-kip single-axle equivalent = 0.041 (Table 4.1) 22-kip single-axle equivalent = 2.090 (Table 4.1)

This gives an 18-kip ESAL total of 2.131 for single-unit trucks. For tractor semi-trailer trucks,

10-kip single-axle equivalent = 0.102 (Table 4.1) 16-kip tandem-axle equivalent = 0.057 (Table 4.2)

44-kip triple-axle equivalent = 0.769 (Table 4.3) This gives an 18-kip ESAL total of 0.928 for tractor semi-trailer trucks. Note the comparatively small effect of cars and other light vehicles in terms of the 18-kip ESAL. This small effect underscores the nonlinear relationship between axle loads and pavement damage. For example, from Table 4.2 with SN = 4, a 36-kip single-axle load has 14.4 times the impact on pavement as an 18-kip single-axle load (twice the weight has 14.4 times the impact).

4.4 Traditional AASHTO Flexible-Pavement Design Procedure 121

Given the computed 18-kip ESAL, the daily traffic on this highway produces an 18-kip ESAL total of 2467.8 (0.0004 × 30,000 + 2.131 × 1000 + 0.928 × 350). Traffic (total axle accumulations) over the 10-year design period will be

2467.8 × 365 × 10 = 9,007,470 18-kip ESAL

With an initial PSI of 4.2 and a TSI of 2.5, ΔPSI = 1.7. Solving Eq. 4.1 for SN (using an equation solver on a calculator or computer) with ZR = −1.645 (which corresponds to R = 95%, as shown in Table 4.4) gives SN = 3.94 (Fig. 4.5 can also be used to arrive at an approximate solution for SN). Note that this is very close to the value that was assumed (SN = 4.0) to get the load equivalency factors from Tables 4.1, 4.2, and 4.3. If Eq. 4.1 gave SN = 5, we would go back and recompute total axle accumulations using the SN of 5 to read the axle-load equivalency factors in Tables 4.1, 4.2, and 4.3. Usually one iteration of this type is all that is needed. Later, Examples 4.3 and 4.5 will demonstrate this type of iteration.

Given that SN = 3.94, Eq. 4.3 can be applied with a1 = 0.44 (surface course, hot- mix asphalt, Table 4.5), a2 = 0.20 (base course, soil cement, Table 4.5), and a3 = 0.11 (subbase, crushed stone, Table 4.5), M2 = 1.0 (given), M3 = 1.0 (given), D1 = 4.0 inches (given), and D3 = 10.0 inches (given). We have

SN = a1D1 + a2D2M2 + a3D3M3

3.94 = 0.44(4) + 0.20D2(1.0) + 0.11(10.0)(1.0) Solving for D2 gives D2 = 5.4 inches. Using D2 = 5.5 inches would be a conservative estimate and allow for variations in construction. Rounding up to the nearest 0.5 inch is a safe practice.

EXAMPLE 4.2 FLEXIBLE-PAVEMENT DESIGN—RELIABILITY ASSESSMENT

A flexible pavement is constructed with 4 inches of HMA wearing surface, 8 inches of emulsion/aggregate-bituminous base, and 8 inches of crushed stone subbase. The subgrade has a soil resilient modulus of 10,000 lb/in2, and M2 and M3 are equal to 1.0 for the materials in the pavement structure. The overall standard deviation is 0.5, the initial PSI is 4.5, and the TSI is 2.5. The daily traffic has 1080 20-kip single axles, 400 24-kip single axles, and 680 40-kip tandem axles. How many years would you estimate this pavement would last (how long before its PSI drops below a TSI of 2.5) if you wanted to be 90% confident that your estimate was not too high, and if you wanted to be 99% confident that your estimate was not too high?

SOLUTION

The pavement’s SN is determined from Eq. 4.3, using Table 4.5 to find a1 = 0.44, a2 = 0.30 and a3 = 0.11, and with D1 = 4, D2 = 8, D3 = 8, M2 = M3 = 1.0 (all given) as

SN = a1D1 + a2D2M2 + a3D3M3

SN = 0.44(4) + 0.30(8)(1.0) + 0.11(8.0)(1.0) = 5.04

For the daily axle loads, the equivalency factors (reading axle equivalents from Tables 4.1 and 4.2 while using SN = 5, which is very close to the 5.04 computed above) are

122 Chapter 4 Pavement Design

20-kip single-axle equivalent = 1.51 (Table 4.1) 24-kip single-axle equivalent = 3.03 (Table 4.1)

40-kip tandem-axle equivalent = 2.08 (Table 4.2) Thus the total daily 18-kip ESAL is

Daily W18 = 1.51(1080) + 3.03(400) + 2.08(680) = 4257.2 18-kip ESAL

Applying Eq. 4.1, with So = 0.5, SN = 5.04, ΔPSI = 2.0 (4.5 − 2.5), and MR = 10,000 lb/in2, we find that at R = 90% (ZR = −1.282 for purposes of Eq. 4.1, as shown in Table 4.4), W18 is 26,128,077. Therefore, the number of years is

26,128,077 years =

365 × 4257.2 =16.82 years

Similarly, with R = 99% (ZR = −2.326 for purposes of Eq. 4.1, as shown in Table 4.4), W18 is 7,854,299, so the number of years is

7,854,299 years =

365 × 4257.2 =5.05 years

These results show that one can be 99% confident that the pavement will last (have a PSI above 2.5) at least 5.05 years, and one can be 90% confident that it will have a PSI above 2.5 for 16.82 years. This example demonstrates the large impact that the chosen reliability value can have on pavement design.

4.5 PAVEMENT SYSTEM DESIGN: PRINCIPLES FOR RIGID PAVEMENTS

Rigid pavements distribute wheel loads by the beam action of the PCC slab, which is made of a material that has a high modulus of elasticity, on the order of 4 to 5 million lb/in2. This beam action (see Fig. 4.6) distributes the wheel loads over a large area of the pavement, thus reducing the high stresses experienced at the surface of the pavement to a level that is acceptable to the subgrade soil.

Figure 4.6 Beam action of a rigid pavement.

4.6 Traditional AASHTO Rigid-Pavement Design Procedure 123

4.6 TRADITIONAL AASHTO RIGID-PAVEMENT DESIGN PROCEDURE

The design procedure for rigid pavements presented in the AASHTO design guide is also based on the field results of the AASHO Road Test. The AASHTO design procedure is applicable to jointed-plain concrete pavements, jointed-reinforced concrete pavements, and continuously-reinforced concrete pavements. Jointed- plain concrete pavements (JPCP) do not have slab-reinforcing material and can have doweled joints (steel bars to transfer loads between slabs as shown in Fig. 4.2) or undoweled joints. The traverse joints between slabs are spaced at about 10 to 13 ft. Jointed reinforced concrete pavements (JRCP) have steel reinforced slabs with joints that are 40 ft or more apart. Finally, continuously- reinforced concrete pavements (CRCP) do not have traverse expansion/ contraction joints, necessitating the use of extensive steel-bar reinforcement in the slab. The idea with both jointed-reinforced and continuously-reinforced pavements is to permit slab cracking but to provide sufficient slab reinforcement to hold the cracks tightly together to ensure load transfer. It is important to note that faulting, which is a distress characterized by different slab elevations, was not a failure consideration in the AASHO Road Test, and thus the design of undoweled joints must be checked with a procedure other than that presented here (more information on faulting is provided in Section 4.7.5).

The design procedure for rigid pavements is based on a selected reduction in serviceability and is similar to the procedure for flexible pavements. However, instead of measuring pavement strength by using a SN, the thickness of the PCC slab is the measure of strength. The regression equation that is used to determine the thickness of a rigid-pavement PCC slab is

( ) [ ]

( )

( ) ( ){ }

  −  Δ

 × +     ′ −  −   −   

10 18 o 10

10

8.467

0.75

10 0.250.75

log = +7.35 log + 1 0.06

log PSI / 3.0 +

1+ 1.624 10 1

1.132 + 4.22 0.32TSI log

215.63 18.42

R

c d

c

W Z S D

D

S C D

J D E k

(4.4)

where

W18 = 18-kip−equivalent single-axle loads,

TSI = Pavement’s terminal serviceability index,

ZR = Reliability (z-statistic from the standard normal curve),

S′c = Concrete modulus of rupture in lb/in2,

So = Overall standard deviation of traffic,

Cd = Drainage coefficient, J = Load transfer coefficient,

D = PCC slab thickness in inches,

Ec = Concrete modulus of elasticity in lb/ in2, and

ΔPSI = Loss in serviceability from the time when the pavement is new until it reaches its TSI,

k = Modulus of subgrade reaction in lb/in3.

124 Chapter 4 Pavement Design

A graphic solution to Eq. 4.4 is shown in Figs. 4.7 and 4.8. The terms used in Eq. 4.4 and Figs. 4.8 and 4.9 are defined as follows: W18 The 18-kip–equivalent single-axle load is the same concept as discussed for

the flexible-pavement design procedure. However, instead of being a function of the SN, this value is a function of slab thickness. The axle- load equivalency factors used in rigid-pavement design are presented in Tables 4.6 (for single axles), 4.7 (for tandem axles), and 4.8 (for triple axles).

ZR As in flexible-pavement design, the reliability, ZR, is defined as the probability that serviceability will be maintained at adequate levels from a user’s point of view throughout the design life of the facility (the PSI will stay above the TSI). In the rigid-pavement–design nomograph (Figs. 4.7 and 4.8), the probabilities (in percent) are used directly (instead of ZR as in Eq. 4.4), and these percent probabilities are denoted R (see Table 4.4, which still applies).

So As in flexible-pavement design, the overall standard deviation, So, takes into account designers’ inability to accurately estimate future 18-kip– equivalent axle loads and the statistical error in the equations resulting from variability in materials and construction practices.

TSI The pavement’s terminal serviceability index, TSI, is the point at which the pavement can no longer perform in a serviceable manner, as discussed previously for the flexible-pavement design procedure.

ΔPSI The amount of serviceability loss, ΔPSI, over the life of the pavement is the difference between the initial PSI and the TSI, as discussed for the flexible-pavement design procedure.

S′c The concrete modulus of rupture, S′c, is a measure of the tensile strength of the concrete and is determined by loading a beam specimen, at the third points, to failure. The test method is ASTM C78, Flexural Strength of Concrete. Because concrete gains strength with age, the average 28-day strength is used for design purposes. Typical values are 500 to 1200 lb/in2.

Cd The drainage coefficient, Cd, is slightly different from the value used in flexible-pavement design. In rigid-pavement design, it accounts for the drainage characteristics of the subgrade. A value of 1.0 for the drainage coefficient represents a material with good drainage characteristics (such as a sandy material). Soils with less-than-ideal drainage characteristics will have drainage coefficients less than 1.0.

4.6 Traditional AASHTO Rigid-Pavement Design Procedure 125

J The load transfer coefficient, J, is a factor that is used to account for the ability of pavement to transfer a load from one PCC slab to another across the slab joints. Many rigid pavements have dowel bars across the joints to transfer loads between slabs. Pavements with dowel bars at the joints are typically designed with a J value of 3.2.

Table 4.6 Axle-Load Equivalency Factors for Rigid Pavements, Single Axles, and TSI = 2.5

Axle load

(kips)

Slab thickness, D (inches)

6 7 8 9 10 11 12 13 14

2 0.0002 0.0002 0.0002 0.0002 0.0002 0.0002 0.0002 0.0002 0.0002 4 0.003 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 6 0.012 0.011 0.010 0.010 0.010 0.010 0.010 0.010 0.010 8 0.039 0.035 0.033 0.032 0.032 0.032 0.032 0.032 0.032

10 0.097 0.089 0.084 0.082 0.081 0.080 0.080 0.080 0.080 12 0.203 0.189 0.181 0.176 0.175 0.174 0.174 0.174 0.173 14 0.376 0.360 0.347 0.341 0.338 0.337 0.336 0.336 0.336 16 0.634 0.623 0.610 0.604 0.601 0.599 0.599 0.599 0.598 18 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 20 1.51 1.52 1.55 1.57 1.58 1.58 1.59 1.59 1.59 22 2.21 2.20 2.28 2.34 2.38 2.40 2.41 2.41 2.41 24 3.16 3.10 3.22 3.36 3.45 3.50 3.53 3.54 3.55 26 4.41 4.26 4.42 4.67 4.85 4.95 5.01 5.04 5.05 28 6.05 5.76 5.92 6.29 6.61 6.81 6.92 6.98 7.01 30 8.16 7.67 7.79 8.28 8.79 9.14 9.35 9.46 9.52 32 10.8 10.1 10.1 10.7 11.4 12.0 12.3 12.6 12.7 34 14.1 13.0 12.9 13.6 14.6 15.4 16.0 16.4 16.5 36 18.2 16.7 16.4 17.1 18.3 19.5 20.4 21.0 21.3 38 23.1 21.1 20.6 21.3 22.7 24.3 25.6 26.4 27.0 40 29.1 26.5 25.7 26.3 27.9 29.9 31.6 32.9 33.7 42 36.2 32.9 31.7 32.2 34.0 36.3 38.7 40.4 41.6 44 44.6 40.4 38.8 39.2 41.0 43.8 46.7 49.1 50.8 46 54.5 49.3 47.1 47.3 49.2 52.3 55.9 59.0 61.4 48 66.1 59.7 56.9 56.8 58.7 62.1 66.3 70.3 73.4 50 79.4 71.7 68.2 67.8 69.6 73.3 78.1 83.0 87.1

Source: AASHTO Guide for Design of Pavement Structures, The American Association of State Highway and Transportation Officials, Washington, DC, 1993. Used by permission.

126 Chapter 4 Pavement Design

Table 4.7 Axle-Load Equivalency Factors for Rigid Pavements, Tandem Axles, and TSI = 2.5

Axle load

(kips)

Slab thickness, D (inches)

6 7 8 9 10 11 12 13 14

2 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 4 0.0006 0.0006 0.0005 0.0005 0.0005 0.0005 0.0005 0.0005 0.0005 6 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002 8 0.007 0.006 0.006 0.005 0.005 0.005 0.005 0.005 0.005

10 0.015 0.014 0.013 0.013 0.012 0.012 0.012 0.012 0.012 12 0.031 0.028 0.026 0.026 0.025 0.025 0.025 0.025 0.025 14 0.057 0.052 0.049 0.048 0.047 0.047 0.047 0.047 0.047 16 0.097 0.089 0.084 0.082 0.081 0.081 0.080 0.080 0.080 18 0.155 0.143 0.136 0.133 0.132 0.131 0.131 0.131 0.131 20 0.234 0.220 0.211 0.206 0.204 0.203 0.203 0.203 0.203 22 0.340 0.325 0.313 0.308 0.305 0.304 0.303 0.303 0.303 24 0.475 0.462 0.450 0.444 0.441 0.440 0.439 0.439 0.439 26 0.644 0.637 0.627 0.622 0.620 0.619 0.618 0.618 0.618 28 0.855 0.854 0.852 0.850 0.850 0.850 0.849 0.849 0.849 30 1.11 1.12 1.13 1.14 1.14 1.14 1.14 1.14 1.14 32 1.43 1.44 1.47 1.49 1.50 1.51 1.51 1.51 1.51 34 1.82 1.82 1.87 1.92 1.95 1.96 1.97 1.97 1.97 36 2.29 2.27 2.35 2.43 2.48 2.51 2.52 2.52 2.53 38 2.85 2.80 2.91 3.03 3.12 3.16 3.18 3.20 3.20 40 3.52 3.42 3.55 3.74 3.87 3.94 3.98 4.00 4.01 42 4.32 4.16 4.30 4.55 4.74 4.86 4.91 4.95 4.96 44 5.26 5.01 5.16 5.48 5.75 5.92 6.01 6.06 6.09 46 6.36 6.01 6.14 6.53 6.90 7.14 7.28 7.36 7.40 48 7.64 7.16 7.27 7.73 8.21 8.55 8.75 8.86 8.92 50 9.11 8.50 8.55 9.07 9.68 10.14 10.42 10.58 10.66 52 10.8 10.0 10.0 10.6 11.3 11.9 12.3 12.5 12.7 54 12.8 11.8 11.7 12.3 13.2 13.9 14.5 14.8 14.9 56 15.0 13.8 13.6 14.2 15.2 16.2 16.8 17.3 17.5 58 17.5 16.0 15.7 16.3 17.5 18.6 19.5 20.1 20.4 60 20.3 18.5 18.1 18.7 20.0 21.4 22.5 23.2 23.6 63 23.5 21.4 20.8 21.4 22.8 24.4 25.7 26.7 27.3 64 27.0 24.6 23.8 24.4 25.8 27.7 29.3 30.5 31.3 66 31.0 28.1 27.1 27.6 29.2 31.3 33.2 34.7 35.7 68 35.4 32.1 30.9 31.3 32.9 35.2 37.5 39.3 40.5 70 40.3 36.5 35.0 35.3 37.0 39.5 42.1 44.3 45.9 72 45.7 41.4 39.6 39.8 41.5 44.2 47.2 49.8 51.7 74 51.7 46.7 44.6 44.7 46.4 49.3 52.7 55.7 58.0 76 58.3 52.6 50.2 50.1 51.8 54.9 58.6 62.1 64.8 78 65.5 59.1 56.3 56.1 57.7 60.9 65.0 69.0 72.3 80 73.4 66.2 62.9 62.5 64.2 67.5 71.9 76.4 80.2 82 82.0 73.9 70.2 69.6 71.2 74.7 79.4 84.4 88.8 84 91.4 82.4 78.1 77.3 78.9 82.4 87.4 93.0 98.1 86 102.0 92.0 87.0 86.0 87.0 91.0 96.0 102.0 108.0 88 113.0 102.0 96.0 95.0 96.0 100.0 105.0 112.0 119.0 90 125.0 112.0 106.0 105.0 106.0 110.0 115.0 123.0 130.0

Source: AASHTO Guide for Design of Pavement Structures, The American Association of State Highway and Transportation Officials, Washington, DC, 1993. Used by permission.

4.6 Traditional AASHTO Rigid-Pavement Design Procedure 127

Table 4.8 Axle-Load Equivalency Factors for Rigid Pavements, Triple Axles, and TSI = 2.5

Axle load

(kips)

Slab thickness, D (inches)

6 7 8 9 10 11 12 13 14

2 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 4 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 6 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 8 0.003 0.002 0.002 0.002 0.002 0.002 0.002 0.002 0.002

10 0.006 0.005 0.005 0.005 0.005 0.005 0.005 0.005 0.005 12 0.011 0.010 0.010 0.009 0.009 0.009 0.009 0.009 0.009 14 0.020 0.018 0.017 0.017 0.016 0.016 0.016 0.016 0.016 16 0.033 0.030 0.029 0.028 0.027 0.027 0.027 0.027 0.027 18 0.053 0.048 0.045 0.044 0.044 0.043 0.043 0.043 0.043 20 0.080 0.073 0.069 0.067 0.066 0.066 0.066 0.066 0.066 22 0.116 0.107 0.101 0.099 0.098 0.097 0.097 0.097 0.097 24 0.163 0.151 0.144 0.141 0.139 0.139 0.138 0.138 0.138 26 0.222 0.209 0.200 0.195 0.194 0.193 0.192 0.192 0.192 28 0.295 0.281 0.271 0.265 0.263 0.262 0.262 0.262 0.262 30 0.384 0.371 0.359 0.354 0.351 0.350 0.349 0.349 0.349 32 0.490 0.480 0.468 0.463 0.460 0.459 0.458 0.458 0.458 34 0.616 0.609 0.601 0.596 0.594 0.593 0.592 0.592 0.592 36 0.765 0.762 0.759 0.757 0.756 0.755 0.755 0.755 0.755 38 0.939 0.941 0.946 0.948 0.950 0.951 0.951 0.951 0.951 40 1.14 1.15 1.16 1.17 1.18 1.18 1.18 1.18 1.18 42 1.38 1.38 1.41 1.44 1.45 1.46 1.46 1.46 1.46 44 1.65 1.65 1.70 1.74 1.77 1.78 1.78 1.78 1.78 46 1.97 1.96 2.03 2.09 2.13 2.15 2.16 2.16 2.16 48 2.34 2.31 2.40 2.49 2.55 2.58 2.59 2.60 2.60 50 2.76 2.71 2.81 2.94 3.02 3.07 3.09 3.10 3.11 52 3.24 3.15 3.27 3.44 3.56 3.62 3.66 3.68 3.68 54 3.79 3.66 3.79 4.00 4.16 4.26 4.30 4.33 4.34 56 4.41 4.23 4.37 4.63 4.84 4.97 5.03 5.07 5.09 58 5.12 4.87 5.00 5.32 5.59 5.76 5.85 5.90 5.93 60 5.91 5.59 5.71 6.08 6.42 6.64 6.77 6.84 6.87 63 6.80 6.39 6.50 6.91 7.33 7.62 7.79 7.88 7.93 64 7.79 7.29 7.37 7.82 8.33 8.70 8.92 9.04 9.11 66 8.90 8.28 8.33 8.83 9.42 9.88 10.17 10.33 10.42 68 10.1 9.4 9.4 9.9 10.6 11.2 11.5 11.7 11.9 70 11.5 10.6 10.6 11.1 11.9 12.6 13.0 13.3 13.5 72 13.0 12.0 11.8 12.4 13.3 14.1 14.7 15.0 15.2 74 14.6 13.5 13.2 13.8 14.8 15.8 16.5 16.9 17.1 76 16.5 15.1 14.8 15.4 16.5 17.6 18.4 18.9 19.2 78 18.5 16.9 16.5 17.1 18.2 19.5 20.5 21.1 21.5 80 20.6 18.8 18.3 18.9 20.2 21.6 22.7 23.5 24.0 82 23.0 21.0 20.3 20.9 22.2 23.8 25.2 26.1 26.7 84 25.6 23.3 22.5 23.1 24.5 26.2 27.8 28.9 29.6 86 28.4 25.8 24.9 25.4 26.9 28.8 30.5 31.9 32.8 88 31.5 28.6 27.5 27.9 29.4 31.5 33.5 35.1 36.1 90 34.8 31.5 30.3 30.7 32.2 34.4 36.7 38.5 39.8

Source: AASHTO Guide for Design of Pavement Structures, The American Association of State Highway and Transportation Officials, Washington, DC, 1993. Used by permission.

128 Chapter 4 Pavement Design

Ec The concrete modulus of elasticity, Ec, is derived from the stress-strain curve as taken in the elastic region. The modulus of elasticity is also known as Young’s modulus. Typical values of Ec for PCC are between 3 and 7 million lb/in2.

k The modulus of subgrade reaction, k, depends upon several different factors, including the moisture content and density of the soil. It should be noted that most highway agencies do not perform testing to measure the k value of the soil. At best, the agency will have a CBR value for the subgrade. Typical values for k range from 100 to 800 lb/in3. Table 4.9 indicates the relationship between CBR and k values.

Table 4.9 Relationship between California Bearing Ratio (CBR) and Modulus of Subgrade Reaction, k

CBR k, lb/in3

2 100

10 200 20 250

25 290

40 420 50 500

75 680

100 800

EXAMPLE 4.3 RIGID-PAVEMENT DESIGN SLAB THICKNESS DETERMINATION

A rigid pavement is to be designed to provide a service life of 20 years and has an initial PSI of 4.4 and a TSI of 2.5. The modulus of subgrade reaction is determined to be 300 lb/in3. For design, the daily car, pickup truck, and light van traffic is 20,000; and the daily truck traffic consists of 200 passes of single-unit trucks with single and tandem axles, and 410 passes of tractor semi-trailer trucks with single, tandem, and triple axles. The axle weights are

cars, pickups, light vans = two 2000-lb single axles

single-unit trucks = 10,000-lb steering, single axle

= 22,000-lb drive, tandem axle

tractor semi-trailer trucks = 12,000-lb steering, single axle

= 18,000-lb drive, tandem axle

= 50,000-lb trailer, triple axle

Reliability is 95%, the overall standard deviation is 0.45, the concrete’s modulus of elasticity is 4.5 million lb/in2, the concrete’s modulus of rupture is 900 lb/in2, the load transfer coefficient is 3.2, and the drainage coefficient is 1.0. Determine the required slab thickness.

129

Figure 4.7 Segment 1 of the design chart for rigid pavement based on the use of mean values for each input variable. Redrawn from AASHTO Guide for Design of Pavement Structures, The American Association of State Highway and Transportation Officials, Washington, DC, 1993. Used by permission.

130 Chapter 4 Pavement Design

Figure 4.8 Segment 2 of the design chart for rigid pavements based on the use of mean values for each input variable. Redrawn from AASHTO Guide for Design of Pavement Structures, The American Association of State Highway and Transportation Officials, Washington, DC, 1993. Used by permission.

4.6 Traditional AASHTO Flexible-Pavement Design Procedure 131

SOLUTION

Because the axle-load equivalency factors presented in Tables 4.6, 4.7, and 4.8 are a function of the slab thickness (D), we have to assume a D value to start the problem (later we will arrive at a slab thickness and check to make sure that it is consistent with our assumed value). A typical assumption is to let D = 10 inches. Given this, the 18- kip–equivalent single-axle load (18-kip ESAL) for cars, pickups, and light vans is

2-kip single-axle equivalent = 0.0002 (Table 4.6)

This gives an 18-kip ESAL total of 0.0004 for each vehicle. For single-unit trucks,

10-kip single-axle equivalent = 0.081 (Table 4.6)

22-kip tandem-axle equivalent = 0.305 (Table 4.7)

This gives an 18-kip ESAL total of 0.386 for single-unit trucks. For tractor semi-trailer trucks,

12-kip single-axle equivalent = 0.175 (Table 4.6)

18-kip tandem-axle equivalent = 0.132 (Table 4.7)

50-kip triple-axle equivalent = 3.020 (Table 4.8)

This gives an 18-kip ESAL total of 3.327 for tractor semi-trailer trucks. Given the computed 18-kip ESAL, the daily traffic on this highway produces an

18-kip ESAL total of 1449.27 (0.0004 × 20,000 + 0.386 × 200 + 3.327 × 410). Traffic (total axle accumulations) over the 20-year design period will be

1449.27 × 365 × 20 = 10,579,671 18-kip ESAL

With an initial PSI of 4.4 and a TSI of 2.5, ΔPSI = 1.9. Solving Eq. 4.4 for D (using an equation solver on a calculator or computer) with ZR = −1.645 (which corresponds to R = 95%, as shown in Table 4.4) gives D = 9.21 inches. (Figs. 4.7 and 4.8 can also be used to arrive at an approximate solution for D.) Note that this value differs from the slab thickness assumed to derive the axle-load equivalency factors. Recomputing the axle-load equivalency factors with D = 9 inches (for Tables 4.6, 4.7, and 4.8) for cars, pickups, and light vans gives

2-kip single-axle equivalent = 0.0002 (Table 4.6)

This gives an 18-kip ESAL total of 0.0004 (same as before) for each vehicle. For single- unit trucks,

10-kip single-axle equivalent = 0.082 (Table 4.6)

22-kip tandem-axle equivalent = 0.308 (Table 4.7)

This gives an 18-kip ESAL total of 0.390 (up from 0.386) for single-unit trucks. For tractor semi-trailer trucks,

12-kip single-axle equivalent = 0.176 (Table 4.6)

18-kip tandem-axle equivalent = 0.133 (Table 4.7)

50-kip triple-axle equivalent = 2.940 (Table 4.8)

132 Chapter 4 Pavement Design

This gives an 18-kip ESAL total of 3.249 (down from 3.327) for tractor semi-trailer trucks. Given the computed 18-kip ESAL, the daily traffic on this highway produces an 18-kip ESAL of 1418.09 (0.0004 × 20,000 + 0.390 × 200 + 3.249 × 410). Traffic (total axle accumulations) over the 20-year design period will be

1418.09 × 365 × 20 = 10,352,057 18-kip ESAL

Again, solving Eq. 4.4 for D gives D = 9.17 inches. (Figs. 4.7 and 4.8 can also be used to arrive at an approximate solution for D.) This is very close to the assumed D = 9.0 inches and is only a minor change from the 9.21 inches previously obtained. To be conservative, we would round up to the nearest 0.5 inch and make the slab 9.5 inches.

4.7 DESIGN-LANE LOADS

The final point to be covered with regard to pavement design (for both flexible and rigid pavements) relates to the case where there are multiple lanes of a highway (such as an interstate) in one direction. Because traffic tends to be distributed among the lanes, in some instances, the pavement can be designed using a fraction of the total directional W18. However, because traffic tends to concentrate in the right lane (particularly heavy vehicles), this fraction is not as simple as dividing W18 by the number of lanes. In equation form,

design-lane W18 = PDL × directional W18 (4.5)

where

W18 = 18-kip–equivalent single-axle load (ESAL) and

PDL = proportion of directional W18 assumed to be in the design lane.

AASHTO-recommended values for PDL are given in Table 4.10.

As an example, suppose the computed directional W18 is an 18-kip ESAL of 10,000,000 and there are three lanes in the direction of travel. If the highway is conservatively designed, Table 4.10 shows that 80% of the axle loads can be assumed to be in the design lane (PDL3-lanes = 0.8). So the design W18 would be 8,000,000 (0.8 × 10,000,000), and this value would be used in the equations and nomographs. This design procedure applies to both flexible and rigid pavements.

4.7 Design-Lane Loads 133

Table 4.10 Proportion of Directional W18 Assumed to Be in the Design Lane

Number of directional lanes Proportion of directional W18

in the design lane (PDL)

1 1.00 2 0.80–1.00

3 0.60–0.80

4 0.50–0.75

EXAMPLE 4.4 RIGID-PAVEMENT DESIGN WITH TRAFFIC DISTRIBUTION BY LANE

In 1996, a rigid pavement on a northbound section of interstate highway was designed with a 12-inch PCC slab, an Ec of 6 × 106 lb/in2, a concrete modulus of rupture of 800 lb/in2, a load transfer coefficient of 3.0, an initial PSI of 4.5, and a TSI of 2.5. The overall standard deviation was 0.45, the modulus of subgrade reaction was 190 lb/in3, and a reliability of 95% was used along with a drainage coefficient of 1.0. The pavement was designed for a 20-year life, and traffic was assumed to be composed entirely of tractor semi-trailer trucks with one 16-kip single axle, one 20-kip single axle, and one 35-kip tandem axle (the effect of all other vehicles was ignored). The interstate has four northbound lanes and was conservatively designed. How many tractor semi-trailer trucks, per day, were assumed to be traveling in the northbound direction?

SOLUTION

Given that the slab thickness D is 12 inches, for the tractor semi-trailer trucks we have

16-kip single-axle equivalent = 0.599 (Table 4.6)

20-kip single-axle equivalent = 1.590 (Table 4.6)

35-kip tandem-axle equivalent = 2.245 (Table 4.7) Note that the value of 2.245 for the 35,000-lb tandem-axle linear interpolation uses 34-kip and 36-kip values [(1.97 + 2.52)/2]. Summing these axle equivalents gives 4.434 18-kip ESAL per truck.

With an initial PSI of 4.5 and a TSI of 2.5, ΔPSI = 2.0. Solving Eq. 4.4 for W18 with ZR = −1.645 (which corresponds to R = 95% as shown in Table 4.4) gives W18 = 39,740,309 18-kip ESAL (Figs. 4.7 and 4.8 can also be used to arrive at an approximate solution for W18). Thus, the total daily truck traffic in the design lane is

39,740,309 18-kip ESAL traffic =

365 days / year × 20 years × 4.434 18-kip ESAL / truck

=1227.76 trucks / day

To determine the total directional volume (total number of northbound trucks), we note from Table 4.10 that the PDL for a conservative design on a four-lane highway is 0.75, and the application of Eq. 4.5 gives

134 Chapter 4 Pavement Design

18 18

4-lanes

design-lane directional =

1227.76 =

0.75 =1637.01 trucks / day

W W

PDL

EXAMPLE 4.5 FLEXIBLE AND RIGID PAVEMENT DESIGN-LIFE EQUIVALENCE

A new flexible pavement was designed for four lanes of traffic (conservatively designed for load distribution among the lanes). The design is for a total directional daily traffic of 967 10-kip single axles and 1935 30-kip tandem axles. The pavement has an 8-in hot- mix asphaltic (HMA) surface, 10-in dense-graded crushed stone base and a 10-in crushed stone subbase (the drainage coefficients are 0.9 for the base and 0.78 for the subbase). The soil CBR is 2, the reliability used was 95%, the overall standard deviation was 0.4, initial PSI was 4.5, and the TSI was 2.5. Determine the required slab thickness for a rigid pavement designed to last the same number of years as the flexible pavement, but with only three lanes (instead of four) in the design direction (again, conservatively designing for the load distribution among the lanes). The design is to use the same truck traffic, reliability, soil, initial PSI, TSI, and overall standard deviations as the flexible pavement. In addition, the rigid pavement is to have a modulus of rupture of 800 lb/in2, a concrete modulus of elasticity of 5.5 × 106 lb/in2, a load transfer coefficient of 3.0, and a drainage coefficient of 0.9.

SOLUTION

We will first need to determine the design life of the existing flexible pavement and then use this value to determine the required slab thickness of the rigid pavement.

The flexible pavement’s SN is determined from Eq. 4.3, using Table 4.5 to find a1 = 0.44, a2 = 0.18 and a3 = 0.11, and with D1 = 8, D2 = 10, D3 = 10, M2 = 0.9 and M3 = 0.78 (all given) as

SN = a1D1 + a2D2M2 + a3D3M3

SN = 0.44(8) + 0.18(10)(0.9) + 0.11(10)(0.9) = 5.998 ≈ 6 With a CBR = 2 (given), Eq. 4.2 is applied to give MR = 3000 (1500 × 2). For other elements required to solve Eq. 4.1 (or alternatively Fig. 4.5), ΔPSI = 2 (initial PSI of 4.5 minus the TSI of 2.5), So = 0.4 (given), and ZR = −1.645 (which corresponds to R = 95%, as shown in Table 4.4). With these values, applying Eq. 4.1 (or using Fig. 4.5) gives W18 = 5,703,439.

For the daily axle loads, the equivalency factors (reading axle equivalents from Tables 4.1 and 4.2 while using SN = 6) are

10-kip single-axle equivalent = 0.08 (Table 4.1)

30-kip tandem-axle equivalent = 0.633 (Table 4.2)

4.7 Design-Lane Loads 135

Thus the total daily 18-kip ESAL is

Daily W18 = 0.08(967) + 0.633(1935) = 1302.215 18-kip ESAL/day

From Table 4.10, the PDL for a conservative design on a four-lane highway is 0.75, so applying Eq. 4.4 gives

18 18 18design-lane = × directional = 976.66 / dayW PDL W W or 356,481 W18/year (976.66 × 365). So the design life of the flexible pavement is:

18

18

design design life (in years)

daily

5,703,439 356,481

16 years

W

W

=

=

For the rigid pavement, we have that all of the design parameters are the same and are given S′c = 800 lb/in2, Ec = 5.5 × 106 lb/in2, J = 3.0, Cd = 0.9. In addition, the modulus of subgrade reaction, k, is 100 lb/in3 from Table 4.9 (with CBR = 2). Note that the accumulated W18 for the rigid pavement will not be the same as the daily axle loads for the flexible pavement because the load equivalency factors will be different. To determine the accumulated W18 for the rigid pavement, we assume a slab thickness of 10 inches (D = 10 inches) and read axle equivalents from Tables 4.6 and 4.7 as

10-kip single-axle equivalent = 0.08 (Table 4.6)

30-kip tandem-axle equivalent = 1.14 (Table 4.7) Thus the total daily 18-kip ESAL is

Daily W18 = 0.08(967) + 1.14(1935) = 2283.26 18-kip ESAL/day

From Table 4.10, the PDL for a conservative design on a three-lane highway is 0.8, so applying Eq. 4.4 gives a required slab thickness of 10.28 inches (Figs. 4.7 and 4.8 can also be used to arrive at an approximate solution for slab thickness). For design, this can be rounded up to 10.5 inches. It can be shown by the reader that going back to the load equivalency factors and assuming 10.5 inches instead of the previously assumed 10 inches will result in the same, correct slab-thickness solution of 10.5 inches.

136 Chapter 4 Pavement Design

EXAMPLE 4.6 FLEXIBLE AND RIGID PAVEMENT DESIGN-LIFE COMPARISON

A roadway is determined to have 400 18-kip single axles, 200 24-kip tandem axles and 100 40-kip triple axles per day. The subgrade CBR is 2 and the roadway pavement is designed for an overall standard deviation of 0.4, a reliability of 99% and the initial PSI is 4.5 and the TSI is 2.5. One newly constructed section of this roadway is a rigid pavement designed with a 9-inch slab with a modulus of rupture of 700 lb/in2, a modulus of elasticity of 4.0 × 106 lb/in2, and a joint transfer coefficient of 3.0. Another newly constructed section of the same roadway is a flexible pavement with a 5-in HMA surface, 10-in dense-graded crushed stone base and a 9-in crushed-stone subbase. If the roadway has four lanes in each direction and is conservatively designed, which of the pavement sections will last longer and by how many years (all drainage coefficients are 1.0)?

SOLUTION

We first determine the total design W18 for the flexible pavement. To do this, the flexible pavement’s SN is computed from Eq. 4.3, using Table 4.5 to find a1 = 0.44, a2 = 0.18 and a3 = 0.11, and with D1 = 5, D2 = 10, D3 = 9, M2 = 1.0 and M3 = 1.0 (all given) as

SN = a1D1 + a2D2M2 + a3D3M3

SN = 0.44(5) + 0.18(10)(1.0) + 0.11(9)(1.0) = 4.99 ≈ 5 With a CBR = 2 (given), Eq. 4.2 is applied to give MR = 3000 (1500 × 2). For other elements required to solve Eq. 4.1 (or alternatively Fig. 4.5), So = 0.4 (given), ZR = −2.326 (which corresponds to R = 99%, as shown in Table 4.4), and ΔPSI = 2 (initial PSI of 4.5 minus the TSI of 2.5). With these values, applying Eq. 4.1 (or using Fig. 4.5) gives W18 = 775,133. To determine the total truck traffic, the equivalency factors (reading axle equivalents from Tables 4.1, 4.2 and 4.3 while using SN = 5) are

18-kip single-axle equivalent = 1.0 (Table 4.1)

24-kip tandem-axle equivalent = 0.260 (Table 4.2)

40-kip triple-axle equivalent = 0.487 (Table 4.3) which gives a total W18 of 500.7 18-kip ESAL/day (1.0 × 400 + 0.260 × 200 + 0.487 × 100). From Table 4.10, the PDL for a conservative design on a four-lane highway is 0.75, so applying Eq. 4.4 gives the design-lane W18 = 375.53 18-kip ESAL/day (0.75 × 500.7). So the design life for the flexible pavement is

18

18

total design-lane design life (in years) =

daily design-lane ×365 days / yr

775,133 =

375.53×365 = 5.655 years

W

W

For the rigid pavement, we are given a slab thickness of 9 inches (D = 9 inches) S′c = 900 lb/in2, Ec = 4.0 × 106 lb/in2, J = 3.0, Cd = 1.0 and the modulus of subgrade reaction,

4.8 Measuring Pavement Quality and Performance 137

k, is 100 lb/in3 from Table 4.9 (with CBR = 2), and all other design parameters are the same as those for the flexible pavement. With these values, applying Eq. 4.4 (or using Figs. 4.7 and 4.8) gives W18 = 2,364,522. To determine the total truck traffic, the equivalency factors (reading axle equivalents from Tables 4.7, 4.8, and 4.9 while using D = 9) are

18-kip single-axle equivalent = 1.0 (Table 4.6)

24-kip tandem-axle equivalent = 0.444 (Table 4.7)

40-kip triple-axle equivalent = 1.17 (Table 4.8) which gives a total W18 of 605.8 18-kip ESAL/day (1.0 × 400 + 0.444 × 200 + 1.17 × 100). As in the flexible-pavement case, from Table 4.10, the PDL for a conservative design on a four-lane highway is 0.75, so applying Eq. 4.4 gives the design-lane W18 = 454.35 18-kip ESAL/day (0.75 × 605.8). So the design life for the rigid pavement is

18

18

total design-lane design life (in years) =

daily design-lane ×365 days / yr

2,364,522 =

454.35×365

=14.258 years

W

W

So the rigid pavement will last 8.603 years longer (14.258 – 5.655).

4.8 MEASURING PAVEMENT QUALITY AND PERFORMANCE

The design procedure for pavements originally focused on the PSI as a measure of pavement quality. However, the PSI is based on the opinions of a panel of experts (as discussed in Section 4.4.1), which can introduce some variability into their determination. As a result, efforts have been undertaken to develop quantitative measures of pavement condition that provide additional insights into pavement quality and performance and that correlate with the traditional pavement serviceability index. Some factors that are regularly measured by highway pavement agencies now include the International Roughness Index, friction measurements, and rut depth.

4.8.1 International Roughness Index The International Roughness Index (IRI) has become the most popular measure for evaluating the condition of pavements. The IRI evolved out of a study commissioned by the World Bank [Sayers et al., 1986] to establish uniformity of the physical measurement of pavement roughness. The IRI is determined by measuring vertical movements in a standardized vehicle’s suspension per unit length of roadway. Units of IRI are reported in inches per mile (in/mi). The higher the value of the IRI, the rougher the road. To get some sense for how the IRI relates to pavement condition assessments and PSI, Tables 4.11 and 4.12 provide IRI and PSI values corresponding to what is considered poor, mediocre,

138 Chapter 4 Pavement Design

fair, good, and very good for Interstate and non-Interstate highways [Federal Highway Administration, 2006]. Note that, due to the higher design standards and performance expectations, interstate highways are held to a higher standard for fair, mediocre, and poor pavement assessments.

4.8.2 Friction Measurements Another important measurement of pavement performance is the surface friction. This is critical because low friction values can increase stopping distances and the probability of accidents. Given the variability of pavement surfaces, weather conditions, and tire characteristics, determining pavement friction over the range of possible values is not an easy task. To estimate friction, a standardized test is conducted under wet conditions using either a treaded or smooth tire. Although other speeds are sometimes used, the standard test is generally conducted at 40 mi/h using a friction-testing trailer in which the wheel is locked on the wetted road surface, and the torque developed from this wheel locking is used to measure a friction number. The friction number resulting from this test gives an approximation of the coefficient of road adhesion under wet conditions (as shown in Table 2.4) and is multiplied by 100 to produce a value between 0 and 100. The friction number with a treaded tire (FNt) attempts to measure pavement microtexture, which is a function of the aggregate quality and composition. The friction number with a smooth tire (FNs) provides a measure of pavement macrotexture, which is critical in providing a water drainage escape path between the pavement and tire.

A number of factors influence the friction number, such as changes in traffic volumes or traffic composition, surface age (friction has been found to increase quickly after construction, then as time passes, to level off and eventually decline), seasonal changes (in northern states, the friction number tends to be highest in the spring and lowest in the fall), and speed (the measured value tends to decrease as the test speed increases). Table 4.11 Relationship between the International Roughness Index (IRI) and Perceptions of Pavement Quality for Interstate Highways

Pavement condition

Present Serviceability

Index

Measured International Roughness Index

in/mi

Very good ≥ 4.0 < 60 Good 3.5 – 3.9 61–94 Fair 3.1 – 3.4 95–119

Mediocre 2.6 – 3.0 120–170 Poor ≤ 2.5 > 170

4.8 Measuring Pavement Quality and Performance 139

Table 4.12 Relationship between the International Roughness Index (IRI) and Perceptions of Pavement Quality for Non-Interstate Highways

Pavement condition

Present Serviceability

Index

Measured International Roughness Index

in/mi

Very good ≥ 4.0 < 60 Good 3.5 – 3.9 61–94 Fair 3.1 – 3.4 95–170

Mediocre 2.6 – 3.0 171–220 Poor ≤ 2.5 > 220

Also, the friction number measured with the treaded tire tends to be greater than that measured with the smooth tire (usually by a value of about 20), but the difference decreases as the surface texture becomes rougher [Li et al., 2003].

In terms of safety, the amount of friction needed to minimize safety-related problems depends on prevailing traffic and geometric conditions. Guidelines used by some states suggest that values of FNt < 30 or FNs < 15 indicate that poor friction may be contributing to wet-weather accidents. Other state agencies have simply put in place guidelines for minimum friction requirements. For example, in Indiana, the minimum friction value is based on the smooth tire test at 40 mi/h, and a pavement with FNs < 20 is considered in need of surfacing work to improve friction (generally resurfacing).

4.8.3 Rut Depth

Rut depth, which is a measure of pavement surface deformation in the wheel paths, can affect roadway safety because the ruts accumulate water and increase the possibility of vehicle hydroplaning (which results in the tire skimming over a film of water, greatly reducing braking and steering effectiveness). Because of its potential impact on vehicle control, rut depths are regularly measured on many highways to determine if pavement rutting has reached critical values that would require resurfacing or other pavement treatments. Virtually all states measure rut depth using automated equipment that seeks to determine the difference in surface elevation of the pavement in the wheel path relative to the pavement that is not in the wheel path. The critical values of rut depth can vary from one highway agency to the next. Usually, rut depths are considered unacceptably high when their values reach between 0.5 and 1.0 inches, indicating that corrective action is warranted.

4.8.4 Cracking

For flexible pavements, four types of cracking are usually monitored: longitudinal-fatigue cracking, transverse cracking, alligator cracking, and reflection cracking. Longitudinal-fatigue cracking is a surface-down cracking that occurs due to material fatigue in the wheel path. Such cracking can accelerate over time and require significant repairs to protect against water penetration into the flexible pavement structure. Transverse cracking is generally the result of low temperatures that cause fractures across the traffic lanes (resulting in an increase in pavement roughness). Alligator-fatigue cracking is a consequence of material fatigue in the wheel path, generally starting from the

140 Chapter 4 Pavement Design

bottom of the asphalt layer. Such material fatigue creates a patch of connected cracks that resembles the skin of an alligator (as with other types of cracks, these can accelerate quickly over time and generate the need for maintenance to protect the integrity of the pavement structure). Finally, reflection cracking occurs when HMA overlays are placed over exiting pavement structures that had alligator-fatigue cracking, or other indications of pavement distress, and these old distresses manifest themselves in new distresses in the overlay. This results in surface cracking that increases surface roughness and the need for maintenance to protect water intrusion into the pavement structure.

For rigid pavements, transverse cracking is a common measure of pavement distress. Such cracking can be the result of slab fatigue and can be initiated either at the surface or base of the slab. The spacing and width of transverse cracks, and the potential impact of severe cracking on the structural integrity of the pavement, are critical measures of rigid-pavement distress.

4.8.5 Faulting For traditional JPCP (Jointed Plain Concrete Pavements) rigid pavements, joint faulting (characterized by different slab elevations) is a critical measure of pavement distress. Faulting is an indicator of erosion or fatigue of the layers beneath the slab and reflects a failure of the load-transfer ability of the pavement between adjacent slabs. Faulting is associated with increased roughness and will be reflected in International Roughness Index measurements.

4.8.6 Punchouts For Continuously Reinforced Concrete Pavements (CRCP) rigid pavements (those built without expansion/contraction joints), fatigue damage at the top of the slab is often measured by punchouts, which occur when the close spacing of transverse cracks cause in high tensile stresses that result in portions of the slab being broken into pieces. Punchouts are associated with increased roughness and are reflected in International Roughness Index measurements.

4.9 MECHANISTIC-EMPIRICAL PAVEMENT DESIGN

The Mechanistic-Empirical Pavement Design Guide (AASHTO 2008) is one of the more recent tools for the design and rehabilitation of pavement structures. The Mechanistic-Empirical Pavement Design Guide was developed to improve on the traditional pavement design procedures presented earlier in this chapter (AASHTO Guide for Design of Pavement Structures, 1993) by providing the ability to predict multiple pavement-performance measures (such as rut depth, various types of cracking, joint faulting, International Roughness Index, etc.) and providing a direct link among pavement elements (materials, structural design, construction, traffic, climate and pavement management practices).

Unlike the traditional pavement-design procedures presented earlier in this chapter, the Mechanistic-Empirical Pavement Design Guide is quite complex and must be done using a software package (the software package is referred to simply as MEPDG, standing for Mechanistic-Empirical Pavement Design Guide). The

4.9 Mechanistic-Empirical Pavement Design 141

design of pavements with MEPDG is an iterative process that can be summarized as follows:

1. The design engineer first selects a pavement structure (layer thicknesses, etc.), often using the traditional AASHTO approach (AASHTO Guide for Design of Pavement Structures, 1993).

2. Various inputs needed for MEPDG pavement assessment are then gathered and classified in the following broad topic groupings (please note that this is a much more time-intensive effort than the traditional AASHTO pavement design approach demonstrated earlier in this chapter):

a. General project information. For this, factors such as base/subbase construction month, pavement construction month, and month that the pavement is open to traffic are needed because these factors can affect pavement-performance criteria.

b. Design criteria and reliability level. For design criteria, the level of tolerable distress such as cracking, faulting, International Roughness Index are needed (these criteria roughly replace the TSI in the traditional AASHTO pavement design). The reliability level needed is similar to that currently used in the traditional AASHTO process.

c. Site conditions and factors. Here, information is needed on truck traffic (including axle-load distributions, speed limit to account for the effect of truck speed on pavement distress, and monthly and hourly distributions of truck-travel), climate (including hourly temperature, precipitation, wind speed, relative humidity, and cloud cover), and detailed soil information (strength, variability, etc.).

d. Material properties. Detailed information on new-pavement material properties is needed. This information is along the lines of the structural coefficient values and concrete-strength measurements used in the traditional AASHTO pavement design (although at a significantly higher level of detail).

3. With the above, the MEPDG software can then be run and software outputs will include calculated changes in pavement layer properties, various distresses (such as rut depth, cracking, and faulting), and the International Roughness Index over the design life of the pavement. The designer can then determine if the criteria for a successful pavement design have been met (critical distresses do not cross values that can be considered a failure of the pavement over its design life). If these criteria are not met, the pavement design is altered and the process is continued until an acceptable pavement design is achieved.

Currently, the use of the mechanistic-empirical pavement design process and the MEPDG software is increasing; however, many highway and transportation agencies still use the traditional AASHTO pavement-design approach (AASHTO Guide for Design of Pavement Structures, 1993).

142 Chapter 4 Pavement Design

4.10 PRACTICE PROBLEMS

PRACTICE PROBLEM 4.1

FLEXIBLE-PAVEMENT DESIGN LIFE AND NUMBER OF LANES

The initial design of northbound direction on a highway consists of a two-lane flexible pavement that has a SN of 5 and is designed for a subgrade CBR of 6, an overall standard deviation of 0.35, and a reliability of 90%. The initial PSI is 4.6 and the final PSI is 2.5. The pavement is designed for 3000 passes per day of a truck that has a 12,000 lb single axle and two 32,000 lb tandem axles. If the pavement is conservatively designed, how many years longer will the pavement last if a third lane is added.

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

This problem shows the effect that the number of lanes can have on the design life of a pavement. Although the number of axle loads may not change when lanes are added or subtracted, the proportion of the axle loads in the design lane will change because axle loads will be distributed over more lanes and fewer will be in the design lanes (as shown in Table 4.10).

To solve the problem, we must first determine the design-lane W18 used in the initial design of the pavement by applying Eq. 4.1 (or by using Fig. 4.5). To apply Eq. 4.1, the soil resilient modulus is determined from Eq. 4.2 as,

MR = 1500 × CBR = 1500 × = , the ΔPSI is determined as,

ΔPSI = initial PSI − final PSI (TSI) = − = , and ZR is determined from Table 4.4 to be with R = 90% (given in the problem). With these values, and the other values given in the problem, solving Eq. 4.1 for W18 (which is the design-lane value used in the existing pavement design) gives,

( ) [ ]

( )

( )

( )

o 10

5.19

10

5.19

= 9.36 log SN 1

PSI / 2.7

0.40 1094 SN 1

2.32 8.07

= 9.36 log 0.20

/ 2.7

0.40 1094 1

2.32

R

R

W Z S

M

 + + −  Δ

+  + +  

+ −

 × + + −  

   +

 + +  

+

10

8.07

log = , so = 33,142,926W W

−

4.10 Practice Problems 143

Initially, the pavement is designed for two lanes in the design direction and, since the problem states that it is conservatively designed, this means that the PDL2-lanes used in Eq. 4.5 is 1.00 from Table 4.10. With the third lane added in the design direction, the PDL3-lanes will become (as shown in Table 4.10 for a conservative design).

To determine the 18-kip–equivalent single-axle loads (EASL) per day, which will be needed to estimate how much longer the pavement will last with the additional lane, the equivalency factors (reading axle equivalents from Tables 4.1 and 4.2 while using SN = 5) are,

12-kip single-axle equivalent = (Table 4.1)

32-kip tandem-axle equivalent = (Table 4.2).

Each truck has one 12-kip single axle and two 32-kip tandem axles so the equivalent axle loads per truck is,

+ 2 × = 1.903 18-kip ESAL/truck. So the total daily 18-kip ESAL is

daily W18 = × trucks/day = 18-kip ESAL/day.

With this, the number of years the northbound two-lane road would last is,

18 2-lanes

2-lanes

as designed years = =

× 365 days / year × 18-kip EASL / day ×365 ×

= years.

W

PDL

Similarly, the number of years the northbound three-lane road would last is,

18 3-lanes

3-lanes

as designed years = =

× 365 days / year × 18-kip EASL / day ×365 ×

= years.

W

PDL

So the pavement will last,

− −3-lanes 2-lanesyears years = = 3.98 years longer

144 Chapter 4 Pavement Design

PRACTICE PROBLEM 4.2

RIGID PAVEMENT DESIGN LIFE AND RELIABILITY

For the conditions in Practice Problem 4.1 above, suppose a rigid pavement was designed with a 12-inch slab (modulus of rupture of 800 lb/in2, a modulus of elasticity of 6 million lb/in2, a joint transfer coefficient of 3.2, and a drainage coefficient of 1.0) for the two-lane northbound direction described in Practice Problem 4.1 (all other traffic and PSI information is the same as in that problem). How confident would you be that the pavement would have a PSI above 2.5 after 30 years?

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

The intent of this problem is to show how the concept of reliability is used in pavement design, and also to underscore the fact that load equivalency factors are not the same for flexible- and rigid-pavement designs even when the axle loads are the same.

To solve this problem, we will need to apply Eq. 4.4 and solve for the reliability z-statistic ZR, and then use Table 4.4 to get the percent probability R which will tell us how confident we would be that the pavement would have a PSI above 2.5 after 30 years. To apply Eq. 4.4, many of the terms are given as part of the problem statement, and others will be the same as those in Practice Problem 4.1 (ΔPSI = 2.1, CBR = 6, etc.). The terms that need to be determined are W18 (which will differ since the pavement is now rigid instead of flexible), and the modulus of subgrade reaction k which is a term not used in flexible-pavement design.

To determine W18, we start by computing the 18-kip–equivalent single-axle loads (EASL) per day (noting that these are different than the ones previously determined for the flexible pavement in Practice Problem 4.1) from the equivalency factors (reading axle equivalents from Tables 4.6 and 4.7 while using D = 12 inches) are

12-kip single-axle equivalent = (Table 4.6)

32-kip tandem-axle equivalent = (Table 4.7).

As stated in Practice Problem 4.1, each truck has one 12-kip single axle and two 32-kip tandem axles so the equivalent axle loads per truck is,

+ 2 × = 3.194 18-kip ESAL/truck. Note that this value is substantially larger than the 1.903 18-kip ESAL/truck found for the flexible-pavement design in Practice Problem 4.1. With this, after 30 years, the design-lane W18 is,

design-lane W18 = PDL2-lanes × design life in years × days/year × trucks/day × ESAL/truck design-lane W18 = × × 365 × × design-lane W18 = 104,922,900.

The modulus of subgrade reaction k is determined to be lb/in3 by interpolation between CBR values of 2 and 10 in Table 4.9. With all values needed for the application of Eq. 4.4 now known, Eq. 4.4 can be rearranged to solve for the reliability z-statistic ZR,

4.10 Practice Problems 145

( )

[ ] ( )

( ) ( ){ }

10 18 10

10

8.467 1 o

0.75

10 0.250.75

log 7.35 log + 1 0.06

log PSI / 3.0 1+ 1.624×10 + 1

1.132 4.22 0.32TSI log

215.63 18.42

R

c d

c

W D

DZ S

S C D

D E k

−

  − −      Δ −  = ×     

   ′ −   − −       × −     

( )

( )

( ) ( )

10 10

10

8.467

0.75

10 0.250.75

log 7.35 log 1 0.06

log / 3.0 1 1.624 10 1

1.132 4.22 0.32 log

215.63 18.42

 − + − 

   −

 + × +=      × −  − − ×

  × −    

1

−

               ×                   

Solving this equation gives ZR = 0.613. With this value, going to Table 4.4 we see the probability R is 73%. Thus we are 73% confident that the pavement will have a PSI above 2.5 after 30 years.

PRACTICE PROBLEM 4.3

RIGID PAVEMENT DESIGN LIFE AND INITIAL SERVICEABILITY

A westbound direction of a rigid pavement consists of three lanes of traffic and it is conservatively designed. The pavement is 10 inches thick, has a joint transfer coefficient of 2.8, reliability of 90%, overall standard deviation of 0.4, subgrade CBR of 55, modulus of elasticity of 5,000,000 lb/in2, modulus of rupture of 850 lb/in2, and a drainage coefficient of 0.9. The westbound direction is designed for daily traffic of 200,000 2-kip single axles, 10,000 12-kip single axles, and 20,000 28-kip tandem axles. If the TSI is 2.5, what is the minimum initial serviceability index for the pavement to have a 20-year design life?

SOLUTION

This problem illustrates how the initial serviceability index (PSI) affects the design life of the pavement.

To begin, the design-lane W18 needs to be determined. This is done by first converting existing axle loads into an 18-kip–equivalent single-axle load (ESAL). Reading axle equivalents from Tables 4.6 and 4.7 while using D = 10 (inches) gives,

2-kip single-axle equivalent = (Table 4.6)

12-kip single-axle equivalent = (Table 4.6)

24-kip tandem-axle equivalent = (Table 4.7).

146 Chapter 4 Pavement Design

So,

( ) ( ) ( ) ( ) ( ) ( )

18

2-kipsingle-axleequivalent 2-kipsingle-axle loadsper day

directional 12-kipsingle-axleequivalent 12-kipsingle-axle loadsper day

24-kiptandem-axleequivalent 24-kiptandam-axle loadsper day

W

 ×  

= + ×   + ×   20 years 365 days/year

200,000

= 20 years 365 days/year = 38,872,500

× ×

 ×   + × × ×    + ×  

With PDL = for a conservatively designed highway with three directional lanes (see Table 4.10), Eq. 4.5 is applied for the design-lane W18 as,

design-lane W18 = PDL3-lanes × directional W18 design-lane W18 = × = 31,098,000.

With a CBR of 55 (given), Table 4.9 shows that the modulus of subgrade reaction is, by interpolation, lb/in3. And, with a given reliability of 90% (R = 90%), Table 4.4 shows ZR = 1.282. With these values (and other values provided in the problem description), solving Eq. 4.4 for PSI (using an equation solver on a calculator or computer, or using Figs. 4.7 and 4.8 to arrive at an approximate solution for PSI) we find that PSI = . With this, the minimum initial serviceability index for the pavement (initial PSI) can be determined from,

PSI = initial PSI − TSI

so, initial PSI = TSI + PSI = + = 4.334.

PRACTICE PROBLEM 4.4

FLEXIBLE PAVEMENT AND SOLVING FOR UNKOWN DESIGN LIFE

A northbound direction on an interstate highway consists of four lanes of flexible pavement. It is conservatively designed with 10-inch sand-mix asphaltic surface, 10-crushed stone base, and a 10-inch crushed stone subbase (all drainage coefficients are 1.0). The subgrade CBR is 3, the overall standard deviation is 0.4, and the reliability is 95%. The initial PSI is 4.6 and the final PSI is 2.5. Daily total northbound road traffic consists of 51,000 cars (each with two 2-kip single axles), 840 buses (each with two 20-kip single axles), and 1000 heavy trucks (each with one 12-kip single axle and two 34-kip tandem axles). How many years was this pavement designed to last?

SOLUTION

This problem improves the reader’s understanding of the relationship between various pavement characteristics and design life by formulating the problem as one where the analysis solves for design life, instead the more conventional approach where the

4.10 Practice Problems 147

analyst designs the pavement with a given design life. By looking at the pavement problem in this way, the reader can gain a better understanding of the pavement-design process.

The number of years the pavement will last is equal to the W18 used to design the pavement (the W18 used in Eq. 4.1, which is the total W18 in the design lane until pavement reaches the TSI) divided by the W18 accumulated annually (which is determined from annual axle loads). To determine the W18 from Eq. 4.1, we apply Eq. 4.2 to get,

MR = 1500 × CBR = 1500 × = From Table 4.4 with R = 95%, ZR = , and

ΔPSI = initial PSI − TSI = − = Next, the SN of the existing pavement is computed. This is done by first determining the structural coefficients of the flexible-pavement layers, which are:

sand-mix asphalt surface a1 = (Table 4.5)

crushed stone base a2 = (Table 4.5)

crushed stone subbase a3 = (Table 4.5).

With these structural coefficients, and the layer depths and drainage coefficients given in the problem, Eq. 4.3 can be applied to give,

SN = a1D1 + a2D2M2 + a3D3M3

SN = × + × × 1 + × 10 × = 6.0 With these values, Eq. 4.1 is solved as,

( ) [ ]

( )

( )

( )

 + −  Δ

+  + +  

+ −

  −  

   

 +  

+ −

o 10

5.19

10

10

5.19

= 9.36 log SN + 1

PSI / 2.7

0.40 1094 SN 1

2.32 8.07

= × + 9.36 log + 1 0.20

log / 2.7 +

0.40 + 1094 1

2.32

R

R

W Z S

M

8.07

= , so = 16,303,758W W

148 Chapter 4 Pavement Design

Now, to determine the accumulated W18 per day in the design lane, the directional daily 18-kip–equivalent single-axle loads (ESALs) need to be computed. With SN = 6, and the four axle type/weight combinations given in the problem, we have,

2-kip single-axle equivalent = (Table 4.1)

12-kip single-axle equivalent = (Table 4.1).

20-kip single-axle equivalent = (Table 4.1)

34-kip tandem-axle equivalent = (Table 4.2).

With 51,000 cars per day having two 2-kip single axles each, 840 buses per day having two 20-kip single axles each, and 1,000 heavy trucks per day having one 12-kip single axle and two 34-kip tandem axles we have,

( ) ( ) ( ) ( ) ( ) ( )18

2-kip single-axle equivalent 2-kip single-axle loads per day

12-kip single-axle equivalent 12-kip single-axle loads per day daily directional

20-kip single-axle equivalent 20-kip single-axle loads per day

34-kip tande

W

×

+ × =

+ ×

+( ) ( )m-axle equivalent 34-kip tandam-axle loads per day

= 102,000 1,000 1,680

           ×  

× + × + × + ×

= 4960.4 18-kip EASL per day. To calculate the daily design-lane W18, Eq. 4.5 is written as,

daily design-lane W18 = PDL4-lanes × daily directional W18. From Table 4.10, with a conservative design, PDL4-lanes = , so

daily design-lane W18 = × = 3720.3 18-kip EASL per day or 1,357,909.5 18-kip EASL per year (3720.3 × 365).

With these values, the number of years the pavement will last is given by,

18

18

as designed years = = = years.

per year

W W

References 149

NOMENCLATURE FOR CHAPTER 4

a1, a2, a3 structural-layer coefficients for wearing surface, base, and subbase

Cd drainage coefficient for rigid-pavement design

CBR California bearing ratio D slab thickness, AASHTO design

equation D1, D2, D3 structural-layer thicknesses for wearing

surface, base, and subbase Ec concrete modulus of elasticity FNs friction number with a smooth tire FNt friction number with a treaded tire IRI International Roughness Index k modulus of subgrade reaction M2, M3 drainage coefficients for base and

subbase

MR soil resilient modulus PDL proportion of directional W18 assumed

in the design lane PSI present serviceability index R reliability

′ concrete modulus of rupture

So overall standard deviation for AASHTO design equations structural number

SN structural number TSI terminal serviceability index W18 18-kip (80.1-kN)–equivalent single-axle

load ZR reliability for AASHTO design

equations ΔPSI change in the present serviceability

index

REFERENCES

AASHTO (American Association of State Highway and Transportation Officials). AASHTO Guide for Design of Pavement Structures. Washington, DC: AASHTO, 1993.

AASHTO (American Association of State Highway and Transportation Officials). Mechanistic- Empirical Pavement Design Guide—A Manual of Practice. Washington, DC: AASHTO, 2008.

Carey, W., and P. Irick. The Pavement Serviceability-Performance Concept. Highway Research Board Special Report 61E, AASHO Road Test, 1962.

Federal Highway Administration. “2006 Status of the Nation’s Highways, Bridges, and Transit: Conditions and Performance; Report to

Congress”, United States Department of Transportation, Washington, DC, 2007.

Li, S., S. Noureldin and K. Zhu. Upgrading the INDOT Pavement Friction Testing Program. Final report FHWA/IN/JTRP-2003/23, West Lafayette, Indiana, 2003.

Sayers, M., T. Gillespie, and W. Paterson. Guidelines for the Conducting and Calibrating Road Roughness Measurements. World Bank Technical Paper No. 46. The World Bank, Washington, DC, 1986.

Yoder, E. J., and M. W. Witczak. Principles of Pavement Design, 2nd ed. New York: Wiley, 1991.

P-15

CHAPTER 5 PROBLEMS

Traffic Stream Parameters and Basic Traffic Stream Models (Sections 5.2–5.3)

5.1 Assume that you are observing traffic in a single lane of a highway at a specific location. You measure the average headway and average spacing of passing vehicles as 3.2 seconds and 165 ft, respectively. Calculate the flow, average speed, and density of the traffic stream in this lane.

5.2 Assume that you are an observer standing at a point along a three-lane roadway. All vehicles in lane 1 are traveling at 30 mi/h, all vehicles in lane 2 are traveling at 45 mi/h, and all vehicles in lane 3 are traveling at 60 mi/h. There is also a constant spacing of 0.5 mile between vehicles. If you collect spot speed data for all vehicles as they cross your observation point, for 30 minutes, what will be the time-mean speed and space-mean speed for this traffic stream?

5.3 Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h, 185 mi/h, and 180 mi/h, respectively. Assume that you are an observer standing at a point on the track for a period of 30 minutes and are recording the instantaneous speed of each vehicle as it crosses your point. What is the time-mean speed and space-mean speed for these vehicles for this time period? (Note: Be careful with rounding.)

5.4 For Problem 5.3, calculate the space-mean speed assuming that you were given only an aerial photo of the circling race cars and the constant travel speed of each of the vehicles.

5.5 On a specific westbound section of highway, studies show that the speed–density relationship is

   = −      

3.5

1f f

k u u

k

It is known that the capacity is 4200 veh/h and the jam density is 210 veh/mi. What is the space-mean speed of the traffic at capacity, and what is the free- flow speed?

5.6 A section of highway has the following flow– density relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the capacity of the highway section, the speed at capacity, and the density when the highway is at one-quarter of its capacity?

Models of Traffic Flow (Section 5.4)

5.7 An observer has determined that the time headways between successive vehicles on a section of highway are exponentially distributed and that 65% of the headways between vehicles are 9 seconds or greater. If the observer decides to count traffic in 30- second time intervals, estimate the probability of the observer counting exactly four vehicles in an interval.

5.8 At a specified point on a highway, vehicles are known to arrive according to a Poisson process. Vehicles are counted in 20-second intervals, and vehicle counts are taken in 120 of these time intervals. It is noted that no cars arrive in 18 of these 120 intervals. Approximate the number of these 120 intervals in which exactly three cars arrive.

5.9 Consider the conditions in Practice Problem 5.2. How short would the driver reaction times of oncoming vehicles have to be for the probability of an accident to equal 0.20?

Queuing Theory and Traffic Flow Analysis (Section 5.5)

5.10 Vehicles arrive at a single toll booth beginning at 8:00 A.M. They arrive and depart according to a uniform deterministic distribution. However, the toll booth does not open until 8:10 A.M. The average arrival rate is 8 veh/min, and the average departure rate is 10 veh/min. Assuming D/D/1 queuing, when does the initial queue clear and what are the total delay, the average delay per vehicle, longest queue length (in vehicles), and the wait time of the 100th vehicle to arrive (assuming first-in-first-out)?

5.11 Vehicles begin to arrive at a park entrance at 7:45 A.M. at a constant rate of six per minute and at a constant rate of four vehicles per minute from 8:00 A.M. on. The park opens at 8:00 A.M. and the manager wants to set the departure rate so that the average delay per vehicle is no greater than 9 minutes (measured from the time of the first arrival until the total queue clears). Assuming D/D/1 queuing, what is the minimum departure rate needed to achieve this?

5.12 A toll booth on a turnpike is open from 8:00 A.M. to 12 midnight. Vehicles start arriving at 7:45 A.M. at a uniform deterministic rate of six per minute until 8:15 A.M. and from then on at two per minute. If vehicles are processed at a uniform deterministic rate of six per minute, determine when the queue will dissipate, the total delay, the maximum

P-16 Chapter 5 Problems

queue length (in vehicles), the longest vehicle delay under FIFO, and the longest vehicle delay under LIFO.

5.13 Vehicles begin to arrive at a parking lot at 6:00 A.M. at a rate of eight per minute. Due to an accident on the access highway, no vehicles arrive from 6:20 to 6:30 A.M. From 6:30 A.M. on, vehicles arrive at a rate of two per minute. The parking-lot attendant processes incoming vehicles (collects parking fees) at a rate of four per minute throughout the day. Assuming D/D/1 queuing, determine total vehicle delay.

5.14 Vehicles begin to arrive at a toll booth at eight vehicles per minute from 9 A.M. to 10 A.M. The booth opens at 9:10 A.M. and services at a rate of 10 vehicles per minute until 9:40 A.M. From 9:40 A.M. until 10 A.M. the service rate is six vehicles per minute. Assuming D/D/1 queuing, what is the total vehicle delay from 9 A.M. to 10 A.M. assuming D/D/1 queuing?

5.15 The arrival rate at a parking lot is 6 veh/min. Vehicles start arriving at 6:00 P.M., and when the queue reaches 36 vehicles, service begins. If company policy is that total vehicle delay should be equal to 500 veh-min, what is the departure rate? (Assume D/D/1 queuing and a constant service rate.)

5.16 At 8:00 A.M. there are 10 vehicles in a queue at a toll booth and vehicles are arriving at a rate of λ(t) = 6.9 − 0.2t. Beginning at 8 A.M., vehicles are being serviced at a rate of μ(t) = 2.1 + 0.3t [λ(t) and μ(t) are in vehicles per minute and t is in minutes after 8:00 A.M.]. Assuming D/D/1 queuing, what is the maximum queue length, and what would the total delay be from 8:00 A.M. until the queue clears?

5.17 At the end of a sporting event, vehicles begin leaving a parking lot at λ(t) = 12 − 0.25t and vehicles are processed at μ(t) = 2.5 + 0.5t [t is in minutes and λ(t) and μ(t) are in vehicles per minute]. Assuming D/D/1 queuing, determine the total vehicle delay, longest queue, and the wait time of the 50th vehicle to arrive assuming first-in-first-out and D/D/1 queuing.

5.18 Vehicles arrive at a single park-entrance booth where a brochure is distributed. At 8 A.M., there are 20 vehicles in the queue and vehicles continue to arrive at the deterministic rate of λ(t) = 4.2 − 0.1t, where λ(t) is in vehicles per minute and t is in minutes after 8:00 A.M. From 8 A.M. until 8:10 A.M., vehicles are served at a constant deterministic rate of three per minute. Starting at 8:10 A.M., another brochure-distributing person is added and the brochure-service rate increases to six per minute (still

at a single booth). Assuming D/D/1 queuing, determine the longest queue, the total delay from 8 A.M. until the queue dissipates; and the wait time of the 40th vehicle to arrive.

5.19 Vehicles arrive at a single toll booth beginning at 7:00 A.M. at a rate of 8 veh/min. Service also starts at 7:00 A.M. at a rate of μ(t) = 6 + 0.2t where μ(t) is in vehicles per minute and t is in minutes after 7:00 A.M. Assuming D/D/1 queuing, determine when the queue will clear, the total delay, and the maximum queue length in vehicles.

5.20 Vehicles begin arriving at a single toll-road booth at 8:00 A.M. at a time-dependent deterministic rate of λ(t) = 2 + 0.1t [with λ(t) in veh/min and t in minutes]. At 8:07 A.M., the toll booth opens and vehicles are serviced at a constant deterministic rate of 6 veh/min. Assuming D/D/1 queuing, what is the average delay per vehicle from 8:00 A.M. until the initial queue clears and what is the delay of the 20th vehicle to arrive?

5.21 Vehicles begin to arrive at a toll booth at 8:50 A.M. with an arrival rate of λ(t) = 4.1 + 0.01t [with t in minutes and λ(t) in vehicles per minute]. The toll booth opens at 9:00 A.M. and processes vehicles at a rate of 12 per minute throughout the day. Assuming D/D/1 queuing, when will the queue dissipate and what will be the total vehicle delay?

5.22 Vehicles begin to arrive at a toll booth at 7:50 A.M. with an arrival rate of λ(t) = 5.2 − 0.01t [with t in minutes after 7:50 A.M. and λ in vehicles per minute]. The toll booth opens at 8:00 A.M. and serves vehicles at a rate of μ(t) = 3.3 + 2.4t (with t in minutes after 8:00 A.M. and μ in vehicles per minute). Once the service rate reaches 10 veh/min, it stays at that level for the rest of the day. If queuing is D/D/1, when will the queue that formed at 7:50 A.M. be cleared?

5.23 Vehicles arrive at a freeway on-ramp meter at a constant rate of six per minute starting at 6:00 A.M. Service begins at 6:00 A.M. such that μ(t) = 2 + 0.5t, where μ(t) is in veh/min and t is in minutes after 6:00 A.M. What is the total delay and the maximum queue length (in vehicles)?

5.24 Vehicles arrive at a toll booth according to the function λ(t) = 5.2 − 0.20t, where λ(t) is in vehicles per minute and t is in minutes. The toll booth operator processes one vehicle every 20 seconds. Determine total delay, maximum queue length, and the time that the 20th vehicle to arrive waits from its arrival to its departure.

Chapter 5 Problems P-17

5.25 There are 10 vehicles in a queue when an attendant opens a toll booth. Vehicles arrive at the booth at a rate of four per minute. The attendant opens the booth and improves the service rate over time following the function μ(t) = 1.1 + 0.30t, where μ(t) is in vehicles per minute and t is in minutes. When will the queue clear, what is the total delay, and what is the maximum queue length?

5.26 Vehicles begin to arrive at a parking lot at 6:00 A.M. with an arrival rate function (in vehicles per minute) of λ(t) = 1.2 + 0.3t, where t is in minutes. At 6:10 A.M., the parking lot opens and processes vehicles at a rate of 12 per minute. What is the total delay and the maximum queue length?

5.27 At a parking lot, vehicles arrive according to a Poisson process and are processed (parking fee collected) at a uniform deterministic rate at a single station. The mean arrival rate is 4.2 veh/min and the processing rate is 5 veh/min. Determine the average length of queue, the average time spent in the system, and the average waiting time in the queue.

5.28 Consider the parking lot and conditions described in Problem 5.27. If the rate at which vehicles are processed became exponentially distributed (instead of deterministic) with a mean processing rate of 5 veh/min, what would be the average length of queue, the average time spent in the system, and the average waiting time in the queue?

5.29 Vehicles arrive at a toll booth with a mean arrival rate of 3 veh/min (the time between arrivals is exponentially distributed). The toll booth operator processes vehicles (collects tolls) at a uniform deterministic rate of one every 15 seconds. What is the average length of queue, the average time spent in the system, and the average waiting time in the queue?

5.30 A business owner decides to pass out free transistor radios (along with a promotional brochure) at a booth in a parking lot. The owner begins giving the radios away at 9:15 A.M. and continues until 10:00 A.M. Vehicles start arriving for the radios at 8:45 A.M. at a uniform deterministic rate of four per minute and continue to arrive at this rate until 9:15 A.M. From 9:15 to 10:00 A.M., the arrival rate becomes 8 per minute. The radios and brochures are distributed at a uniform deterministic rate of 11 cars per minute over the 45-minute time period. Determine total delay, maximum queue length, and longest vehicle delay assuming FIFO and LIFO.

5.31 Consider the conditions described in Problem 5.30. Suppose that the owner decides to accelerate

the radio-brochure distribution rate (in veh/min) so that the queue that forms will be cleared by 9:45 A.M. What would this new distribution rate be?

5.32 A ferryboat queuing lane holds 40 vehicles. If vehicles are processed (tolls collected) at a uniform deterministic rate of five vehicles per minute and processing begins when the lane reaches capacity, what is the uniform deterministic arrival rate if the vehicle queue is cleared 35 minutes after vehicles begin to arrive?

5.33 At a toll booth, vehicles arrive and are processed (tolls collected) at uniform deterministic rates λ and μ, respectively. The arrival rate is 3 veh/min. Processing begins 15 minutes after the arrival of the first vehicle, and the queue dissipates t minutes after the arrival of the first vehicle. Letting the number of vehicles that must actually wait in a queue be x, develop an expression for determining processing rates in terms of x.

5.34 Vehicles arrive at a recreational park booth at a uniform deterministic rate of 5 veh/min. If uniform deterministic processing of vehicles (collecting of fees) begins 20 minutes after the first arrival and the total delay is 3200 veh-min, how long after the arrival of the first vehicle will it take for the queue to be cleared?

5.35 Trucks begin to arrive at a truck weigh station (with a single scale) at 6:00 A.M. at a deterministic but time-varying rate of λ(t) = 4.3 − 0.22t [λ(t) is in veh/min and t is in minutes]. The departure rate is a constant 2 veh/min (time to weigh a truck is 30 seconds). When will the queue that forms be cleared, what will be the total delay, and what will be the maximum queue length?

5.36 Commercial trucks begin to arrive at a seaport entry plaza at 7:50 A.M., at the rate of λ(t) = 6.3 – 0.25t [λ(t) is in veh/min and t is in minutes]. The plaza opens at 8:00 A.M. For the first 10 minutes, one processing booth is open. After the first 10 minutes until the queue clears, two processing booths are open. Each booth processes trucks at a uniform rate of two per minute. What is the average delay per vehicle, the maximum queue length, and the average queue length?

5.37 Vehicles begin to arrive at a remote parking lot after the start of a major sporting event. They are arriving at a deterministic but time-varying rate of λ(t) = 3.3 − 0.1t [λ(t) is in veh/min and t is in minutes]. The parking-lot attendant processes vehicles (assigns spaces and collects fees) at a deterministic rate at a single station. A queue

P-18 Chapter 5 Problems

exceeding four vehicles will back up onto a congested street, and is to be avoided. How many vehicles per minute must the attendant process to ensure that the queue does not exceed four vehicles?

5.38 A truck weighing station has a single scale. The time between truck arrivals at the station is exponentially distributed with a mean arrival rate of 1.6 veh/min. The time it takes vehicles to be weighed is exponentially distributed with a mean rate of 2.1 veh/min. When more than five trucks are in the system, the queue backs up onto the highway and interferes with through traffic. What is the probability that the number of trucks in the system will exceed 5?

5.39 Consider the convenience store described in Example 5.14.The owner is concerned about customers not finding an available parking space when they arrive during the busiest hour. How many spaces must be provided for there to be less than a 1% chance of an arriving customer not finding an open parking space?

5.40 Vehicles arrive at a toll bridge at a rate of 420 veh/h (the time between arrivals is exponentially distributed). Two toll booths are open and each can process arrivals (collect tolls) at a mean rate of 12 seconds per vehicle (the processing time is also exponentially distributed). What is the total time spent in the system by all vehicles in a 1-hour period?

5.41 Vehicles leave an airport parking facility (arrive at parking fee collection booths) at a rate of 500 veh/h (the time between arrivals is exponentially distributed). The parking facility has a policy that the average time a patron spends in a queue waiting to pay for parking is not to exceed 5 seconds. If the time required to pay for parking is exponentially distributed with a mean of 15 seconds, what is the smallest number of payment processing booths that must be open to keep the average time spent in a queue below 5 seconds?

5.42 Vehicles begin to arrive at a parking lot at 7:45 A.M. at a constant rate of 4 veh/min and continue to arrive at that rate throughout the day. The parking lot opens at 8:00 A.M. and vehicles are processed at a constant rate of one vehicle every 10 seconds. Assuming D/D/1 queuing, what is the longest queue, the queue at 8:15 A.M., and the average delay per vehicle from 7:45 A.M. until the queue clears?

5.43 Vehicles arrive at a toll both starting at 7:00 A.M. at a rate of λ(t) = 5.1 – 0.05t [with λ(t) in veh/min and t in minutes after 7:00 A.M.]. The first operator processes cars at a rate of 3 veh/minute 7:00 A.M. until 7:15 A.M when the person leaves because of illness.

From 7:15 A.M to 7:25 A.M, no one is at the toll booth but a new operator arrives at 7:25 A.M and processes at a rate of μ(t) = 8 + 0.3t [with μ(t) in veh/min and t in minutes after 7:25 A.M.]. Assuming D/D/1 queuing, what is the maximum queue length (in vehicles) and the average delay per vehicle?

5.44 Vehicles arrive at a toll both starting at 5:55 A.M. at a rate of λ(t) = 4.0 – 0.05t [with λ(t) in veh/min and t in minutes after 5:55 A.M.]. The toll booth opens at 6:00 A.M. processes vehicles at a rate of μ(t) = 2 + 0.2t [with μ(t) in veh/min and t in minutes after 6:00 A.M.] until a rate of 4.5 veh/min is reached and the service rate remains at 4.5 veh/min until the queue clears. Assuming D/D/1 queuing, when does the queue clear and what is the average delay per vehicle?

5.45 Vehicles begin to arrive at an amusement park entrance at 8:00 A.M. at a rate of 1000veh/h. Some of these vehicles have electronic identifiers that allow them to enter the park immediately, beginning at 8:00 A.M., without stopping (they are billed remotely). All vehicles without such identifiers stop at a single processing booth, but they wait in line until it opens at 8:10 A.M. Once open, the operator processes vehicles at μ(t) = 8 + 0.5t [where μ(t) is in vehicles per minute and t is in minutes after 8:10 A.M.]. An observer notes that at 8:25 there are exactly 20 vehicles in the queue. What percent of arriving vehicles have electronic identifiers and what is the total delay (from the 8:00 A.M. until the queue clears) for those vehicles without the electronic identifiers (assume D/D/1 queuing)?

5.46 At 9:00 A.M., there are 22 vehicles in a queue at a toll booth. Starting at 9:00 A.M., vehicles are arriving at a rate of λ(t) = 9.2 – 0.3t [where λ(t) is in vehicles per minute and t is in minutes after 9:00 A.M.] and vehicles are being serviced at a constant rate of four vehicles per minute. What is the maximum queue length, and what would the total delay be from 9:00 A.M. until the queue clears?

5.47 A transportation engineering midterm exam starts at noon and students start completing a the exam at 1:05 P.M, and the rate they complete the exam and hand it in to the instructor at the front of the class is given by the function λ(t) = 1.1 + 1.2t [with λ(t) in students/minute and t is in minutes after 1:05 P.M]. The professor takes 10 seconds to process each student (record exam completion). There are 137 students in the class. Assuming D/D/1 queuing, when will the last student have their exam processed, what is the total student delay in handing in the exams, and what is the longest student queue?

Chapter 5 Problems P-19

Traffic Analysis at Highway Bottlenecks (Section 5.6)

5.48 A freeway with two northbound lanes is shut down because of an accident. At the time of the accident, the traffic flow rate is 1200 vehicles per hour per lane and the flow remains at this level. The capacity of the freeway is 2200 vehicles per hour per lane when not impacted by an accident. The freeway is shut down completely for 20 minutes after the accident and then one lane is open for 20 minutes and finally both lanes are opened (40 minutes after the accident). What is the average delay per vehicle resulting from the accident (assuming D/D/1 queuing)?

5.49 A four-lane highway has a normal capacity of 1800 vehicles per hour per lane. In the southbound direction, a vehicle disablement on the roadway shoulder occurs at 4:30 P.M. Due to rubbernecking, the capacity in the southbound direction is reduced to 1200 veh/h/lane at this time. At 4:45 P.M., the disabled vehicle is removed from the shoulder and the capacity increases to 1500 veh/h/lane. At 5:00 P.M., the roadway capacity returns to its full value of 1800 veh/h/lane. From 4:30 P.M., until the queue clears the traffic flow rate in the southbound direction is 1600 veh/h/lane. What is the average delay per vehicle, the maximum queue length, and the average queue length in the southbound direction resulting from the incident (assuming D/D/1 queuing)?

Multiple Choice Problems (Multiple Sections)

5.50 Five minivans and three trucks are traveling on a 3.0 mile circular track and complete a full lap in 98.0, 108.0, 113.0, 108.0, 102.0, 101.0, 85.0, and 95 seconds, respectively. Assuming that all the vehicles are traveling at constant speeds, what is the time- mean speed of the minivans? Pay attention to rounding.

a) 102.332 mi/h b) 107.417 mi/h c) 102.079 mi/h d) 102.400 mi/h

5.51 Vehicles arrive at an intersection at a rate of 400 veh/h according to a Poisson distribution. What is the probability that more than five vehicles will arrive in a 1-minute interval?

a) 0.7944 b) 0.6560 c) 0.6547 d) 0.1552

5.52 In studying of traffic flow at a highway toll booth over a course of 60 minutes, it is determined that the arrival and departure rates are deterministic, but not uniform. The arrival rate is found to vary according to the function λ(t) = 1.8 + 0.25t – 0.0030t2. The departure rate function is μ(t) = 1.4 + 0.11t. In both of these functions, t is in minutes after the beginning of the observation and λ(t) and μ(t) are in vehicles per minute. At what time does the maximum queue length occur?

a) 49.4 min b) 2.7 min c) 19.4 min d) 60.0 min

5.53 A theme park has a single entrance gate where visitors must stop and pay for parking. The average arrival rate during the peak hour is 150 veh/h and is Poisson distributed. It takes, on average, 20 seconds per vehicle (exponentially distributed) to pay for parking. What is the average waiting time for this queuing system?

a) 4.167 min/veh b) 2.0 min/veh c) 1.667 min/veh d) 0.833 min/veh

5.54 At an impaired driver checkpoint, the time required to conduct the impairment test varies (according to an exponential distribution) depending on the compliance of the driver, but takes 60 seconds on average. If an average of 30 vehicles per hour arrive (according to a Poisson distribution) at the checkpoint, determine the average time spent in the system.

a) 0.033 min/veh b) 1.5 min/veh c) 1.0 min/veh d) 2.0 min/veh

5.55 A toll road with three toll booths has an average arrival rate of 850 veh/h and drivers take an average of 12 seconds to pay their tolls. If the arrival and departure times are determined to be exponentially distributed, how would the probability of waiting in a queue change if a fourth toll both were opened?

a) 0.088 b) 0.534 c) 0.313 d) 0.847

151

Chapter 5

Fundamentals of Traffic Flow and Queuing Theory

5.1 INTRODUCTION

While the primary function of a highway is mobility (measured by various performance measures such as vehicle speeds), safety must also play a prominent role. Many of the safety-related aspects of highway design were discussed in Chapter 3. Starting with the current chapter, emphasis will now turn to understanding elements of highway traffic flow and queuing, and ultimately on to other measures of highway performance and mobility (Chapters 6 and 7).

The analysis of vehicle traffic (including traffic flow and queuing theory) provides the basis for measuring the operating performance of highways. In undertaking such an analysis, the various dimensions of traffic, such as number of vehicles per unit time (flow), vehicle types, vehicle speeds, and the variation in traffic flow over time, must be addressed because they all influence highway design (the selection of the number of lanes, pavement types, and geometric design) and highway operations (selection of traffic control devices, including signs, markings, and traffic signals), both of which impact the performance of the highway. In light of this, it is important for the analysis of traffic to begin with theoretically consistent quantitative techniques that can be used to model traffic flow, speed, and temporal fluctuations. The intent of this chapter is to focus on models of traffic flow and queuing, thus providing the groundwork for quantifying measures of performance (and levels of service, which will be discussed in Chapters 6 and 7).

5.2 TRAFFIC STREAM PARAMETERS

Traffic streams (the flow of vehicles on a highway) can be characterized by a number of different operational performance measures. Before commencing a discussion of the specific measures, it is important to provide definitions for the contexts in which these measures apply. A traffic stream that operates free from the influence of such traffic control devices as signals and stop signs is classified as uninterrupted flow. This type of traffic flow is influenced primarily by roadway characteristics and the interactions of the vehicles in the traffic stream. Freeways, multilane highways, and two-lane highways often operate under uninterrupted flow conditions. Traffic streams that operate under the influence of signals and stop signs are classified as interrupted flow. Although all the

152 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

concepts in this chapter are generally applicable to both types of flow, there are some additional complexities involved with the analysis of traffic flow at signalized and unsignalized intersections. Chapter 7 will address the additional complexities relating to the analysis of traffic flow at signalized intersections. For details on the analysis of traffic flow at unsignalized intersections, refer to other sources [Transportation Research Board 1975, 2016]. It should be noted that environmental conditions (day vs. night, sunny vs. rainy, etc.) can also affect the flow of traffic, but these issues are beyond the scope of this book.

5.2.1 Traffic Flow, Speed, and Density Traffic flow, speed, and density are variables that form the underpinnings of traffic analysis. To begin the study of these variables, the basic definitions of traffic flow, speed, and density must be presented. Traffic flow is defined as

= n

q t

(5.1)

where

q = traffic flow in vehicles per unit time, n = number of vehicles passing some designated roadway point during time t, and t = duration of time interval.

Flow is often measured over the course of an hour, in which case the resulting value is typically referred to as volume. Thus, when the term “volume” is used, it is generally understood that the corresponding value is in units of vehicles per hour (veh/h). The definition of flow is more generalized to account for the measurement of vehicles over any period of time. In practice, the analysis flow rate is usually based on the peak 15-minute flow within the hour of interest. This aspect will be described in more detail in Chapter 6.

Aside from the total number of vehicles passing a point in some time interval, the amount of time between the passing of successive vehicles (or time between the arrival of successive vehicles) is also of interest. The time between the passage of the front bumpers of successive vehicles, at some designated highway point, is known as the time headway. The time headway is related to t, as defined in Eq. 5.1, by

=

= 1

n

i i

t h (5.2)

where

t = duration of time interval,

hi = time headway of the ith vehicle (the elapsed time between the arrivals of vehicles i and i − 1), and

n = number of measured vehicle time headways at some designated roadway point.

5.2 Traffic Stream Parameters 153

Substituting Eq. 5.2 into Eq. 5.1 gives

1

n

i i =

n q =

h (5.3)

or

1

q = h

(5.4)

where h = average time headway ( ∑ hi / n) in unit time per vehicle. The importance of time headways in traffic analysis will be given additional attention in later sections of this chapter.

The average traffic speed is defined in two ways. The first is the arithmetic mean of the vehicle speeds observed at some designated point along the roadway. This is referred to as the time-mean speed and is expressed as

1

n

i i

t

u

u n

== 

(5.5)

where

tu = time-mean speed in unit distance per unit time,

ui = spot speed (the speed of the vehicle at the designated point on the highway, as might be obtained using a radar gun) of the ith vehicle, and

n = number of measured vehicle spot speeds.

The second definition of speed is more useful in the context of traffic analysis

and is determined on the basis of the time necessary for a vehicle to travel some known length of roadway. This measure of average traffic speed is referred to as the space-mean speed and is expressed as (assuming that the travel time for all vehicles is measured over the same length of roadway)

s l

u t

= (5.6)

where

su = space-mean speed in unit distance per unit time,

l = length of roadway used for travel time measurement of vehicles, and

t = average vehicle travel time, defined as

1

1 n

i i

t t n =

=  (5.7)

154 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

where

ti = time necessary for vehicle i to travel a roadway section of length l, and

n = number of measured vehicle travel times.

Substituting Eq. 5.7 into Eq. 5.6 yields

=

=

 1

1 s n

i i

l u

t n

(5.8)

or

( )=

=       

 1

1

1 1 s n

ii

u

n l t

(5.9)

which is the harmonic mean of speed (space-mean speed). Space-mean speed is the speed variable used in traffic models.

EXAMPLE 5.1 TIME- AND SPACE-MEAN SPEEDS

The speeds of five vehicles were measured (with radar) at the midpoint of a 0.5-mile section of roadway. The speeds for vehicles 1, 2, 3, 4, and 5 were 44, 42, 51, 49, and 46 mi/h, respectively. Assuming that all vehicles were traveling at constant speed over this roadway section, calculate the time-mean and space-mean speeds.

SOLUTION

For the time-mean speed, Eq. 5.5 is applied, giving

1

44 42 51 49 46 5

46.4 mi/h

n

i i

t

u

u n

==

+ + + +=

=



For the space-mean speed, Eq. 5.9 will be applied. This equation is based on travel time; however, because it is known that the vehicles were traveling at constant speed, we can rearrange this equation to utilize the measured speed, knowing that distance, l, divided by travel time, t, is equal to speed ( = ):

( ) ==

= =             

 11

1 1

1 11 1 s nn

ii ii

u

n un l t

5.2 Traffic Stream Parameters 155

=  + + + +  

=

=

1 1 1 1 1 1 1 5 44 42 51 49 46

1 0.02166 46.17 mi/h

Note that the space-mean speed will always be lower than the time-mean speed, unless all vehicles are traveling at exactly the same speed, in which case the two measures will be equal.

Finally, traffic density is defined as

n

k l

= (5.10)

where

k = traffic density in vehicles per unit distance, n = number of vehicles occupying some length of roadway at some specified time,

and l = length of roadway.

The density can also be related to the individual spacing between successive vehicles (measured from front bumper to front bumper). The roadway length, l, in Eq. 5.10 can be defined as

1

n

i i

l s =

= (5.11)

where

si = spacing of the ith vehicle (the distance between vehicles i and i − 1, measured from front bumper to front bumper), and

n = number of measured vehicle spacings.

Substituting Eq. 5.11 into Eq. 5.10 gives

1

n

i i

n k

s =

=

 (5.12)

or

1

k s

= (5.13)

where s = average spacing Σ( / )is n in unit distance per vehicle.

156 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

Time headway and spacing are referred to as microscopic measures because they describe characteristics specific to individual pairs of vehicles within the traffic stream. Measures that describe the traffic stream as a whole, such as flow, average speed, and density, are referred to as macroscopic measures. As indicated by the preceding equations, the microscopic measures can be aggregated and related to the macroscopic measures.

Based on the definitions presented, a simple identity provides the basic relationship among traffic flow, speed (space-mean), and density (denoting space-mean speed, su as simply u for notational convenience):

q uk= (5.14)

where

q = flow, typically in units of veh/h, u = speed (space-mean speed), typically in units of mi/h, and k = density, typically in units of veh/mi.

EXAMPLE 5.2 SPEED, FLOW, AND DENSITY

Vehicle time headways and spacings were measured at a point along a highway, from a single lane, over the course of an hour. The average values were calculated as 2.5 s/veh for headway and 200 ft/veh for spacing. Calculate the average speed of the traffic.

SOLUTION

To calculate the average speed of the traffic, the fundamental relationship in Eq. 5.14 is used. To begin, the flow and density need to be calculated from the headway and spacing data. Flow is determined from Eq. 5.4 as

=

=

1 2.5 s/veh

0.40 veh/s

q

or, because the data were collected for an hour,

0.40 veh/s 3600 s/h

1440 veh/h

q = × =

Density is determined from Eq. 5.13 as

1 200 ft/veh

0.005 veh/ft

k =

=

or, applying this spacing over the course of one mile,

0.005 veh/ft 5280 ft/mi

26.4 veh/mi

k = × =

5.3 Basic Traffic Stream Models 157

Now applying Eq. 5.14, after rearranging to solve for speed, gives

1440 veh/h

26.4 veh/mi 54.5 mi/h

q u

k =

=

=

Note that the average speed of traffic can be determined directly from the average headway and spacing values, as follows:

=

=

=

200 ft/veh

2.5 s/veh 80 ft/s (54.5 mi/h)

s u

h

5.3 BASIC TRAFFIC STREAM MODELS

While the preceding definitions and relationships provide the basis for the measurement and calculation of traffic stream parameters, it is also essential to understand the interaction of the individual macroscopic measures in order to fully analyze the operational performance of traffic streams. The models that describe these interactions are discussed in the following sections, and it will be shown that Eq. 5.14 serves the important function of linking specific models of traffic into a consistent, generalized model.

5.3.1 Speed-Density Model The most intuitive starting point for developing a consistent, generalized traffic model is to focus on the relationship between speed and density. To begin, consider a section of highway with only a single vehicle on it. Under these conditions, the density (veh/mi) will be very low and the driver will be able to travel freely at a speed close to the design speed of the highway. This speed is referred to as the free-flow speed because vehicle speed is not inhibited by the presence of other vehicles. As more and more vehicles begin to use a section of highway, the traffic density will increase and the average operating speed of vehicles will decline from the free-flow value as drivers slow to allow for the maneuvers of other vehicles. Eventually, the highway section will become so congested (will have such a high density) that the traffic will come to a stop (u = 0), and the density will be determined by the length of the vehicles and the spaces that drivers leave between them. This high-density condition is referred to as the jam density.

One possible representation of the process described above is the linear relationship shown in Fig. 5.1. Mathematically, such a relationship can be expressed as

158 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

 

= −     1f

j

k u u

k (5.15)

where

u = space-mean speed in mi/h, uf = free-flow speed in mi/h, k = density in veh/mi, and kj = jam density in veh/mi.

The advantage of using a linear representation of the speed–density relationship is that it provides a basic insight into the relationships among traffic flow, speed, and density interactions without clouding these insights by the additional complexity that a nonlinear speed–density relationship introduces. However, it is important to note that field studies have shown that the speed–density relationship tends to be nonlinear at low densities and high densities (those that approach the jam density). In fact, the overall speed–density relationship is better represented by three relationships: (1) a nonlinear relationship at low densities that has speed slowly declining from the free-flow value, (2) a linear relationship over the large medium-density region (speed declining linearly with density as shown in Eq. 5.15), and (3) a nonlinear relationship near the jam density as the speed asymptotically approaches zero with increasing density. For the purposes of exposition, we present only traffic stream models that are based on the assumption of a linear speed–density relationship. Examples of nonlinear speed–density relationships are provided elsewhere [Pipes, 1967; Drew, 1965].

Figure 5.1 Illustration of a typical linear speed–density relationship.

5.3 Basic Traffic Stream Models 159

5.3.2 Flow-Density Model Using the assumption of a linear speed–density relationship as shown in Eq. 5.15, a parabolic flow-density model can be obtained by substituting Eq. 5.15 into Eq. 5.14:

 

= −    

2

f j

k q u k

k (5.16)

where all terms are as defined previously. The general form of Eq. 5.16 is shown in Fig. 5.2. Note in this figure that the

maximum flow rate, qcap, represents the highest rate of traffic flow that the highway is capable of handling. This is referred to as the traffic flow at capacity, or simply the capacity of the roadway. The traffic density that corresponds to this capacity flow rate is kcap, and the corresponding speed is ucap. Equations for qcap, kcap, and ucap can be derived by differentiating Eq. 5.16, because at maximum flow

 

= − =    

2 1 0f

j

dq k u

dk k (5.17)

and because the free-flow speed (uf) is not equal to zero,

2

j cap

k k = (5.18)

Substituting Eq. 5.18 into Eq. 5.15 gives

  = −  

 

=

1 2

2

j cap f

j

f

k u u

k

u

(5.19)

Figure 5.2 Illustration of the parabolic flow–density relationship.

160 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

and using Eq. 5.18 and Eq. 5.19 in Eq. 5.14 gives

4

cap cap cap

f j

q u k

u k

=

= (5.20)

5.3.3 Speed-Flow Model Again returning to the linear speed–density model (Eq. 5.15), a corresponding speed-flow model can be developed by rearranging Eq. 5.15 to

 

= −     1j

f

u k k

u (5.21)

and by substituting Eq. 5.21 into Eq. 5.14,

 

= −    

2

j f

u q k u

u (5.22)

The speed-flow model defined by Eq. 5.22 again gives a parabolic function, as shown in Fig. 5.3. Note that Fig. 5.3 shows that two speeds are possible for flows, q, up to the highway’s capacity, qcap (this follows from the two densities possible for given flows as shown in Fig. 5.2). It is desirable, for any given flow, to keep the average space-mean speed on the upper portion of the speed-flow curve (above ucap). When speeds drop below ucap, traffic is in a highly congested and unstable condition.

All of the flow, speed, and density relationships and their interactions are graphically represented in Fig. 5.4.

Figure 5.3 Illustration of the parabolic speed–flow relationship.

5.3 Basic Traffic Stream Models 161

Figure 5.4 Flow–density, speed–density, and speed–flow relationships (assuming a linear speed–density model).

EXAMPLE 5.3 APPLICATION OF SPEED–FLOW–DENSITY RELATIONSHIPS

A section of highway is known to have a free-flow speed of 55 mi/h and a capacity of 3300 veh/h. In a given hour, 2100 vehicles were counted at a specified point along this highway section. If the linear speed–density relationship shown in Eq. 5.15 applies, what would you estimate the space-mean speed of these 2100 vehicles to be?

SOLUTION

The jam density is first determined from Eq. 5.20 as

4

4 3300 55

240.0 veh/mi

cap j

f

q k

u =

× =

=

Rearranging Eq. 5.22 to solve for u,

− + =2 0j j

f

k u k u q

u

Substituting,

2240.0 240.0 2100 0

55 u u − + =

which gives u = 44.08 mi/h or 10.92 mi/h. Both of these speeds are feasible, as shown in Fig. 5.3.

162 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

5.4 MODELS OF TRAFFIC FLOW

With the basic relationships among traffic flow, speed, and density formalized, attention can now be directed toward a more microscopic view of traffic flow. That is, instead of simply modeling the number of vehicles passing a specified point on a highway in some time interval, there is considerable analytic value in modeling the time between the arrivals of successive vehicles (the concept of vehicle time headway presented earlier). The most simplistic approach to vehicle arrival modeling is to assume that all vehicles are equally or uniformly spaced. This results in what is termed a deterministic, uniform arrival pattern. Under this assumption, if the traffic flow is 360 veh/h, the number of vehicles arriving in any 5-minute time interval is 30 and the headway between all vehicles is 10 seconds (because h will equal 3600/q). However, actual observations show that such uniformity of traffic flow is not always realistic because some 5-minute intervals are likely to have more or less traffic flow than other 5-minute intervals. Thus, a representation of vehicle arrivals that goes beyond the deterministic, uniform assumption is often warranted.

5.4.1 Poisson Model Models that account for the nonuniformity of flow are derived by assuming that the pattern of vehicle arrivals (at a specified point) corresponds to some random process. The problem then becomes one of selecting a probability distribution that is a reasonable representation of observed traffic arrival patterns. An example of such a distribution is the Poisson distribution (the limitations of which will be discussed later), which is expressed as

( )

( ) !

n tt e P n

n

λλ −

= (5.23)

where P(n) = probability of having n vehicles arrive in time t,

λ = average vehicle flow or arrival rate in vehicles per unit time,

t = duration of the time interval over which vehicles are counted, and e = base of the natural logarithm (e = 2.718).

EXAMPLE 5.4 VEHICLE ARRIVALS AS A POISSON PROCESS

An observer counts 360 veh/h at a specific highway location. Assuming that the arrival of vehicles at this highway location is Poisson distributed, estimate the probabilities of having 0, 1, 2, 3, 4, and 5 or more vehicles arriving over a 20-second time interval.

SOLUTION

The average arrival rate, λ, is 360 veh/h, or 0.1 vehicles per second (veh/s). Using this in Eq. 5.23 with t = 20 seconds, the probabilities of having exactly 0, 1, 2, 3, and 4 vehicles arrive are

5.4 Models of Traffic Flow 163

( ) ( ) ( )0 0.1 200.1 20 0 0.135

0!

e P

−× = =

( ) ( ) ( )1 0.1 200.1 20 1 0.271

1!

e P

−× = =

( ) ( ) ( )2 0.1 200.1 20 2 0.271

2!

e P

−× = =

( ) ( ) ( )3 0.1 200.1 20 3 0.180

3!

e P

−× = =

( ) ( ) ( )4 0.1 200.1 20 4 0.090

4!

e P

−× = =

For five or more vehicles,

( ) ( )5 1 5

1 0.135 0.271 0.271 0.180 0.090

0.053

P n P n≥ = − < = − − − − − =

A histogram of these probabilities is shown in Fig. 5.5.

Figure 5.5 Histogram of the Poisson distribution for λ = 0.1 vehicles per second.

164 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

EXAMPLE 5.5 VEHICLE ARRIVALS AS A POISSON PROCESS WITH DETAILED VEHICLE-ARRIVAL DATA

Traffic data are collected in 60-second intervals at a specific highway location as shown in Table 5.1. Assuming the traffic arrivals are Poisson distributed and continue at the same rate as that observed in the 15 time periods shown, what is the probability that six or more vehicles will arrive in each of the next three 60-second time intervals (12:15 P.M. to 12:16 P.M., 12:16 P.M. to 12:17 P.M., and 12:17 P.M. to 12:18 P.M.)?

Table 5.1 Observed Traffic Data for Example 5.5

Time period Observed number of vehicles

12:00 P.M. to 12:01 P.M. 3

12:01 P.M. to 12:02 P.M. 5 12:02 P.M. to 12:03 P.M. 4 12:03 P.M. to 12:04 P.M. 10 12:04 P.M. to 12:05 P.M. 7 12:05 P.M. to 12:06 P.M. 4 12:06 P.M. to 12:07 P.M. 8 12:07 P.M. to 12:08 P.M. 11 12:08 P.M. to 12:09 P.M. 9 12:09 P.M. to 12:10 P.M. 5 12:10 P.M. to 12:11 P.M. 3 12:11 P.M. to 12:12 P.M. 10 12:12 P.M. to 12:13 P.M. 9 12:13 P.M. to 12:14 P.M. 7 12:14 P.M. to 12:15 P.M. 6

SOLUTION

Table 5.1 shows that a total of 101 vehicles arrive in the 15-minute period from 12:00 P.M. to 12:15 P.M. Thus, the average arrival rate, λ, is 0.112 veh/s (101/900). As in Example 5.4, Eq. 5.23 is applied to find the probabilities of exactly 0, 1, 2, 3, 4, and 5 vehicles arriving.

Applying Eq. 5.23, with λ = 0.112 veh/s and t = 60 seconds, the probabilities of having 0, 1, 2, 3, 4, and 5 vehicles arriving in a 60-second time interval are (using λt = 6.733)

( ) ( )

( )

( )

( )

0 6.733

1 6.733

2 6.733

3 6.733

6.733 0 0.0012

0!

6.733 (1) 0.008

1!

6.733 (2) 0.027

2!

6.733 (3) 0.0606

3!

e P

e P

e P

e P

−

−

−

−

= =

= =

= =

= =

5.4 Models of Traffic Flow 165

( )

( )

4 6.733

5 6.733

6.733 (4) 0.102

4!

6.733 (5) 0.137

5!

e P

e P

−

−

= =

= =

The summation of these probabilities is the probability that 0 to 5 vehicles will arrive in any given 60-second time interval, which is

5

0

( 5) ( )

0.0012 0.008 0.027 0.0606 0.102 0.137

0.3358

i

P n P n =

≤ =

= + + + + + =



So 1 minus P(n ≤ 5) is the probability that six or more vehicles will arrive in any 60-second time interval, which is

( )6 1 ( 5)

1 0.3358

0.6642

P n P n≥ = − ≤ = − =

The probability that six or more vehicles will arrive in three successive time intervals (t1, t2, and t3) is simply the product of probabilities, which is

( ) ( )

( )

3

1

3

6 for three successive time periods 6

0.6642

0.293

it

P n P n

=

≥ = ≥

= =

∏

The assumption of Poisson vehicle arrivals also implies a distribution of the

time intervals between the arrivals of successive vehicles (time headway). To show this, note that the average arrival rate is

3600

q λ = (5.24)

where

λ = average vehicle arrival rate in veh/s,

q = flow in veh/h, and 3600 = number of seconds per hour.

Substituting Eq. 5.24 into Eq. 5.23 gives

( ) ( ) 36003600

!

n qtqt e P n

n

−

= (5.25)

166 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

Note that the probability of having no vehicles arrive in a time interval of length t, P(0), is equivalent to the probability of a vehicle headway, h, being greater than or equal to the time interval t. So from Eq. 5.25,

( ) ( )

3600

0 qt

P P h t

e−

= ≥

= (5.26)

This distribution of vehicle headways is known as the negative exponential distribution and is often simply referred to as the exponential distribution.

EXAMPLE 5.6 HEADWAYS AND THE NEGATIVE EXPONENTIAL DISTRIBUTION

Consider the traffic situation in Example 5.4 (360 veh/h). Again assume that the vehicle arrivals are Poisson distributed. What is the probability that the headway between successive vehicles will be less than 8 seconds, and what is the probability that the headway between successive vehicles will be between 8 and 10 seconds?

SOLUTION

By definition, P(h < t) = 1 − P(h ≥ t). This expression gives the probability that the headway will be less than 8 seconds as

( ) ( )

3600

360 8 3600

1

1

0.551

qtP h t e

e

−

−

< = −

= − =

To determine the probability that the headway will be between 8 and 10 seconds, compute the probability that the headway will be greater than or equal to 10 seconds:

( ) ( )

3600

360 10 3600

0.368

qtP h t e

e

−

−

≥ =

= =

So the probability that the headway will be between 8 and 10 seconds is 0.081

(1 − 0.551 − 0.368).

To help in visualizing the shape of the exponential distribution, Fig. 5.6

shows the probability distribution implied by Eq. 5.26, with the flow, q, equal to 360 veh/h as in Example 5.4.

5.4.2 Limitations of the Poisson Model Empirical observations have shown that the assumption of Poisson-distributed traffic arrivals is most realistic in lightly congested traffic conditions. As traffic flows become heavily congested or when traffic signals cause cyclical traffic stream disturbances, other distributions of traffic flow become more appropriate. The primary limitation of the Poisson model of vehicle arrivals is

5.5 Queuing Theory and Traffic Flow Analysis 167

Figure 5.6 Exponentially distributed probabilities of headways greater than or equal to t, with q = 360 veh/h.

the constraint imposed by the Poisson distribution that the mean of period observations equals the variance. For example, the mean of period-observed traffic in Example 5.5 is 6.733 and the corresponding variance, σ2, is 7.210. Because these two values are close, the Poisson model was appropriate for this example. If the variance is significantly greater than the mean, the data are said to be overdispersed, and if the variance is significantly less than the mean, the data are said to be underdispersed. In either case the Poisson distribution is no longer appropriate, and another distribution should be used. Such distributions are discussed in detail in more specialized sources [Transportation Research Board, 1975; Poch and Mannering, 1996; Lord and Mannering, 2010].

5.5 QUEUING THEORY AND TRAFFIC FLOW ANALYSIS

The formation of traffic queues during congested periods is a source of considerable delay and results in a loss of highway performance. Under extreme conditions, queuing delay can account for 90% or more of a motorist’s total trip travel time. Given this, it is essential in traffic analysis to develop a clear understanding of the characteristics of queue formation and dissipation along with mathematical formulations that can predict queuing-related elements.

As is well known, the problem of queuing is not unique to traffic analysis. Many non-transportation fields, such as the design and operation of industrial plants, retail stores, service-oriented industries, and computer networks, must also give serious consideration to the problem of queuing. The impact of queues on performance and productivity in manufacturing, retailing, and other fields has led to numerous theories of queuing behavior (the process by which queues form and dissipate). As will be shown, the models of traffic flow presented earlier (uniform, deterministic arrivals and Poisson arrivals) will form the basis for studying traffic queues within the more general context of queuing theory.

5.5.1 Dimensions of Queuing Models The purpose of traffic queuing models is to provide a means to estimate important measures of highway performance, including vehicle delay and traffic queue lengths. Such estimates are critical to roadway design (the required length

168 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

of left-turn bays and the number of lanes at intersections) and traffic operations control, including the timing of traffic signals at intersections.

Queuing models are derived from underlying assumptions regarding arrival patterns, departure characteristics, and queue disciplines. Traffic arrival patterns were explored in Section 5.4, where, given an average vehicle arrival rate (λ), two possible distributions of the time between the arrival of successive vehicles were considered:

1. Equal time intervals (derived from the assumption of uniform, deterministic arrivals)

2. Exponentially distributed time intervals (derived from the assumption of Poisson-distributed arrivals)

In addition to vehicle arrival assumptions, the derivation of traffic queuing models requires assumptions relating to vehicle departure characteristics. Of particular interest is the distribution of the amount of time it takes a vehicle to depart—for example, the time to pass through an intersection at the beginning of a green signal, the time required to pay a toll at a toll booth, or the time a driver takes before deciding to proceed after stopping at a stop sign. As was the case for arrival patterns, given an average vehicle departure rate (denoted as μ, in vehicles per unit time), the assumption of a deterministic or exponential distribution of departure times is appropriate.

Another important aspect of queuing models is the number of available departure channels. For most traffic applications only one departure channel will exist, such as a highway lane or group of lanes passing through an intersection. However, multiple departure channels are encountered in some traffic applications, such as at toll booths on turnpikes and at entrances to bridges.

The final necessary assumption relates to the queue discipline. In this regard, two options have been popularized in the development of queuing models: first- in, first-out (FIFO), indicating that the first vehicle to arrive is the first to depart, and last-in, first-out (LIFO), indicating that the last vehicle to arrive is the first to depart. For virtually all traffic-oriented queues, the FIFO queuing discipline is the more appropriate of the two.

Queuing models are often identified by three alphanumeric values. The first value indicates the arrival rate assumption, the second value gives the departure rate assumption, and the third value indicates the number of departure channels. For traffic arrival and departure assumptions, the uniform, deterministic distribution is denoted D and the exponential distribution is denoted M. Thus a D/D/1 queuing model assumes deterministic arrivals and departures with one departure channel. Similarly, an M/D/1 queuing model assumes exponentially distributed arrival times, deterministic departure times, and one departure channel.

5.5.2 D/D/1 Queuing The case of deterministic arrivals and departures with one departure channel (D/D/1 queue) is an excellent starting point in understanding queuing models because of its simplicity. The D/D/1 queue lends itself to an intuitive graphical or mathematical solution that is best illustrated by an example.

5.5 Queuing Theory and Traffic Flow Analysis 169

EXAMPLE 5.7 D/D/1 QUEUING WITH CONSTANT ARRIVAL AND DEPARTURE RATES

Vehicles arrive at an entrance to a recreational park. There is a single gate (at which all vehicles must stop), where a park attendant distributes a free brochure. The park opens at 8:00 A.M., at which time vehicles begin to arrive at a rate of 480 veh/h. After 20 minutes, the arrival flow rate declines to 120 veh/h, and it continues at that level for the remainder of the day. If the time required to distribute the brochure is 15 seconds, and assuming D/D/1 queuing, describe the operational characteristics of the queue.

SOLUTION

Begin by putting arrival and departure rates into common units of vehicles per minute:

480 veh/h 8 veh/min for 20 min

60 min/h 120 veh/h

2 veh/min for 20 min 60 min/h 60 s/min

4 veh/min for all 15 s/veh

t

t

t

λ

λ

μ

= = ≤

= = >

= =

Equations for the total number of vehicles that have arrived and departed up to a specified time, t, can now be written. Define t as the number of minutes after the start of the queuing process (in this case the number of minutes after 8:00 A.M.). The total number of vehicle arrivals at time t is equal to

≤8 for 20 mint t

and

( )160 2 20 for 20 m in t t + − >

Similarly, the number of vehicle departures is

4 for all t t

The preceding equations can be illustrated graphically as shown in Fig. 5.7. When the arrival curve is above the departure curve, a queue condition exists. The point at which the arrival curve meets the departure curve is the moment when the queue dissipates (no more queue exists). In this example, the point of queue dissipation can be determined graphically by inspection of Fig. 5.7, or analytically by equating appropriate arrival and departure equations, that is,

( )160 2 20 4 t t+ − =

Solving for t gives t = 60 minutes. Thus the queue that began to form at 8:00 A.M. will dissipate 60 minutes later (9:00 A.M.), at which time 240 vehicles will have arrived and departed (4 veh/min × 60 min).

Another aspect of interest is individual vehicle delay. Under the assumption of a FIFO queuing discipline, the delay of an individual vehicle is given by the horizontal distance between arrival and departure curves starting from the time of the vehicle’s arrival in the queue. So, by inspection of Fig. 5.7, the 160th vehicle to arrive will have the longest delay, 20 minutes (the longest horizontal distance between arrival and departure curves), and vehicles arriving after the 239th vehicle will encounter no queue delay because the queue will have dissipated and the departure rate will continue to

170 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

exceed the arrival rate. It follows that with the LIFO queuing discipline, the first vehicle to arrive would have to wait until the entire queue clears (60 minutes of delay).

The total length of the queue at a specified time, expressed as the number of vehicles, is given by the vertical distance between arrival and departure curves at that time. For example, at 10 minutes after the start of the queuing process (8:10 A.M.) the queue is 40 vehicles long, and the longest queue (longest vertical distance between arrival and departure curves) will occur at t = 20 minutes and is 80 vehicles long (see Fig. 5.7).

Total vehicle delay, defined as the summation of the delays for the individual vehicles, is given by the total area between the arrival and departure curves (see Fig. 5.7) and, in this case, is in units of vehicle-minutes. In this example, the area between the arrival and departure curves can be determined by summing triangular areas, giving total delay, Dt, as

1 1 (80 20) (80 40)

2 2 2400 veh-min

tD = × + ×

=

Finally, because 240 vehicles encounter queuing delay (as previously determined), the average delay per vehicle is 10 minutes (2400 veh-min/240 veh), and the average queue length is 40 vehicles (2400 veh-min/60 min).

Figure 5.7 D/D/1 queuing diagram for Example 5.7.

5.5 Queuing Theory and Traffic Flow Analysis 171

EXAMPLE 5.8 D/D/1 QUEUING WITH TIME-VARYING ARRIVAL RATE AND CONSTANT DEPARTURE RATE

Vehicles start arriving at an entrance to a national park at 5:45 A.M. and a park booth collecting entrance fees opens at 6:00 A.M. and process vehicles at a deterministic rate of 20 veh/min. It is estimated that 20% of the arriving vehicles have priority passes that allow them to access a separate processing booth that also opens at 6:00 A.M. but processes vehicles at a slower deterministic rate of 15 veh/min. The deterministic arrival rate of vehicles is a function of time such that λ(t) = 17.2 − 0.2t where t is in minutes after 5:45 A.M. Determine the average delay per vehicle for the non–priority- pass vehicles and the priority-pass vehicles (starting with their arrival at 6:45 A.M.) until their respective queues clear, assuming D/D/1 queuing.

SOLUTION

This problem has time-dependent deterministic arrivals and departure rates that do not vary over time. Begin by computing the arrivals of non–priority-pass vehicles. Because 20% of all arrivals are priority-pass vehicles, the non–priority-pass vehicle arrivals will be (starting at 5:45 A.M.):

( ) 0

2 2

2 2

2

= 17.2 0.2

17.2 0.1 0.2 17.2 0.1

17.2 0.1 3.44 0.02

13.76 0.08

t

t dt

t t t t

t t t t

t t

−

= − − −

= − − +

= −



The time required to clear the queue of the non–priority-pass vehicles (with a departure rate of 20 veh/min) is determined as:

2

2

13.76 0.08 20( 15)

0.08 6.24 300 0

33.60min

t t t

t t

t

− = −

− − + = =

At t = 33.60 minutes, the total number of non–priority-pass vehicle arrivals is

2

2

13.76 0.08

13.76(33.60) 0.08(33.60)

372.02 veh

t t= −

= − =

So the total delay for the non–priority-pass vehicles, Dnp will be the area under the arrival function minus the area under the departure function, which will be a simple triangle with a height of 372.02 vehicles and a base of 18.06 min (33.60 − 15):

172 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

( ) ( ) 33.60

2

0

2 3

1 = 13.76 0.08 33.60 15 372.02

2

6.88(33.60) 0.0267(33.60) 3459.786

3294.65 veh-min

npD t t dt

− − −

= − − =



So the average delay per vehicle for the non–priority-pass vehicles is 8.86 min (3294.65/372.02).

For the priority-pass vehicles, the arrival function is 20% of the total vehicle arrival function or,

( ) 0

2

2

=0.2 17.2 0.2

0.2 17.2 0.1

3.44 0.02

t

t dt

t t

t t

−

= −

= −



The time required to clear the queue of the priority-pass vehicles (with a departure rate of 15 veh/min) is:

2

2

3.44 0.02 15( 15)

0.02 11.56 225 0

18.84min

t t t

t t

t

− = −

− − + = =

At t = 18.84 minutes, the total number of priority-pass vehicle arrivals is

2

2

3.44 0.02

3.44(18.84) 0.02(18.84)

57.71 veh

t t= −

= − =

So the total delay for the priority-pass vehicles, Dp will again be the area under the arrival function minus the area under the departure function, which will be a simple triangle with a height of 57.71 vehicles and a base of 3.84 min (18.84 − 15):

( )( )− − −

= − − =

 18.84

2

0

2 3

1 = 3.44 0.02 18.84 15 57.71

2

1.72(18.84) 0.0067(18.84) 110.803

454.90 veh-min

pD t t dt

So the average delay per vehicle for the priority-pass vehicles is 7.88 min (454.9/57.71). Thus the priority pass saves an average of 0.98 min, or about 59 s.

5.5 Queuing Theory and Traffic Flow Analysis 173

EXAMPLE 5.9 D/D/1 QUEUING WITH TIME-VARYING ARRIVAL AND DEPARTURE RATES

After observing arrivals and departures at a highway toll booth over a 60-minute time period, an observer notes that the arrival and departure rates (or service rates) are deterministic, but instead of being uniform, they change over time according to a known function. The arrival rate is given by the function λ(t) = 2.2 + 0.17t − 0.0032t2, and the departure rate is given by μ(t) = 1.2 + 0.07t, where t is in minutes after the beginning of the observation period and λ(t) and μ(t) are in vehicles per minute. Determine the total vehicle delay at the toll booth and the longest queue, assuming D/D/1 queuing.

SOLUTION

Note that this problem is an example of a time-dependent deterministic queue because the deterministic arrival and departure rates change over time. Begin by computing the time to queue dissipation by equating vehicle arrivals and departures:

2

0 0

2 3 2

3 2

2.2 0.17 0.0032 1.2 0.07

2.2 0.085 0.00107 1.2 0.035

0.00107 0.05 0

t t

t t dt t dt

t t t t t

t t t

+ − = +

+ − = +

− + + =

 

This gives t = 61.8 minutes. Therefore, the total vehicle delay (the area between the arrival and departure functions) is

( ) ( ) ( )

61.8 61.8 2 3 2

0 0

61.8 2 3 4 2 3

0

4 3 2

2.2 0.085 0.00107 1.2 0.035

1.1 0.0283 0.0002675 0.6 0.0117

0.0002675 61.8 0.0166 61.8 0.5 61.8

1925.8 veh-min

|

tD t t t dt t t dt

t t t t t

= + − − +

= + − − −

= − + + =

 

The queue length (in vehicles) at any time t is given by the function

( ) 2

0 0

3 2

2.2 0.17 0.0032 1.2 0.07

0.00107 0.05

t t

Q t t t dt t dt

t t t

= + − − +

= − + +

 

Solving for the time at which the maximum queue length occurs yields

( ) 20.00321 0.1 1 0

39.12 min

dQ t t t

dt t

= − + + =

=

174 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

Substituting with t = 39.12 minutes gives the maximum queue length:

( )

( ) ( )

39.12 3 2

0

3 2

39.12 0.00107 0.05

0.00107 39.12 0.05 39.12 39.12

51.58 veh

|Q t t t

= − + +

= − + + =

EXAMPLE 5.10 DETERMINING A REQUIRED DEPARTURE RATE

A parking garage has a single processing booth where cars pay for parking. The garage opens at 6:00 A.M. and vehicles start arriving at 6:00 A.M. at a deterministic rate of λ(t) = 6.1 − 0.22t where λ(t) is in vehicles per minute and t is in minutes after 6:00 A.M. What is the minimum constant departure rate (from 6:00 A.M. on) needed to ensure that the queue length does not exceed 10 vehicles?

SOLUTION

Let μ be the unknown departure rate (in veh/min) so that the queue length (in vehicles) at any time t is:

( ) 0 0

2

6.1 0.22

6.1 0.11

t t

Q t t dt dt

t t t

μ

μ

= − −

= − −

 

Using this and solving for the time at which the maximum queue length occurs gives

( ) μ

μ

= − − =

−=

6.1 0.22 0

6.1 0.22

dQ t t

dt

t

Substituting this value of t into the queue-length equation with a maximum Q(t) = 10 vehicles as specified in the problem gives

( ) μ μ μμ− − −     = = − −           

2 6.1 6.1 6.1

10 6.1 0.11 0.22 0.22 0.22

Q t

which gives 2.271μ2 − 27.727μ + 74.566 = 0. Solving for μ gives possible solutions as 8.21 veh/min and 4 veh/min, so the minimum departure rate would be 4 veh/min.

5.5 Queuing Theory and Traffic Flow Analysis 175

5.5.3 M/D/1 Queuing The assumption of exponentially distributed times between the arrivals of successive vehicles (Poisson arrivals) will, in some cases, give a more realistic representation of traffic flow than the assumption of uniformly distributed arrival times. Therefore, the M/D/1 queue (exponentially distributed arrivals, deterministic departures, and one departure channel) has some important applications within the traffic analysis field. Although a graphical solution to an M/D/1 queue is difficult, a mathematical solution is straightforward. Defining a new term (traffic intensity) for the ratio of average arrival to departure rates as

λρ μ

= (5.27)

where

ρ = traffic intensity, unitless,

λ = average arrival rate in vehicles per unit time, and

μ = average departure rate in vehicles per unit time,

and assuming that ρ is less than 1, it can be shown that for an M/D/1 queue the following queuing performance equations apply:

( ) 2

2 1 Q

ρ ρ

= −

(5.28)

( )2 1 w

ρ μ ρ

= −

(5.29)

( ) 2

2 1 t

ρ μ ρ

−= −

(5.30)

where

Q = average length of queue in vehicles,

w = average waiting time in the queue, in unit time per vehicle, t = average time spent in the system (the summation of average waiting time in the

queue and average departure time), in unit time per vehicle, and

other terms are as defined previously. It is important to note that under the assumption that the traffic intensity is

less than 1 (λ < μ), the D/D/1 queue will predict no queue formation. However, a queuing model that is derived based on random arrivals or departures, such as the M/D/1 queuing model, will predict queue formations under such conditions. Also, note that the M/D/1 queuing model presented here is based on steady-state conditions (constant average arrival and departure rates), with randomness arising from the assumed probability distribution of arrivals. This contrasts with

176 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

the time-varying deterministic queuing case, presented in Example 5.9, in which arrival and departure rates changed over time but randomness was not present.

EXAMPLE 5.11 M/D/1 QUEUING: PARK-ENTRANCE APPLICATION

Consider the entrance to the recreational park described in Example 5.7. However, let the average arrival rate be 180 veh/h and Poisson distributed (exponential times between arrivals) over the entire period from park opening time (8:00 A.M.) until closing at dusk. Compute the average length of queue (in vehicles), average waiting time in the queue, and average time spent in the system, assuming M/D/1 queuing.

SOLUTION

Putting arrival and departure rates into common units of vehicles per minute gives

180 veh/h 3 veh/min for all

60 min/h 60 s/min

4 veh/min for all 15 s/veh

t

t

λ

μ

= =

= =

and

3 veh/min 0.75

4 veh/min

λρ μ

= = =

For the average length of queue (in vehicles), Eq. 5.28 is applied:

( )= −

=

20.75 2 1 0.75

1.125 veh

Q

For average waiting time in the queue, Eq. 5.29 gives

( ) ( ) 0.75

2 4 1 0.75

0.375 min/veh

w = −

=

For average time spent in the system [queue time plus departure (service) time], Eq. 5.30 is used:

( ) ( ) 2 0.75

2 4 1 0.75

0.625 min/veh

t −

= −

=

or, alternatively, because the departure (service) time is 1/μ (the 0.25 minutes it takes the park attendant to distribute the brochure),

5.5 Queuing Theory and Traffic Flow Analysis 177

1

1 0.375

4 0.625 min/veh

t w

μ = +

= +

=

5.5.4 M/M/1 Queuing A queuing model that assumes one departure channel and exponentially distributed departure times in addition to exponentially distributed arrival times (an M/M/1 queue) is applicable in some traffic applications. For example, exponentially distributed departure patterns might be a reasonable assumption at a toll booth, where some arriving drivers have the correct toll and can be processed quickly, and others do not have the correct toll, producing a distribution of departures about some mean departure rate. Under standard M/M/1 assumptions, it can be shown that the following queuing performance equations apply (again assuming that ρ is less than 1):

2

1 Q

ρ ρ

= −

(5.31)

( )w λ

μ μ λ =

− (5.32)

1

t μ λ

= −

(5.33)

where

Q = average length of queue in vehicles,

w = average waiting time in the queue, in unit time per vehicle, t = average time spent in the system (w + 1/μ), in unit time per vehicle, and

other terms are as defined previously.

EXAMPLE 5.12 M/M/1 QUEUING: PARKING-LOT APPLICATION

Assume that the park attendant in Examples 5.7 and 5.11 takes an average of 15 seconds to distribute brochures, but the distribution time varies depending on whether park patrons have questions relating to park operating policies. Given an average arrival rate of 180 veh/h as in Example 5.11, compute the average length of queue (in vehicles), average waiting time in the queue, and average time spent in the system, assuming M/M/1 queuing.

178 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

SOLUTION

Using the average arrival rate, departure rate, and traffic intensity as determined in Example 5.11, the average length of queue is (from Eq. 5.31)

20.75 1 0.75 2.25 veh

Q

= −

=

the average waiting time in the queue is (from Eq. 5.32)

( ) 3

4 4 3

0.75 min/veh

w = −

=

and the average time spent in the system is (from Eq. 5.33)

1 4 3

1 min/veh

t = −

=

5.5.5 M/M/N Queuing A more general formulation of the M/M/1 queue is the M/M/N queue, where N is the total number of departure channels. M/M/N queuing is a reasonable assumption at toll booths on turnpikes or at toll bridges, where there is often more than one departure channel available (more than one toll booth open). A parking lot is another example, with N being the number of parking stalls in the lot and the departure rate, μ, being the exponentially distributed times of parking duration. M/M/N queuing is also frequently encountered in non- transportation applications such as checkout lines at retail stores, security checks at airports, and so on.

The following equations describe the operational characteristics of M/M/N queuing. Note that unlike the equations for M/D/1 and M/M/1, which require that the traffic intensity, ρ, be less than 1, the following equations allow ρ to be greater than 1 but apply only when ρ/N (which is called the utilization factor) is less than 1.

( ) ρ ρ

ρ

−

=

= +

− 0 1

0

1

! ! 1

c

c

N n N

cn

P

n N N

(5.34)

0 for !

n

n

P P n N

n ρ

= ≤ (5.35)

5.5 Queuing Theory and Traffic Flow Analysis 179

ρ

−= ≥0 for !

n

n n N

P P n N

N N (5.36)

( ) 1

0

! 1

N

n N

P P

N N N ρ

ρ

+

> = −

(5.37)

where

P0 = probability of having no vehicles in the system, Pn = probability of having n vehicles in the system,

Pn>N = probability of waiting in a queue (the probability that the number of vehicles in the system is greater than the number of departure channels),

n = number of vehicles in the system, N = number of departure channels, nc = departure channel number, and

ρ = traffic intensity (λ/μ).

( )

ρ ρ

+    =  − 

1 0

2

1 ! 1

NP Q

N N N (5.38)

ρ

λ μ += − 1 Q

w (5.39)

Q

t ρ

λ +

= (5.40)

where Q = average length of queue (in vehicles),

w = average waiting time in the queue, in unit time per vehicle, t = average time spent in the system, in unit time per vehicle, and

other terms are as defined previously.

EXAMPLE 5.13 M/M/N QUEUING: TOLL-BRIDGE APPLICATION

At an entrance to a toll bridge, four toll booths are open. Vehicles arrive at the bridge at an average rate of 1200 veh/h, and at the booths, drivers take an average of 10 seconds to pay their tolls. Both the arrival and departure rates can be assumed to be exponentially distributed. How would the average queue length, time in the system, and probability of waiting in a queue change if a fifth toll booth were opened?

SOLUTION

Using the equations for M/M/N queuing, we first compute the four-booth case. Note that μ = 6 veh/min and λ = 20 veh/min, and therefore ρ = 3.333. Also, because ρ/N = 0.833 (which is less than 1), Eqs. 5.34 to 5.40 can be used. The probability of having no vehicles in the system with four booths open (using Eq. 5.34) is

180 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

( ) 0 2 3 4

1

3.333 3.333 3.333 3.333 1

1! 2! 3! 4! 0.1667

0.0213

P

= + + + +

=

The average queue length is (from Eq. 5.38)

( ) ( )    =   

=

5

2

0.0213 3.333 1 4!4 0.1667

3.287 veh

Q

The average time spent in the system is (from Eq. 5.40)

3.333 3.287 20

0.331 min/veh

t

+=

=

And the probability of having to wait in a queue is (from Eq. 5.37)

( ) ( )> =

5 0.0213 3.333

4!4 0.1667

0.548

n NP

=

With a fifth booth open, the probability of having no vehicles in the system is (from Eq. 5.34)

( ) 0 2 3 4 5

1 =

3.333 3.333 3.333 3.333 3.333 1 + + + + +

1! 2! 3! 4! 5! 0.3333

= 0.0318

P

The average queue length is (from Eq. 5.38)

( ) ( )    =   

6

2

0.0318 3.333 1

5!5 0.3333

= 0.654 veh

Q

The average time spent in the system is (from Eq. 5.40)

3.333 0.654 20

0.199 min/veh

t +

=

=

And the probability of having to wait in a queue is (from Eq. 5.37)

( ) ( )

6 0.0318 3.333

5!5 0.3333

0.218

n NP > =

=

5.6 Traffic Analysis at Highway Bottlenecks 181

So opening a fifth booth would reduce the average queue length by 2.633 veh (3.287 −

0.654), the average time in the system by 0.132 min/veh (0.331 − 0.199), and the

probability of waiting in a queue by 0.330 (0.548 − 0.218).

EXAMPLE 5.14 M/M/N QUEUING: PARKING-LOT APPLICATION

A convenience store has four available parking spaces. The owner predicts that the duration of customer shopping (the time that a customer’s vehicle will occupy a parking space) is exponentially distributed with a mean of 6 minutes. The owner knows that in the busiest hour customer arrivals are exponentially distributed with a mean arrival rate of 20 customers per hour. What is the probability that a customer will not find an open parking space when arriving at the store?

SOLUTION

Putting mean arrival and departure rates in common units gives μ = 10 veh/h and λ= 20 veh/h. So ρ = 2.0, and because ρ/N = 0.5 (which is less than 1), Eqs. 5.34 to 5.40 can be used. The probability of having no vehicles in the system with four parking spaces available (using Eq. 5.34) is

( ) 0 2 3 4

1

2 2 2 2 1

1! 2! 3! 4! 0.5

0.1304

P = + + + +

=

Thus the probability of not finding an open parking space upon arrival is (from Eq. 5.37)

( ) ( )

5 0.1304 2

4 !4 0.5

0.087

n NP > =

=

5.6 TRAFFIC ANALYSIS AT HIGHWAY BOTTLENECKS

Some of the most severe congestion problems occur at highway bottlenecks, which are defined as a portion of highway with a lower capacity (qcap) than the incoming section of highway. This reduction in capacity can originate from a number of sources, including a decrease in the number of highway lanes and reduced shoulder widths (which tend to cause drivers to slow and thus effectively reduce highway capacity, as will be discussed in Chapter 6). There are two general types of traffic bottlenecks—those that are recurring and those that are incident induced. Recurring bottlenecks exist where the highway itself limits capacity—for example, by a physical reduction in the number of lanes. Traffic congestion at such bottlenecks results from recurring traffic flows that exceed the vehicle capacity of the highway in the bottleneck area. In contrast, incident-induced bottlenecks occur as a result of vehicle breakdowns or accidents that effectively reduce highway capacity by restricting the through movement of traffic. Because

182 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

incident-induced bottlenecks are unanticipated and temporary in nature, they have features that distinguish them from recurring bottlenecks, such as the possibility that the capacity resulting from an incident-induced bottleneck may change over time. For example, an accident may initially stop traffic flow completely, but as the wreckage is cleared, partial capacity (one lane open) may be provided for a period of time before full capacity is eventually restored. A feature shared by recurring and incident-induced bottlenecks is the adjustment in traffic flow that may occur as travelers choose other routes and/or different trip departure times, to avoid the bottleneck area, in response to visual information or traffic advisories.

The analysis of traffic flow at bottlenecks can be undertaken using the queuing models discussed in Section 5.5. The most intuitive approach to analyzing traffic congestion at bottlenecks is to assume D/D/1 queuing.

EXAMPLE 5.15 D/D/1 QUEUING: HIGHWAY BOTTLENECK APPLICATION

An incident occurs on a freeway that has a capacity in the northbound direction, before the incident, of 4000 veh/h and a constant flow of 2900 veh/h during the morning commute (no adjustments to traffic flow result from the incident). At 8:00 A.M., a traffic accident closes the freeway to all traffic. At 8:12 A.M., the freeway is partially opened with a capacity of 2000 veh/h. Finally, the wreckage is removed, and the freeway is restored to full capacity (4000 veh/h) at 8:31 A.M. Assume D/D/1 queuing to determine time of queue dissipation, longest queue length, total delay, average delay per vehicle, and longest wait of any vehicle (assuming FIFO).

SOLUTION

Let μ be the full-capacity departure rate and μr be the restrictive partial-capacity departure rate. Putting arrival and departure rates in common units of vehicles per minute, we have

μ

μ

λ

= =

= =

= =

4000 veh/h 66.67 veh/min

60 min/h 2000 veh/h

33.33 veh/min 60 min/h

2900 veh/h 48.33 veh/min

60 min/h

r

The arrival rate is constant over the entire time period, and the total number of vehicles is equal to λt, where t is the number of minutes after 8:00 A.M. The total number of departing vehicles is

( ) ( )

0 for 12 min

12 for 12 min 31 min

633.33 31 for 31 min

r

t

t t

t t

μ μ

≤ − < ≤

+ − >

Note that the value of 633.33 in the departure rate function for t > 31 is based on the preceding departure rate function [33¹⁄³ (31 − 12)]. These arrival and departure rates

can be represented graphically as shown in Fig. 5.8. As discussed in Section 5.5, for D/D/1 queuing, the queue will dissipate at the intersection point of the arrival and

5.6 Traffic Analysis at Highway Bottlenecks 183

departure curves, which can be determined as

λt = 633.33 + μ(t – 31) or t = 78.16 min (just after 9:18 A.M.)

At this time a total of 3777.5 vehicles (48.33 × 78.16) will have arrived and departed (for the sake of clarity, fractions of vehicles are used). The longest queue (longest vertical distance between arrival and departure curves) occurs at 8:31 a.m. and is

( ) ( ) ( )

12

48.33 31 33.33 19

865 veh

rQ t t

λ μ= − −

= − =

Total vehicle delay is (using equations for triangular and trapezoidal areas to calculate the total area between the arrival and departure curves)

( )( ) ( )( ) ( )( )

( )( )

= + + −

+ − −

=

1 1 1 12 580 580 1498.33 19 19 633.33

2 2 2 1

1498.33 633.33 78.16 31 2

37,604.2 veh-min

tD

Figure 5.8 D/D/1 queuing diagram for Example 5.15. The average delay per vehicle is 9.95 min (37,604.2/3777.5). The longest wait of any vehicle (the longest horizontal distance between the arrival and departure curves), assuming a FIFO queuing discipline, will be the delay time of the 633.33rd vehicle to arrive. This vehicle will arrive 13.1 minutes (633.33/48.33) after 8:00 A.M. and will depart at 8:31 A.M., being delayed a total of 17.9 min.

184 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

5.7 IMPACT OF AUTONOMOUS VEHICLES

While there has been tremendous progress in recent years with respect to the technology that allows a vehicle to drive autonomously, understanding the impact that such vehicles will have on traffic operations and speed-flow relationships is not well understood.

The traffic stream models of Section 5.3 are theoretical realizations based on empirical observations of traffic flow. Consequently, these models of traffic flow are founded on human driving behavior. This driving behavior, as well as the automobile, has changed significantly since the relationships of Section 5.3 were first proposed. Consequently, traffic stream models based on current observations of traffic flow, while similar to the models of Section 5.3, do consist of some significant differences, as will be shown in Chapter 6.

With the notion of a traffic stream comprised entirely of computer-controlled vehicles, the possibility exists to essentially define the traffic stream relationships as desired to achieve a much more efficient use of existing roadway space.

EXAMPLE 5.16 FREEWAY TRAFFIC FLOW WITH AUTONOMOUS VEHICLES

A new section of freeway is to be utilized exclusively by autonomous vehicles. The freeway control system will operate the traffic stream at a capacity of 3600 vehicles per hour per lane with a vehicle speed of 70 mi/h. For this capacity flow rate and speed, what is the corresponding inter-vehicle headway, spacing, and the density of the traffic stream? Additionally, considering Fig. 5.4, sketch a set of speed–flow–density relationships that could correspond to this autonomous-vehicle traffic stream.

SOLUTION

Applying Eq. 5.4, rearranged to solve for h and substituting a unit conversion factor into the numerator to convert the headway from units of h/veh to units of s/veh, the average headway per lane is

= =3600 s/h 1.0 s/veh

3600 veh/h h

Multiplying speed (with a unit conversion for mi/h to ft/s) by headway gives the inter- vehicle spacing as

s = (70 × 5280/3600) × 1.0 = 102.67 ft/veh

Applying Eq. 5.13, rearranged to solve for k and substituting a unit conversion factor into the numerator to convert the density from units of veh/ft to units of veh/mi, the density per lane is

= =5280 ft/mi 51.4 veh/mi

102.67 ft/veh k

Note that applying Eq. 5.14 provides a check on our previous calculations

q = uk

= 70 × 51.4 3600 veh/h

5.7 Impact of Autonomous Vehicles 185

In considering the speed–flow–density relationships for this section of freeway, it needs to be recognized that the autonomous vehicle stream will be controlled to ensure that there will be no breakdown (i.e., the point at which traffic flow transitions from uncongested to congested conditions). This breakdown point is represented by the apex of the parabolic speed–flow and flow–density relationships shown in Fig. 5.4, and is regarded as the capacity flow rate value. Furthermore, since vehicle speeds and spacing will be precisely regulated, the parabolic shape of the speed–flow and flow–density relationships of Figure 5.4 will be avoided. Currently, the parabolic shapes of the speed–flow and flow–density relationships reflect the phenomenon of drivers reducing their speed as more vehicles use the roadway, and reducing their inter-vehicle spacing, as reflected in the speed–density relationship of Fig. 5.4. With autonomous vehicles, a potential realization of the speed– flow–density relationships is shown in Fig. 5.9. Note that in these relationships, there is no jam density (kj) and speed at capacity and free-flow speed are the same.

Figure 5.9. Flow–density, speed–density, and speed–flow relationships for autonomous vehicle stream.

Due to uncertainties relating to the technologies involved, it is currently impossible to project the traffic-stream parameter values that will ultimately be realized from autonomous vehicle traffic streams. However, there is clearly great potential for such technologies to lead to significant improvements in traffic operations efficiency. Still, it must also be pointed out that other considerations may ultimately place limits on the traffic stream operational improvements realized from autonomous vehicles. For example, fuel consumption and emissions output considerations could result in limits on speed, human tolerances for deceleration and the physics of vehicle braking could result in limits to speed and/or vehicle spacing, and so on.

S p

ee d

( m

i/h )

Flow Rate (veh/h/ln) qcap = 3600

F lo

w R

at e

(v eh

/h /ln

)

Density (veh/mi/ln)

S p

ee d

( m

i/h )

Density (veh/mi/ln) kcap = 51.4

uf = 70

qcap = 3600

186 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

5.8 PRACTICE PROBLEMS

PRACTICE PROBLEM 5.1

CAPACITY AND SPEED–FLOW RELATIONSHIPS

A section of highway has a speed–flow relationship of the form q = au2 + bu, where q is the traffic flow in veh/h, u is the space-mean speed of traffic in mi/h, and a and b are unknown parameters. It is known that at capacity (which is 3100 veh/h) the space- mean speed of traffic is 28 mi/h. Determine the speed when the flow is 1500 veh/h and the free-flow speed.

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

This problem seeks to improve readers’ understanding of the speed–flow relationship (which Chapter 6 will show is critical in determining highway mobility performance) by demonstrating elements of the functional form of the relationship. In looking at the problem statement, to find the free-flow speed and the speed when the flow is 1500 veh/h we need to first find the unknown parameters a and b. With flow and speed at capacity known, we can set the first derivative of q = au2 + bu equal to zero to arrive at capacity values,

2 +

0

so .

cap cap cap

cap

cap

q au bu

dq a b

du

b a

=

= × × + =

= − × ×

With qcap = 3100 veh/h and ucap = 28 mi/h (both given), substituting these values and the expression for b obtained from the first derivative (above) back into q = au2 + bu gives,

( )= × − × × ×

= −

2 +

3.954.

a

a

With this, b can be calculated,

= − × − × =

3.954

221.424.

b

With these values of a and b, the speed with a flow of 1500 veh/h is,

2

2

+

+ .

q au bu

u u

=

= × ×

Note that the equation above, for a given flow, will give two feasible values for speed, one in uncongested flow conditions (the higher speed) and one in congested flow conditions (the lower speed). This is illustrated by the general speed–flow relationship shown in Fig. 5.3. Solving the above equation for speed gives u = 48.12 mi/h in uncongested conditions and u = 7.88 mi/h in congested conditions.

Finally, the free-flow speed is obtained by substituting q = 0 into the speed–flow relationship above, giving u = 56 mi/h.

5.8 Practice Problems 187

PRACTICE PROBLEM 5.2

POISSON ARRIVAL PATTERNS AND VEHICLE TIME HEADWAYS

A vehicle pulls out onto a single-lane highway that has a flow rate of 300 veh/h (Poisson distributed). The driver of the vehicle does not look for oncoming traffic. Road conditions and vehicle speeds on the highway are such that it takes 1.7 seconds for an oncoming vehicle to stop once the brakes are applied. Assuming that a standard driver reaction time of 2.5 seconds (see Chapter 3) and ignoring the length of the vehicles (which will have just a small influence on the results), what is the probability that the vehicle pulling out will get in an accident with oncoming traffic?

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

This problem shows how a Poisson process can be used to arrive at time spacings between vehicles. With 1.7 seconds required for a stop and 2.5 seconds required for driver reaction time (for a total of 4.2 seconds), an accident will occur when the vehicle headway is less than 4.2 seconds. Using Eq. 5.26 with t = 4.2 s and q = 300 veh/h we have,

( ) ( )

3600

( )( ) 3600 .

qtP h t e

P h e

−

−

≥ =

≥ = =

With this, the probability that the time-headway will be less than 4.2 seconds, and thus the probability of an accident, is,

( ) ( )4.2 1

1 = 0.295.

P h P h< = − ≥

= −

PRACTICE PROBLEM 5.3

D/D/1 QUEUING WITH TIME-INVARIANT ARRIVALS AND DEPARTURES Vehicles begin arriving at a closed entrance to a parking lot at a rate of six vehicles per minute for the first 15 minutes and then continue to arrive at two vehicles per minute for times after this initial 15-minute period. The parking-lot entrance opens when the vehicle queue reaches 30 vehicles and the parking-lot attendant services one vehicle every 15 seconds. Assuming D/D/1 queuing, what is the total delay?

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

To begin, a few key points for the points for the D/D/1 queuing diagram need to be determined. First, the number of minutes into the process when parking-lot entrance opens can be calculated as,

number of queued vehicles when open 5 min.

vehicle arrival rate opent = = =

Next, at 15 min into the process the arrival rate changes from 6 veh/min to 2 veh/min. So after the first 15 min total vehicle arrivals are,

arrivals15 = 15 × = 90 veh.

188 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

It is known that the parking-lot attendant services vehicles at a rate of one every 15 sec or, equivalently, 4 veh/min starting 15 min into the process (when parking service begins). So the number of departures after 15 min will be,

departures15 = 10 × = veh, which means that the queue at 15 min into the process is,

queue15 = arrivals15 departures15 = = veh. So the equation for the time after 15 min to queue clearance (t>15) can be determined from,

queue15 + (arrival rate after 15 min)(t>15) = (departure rate after 15 min)(t>15) or,

( ) ( ) 15

15 queue

departure rate after 15 min arrival rate after 15 min

25 min.

t> = −

= = −

So the queue will clear 15 min + 25 min or 40 min after the start of the process. For the total delay, we need the area between arrival and departure curves. At the 15th minute, 90 veh have arrived so the area under the arrival curve is a triangle with a height of 90 veh and a base of 15 min. Also at the 15th minute, 40 veh will have departed so the area under the departure curve is a triangle with a height of 40 veh and but the base is now 10 min (15 5) since departures do not start until 5 min after the first vehicle arrival. The subtraction of these two triangles will give the area between arrival and departure curves up to 15 min.

For the area between arrival and departure curves after 15 min, note that the difference between total vehicle arrivals and departures at 15 min is 50 veh (90 veh 40 veh) which forms the base of a triangular area, and the height of this triangle is the time between 15 min and the 40 min clearance time or t>15 = 25 min.

With these areas, the total delay Dt is,

2 2 2

1,100 veh-min.

tD      × × ×= − +                

=

PRACTICE PROBLEM 5.4

D/D/1 QUEUING WITH TIME-VARYING ARRIVAL AND DEPARTURE RATES The gate entrance to a park opens at 9:00 A.M. At 9:00 A.M. there are 32 vehicles in the queue waiting to enter. Vehicles continue to arrive (from 9:00 A.M. onward) at a rate of λ(t) = 4.2 – 0.05t (with λ(t) in veh/min and t in minutes after 9:00 A.M.). The gate attendant processes vehicles at a rate of μ(t) = 3 + 0.3t [with μ(t) in veh/min and t in minutes after 9:00 A.M.]. Assuming D/D/1 queuing, what is the maximum queue length and total vehicle delay from 9:00 A.M. onward?

5.8 Practice Problems 189

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

To begin, we need to develop equations for arrivals and departures from the equations we are given for arrival and departure rates. If we integrate the arrival and departure rate equations with respect to time, we will have equations for arrivals and departures (in units of vehicles) at any time t (which will be the number of minutes after 9:00 A.M.). To do this for the arrivals we have,

0

= 4.2 0.05 , t

t dt− = × − ×

and for departures we have,

0

= 3 0.3 . t

t dt+ = × + ×

So the queue length at any time Q(t), at any time t after 9:00 A.M., is going to be the arrivals minus the departures from the above functions plus the 32 vehicles waiting at 9:00 A.M.,

( )    = × − × − × + × +       

= − + +

2

32

0.175 1.2 32

Q t

t t

Solving for the time at which the maximum queue length occurs by taking the first derivative of the above equation with respect to time yields

( ) 0

3.43 min.

dQ t t

dt t

= − × + =

=

Substituting with t = 3.43 minutes gives the maximum queue length as

( ) 2

2

0.175 1.2 32

0.175 1.2 32 34.06 veh.

Q t t t

= − + +

= − × + × + =

Next, to determine total delay, we must first determine when the queue will clear. This will happen when the total number of arrivals equals the total number of departures. Equating the arrival and departure equations from above (with the 32 vehicles initially in the queue added to the arrivals), and solving for time t gives,

   + × − × = × + ×       

32

,

which gives 20 32 t t = − × + × + , or t = 17.38 min (time until queue clearance).

With this, the total delay Dt is going to be the area under the arrival curve minus the area under the departure curve (plus the delay of the initial 32 vehicles), so

190 Chapter 5 Fundamentals of Traffic Flow and Queuing Theory

0 0

17.38 2 3 2 3

0

32

32

431.17 veh-min.

|

t t

tD t dt dt

t t t t t

= × + × − × − × + ×

= + × − × − × − ×

=

 

NOMENCLATURE FOR CHAPTER 5

D deterministic arrivals or departures Dt total vehicle delay h vehicle time headway k traffic density kj traffic jam density kcap traffic density at capacity l roadway length M exponentially distributed arrivals or

departures n number of vehicles nc departure channel number N total number of departure channels q traffic flow qcap traffic flow at capacity (maximum traffic

flow) Q length of queue Q average length of queue

Qmax maximum length of queue s vehicle spacing t time t average time spent in the system u space-mean speed (also denoted su ) ui spot speed of vehicle i uf free-flow speed ucap speed at capacity

su space-mean speed (also denoted simply as u)

tu time-mean speed w average time waiting in the queue λ arrival rate μ departure rate ρ traffic intensity

REFERENCES

Drew, D. R. “Deterministic Aspects of Freeway Operations and Control.” Highway Research Record, 99, 1965.

Lord, D., and F. Mannering. “The Statistical Analysis of Crash-Frequency Data: A Review and Assessment of Methodological Alternatives,” Transportation Research Part A, vol. 44, no. 5, 2010.

Pipes, L. A. “Car Following Models and the Fundamental Diagram of Road Traffic.” Transportation Research, vol. 1, no. 1, 1967.

Poch, M., and F. Mannering. “Negative Binomial Analysis of Intersection-Accident Frequencies,” Journal of Transportation Engineering, vol. 122, no. 2, March/April 1996.

Transportation Research Board. Traffic Flow Theory: A Monograph. Special Report 165. Washington, DC: National Research Council, 1975.

Transportation Research Board. Highway Capacity Manual. Washington, DC: National Research Council, 2010.

P-20

CHAPTER 6 PROBLEMS

Freeways (Section 6.4)

6.1 A six-lane freeway (three lanes in each direction) currently operates at maximum LOS C conditions. The lanes are 11 ft wide, the right-side shoulder is 4 ft wide, and there are two ramps within three miles upstream of the segment midpoint and one ramp within three miles downstream of the segment midpoint. The highway is on rolling terrain with 10% heavy vehicles, and the peak-hour factor is 0.90. Determine the hourly volume for these conditions.

6.2 Consider the freeway in Problem 6.1. At one point along this freeway there is a 3.5% upgrade with a directional hourly traffic volume of 5435 vehicles. The heavy vehicle split is 50% single-unit trucks/50% tractor-trailer trucks. If all other conditions are as described in Problem 6.1, how long can this grade be without the freeway LOS dropping to F?

6.3 A four-lane freeway (two lanes in each direction) is located on rolling terrain and has 12-ft lanes, no lateral obstructions within 6 ft of the pavement edges, and there are two ramps within three miles upstream of the segment midpoint and three ramps within three miles downstream of the segment midpoint. A weekday directional peak-hour volume of 1800 vehicles (familiar users) is observed, with 700 arriving in the most congested 15-min period. If a LOS no worse than C is desired, determine the maximum number of heavy vehicles that can be present in the peak-hour traffic stream.

6.4 Consider the freeway and traffic conditions described in Problem 6.3. If 180 of the 1800 vehicles observed in the peak hour were heavy vehicles (assume a 70%/30% SUT/TT split), what would the LOS of this freeway be on a 5-mi, 6% downgrade?

6.5 A six-lane freeway (three lanes in each direction) in a scenic area has a measured FFS of 55 mi/h. There are 7% SUTs and 7% TTs in the traffic stream. One upgrade is 5% and 0.5 mi long. An analyst has determined that the freeway is operating at capacity on this upgrade during the peak hour. If the peak-hour traffic volume is 3900 vehicles, what value for the peak-hour factor was used?

6.6 A freeway is being designed for a location in rolling terrain. The expected FFS is 55 mi/h. During the peak hour, it is expected that there will be a

directional peak-hour volume of 2700 vehicles and 18% heavy vehicles. The PHF is expected to be 0.88. If a LOS no worse than D is desired, determine the necessary number of lanes.

6.7 A segment of four-lane freeway (two lanes in each direction) has a 3% upgrade that is 1500 ft long followed by a 1000-ft 4% upgrade. It has 12-ft lanes and 3-ft shoulders. The directional hourly traffic flow is 2,000 vehicles with 3% SUTs and 3% TTs. The total ramp density for this freeway segment is 2.33 ramps per mile. If the peak-hour factor is 0.90, what is the LOS of this compound-grade freeway segment?

6.8 Consider Example 6.2, in which it was determined that 1537 vehicles could be added to the peak hour before capacity is reached. Assuming level terrain as in Example 6.1, how many passenger cars could be added to the original traffic mix before peak-hour capacity is reached? (Assume only passenger cars are added and that the number of heavy vehicles originally in the traffic stream remains constant.)

6.9 An eight-lane freeway (four lanes in each direction) is on rolling terrain and has 11-ft lanes with a 4-ft right-side shoulder. The total ramp density is 1.5 ramps per mile. The directional peak- hour traffic volume is 5400 vehicles with 11% heavy vehicles. The peak-hour factor is 0.95. It has been decided that heavy vehicles will be banned from the freeway during the peak hour. What will the freeway’s density and LOS be before and after the ban? (Assume that the heavy vehicles are removed and all other traffic attributes are unchanged.)

6.10 A 5% upgrade on a six-lane freeway (three lanes in each direction) is 1.25 mi long. On this segment of freeway, there is 3% SUTs and 7% TTs, and the peak-hour factor is 0.90. The lanes are 12 ft wide, there are no lateral obstructions within 6 ft of the roadway, and the total ramp density is 1.0 ramps per mile. What is the maximum directional peak-hour volume that can be accommodated without exceeding LOS C operating conditions?

6.11 A northbound two-lane (one direction) segment of freeway currently has a measured density of 32.0 pc/mi/ln and a measured FFS of 65 mi/h. The peak-hour factor is 0.95, the traffic stream contains 7% single-unit trucks and 3% tractor-trailer trucks. The freeway segment is on a 3.5% upgrade

Chapter 6 Problems P-21

that is ¾ of a mile long. With the peak-hour factor unchanged, how many trucks (assuming the same ST/TT split) can be added to the peak hour before capacity is reached?

6.12 A northbound freeway segment is on a 4% upgrade from station 430+20 to 450+00 and has two 11-ft wide lanes, a 5-ft right shoulder, and has a ramp density of 1 per mile in the 3 miles before and after station 440+10. The peak-hour factor is 0.9. Northbound traffic during the peak hour is 2550 cars, 300 STs, and 300 TTs. Determine the density and LOS of the freeway segment.

6.13 A westbound freeway segment has four 12-ft wide lanes (one direction), a 4-ft right shoulder, and 4 ramps in the 3 miles before and after the midpoint of the analysis segment. The peak-hour factor is 0.85. Westbound traffic during the peak hour is 4500 passenger cars, 350 single-unit trucks, and 150 tractor-trailer trucks. The freeway segment is on a 3.5% upgrade. What is the maximum length of this grade if no worse than LOS C must be maintained?

6.14 A freeway segment currently has four 11-ft wide lanes and a 3-ft right shoulder in the southbound direction. It is operating in LOS B with a traffic density of 15.6 pc/mi/ln and a peak-hour factor of 0.85. The freeway is on level terrain with 15% heavy vehicles, and currently has no on- or off- ramps within a distance of three miles upstream and downstream of the mid-point of the analysis section. New interchanges are to be constructed that would result in 6 on- and off-ramps within the distance of three miles upstream and downstream of the mid- point of the analysis section and total traffic volume is projected to double, with the same percentage of heavy vehicles, and the peak-hour factor going to 0.95. What would the new LOS and density be after the interchange construction and new traffic conditions?

6.15 A westbound section of freeway currently has three 12-ft wide lanes, a 6-ft right shoulder, and no ramps within 3 miles upstream and downstream of the segment midpoint. It is on rolling terrain with 10% heavy vehicles and is operating at capacity with a peak-hour factor of 0.9. If the road is expanded to four 11-foot lanes with a 2-foot right shoulder, and traffic after the expansion is projected to increase by 10% with the same heavy vehicle percentage and peak-hour factor, what is the new LOS and estimated density?

6.16 A six-mile long freeway segment has 5 lanes in the northbound direction. The lanes are 10-ft wide and there is a 3-ft right shoulder. There are 8 on/off ramps in this section. What is the estimated FFS at the mid-point of this six-mile segment?

Multilane Highways (Section 6.5)

6.17 A multilane highway (two lanes in each direction) is on level terrain. The FFS has been measured at 45 mi/h. The peak-hour directional traffic flow is 1300 vehicles with 8% heavy vehicles. If the peak-hour factor is 0.85, determine the highway’s LOS.

6.18 Consider the multilane highway in Problem 6.17. If the proportion of vehicle types and peak- hour factor remain constant, how many vehicles can be added to the directional traffic flow before capacity is reached?

6.19 A six-lane multilane highway (three lanes in each direction) has a peak-hour factor of 0.90, 11-ft lanes with a 4-ft right-side shoulder, and a two-way left-turn lane in the median. The directional peak- hour traffic flow is 4000 vehicles with 14% SUTs and 6% TTs. What will the LOS of this highway be on a 4% upgrade that is 1.5 miles long if the speed limit is 55 mi/h and there are 15 access points per mile?

6.20 A divided multilane highway in a recreational area has four lanes (two lanes in each direction) and is on rolling terrain. The highway has 10-ft lanes with a 6-ft right-side shoulder and a 3-ft left-side shoulder. The posted speed is 50 mi/h. Previously, there were four access points per mile, but recent development has increased the number of access points to 12 per mile. Before development, the peak- hour factor was 0.95, and the directional hourly volume was 2200 vehicles with 13% heavy vehicles. After development, the peak-hour directional flow is 2600 vehicles with the same vehicle percentages and peak-hour factor. What is the LOS before and after the development?

6.21 A multilane highway has four lanes (two lanes in each direction) and a measured FFS of 55 mi/h. The directional peak-hour volume is 1900 vehicles (the peak-hour factor is 0.80). One upgrade is 5% and is 0.62 mi long. Currently, heavy vehicles are not permitted on the highway, but local authorities are considering allowing heavy vehicles on this upgrade. If this is done, they estimate that 150 heavy vehicles will use the highway during the peak

P-22 Chapter 6 Problems

hour. What would be the LOS before and after the heavy vehicles are allowed on the upgrade (assuming 50% SUTs and 50% TTs.)?

6.22 A four-lane undivided multilane highway (two lanes in each direction) has 11-ft lanes, 4-ft shoulders, and 10 access points per mile. It is determined that the roadway currently operates at capacity with PHF = 0.80. If the highway is on level terrain with 8% heavy vehicles and the speed limit is 55 mi/h, what is the directional hourly volume?

6.23 A new four-lane divided multilane highway (two lanes in each direction) is being planned with 12-ft lanes, 6-ft shoulders on both sides, and a 50- mi/h speed limit. One 3% downgrade is 4.5 mi long, and there will be four access points per mile. The peak-hour directional volume along this grade is estimated to consist of 1800 passenger cars, 140 SUTs, and 60 TTs. If the peak-hour factor is estimated to be 0.85, what LOS will this segment of highway operate under?

6.24 A six-lane divided multilane highway (three lanes in each direction) has a measured FFS of 50 mi/h. It is on rolling terrain with a traffic stream consisting of 6% heavy vehicles. One direction of the highway currently operates at maximum LOS C conditions, and it is known that the highway has PHF = 0.90. How many vehicles can be added to this highway before capacity is reached, assuming the proportion of vehicle types remains the same but the peak-hour factor increases to 0.95?

6.25 A four-lane undivided multilane highway (two lanes in each direction) has 11-ft lanes and 5-ft shoulders. At one point along the highway, there is a 4% upgrade that is 0.62 mi long. There are 15 access points along this grade. The peak-hour traffic volume is 2340 vehicles, with 10% single-unit trucks and 10% tractor-trailer trucks, and 620 of these vehicles arrive in the most congested 15-min period. The posted speed limit is 60 mi/h. To improve the LOS, the local transportation agency is considering reducing the number of access points by blocking some driveways and rerouting their traffic. How many of the 15 access points must be blocked to achieve LOS C?

6.26 A multilane highway, with three lanes in each direction, operates at maximum level of service D conditions during the peak hour with an observed speed of 48.0 mi/h (with 3690 veh/h observed). The highway is on rolling terrain and the traffic stream

consists of cars and some percentage of heavy vehicles. The peak-hour factor is 0.85. If heavy vehicles were banned from the highway and car traffic remained the same, what would be the LOS?

6.27 An undivided multilane highway segment has two 11-ft lanes in the eastbound direction with no shoulders and a 55 mi/h speed limit. This highway segment has 17 access points on a 0.625 mile, 4% upgrade. During the highest 15 minutes of traffic flow within the peak hour, there are 720 cars, 56 single-unit trucks, and 24 tractor-trailer trucks. What are the estimated speed, density, and LOS of this upgrade?

6.28 A multilane undivided highway is on level terrain and has two lanes in the northbound direction. The posted speed limit is 55 mi/h, and there are 11-ft lanes, 3-ft right shoulders, and 8 access points per mile. The highway operates at capacity during the peak hour with 20% heavy vehicles and a peak-hour factor of 0.925. What is the peak-hour demand volume?

6.29 There are two 11-ft lanes in a northbound direction of an undivided multilane highway. The highway has a 50-mi/h speed limit with a 4-ft shoulder. During the peak hour, there are 2,475 total vehicles, with 8% heavy vehicles (even split between STs and TTs). The peak-hour factor is 0.8. One segment of this highway has a +2% grade that is ¾ of a mile long with 9 access points. What is the density and LOS of this highway segment?

Design Traffic Volumes (Section 6.7)

6.30 A four-lane freeway (two lanes in each direction) segment consists of passenger cars only, a peak-hour directional distribution of 0.70, a peak- hour factor of 0.80, and a measured FFS of 70 mi/h. Assuming Fig. 6.8 applies, if the AADT is 30,000 veh/day, determine the LOS for the 10th, 50th, and 100th highest annual hourly volumes.

6.31 A four-lane freeway (two lanes in each direction) operates at capacity during the peak hour. It has 11-ft lanes, 4-ft shoulders, and there are three ramps within three miles upstream of the segment midpoint and four ramps within three miles downstream of the segment midpoint. The freeway is on rolling terrain and has 8% heavy vehicles with a peak-hour factor of 0.85. It is known that 12% of the AADT occurs in the peak hour and that the directional factor is 0.6. What is the freeway’s AADT?

Chapter 6 Problems P-23

6.32 A six-lane multilane highway (three lanes in each direction) currently operates at maximum LOS C conditions. The measured FFS is 55 mi/h. The highway is on rolling terrain with 14% SUTs and 6% TTs, and the peak-hour factor is 0.92. If 17% of all directional traffic occurs during the peak hour, determine the total directional traffic volume.

Multiple Choice Problems (Multiple Sections)

6.33 A six-lane freeway (three lanes in each direction) in rolling terrain has 10-ft lanes and obstructions 4 ft from the right edge of the traveled pavement. There are five ramps within three miles upstream of the segment midpoint and four ramps within three miles downstream of the segment midpoint. A directional peak-hour volume of 2000 vehicles is observed, with 600 vehicles arriving in the highest 15-min flow rate period. The traffic stream contains 18% heavy vehicles. What is the density of the traffic stream?

a) 8.49 pc/mi/ln b) 17.15 pc/mi/ln c) 14.28 pc/mi/ln d) 14.88 pc/mi/ln

6.34 A six-lane freeway (three lanes in each direction) in level terrain has 10-ft lanes and obstructions 5 ft from the right edge. There are zero ramps within three miles upstream of the segment midpoint and one ramp within three miles downstream of the segment midpoint. The traffic stream has a peak-hour factor of 0.84, peak-hour volume of 2500 vehicles, and 4% heavy vehicles. What is the LOS?

a) LOS A b) LOS B c) LOS C d) LOS D

6.35 A four-lane divided multilane highway (two lanes in each direction) in rolling terrain has five access points per mile and 11-ft lanes with a 4-ft shoulder on the right side and 2-ft shoulder on the left. The peak-hour factor is 0.84 with 13% heavy vehicles. If the analysis flow rate is 1250 pc/h/ln, what is the peak-hour volume?

a) 1667 veh/h b) 1859 veh/h c) 1984 veh/h d) 3335 veh/h

6.36 A six-lane undivided multilane highway (three lanes in each direction) has 12-ft lanes with 2-ft shoulders on the right side. There are two access points per mile and the posted speed limit is 50 mi/h. Estimate the free-flow speed (to the nearest 1 mi/h).

a) 50 mi/h b) 43 mi/h c) 48 mi/h d) 52 mi/h

6.37 You are designing a freeway as a passenger-car- only facility, and with ideal roadway characteristics. It is estimated that the freeway will have a traffic demand of 75,000 vehicles per day, a peak-hour factor of 0.88, and a directional distribution of 0.65. Determine the number of lanes (both directions) required to provide at least LOS D using the 60th highest annual hourly volume (see Fig. 6.8).

a) 4 lanes b) 6 lanes c) 8 lanes d) 10 lanes

191

Chapter 6

Highway Capacity and Level-of-Service Analysis

6.1 INTRODUCTION

The underlying objective of traffic analysis is to quantify a roadway’s performance with regard to specified traffic volumes. This performance can be measured in terms of travel delay (as the roadway becomes increasingly congested) as well as other factors. The comparative performance of various roadway segments (which is determined from an analysis of traffic) is important because it can be used as a basis to allocate limited roadway construction and improvement funds. The purpose of this chapter is to apply the elements of uninterrupted traffic flow theory covered in Chapter 5 to the practical field analysis of traffic flow and capacity on freeways, multilane highways, and two-lane highways.

The main challenge of such a process is to adapt the theoretical formulations to the wide range of conditions that occur in the field. These diverse field conditions must be taken into account in a traffic analysis methodology, yet the methodology must remain theoretically consistent. For example, in Chapter 5, capacity (qcap) is simply defined as the highest traffic flow rate that the roadway is capable of supporting. For applied traffic analysis, a consistent and reasonably precise method of determining capacity must be developed within this definition. Because it can readily be shown that the capacity of a roadway segment is a function of factors such as roadway type (freeway, multilane highway, or two-lane highway), free-flow speed (FFS), number of lanes, and widths of lanes and shoulders, the method of capacity determination clearly must account for a wide variety of physical and operational roadway characteristics.

Additionally, recall that Chapter 5 defines traffic flow on the basis of units of vehicles per hour. Two practical issues arise concerning this unit of measure. First, in many cases, vehicular traffic consists of a variety of vehicle types with substantially different performance characteristics. These performance differentials are likely to be magnified by changing roadway geometrics, such as upgrades or downgrades, which have a differential effect on the acceleration and deceleration capabilities of the various types of vehicles; for example, grades have a greater impact on the performance of large trucks (heavy vehicles) than on automobiles. As a result, traffic must be defined not only in terms of vehicles per unit time but also in terms of vehicle composition, because it is clear that a 1500-veh/h traffic flow consisting of 100% automobiles will differ significantly

192 Chapter 6 Highway Capacity and Level-of-Service Analysis

with regard to operating speed and traffic density from a 1500-veh/h traffic flow that consists of 50% automobiles and 50% heavy vehicles.

The other flow-related concern is the temporal distribution of traffic. In practice, the analysis of roadway traffic usually focuses on the most critical condition, which is the most congested hour within a 24-hour daily period (the temporal distribution of traffic is discussed in more detail in Section 6.7). However, within this most congested peak hour, traffic flow is likely to be nonuniform. It is therefore necessary to arrive at some method of defining and measuring the nonuniformity of flow within the peak hour.

To summarize, the objective of applied traffic analysis is to provide a practical method of quantifying the degree of traffic congestion and to relate this to the overall traffic-related performance of the roadway. The following sections of this chapter discuss and demonstrate accepted standards for applied traffic analysis for the three major types of uninterrupted-flow roadways: freeways, multilane highways, and two-lane highways (one lane in each direction).

6.2 LEVEL-OF-SERVICE CONCEPT

The Highway Capacity Manual (HCM), produced by the Transportation Research Board [2016], is a synthesis of the state of the art in methodologies for quantifying traffic operational performance and capacity utilization (congestion level) for a variety of transportation facilities. One of the foundations of the HCM is the concept of level of service (LOS). The LOS represents a qualitative ranking of the traffic operational conditions experienced by users of a facility under specified roadway, traffic, and traffic control (if present) conditions. Current practice designates six levels of service ranging from A to F, with LOS A representing the best operating conditions and LOS F the worst.

A number of operational performance measures, such as speed, flow, and density, can be measured or calculated for any transportation facility. To apply the LOS concept to traffic analysis, it is necessary to select a performance measure that is representative of how motorists actually perceive the quality of service they are receiving on a facility. Motorists tend to evaluate their received quality of service in terms of factors such as speed and travel time, freedom to maneuver, traffic interruptions, and comfort and convenience. Thus, it is important to select a measure that encompasses some or all of these factors. The performance measure that is selected for LOS analysis for a particular transportation facility is referred to as the service measure.

The HCM [Transportation Research Board 2016] defines the LOS categories for freeways and multilane highways as follows:

Level of service A. LOS A represents free-flow conditions (traffic operating at FFS, as defined in Chapter 5). Individual users are virtually unaffected by the presence of others in the traffic stream. Freedom to select speeds and to maneuver within the traffic stream is extremely high. The general level of comfort and convenience provided to drivers is excellent.

6.2 Level-of-Service Concept 193

Level of service B. LOS B also allows speeds at or near FFS, but the presence of other users in the traffic stream begins to be noticeable. Freedom to select speeds is relatively unaffected, but there is a slight decline in the freedom to maneuver within the traffic stream relative to LOS A. Level of service C. LOS C has speeds at or near FFS, but the freedom to maneuver is noticeably restricted (lane changes require careful attention on the part of drivers). The general level of comfort and convenience declines significantly at this level. Disruptions in the traffic stream, such as an incident (for example, vehicular accident or disablement), can result in significant queue formation and vehicular delay. In contrast, the effects of incidents at LOS A or LOS B are minimal, with only minor delay in the immediate vicinity of the event. Level of service D. LOS D represents the conditions where speeds begin to decline slightly with increasing flow. The freedom to maneuver becomes more restricted, and drivers experience reductions in physical and psychological comfort. Incidents can generate lengthy queues because the higher density associated with this LOS provides little space to absorb disruptions in the traffic flow. Level of service E. LOS E represents operating conditions at or near the roadway’s capacity. Even minor disruptions to the traffic stream, such as vehicles entering from a ramp or vehicles changing lanes, can cause delays as other vehicles give way to allow such maneuvers. In general, maneuverability is extremely limited, and drivers experience considerable physical and psychological discomfort. Level of service F. LOS F describes a breakdown in vehicular flow. Queues form quickly behind points in the roadway where the arrival flow rate temporarily exceeds the departure rate, as determined by the roadway’s capacity (see Chapter 5). Such points occur at incidents and on- and off- ramps, where incoming traffic results in capacity being exceeded. Vehicles typically operate at low speeds under these conditions and are often required to come to a complete stop, usually in a cyclic fashion. The cyclic formation and dissipation of queues is a key characterization of LOS F. A visual perspective of the LOS definitions for freeways is provided in Fig. 6.1.

In dealing with LOS, it is important to remember that when the traffic volume is at or near the roadway capacity (which will be shown as a function of the prevailing traffic and physical characteristics of the roadway), the roadway is operating at LOS E. This, however, is not a desirable condition because under LOS E conditions there is considerable driver discomfort, which could increase the likelihood of vehicular crashes and overall delay. In roadway design, the possibility of degradation of LOS to LOS E should be avoided, although this is not always possible due to financial and environmental constraints that may limit the design speed, number of lanes, and other factors affecting roadway capacity.

194 Chapter 6 Highway Capacity and Level-of-Service Analysis

Figure 6.1 Illustration of freeway level of service (A to F).

LOS A LOS B

LOS C LOS D

LOS E LOS F

6.3 Level-of-Service Determination 195

6.3 LEVEL-OF-SERVICE DETERMINATION

There are several steps in a basic LOS determination for an uninterrupted-flow facility. The remainder of this section describes the general details of each step, as applicable to uninterrupted-flow facility analyses. Facility-specific details of these steps are described in the sections that follow.

6.3.1 Base Conditions and Capacity The determination of a roadway’s LOS begins with the specification of base roadway conditions. Recall that in the introduction to this chapter, the effects of vehicle performance and roadway design characteristics on traffic flow were discussed qualitatively. In practice, the effects of such factors on traffic flow are measured quantitatively, relative to the base traffic and roadway design conditions. For uninterrupted-flow roadways, base conditions can be categorized as those relating to roadway conditions, such as lane widths, lateral clearances, access frequency, and terrain; and traffic stream conditions such as the effects of heavy vehicles (large trucks, buses, and RVs). Base conditions are defined as those conditions that represent unrestrictive geometric and traffic conditions. Additionally, base conditions are assumed to consist of favorable environmental conditions (such as dry roadways).

The capacity of a particular roadway segment will be greatest when all roadway and traffic conditions meet or exceed their base values. Empirical studies have identified the values of these base conditions for which the capacity of a roadway segment is maximized. Values in excess of the base conditions will not increase the capacity of the roadway, but values more restrictive than the base conditions will result in a lower capacity. For example, studies have identified a base lane width of 12 ft. That is, lane widths in excess of 12 ft will not result in increased capacity; however, lane widths less than 12 ft will result in a reduction in capacity. Capacity values for base conditions have been determined for all uninterrupted-flow facility types from field studies. It should be noted that for purposes of LOS analysis, capacity is defined not as the absolute maximum flow rate ever observed for a particular facility type, but rather as the maximum flow rate that can be reasonably expected on a recurring basis.

Because all base conditions for a particular roadway type are seldom realized in practice, methods for converting the measured flow rate into an equivalent analysis flow rate in terms of passenger cars for the given traffic conditions and estimating the actual FFS for the given roadway conditions are needed. The following sections describe the procedures for arriving at flow and speed values for given roadway and traffic conditions.

6.3.2 Determine Free-Flow Speed FFS is a term that was introduced in Chapter 5 as the speed of traffic as the traffic density approaches zero. In practice, FFS is governed by roadway design characteristics (horizontal and vertical curves, lane and shoulder widths, and median design), the frequency of access points, the complexity of the driving environment (possible distractions from roadway signs and the like), and posted speed limits.

196 Chapter 6 Highway Capacity and Level-of-Service Analysis

The FFS must be determined given the characteristics of the roadway segment. FFS is the mean speed of traffic as measured when flow rates are low to moderate (specific values are given under the individual sections for each roadway type). Ideally, FFS should be measured directly in the field at the site of interest. However, if this is not possible or feasible, an alternative method can be employed to arrive at an estimate of FFS under the prevailing conditions. This method makes adjustments to a base FFS (BFFS) depending on the physical characteristics of the roadway segment, such as lane width, shoulder width, and access frequency. This method has the same basic structure for the various roadway types, but contains adjustment factors and values appropriate for each roadway type.

6.3.3 Determine Analysis Flow Rate One of the fundamental inputs to a traffic analysis is the actual traffic volume on the roadway, in vehicles per hour, which is given the symbol V. Generally, the highest volume in a 24-hour period (the peak-hour volume) is used for V in traffic analysis computations. However, this hourly volume needs to be adjusted to reflect the temporal variation of traffic demand within the analysis hour and the impacts due to heavy vehicles. To account for these effects, the hourly volume is divided by adjustment factors to obtain an equivalent flow rate in terms of passenger cars per hour (pc/h). Additionally, the flow rate is expressed on a per- lane basis (pc/h/ln) by dividing by the number of lanes in the analysis segment.

6.3.4 Calculate Service Measure(s) and Determine LOS Once the previous steps have been completed, all that remains is to calculate the value of the service measure and then determine the LOS from the service measure value. For freeways and multilane highways, this is a relatively straightforward task. However, for two-lane highways, there are actually two service measures, and the calculation of these and the subsequent LOS determination are more involved.

6.4 BASIC FREEWAY SEGMENTS

A basic freeway segment is defined as a section of a divided roadway having two or more lanes in each direction, full access control, and traffic that is unaffected by merging or diverging movements near ramps. It is important to note that capacity analysis for divided roadways focuses on the traffic flow in one direction only. This is reasonable because the objective is to measure the highest level of congestion. Due to directional imbalance of traffic flows—for example, morning rush hours having higher volumes going toward the central city and evening rush hours having higher volumes going away from the central city— consideration of traffic volumes in both directions is likely to seriously understate the true level of traffic congestion.

6.4.1 Speed versus Flow Rate Relationship Central to the analysis of basic freeway segments is the relationship between average traffic stream speed and flow rate. Chapter 5 introduced a theoretical speed–flow relationship. This relationship was parabolic in form, derived from

6.4 Basic Freeway Segments 197

an assumed linear speed–density relationship. Empirical studies of current traffic stream patterns show a speed–flow relationship, that while similar to the theoretical parabolic relationship, is significantly different in one respect. The parabolic model for speed-flow yields a continuous change in speed with changes in flow rate, whereas the empirical relationship shows that traffic stream speed stays constant up to moderate flow rates, and then starts to decline in a curvilinear manner. The HCM [Transportation Research Board 2016] defines the relationship between average speed and flow rate as follows:

S = FFS pv BP≤ (6.1)

and

( )

( )

2.0

2.0

45 p c

FFS v BP S FFS

c BP

 − −   = −

− < ≤pBP v c (6.2)

where

c = Capacity in pc/h/ln, vp = 15-min passenger car equivalent flow rate in pc/h/ln,

BP = Linear to curvilinear flow rate breakpoint value in pc/h/ln, and

45 = Density at capacity in pc/mi/ln, as shown in Table 6.2. Other terms are as defined previously.

Capacity is given by the following equation:

( )Min 2200 10 50 ,2400c FFS = + −  55 ≤ FFS ≤ 75 (6.3)

Results from this equation for 5-mi/h increments across the valid range of FFS is shown in Table 6.1.

Table 6.1 Relationship between Free-Flow Speed and Capacity on Basic Freeway Segments Source: Transportation Research Board, Highway Capacity Manual, 6th Edition, National Academy of Sciences, Washington, D.C., 2016.

Free-flow speed (mi/h)

Capacity (pc/h/ln)

75 2400

70 2400

65 2350

60 2300

55 2250

The breakpoint, BP, value is given by the following equation:

( )1000 40 75BP FFS = + −  55 ≤ FFS ≤ 75 (6.4)

The speed-flow rate curves resulting from the application of Eqs. 6.1 and 6.2, for 5-mi/h increments across the valid range of FFS are shown in Fig. 6.2.

Table 6.2 provides the LOS criteria corresponding to traffic density, speed, volume-to-capacity ratio, and maximum flow rate. The maximum service flow

198 Chapter 6 Highway Capacity and Level-of-Service Analysis

rate is simply the maximum flow rate, under base conditions, that can be sustained for a given LOS. This value is based on the speed–flow relationship and the maximum density values given in Table 6.2.

It should be noted that the versions of Eqs. 6.1 and 6.2 in the HCM use the variables FFSadj and cadj instead of FFS and c. This is to accommodate additional factors that might reduce the FFS and/or capacity values from those discussed in this chapter. Examples of these types of factors include temporary work zones, adverse weather, incidents, and drivers unfamiliar with the roadway. In these instances, the FFS and c values as calculated from the factors discussed in this chapter would be further adjusted. Discussion of these topics is beyond the scope of this book; therefore, the FFS and c variables are used in lieu of the FFSadj and cadj variables in Eqs. 6.1 and 6.2.

6.4.2 Base Conditions and Capacity The base conditions for a basic freeway segment are defined as [Transportation Research Board 2016]

• 12-ft minimum lane widths • 6-ft minimum right-shoulder clearance between the edge of the travel

lane and objects (utility poles, retaining walls, etc.) that influence driver behavior

• Only passenger cars in the traffic stream • 6-mi or greater interchange ramp spacing • Level terrain (no grades greater than 2%)

These conditions represent a high operating level, with a FFS of 75.4 mi/h or higher.

The capacity, c, for basic freeway segments, in passenger cars per hour per lane (pc/h/ln), is given in Tables 6.1 and 6.2. From Table 6.2, note that, by definition, the upper boundary of LOS E corresponds to the value of capacity and a v/c of 1.0. Other values of v/c for a specific LOS are obtained by simply dividing the maximum flow rate for that LOS by capacity (the maximum flow rate at LOS E).

6.4.3 Service Measure The service measure for basic freeway segments is density. Density, as discussed in Chapter 5, is typically measured in terms of passenger cars per mile per lane (pc/mi/ln) and therefore provides a good measure of the relative mobility of individual vehicles in the traffic stream. A low traffic stream density gives individual vehicles the ability to change lanes and speeds with relative ease, while a high density makes it very difficult for individual vehicles to maneuver within the traffic stream. Thus, traffic density is the primary determinant of freeway LOS.

Recall Eq. 5.14 from Chapter 5:

q = uk (6.5)

6.4 Basic Freeway Segments 199

where q = flow in veh/h, u = speed in mi/h, and k = density in veh/mi.

Density is therefore calculated as flow divided by speed. The following sections will describe how to arrive at flow and speed values for the given roadway and traffic conditions. Once the flow and speed values have been determined according to the given conditions, a density can be calculated and then referenced in Table 6.2 or Fig. 6.2 to arrive at a LOS for the freeway segment. Table 6.2 LOS Criteria for Basic Freeway Segments

Criterion LOS

A B C D E

FFS = 75 mi/h Maximum density (pc/mi/ln) 11 18 26 35 45 Average speed (mi/h) 75.0 73.9 68.3 60.9 53.3 Maximum v/c 0.34 0.55 0.74 0.89 1.00 Maximum flow rate (pc/h/ln) 825 1330 1775 2130 2400

FFS = 70 mi/h Maximum density (pc/mi/ln) 11 18 26 35 45 Average speed (mi/h) 70.0 70.0 66.7 60.4 53.3 Maximum v/c 0.32 0.53 0.72 0.88 1.00 Maximum flow rate (pc/h/ln) 770 1260 1735 2115 2400

FFS = 65 mi/h Maximum density (pc/mi/ln) 11 18 26 35 45 Average speed (mi/h) 65.0 65.0 64.0 58.9 52.2 Maximum v/c 0.30 0.50 0.71 0.88 1.00 Maximum flow rate (pc/h/ln) 715 1170 1665 2060 2350

FFS = 60 mi/h Maximum density (pc/mi/ln) 11 18 26 35 45 Average speed (mi/h) 60.0 60.0 60.0 57.1 51.1 Maximum v/c 0.29 0.47 0.68 0.87 1.00 Maximum flow rate (pc/h/ln) 660 1080 1560 2000 2300

FFS = 55 mi/h Maximum density (pc/mi/ln) 11 18 26 35 45 Average speed (mi/h) 55.0 55.0 55.0 54.7 50.0 Maximum v/c 0.27 0.44 0.64 0.85 1.00 Maximum flow rate (pc/h/ln) 605 990 1430 1915 2250

Note: Maximum flow rate values are rounded to the nearest 5 passenger cars.

200

Figure 6.2 Basic freeway segment speed-flow curves and level-of-service criteria. (pickup from current edition) Reproduced with permission of the Transportation Research Board, Highway Capacity Manual, 6th Edition, Copyright, National Academy of Sciences, Washington, D.C. 2016. Exhibit 12-7, p. 12-11.

0 0

10

20 11 p

c/ m

i/I n

18 p

c/ m

i/I n

26 pc/m i/In

35 pc/mi/In

45 pc/mi/In 30

40

50

60

70

80 75 m/h free-flow speed

70 m/h

65 m/h

60 m/h

55 m/h

LOS A LOS B LOS C LOS D LOS E

LOS F

500 1000 1500

Flow Rate (pc/h/In)

S p

ee d

( m

i/h )

2000 2500

6.4 Basic Freeway Segments 201

6.4.4 Determine Free-Flow Speed For basic freeway segments, FFS is the mean speed of passenger cars operating in flow rates up to 1300 passenger cars per hour per lane (pc/h/ln). If FFS is to be estimated rather than measured, the following equation can be used. It accounts for the roadway characteristics of lane width, right-shoulder lateral clearance, and ramp density.

FFS = 75.4 – fLW – fRLC – 3.22TRD0.84 (6.6)

where

FFS = estimated free-flow speed in mi/h, fLW = adjustment for lane width in mi/h,

fRLC = adjustment for lateral clearance in mi/h, 3.22TRD0.84 = adjustment for total ramp density in mi/h (with TRD in ramps/mi).

The constant value of 75.4 in Eq. 6.6 is considered to be the base free-flow speed (BFFS) and applies to freeways in urban and rural areas. The following sections describe the procedures for estimating the adjustment factor values.

Lane Width Adjustment When lane widths are narrower than the base 12 ft, the adjustment factor fLW is used to reflect the impact on FFS. Such an adjustment is needed because narrow lanes cause traffic to slow as a result of reduced psychological comfort and limits on driver maneuvering and accident avoidance options. Thus, FFS under these conditions is less than the value that would be observed if base lane widths were provided. The adjustment factors used in current practice are presented in Table 6.3. Table 6.3 Adjustment for Lane Width

Lane width (ft) Reduction in free-flow speed, fLW (mi/h)

12 0.0

11 1.9

10 6.6

Reproduced with permission of the Transportation Research Board, Highway Capacity Manual, 6th Edition, Copyright, National Academy of Sciences, Washington, D.C. 2016. Exhibit 12-20, pp. 12–29.

Lateral Clearance Adjustment When obstructions are closer than 6 ft (at the right shoulder) from the traveled pavement, the adjustment factor fRLC is used to reflect the impact on FFS. Again, these conditions lead to reduced psychological comfort for the driver and consequently reduced speeds. An obstruction is a right-side object that can either be continuous (such as a retaining wall or barrier) or periodic (such as light posts or utility poles). Table 6.4 provides corrections for obstructions on the right side of the roadway.

202 Chapter 6 Highway Capacity and Level-of-Service Analysis

Table 6.4 Adjustment for Right-Shoulder Lateral Clearance

Right-shoulder lateral clearance (ft)

Reduction in free-flow speed, fRLC (mi/h), lanes in one direction

2 3 4 ≥5

≥ 6 0.0 0.0 0.0 0.0

5 0.6 0.4 0.2 0.1

4 1.2 0.8 0.4 0.2

3 1.8 1.2 0.6 0.3

2 2.4 1.6 0.8 0.4

1 3.0 2.0 1.0 0.5

0 3.6 2.4 1.2 0.6

Reproduced with permission of the Transportation Research Board, Highway Capacity Manual, 6th Edition, Copyright, National Academy of Sciences, Washington, D.C. 2016. Exhibit 12-21, pp. 12–29.

Total Ramp Density Adjustment Ramp density provides a measure of the impact of merging and diverging traffic on FFS. Total ramp density is the number of on- and off-ramps (in one direction) within a distance of three miles upstream and three miles downstream of the midpoint of the analysis segment, divided by six miles.

6.4.5 Determine Analysis Flow Rate The analysis flow rate is calculated using the following equation:

p HV

V v

PHF N f =

× × (6.7)

where

vp = 15-min passenger car equivalent flow rate in pc/h/ln, V = hourly volume in veh/h,

PHF = peak-hour factor, N = number of lanes, and

fHV = heavy-vehicle adjustment factor. The adjustment factors PHF and fHV are described next.

Peak-Hour Factor As previously mentioned, vehicle arrivals during the period of analysis [typically the highest hourly volume within a 24-h period (peak hour)] will likely be nonuniform. To account for this varying arrival rate, the peak 15-min vehicle arrival rate within the analysis hour is usually used for practical traffic analysis purposes. The peak-hour factor has been developed for this purpose, and is defined as the ratio of the hourly volume to the maximum 15-min flow rate expanded to an hourly volume, as follows:

6.4 Basic Freeway Segments 203

= ×15 4

V PHF

V (6.8)

where

PHF = peak-hour factor, V = hourly volume for hour of analysis,

V15 = maximum 15-min volume within hour of analysis, and

4 = number of 15-min periods per hour.

Equation 6.8 indicates that the further the PHF is from unity, the more peaked or nonuniform the traffic flow is during the hour. For example, consider two roads both of which have a peak-hour volume, V, of 1800 veh/h. The first road has 600 vehicles arriving in the highest 15-min interval, and the second road has 500 vehicles arriving in the highest 15-min interval. The first road has a more nonuniform flow, as indicated by its PHF of 0.75 [1800/(600 × 4)], which is further from unity than the second road’s PHF of 0.90 [1800/(500 × 4)].

Heavy-Vehicle Adjustment Large trucks, buses, and recreational vehicles have performance characteristics (slow acceleration and inferior braking) and dimensions (length, height, and width) that have an adverse effect on roadway capacity. Recall that base conditions stipulate that no heavy vehicles are present in the traffic stream, and when prevailing conditions indicate the presence of such vehicles, the adjustment factor fHV is used to translate the traffic stream from base to prevailing conditions. The fHV correction term is found using a two-step process. The first step is to determine the passenger car equivalent (PCE) for each heavy vehicle in the traffic stream. These values represent the number of passenger cars that would consume the same amount of roadway capacity as a single heavy vehicle. Heavy vehicles are classified as either single-unit trucks (SUTs) or tractor- trailers (TTs). Buses and recreational vehicles are treated as SUTs. The passenger car equivalent values, denoted as ET, are a function of roadway grades because steep grades will tend to magnify the poor performance of heavy vehicles as well as the sight distance problems caused by their larger dimensions (the visibility afforded to drivers in vehicles following heavy vehicles). For segments of freeway that contain a mix of grades, an extended segment analysis can be used as long as no single grade is steep enough or long enough to significantly impact the overall operations of the segment. As a guideline, an extended segment analysis can be used for freeway segments where no single grade that is less than 3% is more than 0.5 mi long, or no single grade that is 3% or greater is longer than 0.25 mi. If an extended segment analysis is used, the terrain must be generally classified according to the following definitions [Transportation Research Board 2016]:

Level terrain. Any combination of horizontal and vertical alignment permitting heavy vehicles to maintain approximately the same speed as passenger cars. This generally includes short grades of no more than 2%.

204 Chapter 6 Highway Capacity and Level-of-Service Analysis

Rolling terrain. Any combination of horizontal and vertical alignment that causes heavy vehicles to reduce their speed substantially below those of passenger cars but does not cause heavy vehicles to operate at their limiting speed [Fnet(V) ≠ 0] for the given terrain for any significant length of time or at frequent intervals due to high grade resistance, as illustrated in Fig. 2.2.

The passenger car equivalency factor values for an extended segment analysis can be obtained from Table 6.5.

Table 6.5 Passenger Car Equivalents (PCEs) for Extended Freeway Segments

Type of terrain

Factor Level Rolling

ET 2.0 3.0

Reproduced with permission of the Transportation Research Board, Highway Capacity Manual, 6th Edition, Copyright, National Academy of Sciences, Washington, D.C. 2016. Exhibit 12-25, pp. 12–35.

Any grade that does not meet the conditions for an extended segment analysis, must be analyzed as a separate segment because of its significant impact on traffic operations. In these cases, grade-specific PCE values must be used. Tables 6.6–6.8 provide these values. These tables correspond to heavy vehicle percentage splits (%SUTs/%TTs) of 30/70, 50/50, 70/30, respectively. These tables assume average weight-to-horsepower ratios of 100 and 150 lb/hp for SUTs and TTs, respectively. Linear interpolation should be used with these tables as necessary.

Note that the equivalency values presented in these tables increase with increasing grade and length of grade, but decrease with increasing heavy vehicle percentage. This decrease with increasing percentage is due to the fact that heavy vehicles tend to group together as their percentages increase on steep, extended grades, thus decreasing their adverse impact on the traffic stream.

Negative grades (downgrades) also have an impact on equivalency factors because the comparatively poor braking characteristics of heavy vehicles have a more deleterious effect on the traffic stream than the level-terrain case. For negative grades, use the level grade PCE values from Tables 6.6–6.8.

Sometimes it is necessary to determine the cumulative effect on traffic operations of several significant grades in succession. For this situation, a distance-weighted average may be used if all grades are less than 4% or the total combined length of the grades is less than 4000 ft. For example, a 2% upgrade for 1000 ft followed immediately by a 3% upgrade for 2000 ft would use the equivalency factor for a 2.67% upgrade [(2 × 1000 + 3 × 2000)/3000] for 3000 ft or 0.568 mi. For information on additional analysis situations involving composite grades, refer to the Highway Capacity Manual [Transportation Research Board, 2016]. These situations include combining two or more successive grades when the grades exceed 4% or the combined length is greater than 4000 ft, determining the length of a grade that starts or ends on a vertical curve, and determining the point of greatest traffic impact in a series of grades (for example, if a long 5% grade were immediately followed by a 2% grade, the end of the 5% grade would be used, as this would be the point of minimum vehicle speed).

6.4 Basic Freeway Segments 205

Table 6.6 Passenger Car Equivalents (ET) for Heavy Vehicles (30% SUTs/70% TTs) on Specific Grades

Grade (%)

Length (mi)

Percentage of heavy vehicles

2 4 5 6 8 10 15 20 ≥25

≤ 0 ≤ 0.125 2.62 2.37 2.30 2.24 2.17 2.12 2.04 1.99 1.97 0.375 2.62 2.37 2.30 2.24 2.17 2.12 2.04 1.99 1.97 0.625 2.62 2.37 2.30 2.24 2.17 2.12 2.04 1.99 1.97 0.875 2.62 2.37 2.30 2.24 2.17 2.12 2.04 1.99 1.97 1.25 2.62 2.37 2.30 2.24 2.17 2.12 2.04 1.99 1.97 ≥ 1.5 2.62 2.37 2.30 2.24 2.17 2.12 2.04 1.99 1.97

2 ≤ 0.125 2.62 2.37 2.30 2.24 2.17 2.12 2.04 1.99 1.97 0.375 3.76 2.96 2.78 2.65 2.48 2.38 2.22 2.14 2.09 0.625 4.47 3.33 3.08 2.91 2.68 2.54 2.34 2.23 2.17 0.875 4.80 3.50 3.22 3.03 2.77 2.61 2.39 2.28 2.21 1.25 5.00 3.60 3.30 3.09 2.83 2.66 2.42 2.30 2.23 ≥ 1.5 5.04 3.62 3.32 3.11 2.84 2.67 2.43 2.31 2.23

2.5 ≤ 0.125 2.62 2.37 2.30 2.24 2.17 2.12 2.04 1.99 1.97 0.375 4.11 3.14 2.93 2.78 2.58 2.46 2.28 2.19 2.13 0.625 5.04 3.62 3.32 3.11 2.84 2.67 2.43 2.31 2.23 0.875 5.48 3.85 3.51 3.27 2.96 2.77 2.50 2.36 2.28 1.25 5.73 3.98 3.61 3.36 3.03 2.83 2.54 2.40 2.31 ≥ 1.5 5.80 4.02 3.64 3.38 3.05 2.84 2.55 2.41 2.32

3.5 ≤ 0.125 2.62 2.37 2.30 2.24 2.17 2.12 2.04 1.99 1.97 0.375 4.88 3.54 3.25 3.05 2.80 2.63 2.41 2.29 2.22 0.625 6.34 4.30 3.87 3.58 3.20 2.97 2.64 2.48 2.38 0.875 7.03 4.66 4.16 3.83 3.39 3.12 2.76 2.57 2.46 1.25 7.44 4.87 4.33 3.97 3.50 3.22 2.82 2.62 2.50 ≥ 1.5 7.53 4.92 4.38 4.01 3.53 3.24 2.84 2.63 2.51

4.5 ≤ 0.125 2.62 2.37 2.30 2.24 2.17 2.12 2.04 1.99 1.97 0.375 5.80 4.02 3.64 3.38 3.05 2.84 2.55 2.41 2.32 0.625 7.90 5.11 4.53 4.14 3.63 3.32 2.90 2.68 2.55 0.875 8.91 5.64 4.96 4.50 3.92 3.56 3.07 2.82 2.67 ≥ 1.0 9.19 5.78 5.08 4.60 3.99 3.62 3.11 2.85 2.70

5.5 ≤ 0.125 2.62 2.37 2.30 2.24 2.17 2.12 2.04 1.99 1.97 0.375 6.87 4.58 4.10 3.77 3.35 3.09 2.73 2.55 2.44 0.625 9.78 6.09 5.33 4.82 4.16 3.76 3.21 2.93 2.77 0.875 11.20 6.83 5.94 5.33 4.56 4.09 3.45 3.12 2.93 ≥ 1.0 11.60 7.04 6.11 5.47 4.67 4.18 3.51 3.17 2.97

≥ 6 ≤ 0.125 2.62 2.37 2.30 2.24 2.17 2.12 2.04 1.99 1.97 0.375 7.48 4.90 4.36 3.99 3.52 3.23 2.83 2.63 2.51 0.625 10.87 6.66 5.79 5.21 4.46 4.01 3.39 3.08 2.89

0.875 12.54 7.54 6.51 5.81 4.94 4.40 3.67 3.30 3.08 ≥ 1.0 13.02 7.78 6.71 5.99 5.07 4.51 3.75 3.37 3.14

Reproduced with permission of the Transportation Research Board, Highway Capacity Manual, 6th Edition, Copyright, National Academy of Sciences, Washington, D.C. 2016. Exhibit 12-26, p. 12–36.

206 Chapter 6 Highway Capacity and Level-of-Service Analysis

Table 6.7 Passenger Car Equivalents (ET) for Heavy Vehicles (50% SUTs/50% TTs) on Specific Grades

Grade (%)

Length (mi)

Percentage of heavy vehicles

2 4 5 6 8 10 15 20 ≥ 25

≤ 0 ≤ 0.125 2.67 2.38 2.31 2.25 2.16 2.11 2.02 1.97 1.93 0.375 2.67 2.38 2.31 2.25 2.16 2.11 2.02 1.97 1.93 0.625 2.67 2.38 2.31 2.25 2.16 2.11 2.02 1.97 1.93 0.875 2.67 2.38 2.31 2.25 2.16 2.11 2.02 1.97 1.93 1.25 2.67 2.38 2.31 2.25 2.16 2.11 2.02 1.97 1.93 ≥ 1.5 2.67 2.38 2.31 2.25 2.16 2.11 2.02 1.97 1.93

2 ≤ 0.125 2.67 2.38 2.31 2.25 2.16 2.11 2.02 1.97 1.93 0.375 3.76 2.95 2.77 2.64 2.47 2.36 2.20 2.11 2.06 0.625 4.32 3.24 3.01 2.84 2.63 2.49 2.29 2.19 2.12 0.875 4.57 3.37 3.11 2.93 2.70 2.55 2.33 2.22 2.15 1.25 4.71 3.45 3.17 2.99 2.74 2.58 2.36 2.24 2.17 ≥ 1.5 4.74 3.47 3.19 3.00 2.75 2.59 2.36 2.24 2.17

2.5 ≤ 0.125 2.67 2.38 2.31 2.25 2.16 2.11 2.02 1.97 1.93 0.375 4.10 3.13 2.92 2.77 2.57 2.44 2.26 2.16 2.10 0.625 4.84 3.52 3.23 3.03 2.77 2.61 2.38 2.26 2.18 0.875 5.17 3.69 3.37 3.15 2.87 2.69 2.43 2.30 2.22 1.25 5.36 3.79 3.45 3.22 2.92 2.73 2.47 2.33 2.24 ≥ 1.5 5.40 3.81 3.47 3.24 2.93 2.74 2.47 2.33 2.25

3.5 ≤ 0.125 2.67 2.38 2.31 2.25 2.16 2.11 2.02 1.97 1.93 0.375 4.89 3.54 3.25 3.05 2.79 2.62 2.39 2.26 2.19 0.625 6.05 4.15 3.75 3.47 3.11 2.89 2.58 2.42 2.32 0.875 6.58 4.43 3.97 3.66 3.26 3.01 2.67 2.49 2.39 1.25 6.88 4.58 4.10 3.77 3.35 3.09 2.72 2.53 2.42 ≥ 1.5 6.95 4.62 4.13 3.80 3.37 3.10 2.73 2.54 2.43

4.5 ≤ 0.125 2.67 2.38 2.31 2.25 2.16 2.11 2.02 1.97 1.93 0.375 5.83 4.03 3.65 3.39 3.05 2.84 2.55 2.39 2.30 0.625 7.53 4.92 4.38 4.01 3.53 3.24 2.83 2.62 2.50 0.875 8.32 5.34 4.72 4.29 3.75 3.42 2.97 2.73 2.59 ≥ 1.0 8.53 5.45 4.81 4.37 3.81 3.47 3.00 2.76 2.62

5.5 ≤ 0.125 2.67 2.38 2.31 2.25 2.16 2.11 2.02 1.97 1.93 0.375 6.97 4.63 4.14 3.81 3.38 3.11 2.74 2.55 2.43 0.625 9.37 5.89 5.16 4.68 4.05 3.67 3.14 2.88 2.72 0.875 10.49 6.48 5.65 5.09 4.37 3.93 3.34 3.03 2.85 ≥ 1.0 10.80 6.64 5.78 5.20 4.46 4.01 3.39 3.08 2.89

≥ 6 ≤ 0.125 2.67 2.38 2.31 2.25 2.16 2.11 2.02 1.97 1.93 0.375 7.64 4.98 4.43 4.05 3.56 3.26 2.85 2.64 2.51 0.625 10.45 6.45 5.63 5.07 4.36 3.92 3.33 3.03 2.85

0.875 11.78 7.16 6.20 5.56 4.74 4.24 3.56 3.22 3.01 ≥ 1.0 12.15 7.35 6.36 5.69 4.85 4.33 3.62 3.27 3.05

Reproduced with permission of the Transportation Research Board, Highway Capacity Manual, 6th Edition, Copyright, National Academy of Sciences, Washington, D.C. 2016. Exhibit 12-27, p. 12–37.

6.4 Basic Freeway Segments 207

Table 6.8 Passenger Car Equivalents (ET) for Heavy Vehicles (70% SUTs/30% TTs) on Specific Grades

Grade (%)

Length (mi)

Percentage of heavy vehicles

2 4 5 6 8 10 15 20 ≥ 25

≤ 0 ≤ 0.125 2.39 2.18 2.12 2.07 2.01 1.96 1.89 1.85 1.83 0.375 2.39 2.18 2.12 2.07 2.01 1.96 1.89 1.85 1.83 0.625 2.39 2.18 2.12 2.07 2.01 1.96 1.89 1.85 1.83 0.875 2.39 2.18 2.12 2.07 2.01 1.96 1.89 1.85 1.83 1.25 2.39 2.18 2.12 2.07 2.01 1.96 1.89 1.85 1.83 ≥ 1.5 2.39 2.18 2.12 2.07 2.01 1.96 1.89 1.85 1.83

2 ≤ 0.125 2.67 2.32 2.23 2.17 2.08 2.03 1.94 1.89 1.86 0.375 3.63 2.82 2.64 2.52 2.35 2.25 2.10 2.02 1.97 0.625 4.12 3.08 2.85 2.69 2.49 2.36 2.18 2.08 2.02 0.875 4.37 3.21 2.96 2.78 2.56 2.42 2.22 2.11 2.05 1.25 4.53 3.29 3.02 2.84 2.60 2.45 2.24 2.13 2.07 ≥ 1.5 4.58 3.31 3.04 2.86 2.61 2.46 2.25 2.14 2.07

2.5 ≤ 0.125 2.75 2.36 2.27 2.20 2.11 2.04 1.95 1.90 1.87 0.375 4.01 3.02 2.80 2.65 2.46 2.33 2.16 2.06 2.01 0.625 4.66 3.35 3.08 2.88 2.64 2.48 2.26 2.15 2.08 0.875 4.99 3.52 3.21 3.00 2.73 2.56 2.32 2.19 2.12 1.25 5.20 3.64 3.30 3.08 2.79 2.60 2.35 2.22 2.14 ≥ 1.5 5.26 3.67 3.33 3.10 2.80 2.62 2.36 2.23 2.15

3.5 ≤ 0.125 2.93 2.45 2.34 2.26 2.16 2.09 1.98 1.92 1.89 0.375 4.86 3.46 3.16 2.96 2.69 2.53 2.30 2.18 2.10 0.625 5.88 3.99 3.59 3.32 2.98 2.76 2.46 2.31 2.22 0.875 6.40 4.26 3.81 3.51 3.12 2.88 2.55 2.38 2.28 1.25 6.74 4.43 3.96 3.63 3.21 2.96 2.60 2.42 2.32 ≥ 1.5 6.83 4.48 3.99 3.66 3.24 2.98 2.62 2.44 2.33

4.5 ≤ 0.125 3.13 2.56 2.43 2.34 2.21 2.13 2.01 1.95 1.91 0.375 5.88 3.99 3.59 3.32 2.98 2.76 2.46 2.31 2.22 0.625 7.35 4.75 4.22 3.85 3.39 3.10 2.71 2.51 2.39 0.875 8.11 5.15 4.54 4.13 3.60 3.27 2.83 2.61 2.47 ≥ 1.0 8.33 5.27 4.63 4.21 3.66 3.33 2.87 2.64 2.50

5.5 ≤ 0.125 3.37 2.69 2.53 2.42 2.28 2.19 2.05 1.98 1.94 0.375 7.09 4.62 4.11 3.76 3.31 3.04 2.66 2.47 2.36 0.625 9.13 5.68 4.97 4.49 3.88 3.51 3.00 2.74 2.59 0.875 10.21 6.24 5.43 4.88 4.18 3.76 3.18 2.89 2.71 ≥ 1.0 10.52 6.41 5.57 5.00 4.27 3.83 3.24 2.93 2.75

≥ 6 ≤ 0.125 3.51 2.76 2.59 2.47 2.32 2.22 2.08 2.00 1.95 0.375 7.78 4.98 4.40 4.01 3.51 3.20 2.78 2.56 2.44 0.625 10.17 6.23 5.42 4.87 4.17 3.75 3.18 2.88 2.71

0.875 11.43 6.88 5.95 5.32 4.53 4.04 3.39 3.06 2.86 ≥ 1.0 11.81 7.08 6.11 5.46 4.64 4.13 3.45 3.11 2.90

Reproduced with permission of the Transportation Research Board, Highway Capacity Manual, 6th Edition, Copyright, National Academy of Sciences, Washington, D.C. 2016. Exhibit 12-28, p. 12–38.

208 Chapter 6 Highway Capacity and Level-of-Service Analysis

It should be noted that utilizing PCE values to accommodate heavy vehicles in the analysis is not appropriate for mountainous terrain situations. Mountainous terrain is considered to be any combination of horizontal and vertical alignment that causes heavy vehicles to operate at their limiting speed for significant distances or at frequent intervals. For this type of situation, a more accurate analysis procedure is provided in the HCM [Transportation Research Board 2016].

Once the appropriate equivalency factors have been obtained, the following equation is applied to arrive at the heavy-vehicle adjustment factor fHV:

( ) 1

1 1HV T T

f P E

= + −

(6.9)

where

fHV = heavy-vehicle adjustment factor, PT = proportion of heavy vehicles in the traffic stream, and ET = passenger car equivalent for heavy vehicles, from Tables 6.5–6.8.

As an example of how the heavy-vehicle adjustment factor is computed, consider a freeway with a 1.25-mi 3.5% upgrade with a traffic stream having 10% heavy vehicles (70% SUT/30% TT split). Table 6.8 must be used because the grade is too steep and long for Table 6.5 to apply. The corresponding equivalency value from Table 6.8 for this roadway is ET = 2.96. Substituting this ET value and PT = 0.1 (for 10% heavy vehicles) into Eq. 6.5 gives fHV = 0.84, or a 16% reduction in effective roadway capacity relative to the base condition of no heavy vehicles in the traffic stream.

6.4.6 Calculate Density and Determine LOS With all the terms in the previous equations defined, these equations can now be applied to determine freeway LOS and capacity. The final step before LOS can be determined is to calculate the density of the traffic stream. The alternative notation to Eq. 6.5 is shown in Eq. 6.10, which will be used in subsequent example problems (for consistency with the Highway Capacity Manual).

pv D

S = (6.10)

where

D = density in pc/mi/ln, vp = flow rate in pc/h/ln, and

S = average speed in mi/h.

The average speed is found by reading it from the y-axis of Fig. 6.2 for the corresponding flow rate (vp) and FFS. Once the density value is calculated, the LOS can be read from Table 6.1 or Fig. 6.2.

Application of the process for determining basic freeway segment capacity and LOS is demonstrated in example problems 6.1 and 6.2.

6.4 Basic Freeway Segments 209

EXAMPLE 6.1 BASIC FREEWAY SEGMENT LOS WITH GENERAL TERRAIN CLASSIFICATION

A six-lane urban freeway (three lanes in each direction) is on level terrain with 11-ft lanes, obstructions 2 ft from the right edge of the traveled pavement, and nine ramps within three miles upstream and three miles downstream of the midpoint of the analysis segment. A directional weekday peak-hour volume of 3000 vehicles is observed, with 810 vehicles arriving in the most congested 15-min period. If the traffic stream has 10% single-unit trucks and 10% tractor-trailer trucks, determine the LOS.

SOLUTION

Determine the FFS according to Eq. 6.6.

FFS = 75.4 – fLW – fRLC − 3.22TRD 0.84

with fLW = 1.9 mi/h (Table 6.3), fRLC = 1.6 mi/h (Table 6.4), and

9 1.5 ramps/mi

6 TRD = =

FFS = 75.4 – 1.9 – 1.6 – 3.22(1.5)0.84 = 67.4 mi/h

Determine the flow rate according to Eq. 6.7:

p HV

V v

PHF N f =

× ×

with

3000 0.926

810 4 PHF = =

×

N = 3 (given), and ET = 2.0 (level terrain, Table 6.5). From Eq. 6.9, we obtain:

( ) 1

0.833 1 0.20 2.0 1HVf = =

+ −

So, from Eq. 6.7,

3000 1296.5 1297 pc/h/ln

0.926 3 0.833pv = = → × ×

To determine the average speed (S), first calculate the breakpoint value for the speed– flow relationship, from Eq. 6.4: ( )1000 40 75 67.4 1304 pc/h/lnBP  = + − =  Since the vp of 1297 is less than the BP value of 1304, S will be equal to the FFS, 67.4 mi/h. In this case, the average speed is still the same as the FFS because the flow rate is low enough such that it is still on the linear/flat part of the speed-flow curve.

210 Chapter 6 Highway Capacity and Level-of-Service Analysis

Now, density can be calculated with Eq. 6.10:

1297 19.2 pc/mi/ln

67.4 D = =

From Table 6.2, it can be seen that this corresponds to LOS C (18.0 [max density for LOS B] < 19.2 < 26.0 [max density for LOS C]). Thus, this freeway segment operates at LOS C. This problem can also be solved graphically by applying Fig. 6.2. Using this figure, draw a vertical line up from 1297 pc/h/ln (on the figure’s x-axis) and find that this line intersects the 67.4 mi/h FFS curve (interpolated between the 65 and 70 mi/h curves) in the LOS C density region (the dashed diagonal lines).

EXAMPLE 6.2 BASIC FREEWAY SEGMENT LOS WITH A SPECIFIC GRADE

Consider the freeway and traffic conditions in Example 6.1. At some point further along the roadway there is a 6% upgrade that is 1.25 mi long. All other characteristics are the same as in Example 6.1. What is the LOS of this portion of the roadway, and how many vehicles can be added before the roadway reaches capacity (assuming that the proportion of vehicle types and the peak-hour factor remain constant)?

SOLUTION

To determine the LOS of this segment of the freeway, we note that all adjustment factors are the same as those in Example 6.1 except fHV, which must now be determined using an equivalency factor, ET, drawn from the specific-grade tables (in this case Table 6.7, for 50%/50% SUT/TT split). From Table 6.7, ET = 3.27, which gives (from Eq. 6.9)

( ) 1

0.688 1 0.20 3.27 1HVf = =

+ −

So,

= = → × × 3000

1569.8 1570 pc/h/ln 0.926 3 0.688pv

In this case, the vp of 1570 is more than the BP value of 1304 (compare to Example 6.1), so S will not be equal to the FFS. Therefore, we must proceed with applying Eq. 6.2. First, calculate the capacity from Eq. 6.3, which gives

( ) = + − = Max 2200 10 67.4 50 ,2400 2374 pc/h/lnc

Substituting the values for FFS, c, vp, and BP into Eq. 6.2 yields

( )

( )

2.0

2.0

2374 67.4 1570 1304

45 67.4 66.5 mi/h

2374 1304 S

 − −   = − =

−

Now, density can be calculated with Eq. 6.10:

1570 23.6 pc/mi/ln

66.5 D = =

6.5 Multilane Highway Segments 211

which still gives LOS C from Table 6.2 (18.0 [max density for LOS B] < 23.6 < 26.0 [max density for LOS C]). However, traffic conditions are still worse on the upgrade section than the level section because the density is higher. To determine how many vehicles can be added before capacity is reached, the hourly volume at capacity must be computed. Recall that capacity corresponds to a volume-to-capacity ratio of 1.0 (the threshold between LOS E and LOS F). For a FFS of 67.4 mi/h, the capacity is 2374 pc/h/ln. Equation 6.7 is rearranged and used to solve for the hourly volume based upon this capacity:

=  = × × × ×

2374 0.926 3 0.688p

HV

V V v

PHF N f

which gives V = 4537 veh/h. This means that about 1537 vehicles (4537 − 3000) can be added during the peak hour before capacity is reached. It should be noted that the assumption that the peak-hour factor will remain constant as the roadway approaches capacity is not very realistic. In practice, it is observed that as a roadway approaches capacity, PHF gets closer to 1. This implies that the flow rate over the peak hour becomes more uniform.

6.5 MULTILANE HIGHWAY SEGMENTS

Multilane highways are similar to freeways in most respects, except for a few key differences:

• Vehicles may enter or leave the roadway at at-grade intersections and driveways (multilane highways do not have full access control).

• Multilane highways may or may not be divided (by a barrier or median separating opposing directions of flow), whereas freeways are always divided.

• Traffic signals may be present. • Design standards (such as design speeds) are sometimes lower than those

for freeways. • The visual setting and development along multilane highways are usually

more distracting to drivers than in the freeway case. Multilane highways usually have four or six lanes (both directions), have posted speed limits between 40 and 65 mi/h, and can have physical medians, medians that are two-way left-turn lanes (TWLTLs), or opposing directional volumes that may not be divided by a median at all. Two examples of multilane highways are shown in Fig. 6.3.

The determination of LOS on multilane highways closely mirrors the procedure for freeways. The main differences lie in the speed–flow relationship and some of the FFS adjustment factors and their values. The procedure we present is valid only for sections of highway that are not significantly influenced by large queue formations and dissipations resulting from traffic signals (this is generally taken as having traffic signals spaced 2.0 mi apart or more), do not have significant on-street parking, do not have bus stops with high usage, and do not have significant pedestrian activity.

212 Chapter 6 Highway Capacity and Level-of-Service Analysis

Figure 6.3 Examples of multilane highways.

6.5.1 Speed versus Flow Rate Relationship Just as for freeways, the speed–flow relationship is central to the analysis of multilane highways. However, unlike freeways, the breakpoint (the transition from a linear to curvilinear flow rate) is not a function of FFS but rather is constant at 1400 pc/h/ln. With this, the HCM [Transportation Research Board 2016] defines the relationship between average speed and flow rate as follows:

S = FFS ≤ 1400pv (6.11)

and

( )

( )

 − −   = −

−

1.31

1.31

1400 45

1400

p c

FFS v S FFS

c < ≤1400 pv c (6.12)

where

c = Capacity in pc/h/ln, vp = 15-min passenger car equivalent flow rate in pc/h/ln,

45 = Density at capacity in pc/mi/ln, as shown in Table 6.10, and

1400 = Linear to curvilinear flow rate breakpoint value in pc/h/ln. Other terms are as defined previously.

Capacity is given by the following equation:

( )Min 1900 20 45 ,2300c FFS = + −  45 ≤ FFS ≤ 70 (6.13)

Results from this equation for 5-mi/h increments across the valid range of FFS are shown in Table 6.9. The speed-flow rate curves resulting from the application of Eqs. 6.11 and 6.12, for 5-mi/h increments across the valid range of FFS are shown in Fig. 6.4. Table 6.10 provides the LOS criteria corresponding to traffic density, speed, volume-to-capacity ratio, and maximum flow rate.

6.5 Multilane Highway Segments 213

Table 6.9 Relationship between Free-Flow Speed and Capacity on Multilane Highway Segments

Source: Transportation Research Board, Highway Capacity Manual, 6th Edition, National Academy of Sciences, Washington, D.C. 2016.

Free-flow speed (mi/h)

Capacity (pc/h/ln)

70 2300 65 2300 60 2200 55 2100 50 2000 45 1900

Table 6.10 LOS Criteria for Multilane Highway Segments

Criterion LOS

A B C D E

FFS = 70 mi/h Maximum density (pc/mi/ln) 11 18 26 35 45 Average speed (mi/h) 70.0 70.0 65.4 58.1 51.1 Maximum v/c 0.33 0.55 0.74 0.88 1.00 Maximum flow rate (pc/h/ln) 770 1260 1700 2035 2300

FFS = 65 mi/h Maximum density (pc/mi/ln) 11 18 26 35 45 Average speed (mi/h) 65.0 65.0 62.7 57.0 51.1 Maximum v/c 0.31 0.51 0.71 0.87 1.00 Maximum flow rate (pc/h/ln) 715 1170 1630 1995 2300

FFS = 60 mi/h Maximum density (pc/mi/ln) 11 18 26 35 45 Average speed (mi/h) 60.0 60.0 59.0 54.1 48.9 Maximum v/c 0.30 0.49 0.70 0.86 1.00 Maximum flow rate (pc/h/ln) 660 1080 1535 1895 2200

FFS = 55 mi/h Maximum density (pc/mi/ln) 11 18 26 35 45 Average speed (mi/h) 55.0 55.0 54.8 51.1 46.7 Maximum v/c 0.29 0.47 0.68 0.85 1.00 Maximum flow rate (pc/h/ln) 605 990 1425 1790 2100

FFS = 50 mi/h Maximum density (pc/mi/ln) 11 18 26 35 45 Average speed (mi/h) 50.0 50.0 50.0 48.0 44.4 Maximum v/c 0.28 0.45 0.65 0.84 1.00 Maximum flow rate (pc/h/ln) 550 900 1300 1680 2000

FFS = 45 mi/h Maximum density (pc/mi/ln) 11 18 26 35 45 Average speed (mi/h) 45.0 45.0 45.0 44.4 42.2 Maximum v/c 0.26 0.43 0.62 0.82 1.00 Maximum flow rate (pc/h/ln) 495 810 1170 1555 1900

Note: Maximum flow rate values are rounded to the nearest 5 passenger cars.

214

Figure 6.4 Multilane highway speed-flow curves and level-of-service criteria. Reproduced with permission of the Transportation Research Board, Highway Capacity Manual, 6th Edition, Copyright, National Academy of Sciences, Washington, D.C. 2016. Exhibit 12-8, p. 12–11.

0 0

10

20

11 p

c/ m

i/I n

18 p

c/ m

i/I n

26 pc/m i/In

35 pc/mi/In

45 pc/mi/In 30

40

50

60

70

80

70 m/h free-flow speed

65 m/h

60 m/h

55 m/h

50 m/h

45 m/h

LOS A LOS B LOS C LOS D LOS E

LOS F

400 800 1200

Flow Rate (pc/h/In)

S p

ee d

( m

i/h )

20001600 2400

6.5 Multilane Highway Segments 215

6.5.2 Base Conditions and Capacity The base conditions for multilane highways are defined as [Transportation Research Board 2016]

• 12-ft minimum lane widths • 12-ft minimum total lateral clearance from roadside objects (right

shoulder and median) in the travel direction • Only passenger cars in the traffic stream • No direct access points along the roadway • Divided highway • Level terrain (no grades greater than 2%) • Free-flow speed of 60 mi/h or more

As was the case with the freeway LOS analysis, adjustments will have to be made when nonbase conditions are encountered.

The capacity, c, for multilane highway segments, in pc/h/ln, is given in Table 6.9. From Table 6.10, note again that these capacity values correspond to the maximum service flow rate at LOS E and a v/c of 1.0.

6.5.3 Service Measure Due to the large degree of similarity between multilane highway and freeway facilities, density is also the service measure (performance measure used for determining LOS) for multilane highways. As can be seen in Table 6.10, the LOS density thresholds for multilane highways are the same as for freeways.

6.5.4 Determine Free-Flow Speed FFS for multilane highways is the mean speed of passenger cars operating in flow rates up to 1400 passenger cars per hour per lane (pc/h/ln). If FFS is to be estimated rather than measured, the following equation can be used, which takes into account the roadway characteristics of lane width, lateral clearance, presence (or lack) of a median, and access frequency:

LW TLC M AFFS BFFS f f f f= − − − − (6.14)

where FFS = estimated free-flow speed in mi/h,

BFFS = estimated free-flow speed, in mi/h, for base conditions, fLW = adjustment for lane width in mi/h, fLC = adjustment for lateral clearance in mi/h, fM = adjustment for median type in mi/h, and fA = adjustment for the number of access points along the roadway in mi/h.

As can be seen, this equation closely resembles Eq. 6.6 in the freeway section. Both include adjustments for lane width and lateral clearance, and the access frequency adjustment is similar to the ramp density adjustment. The main

216 Chapter 6 Highway Capacity and Level-of-Service Analysis

difference is that Eq. 6.14 also includes an adjustment for median type. The presence of a physical barrier or wide separation between opposing flows (such as a TWLTL) will lead to higher FFS than if there is no separation or physical barrier between opposing flows. This adjustment is not included for freeways since, by definition, all freeways are divided.

As for BFFS, many factors can influence the FFS, with the posted speed limit often being a significant one. For multilane highways, research has found that FFS, under base conditions, are about 7 mi/h higher than the speed limit for 40- and 45-mi/h posted-speed-limit roadways, and about 5 mi/h higher for 50- mi/h and higher posted-speed-limit roadways. The following sections describe the procedures for estimating the adjustment factor values.

Lane Width Adjustment The same lane width adjustment factor values are used for multilane highways as are used for freeways. Thus, Table 6.3 should be used for multilane highways as well.

Lateral Clearance Adjustment The adjustment factor for potentially restrictive lateral clearances (fTLC) is determined first by computing the total lateral clearance, which is defined as

TLC = LCR + LCL (6.15)

where

TLC = total lateral clearance in ft, LCR = lateral clearance on the right side of the travel lanes to obstructions

(retaining walls, utility poles, signs, trees, etc.), and LCL = lateral clearance on the left side of the travel lanes to obstructions.

For undivided highways, there is no adjustment for left-side lateral clearance because this is already taken into account in the fM term (thus LCL = 6 ft in Eq. 6.8). If an individual lateral clearance (either left or right side) exceeds 6 ft, 6 ft is used in Eq. 6.15. Finally, highways with TWLTLs are considered to have LCL equal to 6 ft. Once Eq. 6.15 is applied, the value for fTLC can be determined directly from Table 6.11.

Table 6.11 Adjustment for Lateral Clearance *Total lateral clearance is the sum of the lateral clearances of the median (if greater than 6 ft, use 6 ft) and shoulder (if greater than 6 ft, use 6 ft). Therefore, for purposes of analysis, total lateral clearance cannot exceed 12 ft.

Reproduced with permission of the Transportation Research Board, Highway Capacity Manual, 6th Edition, Copyright, National Academy of Sciences, Washington, D.C. 2016. Exhibit 12-22, p. 12-30.

Total lateral

clearance* (ft)

Reduction in free-flow speed, fTLC (mi/h)

Four-lane highways

Six-lane highways

12 0.0 0.0

10 0.4 0.4

8 0.9 0.9

6 1.3 1.3

4 1.8 1.7

2 3.6 2.8

0 5.4 3.9

6.5 Multilane Highway Segments 217

Median Adjustment Values for the adjustment factor for median type, fM, are provided in Table 6.12. This table shows that undivided highways have a FFS that is 1.6 mi/h lower than divided highways (which include those with two-way left-turn lanes).

Table 6.12 Adjustment for Median Type Reproduced with permission of the Transportation Research Board, Highway Capacity Manual, 6th Edition, Copyright, National Academy of Sciences, Washington, D.C. 2016. Exhibit 12-23, p. 12-30.

Median type

Reduction in free-flow speed,

fM (mi/h)

Undivided highways 1.6

Divided highways (including TWLTLs)

0.0

Access Frequency Adjustment The final adjustment factor in Eq. 6.14 is for the number of access points per mile, fA. Access points are defined to include intersections and driveways (on the right side of the highway in the direction being considered) that significantly influence traffic flow, and thus do not generally include driveways to individual residences or service driveways at commercial sites. Adjustment values for access point frequency are provided in Table 6.13.

Table 6.13 Adjustment for Access-Point Frequency Reproduced with permission of the Transportation Research Board, Highway Capacity Manual, 6th Edition, Copyright, National Academy of Sciences, Washington, D.C. 2016. Exhibit 12-24, p. 12-31.

Access points/ mile

Reduction in free-flow speed

fA (mi/h)

0 0.0

10 2.5

20 5.0

30 7.5

≥ 40 10.0

6.5.5 Determine Analysis Flow Rate The analysis flow rate for multilane highways is determined in the same manner as for freeways, using Eq. 6.7 and the remainder of the procedure outlined in Section 6.4.5.

6.5.6 Calculate Density and Determine LOS The procedure for calculating density and determining LOS for multilane highways is essentially the same as for freeways (see Section 6.4.6). Equation 6.10 is applied to arrive at a density. However, different speed-flow curves are used for multilane highways. Once the density value is calculated, the LOS can be determined from Table 6.10 or Fig. 6.4.

218 Chapter 6 Highway Capacity and Level-of-Service Analysis

EXAMPLE 6.3 MULTILANE HIGHWAY FREE-FLOW SPEED

A four-lane undivided highway (two lanes in each direction) has 11-ft lanes, with 4-ft shoulders on the right side. There are seven access points per mile, and the posted speed limit is 50 mi/h. What is the estimated FFS?

SOLUTION

This problem can be solved by direct application of Eq. 6.14 to arrive at an estimated FFS:

= − − − −LW TLC M AFFS BFFS f f f f

with

BFFS = 55 mi/h (assume FFS = posted speed + 5 mi/h), fLW = 1.9 mi/h (Table 6.3), fTLC = 0.4 mi/h (Table 6.11, with TLC = 4 + 6 = 10 from Eq. 6.15, with LCL = 6

ft because the highway is undivided), fM = 1.6 mi/h (Table 6.12), and fA = 1.75 mi/h (Table 6.13, by interpolation).

Substitution gives

55 1.9 0.4 1.6 1.75 49.35 mi/hFFS = − − − − =

which means that the more restrictive roadway characteristics relative to the base conditions result in a reduction in FFS of 5.65 mi/h.

EXAMPLE 6.4 MULTILANE HIGHWAY LOS

A six-lane divided highway (three lanes in each direction) is on rolling terrain with two access points per mile and has 10-ft lanes, with a 5-ft shoulder on the right side and a 3-ft shoulder on the left side. The peak-hour factor is 0.85, and the directional peak- hour volume is 3400 vehicles per hour. There are 7% single-unit trucks and 3% tractor- trailer trucks in the traffic stream. No speed studies are available, but the posted speed limit is 55 mi/h. Determine the LOS.

SOLUTION

We begin by determining FFS by applying Eq. 6.14:

= − − − −LW TLC M AFFS BFFS f f f f

with

BFFS = 60 mi/h (assume FFS = posted speed + 5 mi/h), fLW = 6.6 mi/h (Table 6.3), fTLC = 0.9 mi/h (Table 6.11, with TLC = 5 + 3 = 8 from Eq. 6.15),

fM = 0.0 mi/h (Table 6.12), and fA = 0.5 mi/h (Table 6.13, by interpolation).

6.5 Multilane Highway Segments 219

Substitution gives

= − − − − =60.0 6.6 0.9 0.0 0.5 52.0 mi/hFFS

Next we determine the analysis flow rate using Eq. 6.7:

= × ×p

HV

V v

PHF N f

with

V = 3400 veh/h (given), PHF = 0.85 (given),

N = 3 (given), and ET = 3.0 (Table 6.5).

From Eq. 6.9, we find

( )= = + −

1 0.833

1 0.10 3.0 1HVf

Substitution gives

= = → × × 3400

1600.6 1601 pc/h/ln 0.85 3 0.833pv

To determine the average speed (S), first check the vp value against the BP value. Since 1601 > 1400, S will not be equal to the FFS. Therefore, we must proceed with applying Eq. 6.12. First, calculate the capacity from Eq. 6.13, which gives

( ) = + − = Max 1900 20 52 45 ,2300 2040 pc/h/lnc

Substituting the values for FFS, c, and vp into Eq. 6.12 yields

( )

( )

 − −   = − =

−

1.31

1.31

2040 52 1601 1400

4552 50.5 mi/h 1601 1400

S

Now, density can be calculated with Eq. 6.10:

= =1601 31.7 pc/mi/ln

50.5 D

which gives LOS D from Table 6.10 [26.0 (max density for LOS C) < 31.7 < 35.0 (max density for LOS D)]. This can also be confirmed from Fig. 6.4, for FFS = 52 mi/h (interpolated), where the 1601-pc/h/ln flow rate intersects this curve in the LOS D density region

220 Chapter 6 Highway Capacity and Level-of-Service Analysis

EXAMPLE 6.5 MULTILANE HIGHWAY CAPACITY

A local manufacturer wishes to open a factory near the segment of highway described in Example 6.4. How many heavy vehicles can be added to the peak-hour directional volume before capacity is reached? (Add only heavy vehicles and assume that the PHF remains constant.)

SOLUTION

Note that FFS will remain unchanged at 52 mi/h. The capacity for FFS = 52 mi/h, as calculated in Example 6.4 is 2040 pc/h/ln. The current number of heavy vehicles in the peak-hour traffic stream is 340 (0.10 × 3400). Let us denote the number of new heavy vehicles added as Vnt, so the combination of Eqs. 6.7 and 6.9 gives

( )( ) ( )

+ =

        + + − +   

1

340 1 1

nt p

nt T

nt

V V v

PHF N V

E V V

with

vp = 2040 pc/h/ln (capacity),

V = 3400 veh/h (Example 6.5), PHF = 0.85 (Example 6.4),

N = 3 (Example 6.4), and ET = 3.0 (Example 6.4).

( )( ) ( )

+ =

        + + − +   

3400 2040

1 0.85 3

340 1 3.0 1

3400

nt

nt

nt

V

V V

which gives = 374ntV , which is the number of heavy vehicles that can be added to the

peak-hour volume before capacity is reached.

6.6 Two-Lane Highways 221

6.6 TWO-LANE HIGHWAYS

Two-lane highways are defined as roadways with one lane available in each direction. For level of service (LOS) determination, a key distinction between two- lane highways and the freeways and multilane highways previously discussed is that traffic in both directions must now be considered (previously we considered traffic in one direction only). This is because traffic in an opposing direction has a strong influence on LOS. For example, a high opposing traffic volume limits the opportunity to pass slow-moving vehicles (because such a pass requires the passing vehicle to occupy the opposing lane) and thus forces a lower traffic speed—and, as a consequence, a lower LOS. It also follows that any geometric features that restrict passing sight distance (such as sight distance on horizontal and vertical curves) will have an adverse impact on the LOS. Finally, the type of terrain (grades, horizontal curves) plays a more critical role in LOS calculations, relative to freeways and multilane highways, because of the sometimes limited ability to pass slower-moving vehicles on grades in areas where passing is prohibited due to sight distance restrictions or where opposing traffic does not permit safe passing. Some examples of two-lane highways are shown in Fig. 6.5.

Figure 6.5 Examples of two-lane highways.

(a) (b)

(c) (d)

222 Chapter 6 Highway Capacity and Level-of-Service Analysis

This section provides an overview of the analysis procedure for two-lane highways in the Highway Capacity Manual [Transportation Research Board 2016].

6.6.1 Analysis Concepts The analysis methodology considers several roadway and traffic concepts, described as follows.

Segmentation A two-lane highway is divided into segments for analysis purposes. The ability to pass, lane geometry, grades, lane and shoulder widths, posted speed limits, traffic demands, adjacent land uses, driveways, and other characteristics of the facility should be homogeneous within each analysis segment. Note that segmentation will be different for each direction of the highway because passing zones and other characteristics will start and end in different locations depending on the direction of travel. The results of each segment analysis are then combined into results for the full highway.

The analysis methodology can be used to analyze individual homogeneous segments of a two-lane highway, as well as extended lengths of a two-lane highway that are composed of multiple contiguous segments. The methodology applies to the following segment types:

• Passing Constrained: A segment in which passing in the oncoming lane is either prohibited or is effectively negligible due to geometric or sight distance limitations. Traffic operations in a passing constrained segment are a function of analysis direction flow rate, percent heavy vehicles, horizontal and vertical alignment, and segment length (see Figs. 6.5a and 6.5b).

• Passing Zone: A segment where passing in the oncoming lane is not restricted. However, to be effective in accommodating passing maneuvers, these zones must provide a minimum length of useable passing distance. The effectiveness of a passing zone in improving traffic operations is a function of analysis and opposing direction flow rates, percent heavy vehicles, horizontal and vertical alignment, and segment length (see Fig. 6.5c, direction with centerline skip-striping).

• Passing/Climbing Lane: A segment in which a relatively short length of additional lane is provided in the same travel direction. A passing lane can be effective in dispersing platoons by providing an opportunity for faster vehicles to pass slower vehicles. Passing lanes on significant upgrades (where the speed of large trucks is reduced well below their desired speed) are generally referred to as climbing lanes. The effectiveness of a passing or climbing lane in improving traffic operations is a function of analysis direction flow rate, percent heavy vehicles, horizontal and vertical alignment, and segment length (see Fig. 6.5d).

Access Points Access points are major driveways and side roads where significant traffic enters and/or leaves the two-lane highway within an analysis segment. Access points lower the free-flow speed (FFS) for the segment. By lowering the FFS, access

6.6 Two-Lane Highways 223

points will also indirectly affect the average speed and the follower density. Residential driveways and other low-volume driveways and side roads (generally with average daily traffic below 20 vehicles per day) are generally not counted as access points.

Treatment of Terrain and Heavy Vehicles Unlike other facility analysis methodologies in the HCM, the two-lane highway analysis procedure works directly with traffic stream units of vehicles per hour, rather than passenger cars per hour. That is, passenger car equivalent (PCE) values are not utilized. This approach provides the ability to more accurately represent the performance measure relationships as a function of the more varying horizontal and vertical alignment often encountered on two-lane highways. Additionally, it simplifies the analysis methodology by removing the process of converting units from veh/h to pc/h. The two-lane highway analysis methodology therefore includes the percentage of heavy vehicles as an input to the performance measure estimation models.

In general, a heavy vehicle is defined as any vehicle (or vehicle–trailer unit) with more than four wheels on the ground during normal operation. Heavy vehicles generally consist of large trucks, buses, and recreational vehicles (RVs).

Vertical alignment is accommodated through a classification scheme providing five different levels. Each classification is a function of the segment length and up- or downgrade percentage. The classification levels are based on reductions in heavy-vehicle FFS. On an upgrade, the speed reduction is a function of the effect of grade resistance limiting a vehicle’s acceleration capability. On a downgrade, the speed reduction is a result of trucks downshifting to avoid the potential of a “runaway” acceleration condition. Vertical Alignment Class 1 corresponds to a minimal reduction in FFS, while Vertical Alignment Class 5 corresponds to a crawl-speed condition.

Horizontal alignment is also accommodated through a classification scheme that considers reductions in heavy-vehicle FFS. The horizontal classifications are a function of curve radius and superelevation.

Applicable Performance Measures Three performance measures underpin this analysis methodology: average speed, percent followers, and follower density. Typical agency performance objectives for two-lane highways focus on reduced travel times and reduced pressure to pass (reduced percent and density of followers). The three performance measures were selected to represent the quality of service provided by two-lane highways because they most closely conform to typical agency performance objectives. They also have the advantage of being sensitive to the design and demand features typically considered when planning and designing two-lane highways.

Average Speed Average speed is a key performance measure for two-lane highways. The general relationship between average speed and flow rate is illustrated in Fig. 6.6.

224 Chapter 6 Highway Capacity and Level-of-Service Analysis

Figure 6.6 Average speed versus directional flow rate.

Percent Followers The measure percent followers represents the freedom to maneuver and the comfort and convenience of travel. It can also serve as a proxy for the need to provide passing opportunities. Percent followers is the percentage of vehicles passing a given point on the roadway that are considered to be in a follower state. Being in a follower state is used as a surrogate to indicate a driver’s perception that they are being delayed by a slower driver. Higher values are generally indicative of restrictions in passing (very infrequent passing zones or passing lanes, high volumes in the opposing direction).

A critical headway threshold value of 2.5 s is used to determine whether a vehicle is in a follower state. Thus, any vehicle following another vehicle at a headway of 2.5 s or less is considered to be in a follower state and percent followers is simply the percentage of all vehicles with a headway of 2.5 s or less at a given roadway location. The general relationship between percent followers and flow rate is illustrated in Fig 6.7.

Note that even at a capacity flow rate, 100% followers will not be reached. This is because there will always be some gaps between platoons within the traffic stream as a result of different platoon lead vehicle desired speeds.

Follower Density Follower density is defined as the number of vehicles in a follower state per mile per lane. Mathematically, it is the percent followers multiplied by density. This measure is used as the service measure for two-lane highways because of its sensitivity to both traffic demand and geometric alignment variability. On two-lane highways, it is possible for two roadways to have similar values of density but very different levels of percent followers. Likewise, it is possible for two roadways to have similar percent followers but very different values of density. Therefore, service quality is better represented by considering the combination of follower percentage and density. The general relationship between follower density and flow rate is illustrated in Fig. 6.8.

0 0

10

20

30

40

50

60

70

300 600 900

Flow Rate (veh/h)

A ve

ra g

e S

p ee

d (

m i/h

)

1,200 1,500 1,800

(a) Passing Zone Segment (b) Passing Lane Segment

0 0

10

20

30

40

50

60

70

300 600 900

Flow Rate (veh/h)

A ve

ra g

e S

p ee

d (

m i/h

)

1,200 1,500 1,800

6.6 Two-Lane Highways 225

Figure 6.7 Percent followers versus directional flow rate.

Figure 6.8 Follower density versus directional flow rate.

Analysis Approach The operational performance of a segment, either individually or within a facility, is reported for the end of the segment, rather than corresponding to an aggregated value across the full length of the segment. Thus, the reported values are point estimates of performance and are not necessarily representative of average conditions along an extended length of segment. This approach makes it easier for practitioners to use field-measured point values to directly calculate LOS with the method and to use point values to validate method outputs. Nevertheless, when considering multiple contiguous segments as a facility, the point measure is used as the estimate of the performance of a given segment. From a traveler’s perspective, the conditions at the end of a segment (particularly passing zones) probably factor more heavily into their assessment of the quality of service. For passing lane segments, additional calculations are performed, which is described briefly at the end of this section.

0 200 400 600 800 1,000

Flow Rate (veh/h)

P er

ce n

t Fo

llo w

er s

(% )

1,200 1,400 1,600 1,800 0

10

20

30

40

50

60

70

80

90

100

0 0

5

10

Fo llo

w er

D en

si ty

( fo

llo w

er s/

m i/I

n )

15

20

25

30

200 400 600 800 1,000

Flow Rate (veh/h)

1,200 1,400 1,600 1,800

226 Chapter 6 Highway Capacity and Level-of-Service Analysis

Individual segments can be analyzed with this methodology. Additionally, multiple contiguous two-lane highway segments (in the same direction) may be combined to analyze a longer section (with varying characteristics) as a facility.

While this methodology provides estimation equations for all of the key performance measures, it is also possible to assess LOS through direct field measurement of speed, flow rate, and percent followers at a specific point.

Base Conditions and Capacity The base conditions for two-lane highways are defined as:

• 12-ft minimum lane widths • 6-ft minimum paved shoulder widths • No restrictions on passing in the oncoming lane • Only passenger cars in the traffic stream • No direct access points along the roadway • No impediments to through traffic due to traffic control or turning

vehicles • Level terrain and straight alignment

Flow rates equivalent to capacity are rarely observed on two-lane highways, except possibly in short segments. Because service quality deteriorates at relatively low demand flow rates, most two-lane highways are upgraded before demand approaches capacity. Thus, the consideration of a capacity value for a two-lane highway operations analysis is rarely a concern. Nevertheless, being able to estimate capacity can be important for some planning situations, such as evacuation planning, special event planning, recreational routes, and evaluating the downstream impacts of incident bottlenecks once cleared.

The capacity of nonpassing lane segments, under base conditions, is 1,700 veh/h/ln. A distinction between the capacities of a Passing Zone and a Passing Constrained segment is not made, as the empirical evidence is currently too limited in this regard.

Passing lane segments can significantly improve traffic flow, especially on steep and long grades. However, the merging behavior of vehicles at the end of the passing lane can become problematic at high flow rates. Higher flow rates reduce the average gap between vehicles, which forces drivers to merge into smaller gaps. This behavior creates shockwaves, as following vehicles must decelerate for the merging vehicles. At some point, breakdown is reached, and the performance of the passing lane degrades below that of a nonpassing lane segment. Maximum flow rates for passing lane segments are a function of the heavy vehicle percentage and the vertical classification. For example, the capacity flow rate at the merge point on a vertical class 5 passing lane segment with 25% or higher trucks is approximately 1100 veh/h.

Level of Service If the demand-to-capacity ratio is less than or equal to 1.0, follower density is used as the service measure for all two-lane highways. However, two sets of LOS thresholds are used to account for differences in driver perception between driving on higher-speed versus lower-speed highways.

On higher-speed two-lane highways (≥ 50 mi/h), absolute speed and delay due to passing restrictions are generally both important to motorists. Higher-speed

6.6 Two-Lane Highways 227

two-lane highways are most commonly encountered as inter-city connecting routes. Lower-speed two-lane highways (< 50 mi/h) are typically encountered as intra-city routes and in scenic and rural-developed areas. These highways generally have posted speed limits of 35–45 mi/h and have limited passing opportunities. Thus, for two-lane highways in these areas, high speeds are usually not expected and higher percentages of followers are generally tolerated. Consequently, the follower density thresholds for a given LOS are higher for lower-speed highways than higher-speed highways. The LOS follower density thresholds for high-speed two-lane highways range from 2.0 followers/mi (LOS A/B) to 12.0 followers/mi (LOS E/F). For lower-speed two-lane highways, the thresholds range from 2.5 followers/mi (LOS A/B) to 15.0 followers/mi (LOS E/F).

At LOS A, motorists experience operating speeds near the posted speed limit and little difficulty in passing. Platooning is minimal, and follower density is very low. At LOS E, speeds may still be reasonable, but platooning is significant and follower density is high. Passing, if allowed, is essentially impossible. Conditions for LOS B, C, and D represent gradations between the conditions for LOS A and E. LOS F exists whenever demand flow exceeds the capacity of the segment.

Performance Measure Estimation A summary of the models used to estimate the performance measures used in this analysis methodology is provided in this section. The reader is referred to the HCM for the details on supporting equations, determination of equation coefficient values, and the specific sequence of steps in the analysis procedure.

Average Speed Average speed is computed from Eq. 6.16:

≤ =   − − >  

 

100

0.1 100 1,000

d

p d

d

FFS v

S v FFS m v

(6.16)

where

S = average speed in the analysis direction (mi/h), FFS = free-flow speed (mi/h),

v = flow rate in the analysis direction (veh/h),

m = slope coefficient, and p = power coefficient.

The slope (m) and (p) coefficients are calculated from supporting equations, which include one or more of the following independent variables: analysis direction flow rate, opposing direction flow rate (for passing zone segments), FFS, percent heavy vehicles, segment length, and coefficient values as a function of vertical class (1-5). If horizontal curves are present within the segment alignment, the average speed will be reduced through a supplementary calculation process.

The formulation of Eq. 6.16 allows for the two different shapes of curves as shown in Fig. 6.6—one for passing constrained and passing zone segments and

228 Chapter 6 Highway Capacity and Level-of-Service Analysis

one for passing lane segments. The shape of the speed versus flow rate for passing lane segments (Fig. 6.6b) has a shape more consistent with that for freeways and multilane highways, as expected, since passing lane segments have two lanes and allow for passing without entering the opposing lane.

Percent Followers The percent followers is computed from Eq. 6.17:

(6.17)

where

PF = percent followers in the analysis direction (%), v = flow rate in the analysis direction (veh/h),

m = slope coefficient, and p = power coefficient.

The slope (m) and (p) coefficients are calculated in a similar manner as for average speed.

Follower Density Follower density is computed from Eq. 6.18:

= × 100

dvPF FD

S (6.18)

where

FD = follower density in the analysis direction (followers/mi/ln), and other terms are as defined previously.

EXAMPLE 6.6 TWO-LANE HIGHWAY PASSING CONSTRAINED SEGMENT

A segment of two-lane highway has the following roadway characteristics:

• Passing constrained in the analysis direction • Segment Length (L) = 3,960 ft • Level Terrain (Vertical Class 1) • FFS = 62.7 mi/h (based on posted speed limit of 55 mi/h)

For these conditions, determine the follower density for an analysis direction flow rate (vd) of 800 veh/h (this accounts for the measured hourly volume and the PHF) that consists of only passenger cars (HV% = 0).

6.6 Two-Lane Highways 229

SOLUTION

Applying the full analysis methodology of the Highway Capacity Manual for these roadway characteristics and no heavy vehicles yields a speed–flow relationship as shown in Fig. 6.9. For an analysis direction flow rate of 800 veh/h, the resulting average speed (S) is 59.3 mi/h.

Figure 6.9 Average speed versus directional flow rate for Example 6.6.

Applying the full analysis methodology of the Highway Capacity Manual for these roadway characteristics and no heavy vehicles yields a percent followers-flow relationship as shown in Fig. 6.10. For an analysis direction flow rate of 800 veh/h, the resulting percent followers (PF) is 66.2%.

Figure 6.10 Percent followers versus directional flow rate for Example 6.6.

0 0

10

20

30

40

50

60 59.3

70

300 600 900

Flow Rate (veh/h)

A ve

ra g

e S

p ee

d (

m i/h

)

800 1200 1500 1800

100

90

80

70

P er

ce n

t F

o llo

w er

s (%

)

60

50

40

30

20

10

0 0 300 600 800900

Flow Rate (veh/h)

1200 1500 1800

66.2

230 Chapter 6 Highway Capacity and Level-of-Service Analysis

Substituting the average speed and percent followers values into Eq. 6.18 gives a follower density of:

= × =66.2 800 8.9 followers/mi

100 59.3 FD

This result is also illustrated in Fig. 6.11.

Figure 6.11 Percent follower density versus directional flow rate for Example 6.6.

F o

llo w

er D

en si

ty (

fo llo

w er

s/ m

ile )

50

45

40

35

30

25

20

15

10 8.9

5

0 0 300 600 800900

Flow Rate (veh/h)

1200 1500 1800

6.6 Two-Lane Highways 231

EXAMPLE 6.7 TWO-LANE HIGHWAY PASSING ZONE SEGMENT

A segment of two-lane highway has the following roadway characteristics:

• Passing zone in the analysis direction • Segment Length (L) = 5,280 ft • Vertical Class 2 • FFS = 62.7 mi/h (based on posted speed limit of 55 mi/h)

For these conditions, determine the follower density for an analysis direction flow rate (vd) of 700 veh/h, opposing direction flow rate (vo) of 500 veh/h, and 8% heavy vehicles in the analysis direction.

SOLUTION

Applying the full analysis methodology of the Highway Capacity Manual for these roadway characteristics, opposing direction traffic flow rate of 500 veh/h, and 8% heavy vehicles in the analysis direction yields a speed–flow relationship as shown in Fig. 6.12. For an analysis direction flow rate of 700 veh/h, the resulting average speed (S) is 58.5 mi/h.

Figure 6.12 Average speed versus directional flow rate for Example 6.7.

Applying the full analysis methodology of the Highway Capacity Manual for these roadway characteristics, opposing direction traffic flow rate of 500 veh/h, and 8% heavy vehicles in the analysis direction yields a speed–flow relationship as shown in Fig. 6.13. For an analysis direction flow rate of 700 veh/h, the resulting percent followers (PF) is 60.4%.

A ve

ra g

e S

p ee

d (

m i/h

)

70

60

50

40

30

20

10

0 0 300 600700 900

Flow Rate (veh/h)

1200 1500 1800

58.5

232 Chapter 6 Highway Capacity and Level-of-Service Analysis

Figure 6.13 Percent followers versus directional flow rate for Example 6.7.

Substituting the average speed and percent followers values into Eq. 6.18 gives a follower density of:

= × =60.4 700 7.2 followers/mi

100 58.5 FD

This result is also illustrated in Fig. 6.14.

Figure 6.14 Percent follower density versus directional flow rate for Example 6.7.

P er

ce n

t F

o llo

w er

s (%

)

100

90

80

70

60

50

40

30

20

10

0 0 300 600700 900

Flow Rate (veh/h) 1200 1500 1800

60.4

F o

llo w

er D

en si

ty (

fo llo

w er

s/ m

ile )

50

45

40

35

30

25

20

15

10

6

0 0 300 600

700 900

Flow Rate (veh/h) 1200 1500 1800

7.2

6.7 Design Traffic Volumes 233

Calculations for passing lane segments are more complicated than those for passing constrained and passing zone segments, especially in the context of a two-lane highway facility (multiple contiguous segments analyzed together). Performance improvements in roadway operations can persist well downstream of the end of a passing lane segment, particularly with respect to percent followers. Average speed also improves; however, this improvement is relatively minor and persists for a much shorter distance downstream, relative to follower- related improvements.

The distance downstream of the passing lane segment to which the operational improvements resulting from the passing lane applies is referred to as the effective length of the passing lane. Thus, segments downstream of a passing lane, and within this effective length, will have an adjustment applied to their base follower density result (assuming no upstream passing lane influence), and referred to as the adjusted follower density. For the passing lane segment itself, additional follower density calculations are performed. This is done to account for operating conditions within the passing lane segment to account for the two lanes, in addition to the results for the end of the passing lane immediately downstream of the two lanes merging together.

6.7 DESIGN TRAFFIC VOLUMES

In the preceding sections of this chapter, consideration was given to the determination of LOS, given some hourly volume. However, a procedure for selecting an appropriate hourly volume is needed to compute the LOS and to determine the number of lanes that need to be provided in a new roadway design to achieve some specified LOS. The selection of an appropriate hourly volume is complicated by two issues. First, there is considerable variability in traffic volume by time of day, day of week, time of year, and type of roadway. Figure 6.15 shows such variations in traffic volumes by hour of day and day of week for typical intra-city and inter-city routes. Figure 6.15 gives variations by time of year by comparing monthly percentages of the annual average daily traffic, AADT (in units of vehicles per day and computed as the total yearly traffic volume divided by the number of days in the year). The second concern is an outgrowth of the first: Given the temporal variability in traffic flow, what hourly volume should be used for design and/or analysis? To answer this question, consider the example diagram shown in Fig. 6.16. This figure plots hourly volume (as a proportion of AADT) against the cumulative number of hours that exceed this volume, per year. For example, the highest traffic flow in the year, on this sample roadway, would have an hourly volume of 0.148 × AADT (a volume that is exceeded by zero other hours). Sixty hours in the year would have a volume that exceeds 0.11 × AADT.

234 Chapter 6 Highway Capacity and Level-of-Service Analysis

Figure 6.15 Examples of hourly and daily traffic variations for intra- city and inter-city routes.

Figure 6.16 Example of monthly traffic volume variations for business and recreational access routes.

6.7 Design Traffic Volumes 235

Figure 6.17 Highest 100 hourly volumes over a one-year period for a typical roadway.

In determining the number of lanes that should be provided on a new or

redesigned roadway, it is obvious that using the worst single hour in a year (the hour with the highest traffic flow, which would be 0.148 × AADT from Fig. 6.16) would be a wasteful use of resources because additional lanes would be provided for a relatively rare occurrence. In contrast, if the 100th highest volume is used for design, the design LOS will be exceeded 100 times a year, which will result in considerable driver delay. Clearly, some compromise between the expense of providing additional capacity (such as additional lanes) and the expense of incurring additional driver delay must be made.

A common practice in the United States is to use a design hour-volume (DHV) that is between the 10th and 50th highest-volume hours of the year, depending on the type and location of the roadway (urban freeway, rural/suburban multilane highway, etc.), local traffic data, and engineering judgment. Perhaps the most common hourly volume used for roadway design is the 30th highest of the year. In practice, the K-factor is used to convert annual average daily traffic (AADT) to the 30th highest hourly volume. K is defined as

= DHV AADT

K (6.24)

where

K = factor used to convert annual average daily traffic to a specified annual hourly volume,

DHV = design hour-volume (typically, the 30th highest annual hourly volume), and

AADT = roadway’s annual average daily traffic in veh/day. For example, Fig. 6.16 shows that the K-value corresponding to the 30th highest hourly volume is 0.12. More generally, Ki can be defined as the K-factor corresponding to the ith highest annual hourly volume. Again, for example, the

236 Chapter 6 Highway Capacity and Level-of-Service Analysis

20th highest annual hourly volume would have a K-value, K20, of 0.126, from Fig. 6.16. If K is not subscripted, the 30th highest annual hourly volume is assumed (K = K30).

Finally, in the design and analysis of some highway types (such as freeways and multilane highways), the concern lies with directional traffic flows. Thus, a factor is needed to reflect the proportion of peak-hour traffic volume traveling in the peak direction. This factor is denoted D and is used to arrive at the directional design-hour volume (DDHV) by application of

= × ×DDHV AADTK D (6.25)

where

DDHV = directional design-hour volume,

D = directional distribution factor to reflect the proportion of peak- hour traffic volume traveling in the peak direction, and

Other terms are as defined previously.

EXAMPLE 6.8 DETERMINATION OF REQUIRED NUMBER OF FREEWAY LANES

A freeway is to be designed as a passenger-car–only facility for an AADT of 35,000 vehicles per day. It is estimated that the freeway will have a FFS of 70 mi/h. The peak- hour factor is estimated to be 0.85 with 65% of the peak-hour traffic traveling in the peak direction. Assuming that Fig. 6.16 applies, determine the number of lanes required to provide at least LOS C using the highest annual hourly volume and the 30th highest annual hourly volume.

SOLUTION

By inspection of Fig. 6.16, the highest annual hourly volume has K1 = 0.148. Application of Eq. 6.25 gives

= × × = × × =

1DDHV AADT

0.148 0.65 35,000 3367 veh/h

K D

The next step is to determine the maximum service flow rate that can be accommodated at LOS C for FFS = 70 mi/h. From Table 6.1, we see that this value is 1735 pc/h/ln. Thus, we must provide enough lanes so that the per-lane traffic flow is less than or equal to this value. We can use Eq. 6.7 to find vp, based on an assumed number of lanes. Comparing the calculated value of vp to the maximum service flow rate of 1735 will determine whether we have an adequate number of lanes. Assuming a four-lane freeway (two lanes in each direction), Eq. 6.7 gives

= = × ×

3367 1980.6 pc/h/ln

0.85 2 1.0pv

with

6.8 Practice Problems 237

V = 3367 (DDHV from above), and fHV = 1.0 (no heavy vehicles).

This value is higher than 1735, so we need to provide more lanes. The calculation is repeated, this time with an assumed six-lane freeway (three lanes in each direction):

= = × ×

3367 1320.4 pc/h/ln

0.85 3 1.0pv

Since this value is less than 1735, a six-lane freeway is necessary to provide LOS C operation for the design traffic flow rate. For the 30th highest hourly annual volume, Fig. 6.16 gives K30 = K = 0.12, which when used in Eq. 6.25 gives

= × × = × × =

DDHV AADT

0.12 0.65 35,000 2730 veh/h

K D

Again applying Eq. 6.7, with an assumed four-lane freeway, yields

= = × ×

2730 1605.9 pc/h/ln

0.85 2 1.0pv

This value is less than 1735, so a four-lane freeway (two lanes each direction) is adequate for this design traffic flow rate. This example demonstrates the impact of the chosen design traffic flow rate on roadway design. Only a four-lane freeway is necessary to provide LOS C for the 30th highest annual hourly volume, as opposed to a six-lane freeway needed to satisfy the LOS requirement for the highest annual hourly volume.

6.8 PRACTICE PROBLEMS

PRACTICE PROBLEM 6.1

BASIC FREEWAY SEGMENT LOS WITH A SPECIFIC GRADE

A northbound freeway segment is on a 4% upgrade from station 430 + 60 to 397 + 60 and has three 11 ft-wide lanes, a 5 ft right shoulder and has a ramp density of 1 per mile in the 3 miles before and after the mid-point of this freeway segment. The peak- hour factor is 0.9 and northbound traffic during the peak hour is 2400 vehicles, with 20% heavy vehicles (and a 50% SUT/50% TT mix). Determine the density and LOS of the freeway segment.

238 Chapter 6 Highway Capacity and Level-of-Service Analysis

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

We begin by determining the FFS according to Eq. 6.6,

FFS = 75.4 – fLW – fRLC − 3.22TRD 0.84.

Here, from Table 6.3 fLW = mi/h, from Table 6.4 fRLC = mi/h (Table 6.4), and TRD = 1 ramp/mi (given), so

FFS = 75.4 – – – 3.22(1)0.84 = 69.9 mi/h.

Next, to determine the analysis flow rate vp, we begin noting that the length of this freeway segment is,

430 + 60 minus 397 + 60 = ft. This is 0.625 mi ( /5280) and, because it is on an upgrade that exceeds 3% and is longer the 0.25 mi, extended segment analysis cannot be used.

Using values in Table 6.7 (50% SUTs/50% TTs mix) with 20% heavy vehicles and a section length of 0.625 mi, we find ET = at a 3.5% grade and ET = at a 4.5% grade. Interpolating, we find ET = 2.52 [( + )/2] at a grade of 4%. Applying Eq. 6.9 to arrive at the heavy-vehicle adjustment factor, we get,

( )= = + −

1

1 0.20 1 HVf

So, from Eq. 6.7, the analysis flow rate is,

= = × ×

1159 pc/h/ln

pv .

To determine the average speed (S), calculate the breakpoint (BP) value for the speed– flow relationship, from Eq. 6.4 as,

( ) = + − = 1000 40 75 pc/h/lnBP

Since the vp of 1159 is less than the BP value of , S will be equal to the FFS, mi/h. With this, the density can be calculated using Eq. 6.10 as,

= = 16.6 pc/mi/ln

D .

From Table 6.2, it can be seen that this corresponds to LOS B [ (max density for LOS B) < 16.6 < (max density for LOS B)]. Thus, this freeway segment operates at LOS B.

6.8 Practice Problems 239

PRACTICE PROBLEM 6.2

BASIC FREEWAY SEGMENT LOS WITH A SPECIFIC GRADE AND INCREASING VOLUME

Consider the freeway and conditions in Practice Problem 6.1. Suppose that traffic increases by 50% and that the peak-hour factor increases from 0.9 to 0.95 (as the increase in traffic makes the traffic flow rate more uniform over the peak hour). The proportion of heavy vehicles remains the same as do all other elements of the problem. Determine the new density and LOS of the freeway segment.

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

To determine the new analysis flow rate (vp), note that the free-flow speed (FFS) and heavy-vehicle adjustment factor (fHV) will remain unchanged from Practice Problem 6.1. The new hourly volume will be V = 3600 veh/h (2400 × 1.5) and, with the new PHF = 0.95, the new analysis flow rate from Eq. 6.7 is,

= = × ×

1647 pc/h/ln

pv

As in Practice Problem 6.1, to determine the average speed (S), the breakpoint (BP) value for the speed–flow relationship is calculated from Eq. 6.4 as,

( ) = + − = 1000 40 75 pc/h/ln.BP

Unlike Practice Problem 6.1, the vp of 1647 is greater than the BP value of so S must be computed from Eq. 6.2. To apply Eq. 6.2, we must compute the capacity c (and also check that this value of vp is less than the capacity to ensure Eq. 6.3 can be applied). For capacity, using Eq. 6.3,

( ) = + − ≤ ≤  

Min 2200 10 50 ,2400 55 75,c FFS

which gives c = , so vp ≤ c. Applying Eq. 6.3 gives,

( ) ( )

  − −  

 = − < ≤ −

2.0

2.0

45

pS BP v c

or S = 67.6 mi/h. With this, density is given in Eq. 6.10 as,

= = 24.4 pc/mi/ln

D

From Table 6.2, it can be seen that this density corresponds to LOS C [ (max

density for LOS B) < 26.86 < (max density for LOS C)]. Thus, this freeway segment now operates at LOS C with the increased traffic.

240 Chapter 6 Highway Capacity and Level-of-Service Analysis

PRACTICE PROBLEM 6.3

MULTILANE HIGHWAY SEGMENT LOS WITH A SPECIFIC GRADE

An undivided multilane highway segment has two 11-ft lanes in the eastbound direction with no shoulders and a 55 mi/h speed limit. This highway segment has 40 access points on a 1.25 mile, 2.5% upgrade. During the highest 15 minutes of traffic- flow within the peak hour, there are 755 vehicles and 12% of these are heavy vehicles with a 70%/30% mix of single-unit and tractor-trailer trucks. What are the estimated speed, density, and LOS of upgrade?

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

We begin by determining the FFS according to Eq. 6.14,

= − − − −LW TLC M AFFS BFFS f f f f

In this equation,

BFFS = mi/h (the speed limit pus 5 mi/h),

fLW = mi/h (Table 6.3),

fTLC = mi/h (with TLC = 6 since undivided with 0 ft right shoulder, given),

fM = mi/h (Table 6.12), and

fA = mi/h (40/1.25 × 0.25, since 0.25 mi/h is lost for every access point per mile), (Table 6.13).

With these values, the FFS is computed as,

= − − − − = 47.2 mi/h.FFS

Moving on to the determination of the analysis flow rate vp, we begin noting that the length of this freeway segment is an upgrade of 1.25 mi in length at 2.5% which is a grade less than or equal to 3% that extends for more than 1 mi, so extended segment analysis cannot be used (see discussion in Section 6.5.5). Thus, using values in Table 6.8 (70% SUTs/30% TTs mix) with 12% heavy vehicles and a section length

of 1.25 mi, we find ET = at a 2.5% grade with 10% heavy vehicles, and

ET = at a 2.5% grade with 15% heavy vehicles. Interpolating, we find ET =

[ − ( ) × / ]. Applying Eq. 6.9 to arrive at the heavy-vehicle adjustment factor we get,

( )= = + −

1

1 0.12 1 HVf

Because we are given V15 = 755, Eq. 6.7 can be written as,

× =

× 15 4

p HV

V v

N f .

6.8 Practice Problems 241

So, substituting from Eq. 6.7,

×= = ×

4 1782 pc/h/ln

pv .

As shown in the text (see Eqs. 6.11 and 6.12) and illustrated in Fig. 6.4, the break point (BP) for multilane highways is always 1400 pc/h/ln. Since our vp = 1782 > 1400, Eq. 6.12 needs to be applied to determine the speed, S. We begin by determining the capacity (needed to solve for Speed in Eq. 6.12), using Eq. 6.13,

( )Min 1900 20 45 ,2300 45 70,c FFS = + − ≤ ≤  

which gives c = , so vp ≤ c. Applying Eq. 6.12 gives,

( ) ( )

  − −  

 = − < ≤ −

1.31

1.31

1400

45 1400

1400 pS v c

or S = 44.7 mi/h. With this, density is given in Eq. 6.10 as,

= = 39.9 pc/mi/ln

D

From Table 6.2, it can be seen that this density corresponds to LOS E [ (max

density for LOS D) < 39.9 < (max density for LOS E)]. Thus, this freeway segment now operates at LOS E with the increased traffic.

242 Chapter 6 Highway Capacity and Level-of-Service Analysis

NOMENCLATURE FOR CHAPTER 6

AADT annual average daily traffic BFFS estimated free-flow speed for base

conditions c roadway capacity D density or factor for directional

distribution of traffic DHV design-hour volume DDHV directional design-hour volume ER passenger car equivalent value for

recreational vehicles ET passenger car equivalent value for large

trucks and buses FD follower density (two-lane highways) FFS measured or estimated free-flow speed fA free-flow speed adjustment factor for

access point frequency (multilane highways)

fHV heavy-vehicle adjustment factor fID free-flow speed adjustment factor for

interchange density (freeways) fRLC free-flow speed adjustment factor for

lateral clearance (freeways and multilane highways)

fLW free-flow speed adjustment factor for lane width (freeways and multilane highways)

fM free-flow speed adjustment factor for median type (multilane highways)

Ki factor used to convert AADT to ith highest annual hourly volume

LCL left-side lateral clearance (multilane highways)

LCR right-side lateral clearance (multilane highways)

N number of lanes in one direction PCE passenger car equivalent PHF peak-hour factor PF percent followers (two-lane highways) S average speed TLC total lateral clearance (multilane

highways) v analysis flow rate V hourly volume V15 highest 15-minute volume v/c volume-to-capacity ratio

REFERENCE

Transportation Research Board. Highway Capacity Manual. 6th Edition. National Academy of Sciences, Washington, DC. 2016.

P-24

CHAPTER 7 PROBLEMS

Development of Signal Phasing and Timing Plans (Section 7.4)

7.1 An intersection has a three-timing-stage signal with the movements allowed in each timing stage and corresponding analysis and saturation flow rates shown in Table 7.6. Calculate the sum of the flow ratios for the critical lane groups.

7.2 An intersection has a four-timing-stage signal with the movements allowed in each timing stage and corresponding analysis and saturation flow rates shown in Table 7.7. Calculate the sum of the flow ratios for the critical lane groups.

7.3 The minimum cycle length for an intersection is determined to be 95 seconds. The critical lane group flow ratios were calculated as 0.235, 0.250, 0.170, and 0.125 for timing stages 1–4, respectively. What Xc was used in the determination of this cycle length, assuming a lost time of 5 seconds per timing stage?

7.4 A pretimed four-timing-stage signal has critical lane group flow rates for the first three timing stages of 200, 187, and 210 veh/h (saturation flow rates are 1800 veh/h/ln for all timing stages). The lost time is known to be 4 seconds for each timing stage. If the cycle length is 60 seconds, what is the estimated effective green time of the fourth timing stage?

7.5 A four-timing-stage traffic signal has critical lane group flow ratios of 0.225, 0.175, 0.200, and 0.150. If the lost time per timing stage is 5 seconds and a critical intersection v/c of 0.85 is desired, calculate the minimum cycle length and the timing

stage effective green times such that the lane group v/c ratios are equalized.

7.6 For Problem 7.1, calculate the minimum cycle length and the effective green time for each timing stage (balancing v/c for the critical lane groups). Assume the lost time is 4 seconds per timing stage and a critical intersection v/c of 0.90 is desired.

7.7 For Problem 7.1, calculate the optimal cycle length (Webster’s formulation) and the corresponding effective green times (based on lane group v/c equalization). Assume lost time is 4 seconds per timing stage.

7.8 For Problem 7.2, calculate the minimum cycle length and the effective green time for each timing stage (balancing v/c for the critical movements). Assume the lost time is 4 seconds per timing stage and a critical intersection v/c of 0.95 is desired.

7.9 For Problem 7.2, calculate the optimal cycle length (Webster’s formulation) and the corres- ponding effective green times (based on lane group v/c equalization). Assume lost time is 4 seconds per timing stage.

7.10 Consider Example 7.3. Two additional 12-ft through lanes are added to Vine Street (the street in the intersection shown in Fig. 7.8), one lane in each direction. If the peak-hour traffic volumes are unchanged but the Vine Street left-turn saturation flow rates increase by 100 veh/h because of the added through lanes, what would the revised effective green time, yellow time, and all-red time be for each timing stage? Assume minimum cycle

Table 7.6 Data for Problem 7.1

Timing stage 1 2 3

Allowed movements NB L, SB L NB T/R, SB T/R EB L, WB L EB T/R, WB T/R

Analysis flow rate 330, 365 veh/h 1125, 1075 veh/h 110, 80 veh/h 250, 285 veh/h

Saturation flow rate 1700, 1750 veh/h 3400, 3300 veh/h 650, 600 veh/h 1750, 1800 veh/h

Table 7.7 Data for Problem 7.2

Timing stage 1 2 3 4

Allowed movements EB L, WB L EB T/R, WB T/R SB L, SB T/R NB L, NB T/R

Analysis flow rate 245, 230 veh/h 975, 1030 veh/h 255, 235 veh/h 225, 215 veh/h

Saturation flow rate 1750, 1725 veh/h 3350, 3400 veh/h 1725, 1750 veh/h 1700, 1750 veh/h

Chapter 7 Problems P-25

length and a critical intersection v/c of 0.90 is desired.

7.11 Consider the intersection of Vine and Maple Streets as shown in Fig. 7.8. Suppose Vine Street’s northbound and southbound approaches are both on an 8% upgrade, and the assumed vehicle approach speed is 30 mi/h. What should the yellow and all-red times be?

7.12 Consider Problem 7.10. Calculate the new minimum pedestrian green time, assuming the effective crosswalk width is 6 ft and the maximum number of crossing pedestrians in any timing stage is 20.

Analysis of Traffic at Signalized Intersections (Section 7.5)

7.13 An intersection approach has a saturation flow rate of 1500 veh/h, and vehicles arrive at the approach at the rate of 800 veh/h. The approach is controlled by a pretimed signal with a cycle length of 60 seconds and D/D/1 queuing holds. Local standards dictate that signals should be set such that all approach queues dissipate 10 seconds before the end of the effective green portion of the cycle. Assuming that approach capacity exceeds arrivals, determine the maximum length of effective red that will satisfy local standards.

7.14 An approach to a pretimed signal has 30 seconds of effective red, and D/D/1 queuing holds. The total delay at the approach is 83.33 veh-s/cycle and the saturation flow rate is 1000 veh/h. If the capacity of the approach equals the number of arrivals per cycle, determine the approach flow rate and cycle length.

7.15 An approach to a pretimed signal has 25 seconds of effective green in a 60-second cycle. The approach volume is 500 veh/h and the saturation flow rate is 1400 veh/h. Calculate the average vehicle delay for the approach assuming D/D/1 queuing.

7.16 An observer notes that an approach to a pretimed signal has a maximum of 8 vehicles in a queue in a given cycle. If the saturation flow rate is 1440 veh/h and the effective red time is 40 seconds, how much time will it take this queue to clear after the start of the effective green (assuming that approach capacity exceeds arrivals and D/D/1 queuing applies)?

7.17 An approach to a pretimed signal with a 60- second cycle has 9 vehicles in the queue at the beginning of the effective green. Four of the 9

vehicles in the queue are left over from the previous cycle (at the end of the previous cycle’s effective green). The saturation flow rate of the approach is 1500 veh/h, total delay for the cycle is 5.78 vehicle minutes, and at the end of the effective green there are 2 vehicles left in the queue. Determine the arrival rate assuming that it is unchanged over the duration of the observation period (from the beginning to the end of this 5.78–vehicle-minute delay cycle). (Assume D/D/1 queuing.)

7.18 At the beginning of an effective red, vehicles are arriving at an approach at the rate of 500 veh/h and 16 vehicles are left in the queue from the previous cycle (at the end of the previous cycle’s effective green). However, due to the end of a major sporting event, the arrival rate is continuously increasing at a constant rate of 200 veh/h/min (after 1 minute the arrival rate will be 700 veh/h, after 2 minutes 900 veh/h, etc.). The saturation flow rate is 1800 veh/h, the cycle length is 60 seconds, and the effective green time is 40 seconds. Determine the total vehicle delay on this approach until complete queue clearance. (Assume D/D/1 queuing.)

7.19 The saturation flow rate for an intersection approach is 3600 veh/h. At the beginning of a cycle (effective red) no vehicles are queued. The signal is timed so that when the queue (from the continuously arriving vehicles) is 13 vehicles long, the effective green begins. If the queue dissipates 8 seconds before the end of the cycle and the cycle length is 60 seconds, what is the approach’s arrival rate, assuming D/D/1 queuing?

7.20 The saturation flow rate for a pretimed signalized intersection approach is 1800 veh/h. The cycle length is 80 seconds. It is known that the arrival rate during the effective green is twice the arrival rate during the effective red. During one cycle, there are 2 vehicles in the queue at the beginning of the cycle (the beginning of the effective red) and there are 8 vehicles in the queue at the end of the effective red (the beginning of the effective green). If the queue clears exactly at the end of the effective green and D/D/1 queuing applies, determine the total vehicle delay on the approach in the cycle (in veh-s).

7.21 An approach to a signalized intersection has a saturation flow rate of 1800 veh/h. At the beginning of an effective red, there are 6 vehicles in the queue and vehicles arrive at 900 veh/h. The signal has a 60-second cycle with 25 seconds of effective red. What is the total vehicle delay on the approach after one cycle (assume D/D/1 queuing)?

P-26 Chapter 7 Problems

7.22 An approach to a signalized intersection has a saturation flow rate of 2640 veh/h. For one cycle, the approach has 3 vehicles in queue at the beginning of an effective red, and vehicles arrive at 1064 veh/h. The signal for the approach is timed such that the effective green starts 8 seconds after the approach’s vehicle queue reaches 10 vehicles, and lasts 15 seconds. What is the total vehicle delay on the approach for this signal cycle?

7.23 An approach to a signal has a saturation flow rate of 1800 veh/h. During one 80-second cycle, there are 4 vehicles queued at the beginning of the cycle (the start of the effective red) and 2 vehicles queued at the end of the cycle (the end of the effective green). At the beginning of the effective green there are 10 vehicles in the queue. The arrival rate is constant and the process is D/D/1. If the effective red is known to be less than 40 seconds, what is the total vehicle delay on this approach for this signal cycle?

7.24 Vehicles arrive at an approach to a pretimed signalized intersection. The arrival rate over the cycle is given by the function v(t) = 0.22 + 0.012t [v(t) is in veh/s and t is in seconds]. There are no vehicles in the queue when the cycle (effective red) begins. The cycle length is 60 seconds and the saturation flow rate is 3600 veh/h. Determine the effective green and red times that will allow the queue to clear exactly at the end of the cycle (the end of the effective green), and determine the total vehicle delay for this approach over the cycle (assuming D/D/1 queuing).

7.25 At the start of the effective red at an intersection approach to a pretimed signal, vehicles begin to arrive at a rate of 800 veh/h for the first 40 seconds and 500 veh/h from then on. The approach has a saturation flow rate of 1200 veh/h and an effective green of 20 seconds, and the cycle length is 40 seconds. What is the approach’s total vehicle delay two full cycles after the 800-veh/h arrival rate begins? (Assume D/D/1 queuing.)

7.26 A left-turn movement has a maximum arrival rate of 200 veh/h. The saturation flow of this movement is 1400 veh/h. For this approach, the yellow time is 4 seconds, all-red time is 2 seconds, and total lost time is 3 seconds. The cycle length is 120 seconds. What minimum displayed green time must be provided to ensure that the queue in each cycle clears, and what is the movement’s total delay per cycle and delay per vehicle for this green time? (Assume D/D/1 queuing.)

7.27 Vehicles begin to arrive at a signal approach at a rate of v(t) = 0.2286 + 0.0008t [with v(t) in veh/s and t in seconds] at the beginning of the cycle (the beginning of an effective red) and there are 2 vehicles already in the queue that are left over from the effective green of the previous cycle. The signal is designed so that the effective green starts when there are 10 vehicles in the queue. The saturation flow rate is 1800 veh/h. What is the approach’s total vehicle delay after one cycle (cycles are 60 seconds long) and when will the effective green start in cycle #2 (or, equivalently, how long will the effective red be in cycle #2)? (Assume D/D/1 queuing.)

7.28 At the beginning of a signal approach’s effective red there are 8 vehicles in queue. The arrival rate over the cycle is v(t) = 0.05 + 0.001t [with v(t) in veh/s and t in seconds]. If the saturation flow rate is 1800 veh/h and the cycle length is 80 seconds, what is the minimum effective green time needed for this cycle to have zero vehicles in the approach’s queue when the effective red of the next cycle starts and what is approach’s the total delay in this cycle with this green time. (Assume D/D/1 queuing.)

7.29 A signal approach has 20 seconds of displayed green, 4 seconds of yellow, and 3 seconds of all-red (start-up lost time and clearance times are typical). The cycle length is 60 seconds. At the beginning of an effective red there are no vehicles in queue and vehicles arrive at three-quarters of the saturation flow rate for 30 seconds. Then there is zero flow for 10 seconds, and then one-half of the saturation flow rate from 40 seconds until the end of the cycle. What is the approach’s total delay for this cycle if the saturation flow rate is 1400 veh/h? (Assume D/D/1 queuing.)

7.30 An approach with a saturation flow rate of 1800 veh/h has 3 vehicles in queue at the start of an effective red. For the first cycle, the approach arrival rate is given by the function v(t) = 0.5 – 0.005t [with v(t) in veh/s and t in seconds measured from the beginning of the effective red]. From the second cycle onward (starting at the beginning of the second effective red) vehicles arrive at a fixed rate of 720 veh/h. The approach has 26 seconds of effective red and a 60 second cycle for all cycles. How many cycles will it take to have no vehicles in the queue at the start of an effective red for this approach and what would be the total delay for the approach until this happens? (Assume D/D/1 queuing.)

Chapter 7 Problems P-27

7.31 Vehicles arrive at a signal approach at a rate of v(t) = 0.3 – 0.001t [with v(t) in veh/s and t in seconds measured from the beginning of the effective red of the first cycle]. The signal has a 70-second cycle length with 40 seconds of effective red. The saturation flow rate of the approach is 1800 veh/h. What is the approach’s total vehicle delay after two cycles (when t = 140 seconds) and when will the queue clear (measured from the beginning of the first cycle) during an effective green (that is, the t at which there will no longer be a queue)? (Assume D/D/1 queuing.)

7.32 An approach to a signalized intersection has a green time of 35 seconds, and all-red time of 2 seconds, a yellow time of 3 seconds, and a total lost time of 3 seconds. The arrival rate is v(t) = 0.5 + 0.002t [with v(t) in veh/s and t in seconds measured from the beginning of the effective red of the cycle]. The saturation flow rate of the approach is 3400 veh/h. How long must the cycle length be so that the queue that forms at the beginning of the cycle (effective red) dissipates exactly at the end of the cycle (end of the effective green) and what would be the average delay per vehicle over the cycle? (Assume D/D/1 queuing.)

7.33 Recent computations at an approach to a pretimed-signalized intersection indicate that the volume-to-capacity ratio is 0.8, the saturation flow rate is 1600 veh/h, and the effective green time is 50 seconds. If the uniform delay (assuming D/D/1 queuing) is 11.25 seconds per vehicle, determine the arrival flow rate (in veh/h) and the cycle length.

7.34 An approach to a pretimed signal is timed so that the effective green starts 4 seconds after 13 vehicles are in the queue. The cycle length is 70 seconds and the arrival rate for a specific cycle is v(t) = 0.52 – 0.003t [with v(t) in veh/s and t in seconds measured from the beginning of the effective red of the cycle]. If there are 3 vehicles in the queue at the beginning of a specific cycle (effective red) and the queue clears 4 seconds before the end of the cycle (effective green), what is the saturation flow of the approach and the approach’s total delay over the cycle? (Assume D/D/1 queuing.)

7.35 At an intersection approach, at the beginning of an effective red there are 10 vehicles in a queue. Vehicles arrive at a constant rate of 600 veh/h and the saturation flow of the approach is 1800 veh/h. the cycle length is 60 seconds and the approach gets 30 seconds of effective green. What will the total

delay be (in veh-s) after two cycles at the approach? (Assume D/D/1 queuing.)

7.36 An approach at a signalized intersection with a 70-second cycle gets 25 seconds of displayed green time. Yellow time is 4 seconds and all-red is 2 seconds (lost time is to be determined from standard assumptions). At the beginning of an effective red there are 4 vehicles in the queue and the saturation flow is 818 veh/h. The arrival rate is given as v(t) = 0.17 – 0.002t [with v(t) in veh/s and t in seconds after the beginning of the effective red]. What is the total vehicle delay for this approach at the end of the cycle (until the next effective red) that started with the 4 vehicles queued at the beginning of the effective red? (Assume D/D/1 queuing.)

7.37 At the beginning of an effective red, an approach has 7 vehicles in a queue with vehicles arriving according to the function v(t) = 0.2 + 0.00111t [with v(t) in veh/s and t in seconds after the beginning of the effective red]. The saturation flow is 1800 veh/h, the cycle length is 60 seconds, and the effective red is 20 seconds. What is the total delay for the approach until the queue first clears? (Assume D/D/1 queuing.)

7.38 At one signalized intersection approach, 20 vehicles, on average, arrive during the 30-second effective green time. During the rest of the cycle, 45 vehicles, on average, arrive at the intersection from this approach. The cycle length is 90 seconds. What is the proportion of vehicles arriving on green for this approach?

7.39 Consider Problem 7.38. Assume the saturation flow rate is 8000 veh/h. Determine the average uniform delay for this approach.

7.40 Consider Problems 7.38 and 7.39. Assume that improvements were made to the signal timing such that the PVG for this approach is now 0.45. Also assume that the average number of vehicle arrivals per cycle is still 65 (but the arrivals on green and red will be different from Prob. 7.34). Determine the new average uniform delay for this approach.

7.41 An approach to a pretimed signal has 25 seconds of effective green, a saturation flow rate of 1300 veh/h, and a volume-to-capacity ratio less than 1. If the cycle length is 60 seconds and the overall delay formula (Eq. 7.27) estimates an average delay that is 34 s greater than that estimated by using just the uniform delay formula, determine the vehicle arrival rate. (Assume the signal is isolated and d3 = 0.)

P-28 Chapter 7 Problems

7.42 For Problem 7.6, calculate the northbound average approach delay (using Eq. 7.27) and level of service.

7.43 For Problem 7.6, calculate the southbound average approach delay (using Eq. 7.27) and level of service.

7.44 For Problem 7.6, calculate the westbound average approach delay (using Eq. 7.27) and level of service.

7.45 For Problem 7.6, calculate the eastbound average approach delay (using Eq. 7.27) and level of service.

7.46 For Problem 7.6, calculate the overall intersection average delay (using Eqs. 7.27−7.29) and level of service.

7.47 For Problem 7.8, calculate the northbound average approach delay (using Eq. 7.27) and level of service.

7.48 For Problem 7.8, calculate the southbound average approach delay (using Eq. 7.27) and level of service.

7.49 For Problem 7.8, calculate the westbound average approach delay (using Eq. 7.27) and level of service.

7.50 For Problem 7.8, calculate the eastbound average approach delay (using Eq. 7.27) and level of service.

7.51 For Problem 7.8, calculate the overall intersection average delay (using Eqs. 7.27−7.29) and level of service.

7.52 A new shopping center opens near the intersection of Vine and Maple Streets (the intersection shown in Fig. 7.8). The net effect is to increase the approaching traffic volumes by 10%. Calculate the new level of service for the westbound approach, assuming all else remains the same. Assume other required input values are as in Table 7.5. Use Eq. 7.27 for the delay calculation.

7.53 For Problem 7.52, calculate the new level of service for the northbound approach.

7.54 Calculate the overall intersection level of service for Problem 7.10. Assume other required input values are as in Table 7.5. Use Eqs. 7.27−7.29 for the delay calculations.

7.55 Consider Problem 7.10. How much traffic volume can be added to the southbound approach

(assuming the same turning movement percentage) before LOS D is reached for the approach? Use Eq. 7.27 for the delay calculation.

7.56 Consider Problem 7.10. How much traffic volume must be diverted from the eastbound approach (assuming the same turning movement percentage) to achieve LOS B for the approach? Use Eq. 7.27 for the delay calculation.

Multiple Choice Problems (Multiple Sections)

7.57 A signalized intersection has a cycle length of 70 seconds. For one traffic movement, the displayed all-red time is set to 2 seconds while the displayed yellow time is 5 seconds. The effective red time is 37 seconds and the total lost time per cycle for the movement is 4 seconds. What is the displayed green time for the traffic movement?

a) 30 s b) 31 s c) 33 s d) 65 s

7.58 An isolated pretimed signalized intersection has an approach with a saturation flow rate of 1900 veh/h. For this approach, the displayed red time is 58 seconds, the displayed yellow time is 3 seconds, the all-red time is 2 seconds, the effective green time is 28 seconds, and the total lost time is 4 s/timing stage. What is the average uniform delay per vehicle when the approach flow rate is 550 veh/h?

a) 26.3 s b) 29.7 s c) 30.1 s d) 413.3 s

7.59 An isolated pretimed signalized intersection has an approach with a traffic flow rate of 750 veh/h and a saturation flow rate of 3200 veh/h. This approach is allocated 32 seconds of effective green time. The cycle length is 100 seconds. Determine the average approach delay (using Eq. 7.27).

a) 4.6 s b) 30.2 s c) 34.8 s d) 35.0 s

7.60 Calculate the sum of flow ratios for the critical lane groups for the three-phase timing plan, with traffic and saturation flow rates shown in the following tables.

Chapter 7 Problems P-29

Traffic Flow Rates (veh/h) for Problem 7.60

Timing Stage 1 Timing Stage 2 Timing Stage 3

EB L: 250 EB T/R: 1200 SB L: 75

NB L: 100

WB L: 300 WB T/R: 1350 SB T/R: 420

NB T/R: 425

Saturation Flow Rates (veh/h) for Problem 7.60

Timing Stage 1 Timing Stage 2 Timing Stage 3

EB L: 1800 EB T/R: 3600 SB L: 500

NB L: 525

WB L: 1800 WB T/R: 3600 SB T/R: 1950

NB T/R: 1950

a) 0.950 b) 0.760 c) 0.690 d) 0.622

7.61 A signalized intersection has a sum of critical flow ratios of 0.72 and a total cycle lost time of 12 seconds. Assuming a critical intersection v/c ratio of 0.9, calculate the minimum necessary cycle length.

a) 48.0 s b) 60.0 s c) 82.1 s d) 42.9 s

7.62 A signalized intersection approach has an upgrade of 4%. The total width of the cross street at this intersection is 60 feet. The average vehicle length of approaching traffic is 16 feet. The speed of approaching traffic is 40 mi/h. Determine the sum of the minimum necessary change and clearance intervals.

a) 3.59 s b) 4.96 s c) 4.89 s d) 2.51 s

243

Chapter 7

Traffic Control and Analysis at Signalized Intersections

7.1 INTRODUCTION

Due to conflicting traffic flows, highway intersections are of great concern to traffic engineers. Intersections can be a major source of crashes and vehicle delays (as vehicles yield to avoid conflicts with other vehicles). Most highway intersections are not signalized due to low traffic volumes and adequate sight distances. However, at some point, traffic volumes, accident frequencies and their resulting injury severities, as well as other factors, reach a level that warrants the installation of a traffic signal.

The installation and operation of a traffic signal to control conflicting traffic and pedestrian flows (also referred to as movements) at an intersection has advantages and disadvantages. The advantages include a potential reduction in some types of crashes (particularly angle crashes), provisions for pedestrians to cross the street, provisions for side-street vehicles to enter the traffic stream, provisions for the progressive flow of traffic in a signal-system corridor, and possible improvements in capacity and reductions in delays. However, signals are by no means the perfect solution for delay or potential crash problems at an intersection. A poorly timed signal or one that is not justified can have a negative impact on the operation of the intersection by increasing vehicle delay, increasing the rate of vehicle crashes (particularly rear-end crashes), causing a disruption in traffic progression (adversely impacting the through movement of traffic), and encouraging the use of routes not intended for through traffic (such as routes through residential neighborhoods). Traffic signals are also costly to install, with some basic signal installations costing in excess of $100,000. Therefore, the decision to install a signal must be weighed and studied carefully. To assist transportation engineers in this process, the Federal Highway Administration of the U.S. Department of Transportation publishes the Manual on Uniform Traffic Control Devices (MUTCD) [U.S. Federal Highway Administration, 2009], which contains a section on warrants for the installation of a traffic signal. There are a total of nine warrants, which include consideration of vehicle volumes, pedestrian volumes, school crossings, signal coordination, and crash experience. The reader is referred to the MUTCD for details on these warrants.

Unlike uninterrupted flow, in which vehicle flow is affected only by other vehicles and the highway environment, the introduction of a traffic control

244 Chapter 7 Traffic Control and Analysis at Signalized Intersections

device such as a signal exerts a significant influence on the flow of vehicles. Thus, the analysis of traffic flow at signalized intersections can become very complex. This chapter will make several simplifying assumptions to keep the material at an accessible level.

The chapter begins by providing an overview of the physical elements of intersection configuration and traffic signal control. A basic understanding of these principles provides the foundation for designing intersection geometry and traffic movement sequence plans. This is followed by a presentation of concepts, definitions, and analytical techniques that are used in the design and analysis of signal timing plans at signalized intersections.

7.2 INTERSECTION AND SIGNAL CONTROL CHARACTERISTICS

An intersection is defined as an at-grade crossing of two or more roadways. For analysis, the roadways entering the intersection are segmented into approaches, which are defined by lane groups (groups of one or more lanes). These lane groups are usually based on the allowed movements (left, through, right) within each lane and the sequencing of allowed movements by the traffic signal. The establishment of lane groups will be discussed in more detail in Section 7.4.2.

To illustrate these concepts, note that approach 1 of the intersection depicted in Fig. 7.1 consists of a lane for the exclusive use of left turns, a lane for the exclusive use of right turns, and two lanes for the exclusive use of through movements. Approach 3 is similar to approach 1 but does not include an exclusive right-turn lane; instead, the right turns share the outside lane with the through movements. Because the lanes for the exclusive use of left and right turns are short, they are usually referred to as bays and are intended to hold a limited number of queued vehicles. Queuing analysis can be used to determine the length of bay necessary to prevent queued turning vehicles from overflowing the bay and blocking the through lanes (known as spillover) and/or the length necessary to prevent queued through vehicles from blocking the entrance of the turn bay (known as spillback). Approach 2 consists of a shared through/right- turn lane and an exclusive left-turn lane (not a bay in this case because it has the same length as the adjacent lane). Approach 4 is similar to approach 2, but the inside lane is a shared through/left-turn lane.

From a driver’s perspective, a traffic signal is just a collection of light- emitting devices [usually incandescent bulbs or light-emitting diodes (LEDs)] and lenses that are housed in casings of various configurations (referred to as signal heads) whose purpose is to display red, yellow, and green full circles and/or arrows. Figure 7.2 shows typical configurations of signal heads in the United States. These signal heads are usually mounted to mast arms or wire spanned across the intersection.

The following terminology is commonly used in the design of traffic signal controls.

Indication. The illumination of one or more signal lenses (greens, yellows, reds) indicating an allowed or prohibited traffic movement.

7.2 Intersection and Signal Control Characteristics 245

Figure 7.1 Typical signalized intersection elements.

Figure 7.2 Typical signal head configurations in the United States (G = green; Y = yellow; R = red).

246 Chapter 7 Traffic Control and Analysis at Signalized Intersections

Interval. A period of time during which all signal indications (greens, yellows, reds) remain the same for all approaches. Cycle. One complete sequence (for all approaches) of signal indications (greens, yellows, reds).

Cycle length. The total time for the signal to complete one cycle (given the symbol C and usually expressed in seconds). Green time. The amount of time within a cycle for which a movement or combination of movements receives a green indication (the illumination of a signal lens). This is expressed in seconds and given the symbol G. Yellow time. The amount of time within a cycle for which a movement or combination of movements receives a yellow indication. This is expressed in seconds and given the symbol Y. This time is referred to as the change interval, as it alerts drivers that the signal indication is about to change from green to red. Red time. The amount of time within a cycle for which a movement or combination of movements receives a red indication. This is expressed in seconds and given the symbol R.

All-red time. The time within a cycle in which all approaches have a red indication (expressed in seconds and given the symbol AR). This time is referred to as the clearance interval because it allows vehicles that might have entered at the end of the yellow interval to clear the intersection before the green phase starts for the next conflicting movement(s). This type of interval is becoming increasingly common for safety reasons because the rate of vehicles entering at the end of the yellow and beginning of the red indication has steadily increased in recent years.

Phase. The sum of the displayed green, yellow, and red times for one or more movements that start and stop moving at the same time.

Timing Stage. Consists of the time during which the same combination of movements, across one or more phases, are moving simultaneously. The sum of the timing stage lengths (in seconds) is the cycle length.

The term “movement” is used frequently in the preceding definitions. In

addition to a directional descriptor, such as left, through, or right, a distinction is made by categorizing movements as either protected or permitted.

Protected movement. A movement that has the right-of-way and does not need to yield to conflicting movements, such as opposing vehicle traffic or pedestrians. Through movements, which are always protected, are given a green full circle indication (or in some geometric configurations, a green

7.2 Intersection and Signal Control Characteristics 247

arrow pointing up). Left- or right-turn movements that are protected are given a green arrow indication (pointing either left or right). Permitted movement. A movement that must yield to opposing traffic flow or a conflicting pedestrian movement. This movement is made during gaps (time headways) in opposing traffic and conflicting pedestrian movements. Left- or right-turn movements with a green full circle indication are permitted movements. Left-turning vehicles in this situation must wait for gaps in the opposing through and right-turning traffic before making their turns. Right-turning vehicles must yield to pedestrians in the adjacent crosswalk before making their turns.

To understand how these control characteristics are implemented, it is

useful to analyze the physical implementation of these concepts. The display of the various signal indications (green, yellow, red, protected, permitted) at an intersection is handled by a signal controller (which is typically located in a cabinet next to the intersection). Modern signal controllers are sophisticated pieces of electronic equipment. These controllers, when combined with a method of vehicle detection, offer great flexibility in controlling phase duration and sequence. Traffic signal controllers are designed to operate in one or more of the following modes: pretimed, semi- actuated, or fully actuated. Pretimed. A signal whose timing (cycle length, green time, etc.) is fixed over specified time periods and does not change in response to changes in traffic flow at the intersection. No vehicle detection is necessary with this mode of operation. Semi-actuated. A signal whose timing (cycle length, green time, etc.) is affected when vehicles are detected (by video, pavement-embedded inductance loop detectors, etc.) on some, but not all, approaches. This mode of operation is usually found where a low-volume road intersects a high- volume road, often referred to as the minor and major streets, respectively. In such cases, green time is allocated to the major street until vehicles are detected on the minor street; then the green indication is briefly allocated to the minor street and then returned to the major street. Fully actuated. A signal whose timing (cycle length, green time, etc.) is completely influenced by the traffic volumes, when detected, on all of the approaches. Fully actuated signals are most commonly used at intersections of two major streets and where substantial variations exist in all approach traffic volumes over the course of a day.

7.2.1 Actuated Control Although pretimed signal control does not require the expense of vehicle detection hardware, it results in signal timing that it is not responsive to real-time traffic demands. The fixed-time values of the pretimed signal are based on expected

248 Chapter 7 Traffic Control and Analysis at Signalized Intersections

traffic demands during the time period of interest. However, traffic arrivals can vary significantly from one cycle to the next, as described in Chapter 5. Thus, with fixed timing, a phase may provide excessive green time one cycle (which results in extra delay for the vehicles that move in other phases) and not enough time in another cycle.

Improvements in signal-controller technology over the last 25 years have made possible significant advances in traffic control at signalized intersections. Modern signal controllers are able to accept inputs on traffic demands and utilize this information to adjust the green interval duration and phasing sequence from one cycle to the next.

Actuated control’s ability to respond to changes in traffic demands allows the green time to be reduced for a movement when the arrival rate is lower than normal, the green time to be extended when the arrival rate is higher than normal, or a phase to be skipped altogether if no demand is present for that movement. Thus, an intersection operating under actuated control will almost always result in lower overall delays than one operating under pretimed control.

Vehicle Detection Actuated control requires vehicle detection technology. The most common form of vehicle detection technology is the inductance loop detector (ILD). The ILD, a simple technology that has been in use for several decades now, consists of a loop (or coil) of wire embedded in the pavement through which an electrical current is circulated. This current is monitored by a device that interfaces with the signal controller, and when a vehicle passes over the ILD, the inductance level of the current is altered. When this change in the inductance level is detected by the monitoring device, it sends an input to the signal controller to indicate the presence of a vehicle.

ILDs can take on a variety of shapes, with different shapes having different advantages/disadvantages for detection ability. Furthermore, the sensitivity of the detectors can be tuned. The challenge in tuning the sensitivity is to find the level that allows detection of smaller vehicles (such as motorcycles or bicycles), yet is not so sensitive as to detect objects that are not vehicles.

The limitation of ILD technology is that vehicles can be detected only where the ILDs are placed. This limits the control options because it is prohibitively expensive to implement a large number of ILDs on an intersection approach.

A newer vehicle detection technology that is increasing in popularity at signalized intersections is video imaging processing (VIP). This technology has three main components: video camera, video digitization and processing unit, and signal controller interface. With VIP technology, virtual (software-based) detectors can be placed anywhere within a video camera’s field of view. The video processing unit converts the video from the camera into a digital format, and a software algorithm processes the combination of the digitized field of view and virtual detectors to determine vehicle presence, as well as additional measures that cannot be obtained with ILDs (such as queue length). With the additional measures that can be obtained from VIP systems, a greater number of control strategies are possible. Thus, a greater level of traffic responsiveness can be obtained.

7.2 Intersection and Signal Control Characteristics 249

Typical Phase Operation With the most basic configuration for vehicle detection, signal phases operating under actuated control consist of the following phase periods.

Initial Green. This period provides a practical minimum amount of green time to the traffic movement. Extended Green. After the initial green time expires, the phase enters an extension mode. This extension mode provides a continuation of green time as long as a vehicle arrives (at the detector) within a specified amount of time after the arrival of the previous vehicle.

A maximum green time is also specified for the phase. Thus, even if the arrival rate of vehicles is high enough to continue the extension period indefinitely, the phase will eventually be terminated so that other traffic movements can be served. If the phase terminates in this manner, it is referred to as a max-out condition. If the phase terminates prior to reaching the maximum green time, due to the time gap in successive vehicle arrivals exceeding the extension interval (the time allowed between successive vehicle arrivals before it is assumed no more vehicles are arriving), this is referred to as a gap-out condition.

Initial green periods are intended to serve all, or most, of the vehicle queue that develops during that traffic movement’s red period. Vehicles that join the queue at the start of green, or arrive just after queue clearance, are usually handled during the extension period. Therefore, a large enough green time should be specified so that the regularly expected initial queue length can be served, but not too large that excessive green time is wasted during cycles with much lower than average vehicle arrivals. Additionally, the green time should be at least several seconds greater than the lost time for a phase. Typical minimum green times are on the order of 10 seconds.

The time allowed between successive vehicle arrivals at vehicle detectors before it is assumed that no more vehicles are arriving (the extension interval) ranges from 2 to 4 seconds. Overall phase operations are quite sensitive to the selected value of this extension interval (also commonly referred to as the unit extension). A smaller value, such as 2 seconds, will generally result in a “snappy” operation; that is, the phase will tend to gap out quickly after the initial queue is served. In practice, a small extension interval can often result in a phase gapping out prematurely due to a driver in the queue hesitating (from being distracted or inattentive) in starting up from green, thus increasing delays and driver frustration. A larger extension interval, such as 4 seconds, will generally result in “sluggish” operation and a greater likelihood of a max-out. Although the longer extension interval decreases the likelihood of premature gap out, it can also lead to frustration for the drivers of other movements as they may perceive that the green time for the active phase is lasting an unnecessarily long time.

The basic sequence of events for an actuated phase is as follows: The initial green period is provided; the phase enters the extension period; the extension period continues until the phase either gaps out or maxes out; the yellow and all- red intervals commence and then the phase terminates. This sequence is

250 Chapter 7 Traffic Control and Analysis at Signalized Intersections

illustrated in Fig. 7.3. This is the most basic operation of an actuated phase. Many more elements can be incorporated, depending on the level of traffic responsiveness desired, the level of coordination with other signalized intersections, and the level of vehicle detection.

For a more complete description of actuated control, the reader is referred to the Manual of Traffic Signal Design [Kell and Fullerton, 1998] and the Highway Capacity Manual [Transportation Research Board, 2016].

Figure 7.3 Basic operation of an actuated signal phase.

7.2.2 Signal Controller Operation Most modern signal controllers are designed to operate in what is termed a dual- ring configuration. This configuration allows maximum flexibility for controlling phase duration and sequencing, which is necessary for operating in a fully actuated mode. The dual-ring concept can be best explained with a graphical illustration (see Fig. 7.4a).

In this figure, the three-letter notation in each numbered box refers to direction and movement type. For example, WBL means westbound left and NBT means northbound through. At an intersection with four approaches and separate/exclusive left-turn phases for each approach, a total of eight separate movements are possible. Any right-turn movements are implicitly considered to move with the corresponding through movement. The dual-ring terminology comes from four movements (1–4) being represented on the top ring, and four movements (5–8) being represented on the bottom ring. The movements in each of the numbered boxes must start and end at the same time; thus, the boxes represent phases. The logic behind this configuration is straightforward. Any phase in ring 1 can occur simultaneously with any phase in ring 2 as long as both are on the same side of the barrier (a term used figuratively to separate conflicting traffic movements). For example, the opposing WB and EB left-turn movements (phases 1 and 5) can move simultaneously. However, if no vehicles are detected in the EB left-turn lane, this movement can be skipped and the WB left-turn and WB through movements (phases 1 and 6) can move simultaneously. The same logic applies to movements on the right side of the barrier (phases 3, 4, 7, and 8).

7.2 Intersection and Signal Control Characteristics 251

Furthermore, if the volume of a phase movement subsides sooner than the volume of another phase movement currently moving at the same time, the green time for this lower-volume phase can be terminated and another phase can be initiated, according to the previously described logic. For example, suppose the WB and EB left turns (phases 1 and 5) start to move at the same time, but the WB left-turn (phase 1) volume is considerably larger than the EB left-turn (phase 5) volume. If both phases 1 and 5 received enough green time to satisfy the WB left-turn vehicle demand, this would result in wasted green time for phase 5. With the dual-ring configuration, phase 5 can be terminated before phase 1, and phase 6 can be initiated while phase 1 continues. This results in a more efficient allocation of green time and reduced delay. The phase sequence and phase durations can therefore vary from one cycle to the next at a fully actuated intersection, especially with highly variable approach volumes. Consequently, the cycle length can vary from one cycle to the next. Figure 7.4b illustrates the typical dual-ring phase sequence options as discussed above. Note that phases can be skipped entirely if no traffic demand is present, and other phase options are possible depending on intersection geometry/lane movement assignment.

It must be pointed out that although no two phases are required to either start or terminate at the same time in a dual-ring configuration, all movements on the left side of the barrier must be terminated before any movement on the right side of the barrier can be initiated, and vice versa. This is a safety feature: no movement on the left side of the barrier can be allowed to move simultaneously with any movement on the right side of the barrier, or else conflicting traffic movements will intersect. Movements on the same side of a barrier are commonly referred to as being in the same concurrency group.

252 Chapter 7 Traffic Control and Analysis at Signalized Intersections

Figure 7.4 Dual-ring signal control. (a) Movement/phase-based representation of dual-ring logic. (b) Timing-stage−based representation of dual-ring logic.

Part (a) Adapted from Exhibit 19-2, p. 19-6, Highway Capacity Manual, 6th Edition, Copyright, Transportation Research Board, National Academy of Sciences, Washington, D.C. 2016.

7.3 Traffic Flow Fundamentals for Signalized Intersections 253

7.3 TRAFFIC FLOW FUNDAMENTALS FOR SIGNALIZED INTERSECTIONS

Before presenting the analytical principles and techniques used for signalized intersections, it is important to introduce several key concepts and definitions used in the development of signal timing plans and the analysis of traffic at signalized intersections. Saturation Flow Rate The saturation flow rate is the maximum hourly volume that can pass through an intersection, from a given lane or group of lanes, if that lane (or lanes) were allocated constant green over the course of an hour. Saturation flow rate is given by

3600

s h

= (7.1)

where

s = saturation flow rate in veh/h, h = saturation headway in s/veh, and

3600 = number of seconds per hour. Note the similarity between Eq. 7.1 and Eq. 5.4. The difference is that h in

Eq. 7.1 is a constant minimum headway value maintained for saturated conditions, as opposed to an average headway value as used in Eq. 5.4 (Eq. 7.1 also directly yields units of veh/h for s because of the numerator). The use of the term “saturation” is an important qualifier in this definition, as it implies the presence of constant vehicle demand in measuring the headway. If the measure of interest is simply the traffic flow through the intersection for some period of time, then the appropriate equation would be 5.1 or 5.4.

Research has found that a typical maximum saturation flow rate of 1900 passenger cars per hour per lane (pc/h/ln) is possible at signalized intersections, and this is referred to as the base saturation flow rate. This corresponds to a saturation headway of about 1.9 seconds. Just as in the analysis of uninterrupted flow, a number of roadway and traffic factors can affect the maximum flow rate through an intersection. These factors include lane widths; grades; curbside parking maneuvers; the distribution of traffic among multiple approach lanes; the level of roadside development; bus stops; and the influence of pedestrians, bicycles, and heavy vehicles (since they occupy more roadway space and have poorer acceleration/deceleration capabilities). Additionally, lanes that allow left or right turns usually have lower saturation flow rates because drivers reduce speed to make a turning maneuver (especially heavy vehicles, with their increased turning radii). Furthermore, if a turning movement is permitted rather than protected, its saturation flow rate will be reduced as a result of the turning vehicles yielding to conflicting through and right-turning vehicles (for left turns only), bicycles, and/or pedestrians. All of these factors are accommodated by applying adjustments to the base saturation flow rate. The end result is usually a value less than 1900 pc/h/ln for each approach lane at an intersection, and is referred to as the adjusted saturation flow rate. Additionally, the units are

254 Chapter 7 Traffic Control and Analysis at Signalized Intersections

converted to vehicles per hour per lane (veh/h/ln) due to adjustment of the heavy-vehicle volume with passenger car equivalents (in a manner similar to the procedures of Chapter 6). The process for arriving at an adjusted saturation flow rate by making adjustments to the base saturation flow rate, for the preceding factors, is quite involved and beyond the scope of this book (see the Highway Capacity Manual [Transportation Research Board, 2016]). Of course, saturation flow rates can be measured directly in the field, in which case no further adjustments are necessary. For the rest of this chapter, it should be assumed that the provided saturation flow rates have been adjusted for the given conditions; the term “adjusted” has been dropped just for notational convenience.

Lost Time Due to the traffic signal’s function of continuously alternating the right-of-way between conflicting movements, traffic streams are continuously started and stopped. Every time this happens, a portion of the cycle length is not being completely utilized, which translates to lost time (time that is not effectively serving any movement of traffic). Total lost time is a combination of start-up and clearance lost times. Start-up lost times occur because when a signal indication turns from red to green, drivers in the queue do not instantly start moving at the saturation flow rate; there is an initial lag due to drivers reacting to the change of signal indication and the time it takes for vehicles to accelerate to their saturation-flow speed. This start-up delay results in a portion of the green time for that movement not being completely utilized. This start-up lost time has a typical value of around 2 seconds. This concept is illustrated in Fig. 7.5.

In this figure, note that the headway for the first several vehicles is larger than the saturation headway. The saturation headway is typically reached after the fourth vehicle in the queue. The summation of the amount of headway time greater than the saturation headway for each of the first 4 vehicles yields the total start-up lost time for the movement.

The stopping of a traffic movement also results in lost time. When the signal indication turns from green to yellow, the latter portion of time during the yellow interval is generally not utilized by traffic. Additionally, if there is an all- red interval, this time period is generally not utilized by traffic. These periods of time during the change and clearance intervals that are not effectively used by traffic are referred to as clearance lost time. Typically, the last second of the yellow interval and the entire all-red interval are included in the estimate of clearance lost time. However, for intersections with significant red-light running, the clearance lost time may be negligible.

Start-up and clearance lost times are summed to arrive at a total lost time for the phase, given as

tL = tsl + tcl (7.2)

where tL = total lost time for a movement during a cycle in seconds, tsl = start-up lost time in seconds (typical value of 2 seconds), and tcl = clearance lost time in seconds (typically the last second of the yellow interval

plus the entire all-red interval).

7.3 Traffic Flow Fundamentals for Signalized Intersections 255

Figure 7.5 Concept of saturation headway and lost time.

This amount of time remains fixed, regardless of phase or cycle length. Thus, for shorter cycle lengths, the lost time will comprise a larger percentage of the cycle length and will result in a larger amount of lost time over the course of a day compared with longer cycle lengths. However, longer cycle lengths usually have more phases than shorter cycle lengths, which may result in similar proportions of lost time.

Effective Green and Red Times For analysis purposes, the time during a cycle that is effectively (or not effectively) utilized by traffic must be used rather than the time for which green, yellow, and red signal indications are actually displayed, because they are most likely different. This results in two measures of interest: the effective green time and the effective red time. The effective green time is the time during which a traffic movement is effectively utilizing the intersection. The effective green time for a given movement or phase is calculated as

Lg G Y AR t= + + − (7.3)

where

g = effective green time for a traffic movement in seconds, G = displayed green time for a traffic movement in seconds, Y = displayed yellow time for a traffic movement in seconds,

AR = displayed all-red time in seconds, and tL = total lost time for a movement during a cycle in seconds. The effective red time is the time during which a traffic movement is not

effectively utilizing the intersection. The effective red time for a given movement or phase is calculated as

Lr R t= + (7.4)

256 Chapter 7 Traffic Control and Analysis at Signalized Intersections

where

r = effective red time for a traffic movement in seconds, R = displayed red time for a traffic movement in seconds, and tL = total lost time for a movement during a cycle in seconds.

Alternatively, the effective red time can be calculated as follows, assuming that the cycle length and effective green time have already been determined:

r C g= − (7.5)

where

C = cycle length in seconds, and other terms are as defined previously.

Likewise, the effective green time can be calculated by subtracting the effective red time from the cycle length. Capacity Because movements on an intersection approach do not receive a constant green indication (as assumed in the definition for saturation flow rate), another measure must be defined that accounts for the hourly volume that can be accommodated on an intersection approach given that the approach will receive less than 100% green time. This measure is capacity and is given by

c = s × g/C (7.6)

where

c = capacity (the maximum hourly volume that can pass through an intersection from a lane or group of lanes under prevailing roadway, traffic, and control conditions) in veh/h,

s = saturation flow rate in veh/h, and g/C = ratio of effective green time to cycle length.

7.4 DEVELOPMENT OF A TRAFFIC SIGNAL PHASING AND TIMING PLAN

Assuming the decision to install a traffic signal at an intersection has been made, an appropriate phasing and timing plan must be developed. The development of a traffic signal phasing and timing plan can be complex, particularly if the intersection has multiple-lane approaches and requires protected turning movements (a turn arrow). However, the timing plan analysis can be simplified by dealing with each approach separately. This section provides the basic process and fundamentals needed to develop a phasing and timing plan for an isolated, fixed-time (pretimed) traffic signal. As timing plans become more complex, they simply build on these fundamental principles. The reader is

7.4 Development of a Traffic Signal Phasing and Timing Plan 257

encouraged to review the material in other references to see how actuated and progressive timing plans are developed [Kell and Fullerton, 1998; Transportation Research Board, 2016]. This section describes the basic process that results in the development of a signal phasing and timing plan.

7.4.1 Select Signal Phasing Recall that a timing stage consists of the time during which the same combination of movements, across one or more phases, are moving simultaneously and that the sum of the timing stage lengths is equal to the cycle length. The most basic traffic signal cycle is made up of two timing stages, as shown in Fig. 7.6. In this case, timing stage 1 accommodates the movement of all the eastbound and westbound vehicles, and timing stage 2 accommodates the movement of all the northbound and southbound vehicles. These timing stages will alternate during the continuous operation of the signal. This timing-stage scheme, however, could prove to be very inefficient if one or more of the approaches includes a high left- turn volume.

Given that each approach consists of one lane, vehicles will be delayed behind a left-turning vehicle waiting for a gap in the opposing traffic stream. If the high volume of left turns is present on both the northbound and southbound approaches, for example, each of these approaches could be given a separate timing stage consisting of the left-turn phases. This would be more efficient because left-turning vehicles on these two approaches would not have to wait for gaps in the opposing traffic stream, thus greatly reducing delays for all vehicles. This would result in a three-timing-stage operation, as shown in Fig. 7.6. When phases for movements on opposing approaches are run in a separate timing stage, as in this case, it is referred to as split phasing.

Some common signal timing-stage configurations, along with the applicable dual-ring phase numbers, are shown in Fig. 7.7. In this figure, note that the dashed lines represent permitted movements and the solid lines represent protected movements. When the left turns precede the through and right-turn movements in the timing-stage sequence for an approach, they are referred to as leading left turns. When the left turns follow the through and right-turn movements, they are referred to as lagging left turns. Although not shown in Fig. 7.7, it is also possible for a movement to be protected for a period of time and then permitted for a period of time, or vice versa. This is most commonly seen with left-turn movements, and is referred to as protected plus permitted or permitted plus protected, depending on the sequence.

It is important to remember that there is lost time (start-up and clearance) associated with each timing stage. Thus, with each timing stage added to a cycle, the lost time increases. Although the lost time may be only 3 to 5 seconds per timing stage, the accumulated lost time throughout the day can be significant.

258 Chapter 7 Traffic Control and Analysis at Signalized Intersections

Figure 7.6 Illustration of two-timing and three-timing signal operation.

Figure 7.7 Typical phasing configurations and sequencing.

7.4 Development of a Traffic Signal Phasing and Timing Plan 259

Given this, a point of diminishing returns is reached with the addition of phases as the efficiencies gained by separating traffic movements eventually become outweighed by the inefficiencies of increased lost time. Thus, a primary concern in signal timing is to keep the number of phases to a minimum. Because protected-turn phases add to lost time, they should be used only when warranted. Because of opposing motor vehicle traffic, left-turn movements typically require a protected-turn phase much more often than right turns. There are no nationally established guidelines on when protected left-turn phasing should be used, so local policies and practices should be consulted before a decision is made about whether to provide a protected left-turn phase. In general, decisions on whether to provide a protected left-turn phase are based on one or more of the following factors:

• Volume (just left turn or combination of left turn and opposing volume)

• Delay

• Queuing (spillover)

• Traffic progression

• Opposing traffic speeds

• Geometry (number of left-turn lanes, crossing distance, sight distance)

• Crash experience (which may also be related to any of the above factors) More specific guidance on this issue can be found in several references, including the Highway Capacity Manual [Transportation Research Board, 2016], the Traffic Control Devices Handbook [ITE, 2001], and the Manual of Traffic Signal Design [Kell and Fullerton, 1998].

One of the more common guidelines is the use of the cross product of left- turn volume and opposing through and right-turn volumes. The Highway Capacity Manual offers the following criteria for this guideline: The use of a protected left-turn phase should be considered when the product of left-turning vehicles and opposing traffic volume exceeds 50,000 during the peak hour for one opposing lane, 90,000 for two opposing lanes, or 110,000 for three or more opposing lanes.

EXAMPLE 7.1 DETERMINE LEFT-TURN PHASING

Refer to the intersection shown in Fig. 7.8. Use the cross-product guideline to determine if protected left-turn phases should be provided for any of the approaches.

260 Chapter 7 Traffic Control and Analysis at Signalized Intersections

Figure 7.8 Intersection geometry and peak traffic volumes for example problems.

SOLUTION

There are 250 westbound vehicles that turn left during the peak hour. The product of the westbound left-turning vehicles and the opposing eastbound traffic (right-turn and straight-through vehicles) is 275,000 [250 × (900 + 200)]. There are 300 eastbound vehicles that turn left during the peak hour. The product of the eastbound left-turning vehicles and the opposing westbound traffic (right-turn and straight-through vehicles) is 345,000 [300 × (1000 + 150)]. Because the cross product for each of these approaches is greater than 90,000 (the requirement for two opposing lanes), a protected left-turn phase is suggested for the WB and EB left-turn movements. The NB and SB approaches do not require a separate timing stage for the left-turn movements using this criterion because the cross products for these approaches are less than 50,000 (for one opposing lane). Therefore, a three-timing-stage plan is recommended, as shown in Fig. 7.9.

7.4 Development of a Traffic Signal Phasing and Timing Plan 261

Figure 7.9 Recommended signal phasing plan for the intersection in Example 7.1.

7.4.2 Establish Analysis Lane Groups Each intersection approach is initially treated separately, and the results are later aggregated. Thus, each approach must be subdivided into logical groupings of traffic movements for analysis purposes.

The Highway Capacity Manual [Transportation Research Board, 2016] provides detailed guidelines on this process. Generally, the process consists of placing like movements into movement groups (left-turn, right-turn, and through movements would be identified) and then translating movement groups to lane groups based on the allowable movements from each lane. The only time these two group designations differ is when a shared lane (left-turn and through movements allowed from the same lane) is present on an approach with two or more lanes. When shared lanes are present on an approach with two or more lanes, the Highway Capacity Manual [Transportation Research Board, 2016] employs an iterative procedure to identify the expected distribution of each movement type in each lane, based on the principle that drivers will choose the lane that they perceive will minimize their travel time through the intersection (delay). This procedure is beyond the scope of this book and, as a result, the subsequent example problems and end-of-chapter problems will provide specific lane distributions of traffic as necessary. Consequently, it is only necessary to make reference to lane groups for the remainder of the chapter.

Based on the lane and traffic movement distribution on an approach, lane

groups can be readily determined. The following general guidelines are offered for establishing lane groups [Transportation Research Board, 2016]:

• If an exclusive turn lane (or lanes) is present, it should be treated as a separate lane group.

• Each shared lane on an approach should be treated as a separate lane group.

• Any remaining lanes, which would be exclusive through lanes, should be treated as a separate lane group.

Figure 7.10 shows some typical lane groupings for analysis purposes. Note

that when multiple lanes are combined into a lane group, the subsequent analysis

262 Chapter 7 Traffic Control and Analysis at Signalized Intersections

calculations for this lane group should treat these lanes as a single unit. Because of the aforementioned complexity of dealing with shared lanes, the subsequent example and end-of-chapter problems in this book assume that adjacent exclusive through lanes and shared through/right-turn lanes will be combined into one lane group (such as is done in Example 7.2). This assumption will often introduce only very minimal error when compared to treating the exclusive through lane and shared through/right-turn lane as separate lane groups and does not affect learning the basic concept of aggregating lane group results into overall approach results (this concept is covered later in this chapter).

Figure 7.10 Example lane groupings for analysis (LT = left turn; TH = through; RT = right turn; EXC = exclusive).

Adapted from Exhibit 19-19, p. 19-43, Highway Capacity Manual, 6th Edition. Copyright, Transportation Research Board, National Academy of Sciences, Washington, D.C. 2016.

EXAMPLE 7.2 DETERMINE LANE GROUPS

Determine the lane groups to use for analysis of the Maple and Vine Streets intersection (Fig. 7.8).

SOLUTION

The EB and WB left-turn movements will each be a lane group because they have a separate lane and move in a separate phase from the through/right-turn movements. Likewise, the EB and WB through/right-turn movements proceed together in a separate phase and will therefore be separate lane groups. Although the right turns use only the outside lane, this movement’s impact on the saturation flow rate for the two lanes combined will be determined. The NB and SB left turns will also each be a separate lane group. Even though they move during the same phase as the adjacent through and right-turn movements, these left turns are permitted and will have very different operating characteristics from the through and right-turn movements.

7.4 Development of a Traffic Signal Phasing and Timing Plan 263

Because the through and right-turn movements use the same lane, they will be an individual lane group for both the NB and SB approaches. The recommended lane groups for analysis for each of the approaches are shown in Fig. 7.11.

Figure 7.11 Analysis lane groups for the three-timing-stage design at the intersection of Maple Street and Vine Street.

7.4.3 Calculate Analysis Flow Rates and Adjusted Saturation Flow Rates Just as for the analysis of uninterrupted flow, the hourly traffic volume arriving on each intersection approach must be converted to an analysis flow rate that accounts for the peak 15-minute flow within that hour (typically the peak hour). This is accomplished by calculating the peak-hour factor (PHF) and dividing this into the hourly volume (as shown in Chapter 6), which yields the analysis flow rate.

One note about adjusting for the PHF. With the multiple traffic streams entering an intersection, a separate PHF can be calculated for each approach’s traffic stream. However, adjusting each approach volume by its specific PHF can yield unrealistically high combined analysis volumes, because the different approach volumes usually do not peak during the same 15-minute period. Applying a single PHF determined for the intersection as a whole will result in more reasonable analysis volumes.

The adjustment of the saturation flow rate was discussed in Section 7.3. Again, it is assumed that the approach volumes and saturation flow rates provided in this chapter have already been adjusted.

7.4.4 Determine Critical Lane Groups and Total Cycle Lost Time For any combination of lane group movements during a particular timing stage, one of these lane groups will control the necessary green time for that timing stage. This lane group is referred to as the critical lane group. When the traffic movements of each lane group occur during only one timing stage of the signal cycle, the determination of the critical lane group for each timing stage is straightforward. In this case, the critical lane group for each phase is simply the lane group with the highest ratio of vehicle arrival rate to vehicle departure rate (λ/μ). This quantity is referred to as the flow ratio (which was called the traffic intensity, ρ, in Chapter 5) and is designated v/s (arrival flow rate divided by saturation flow rate). If the allocation of green time for each timing stage is

264 Chapter 7 Traffic Control and Analysis at Signalized Intersections

based on the flow ratio of the critical lane group, then the noncritical lane group movements will be accommodated as well.

As previously discussed, with dual-ring controllers, a wide variety of phasing sequences is possible. The situation where a phase starts in one signal timing stage and continues in the next signal timing stage is referred to as an overlapping phase (see the last two rows of Fig. 7.7). For the case of overlapping phases in a signal cycle, the identification of the critical lane groups is more complex, as the lane group with the highest flow ratio in each timing stage is not necessarily the critical lane group for that timing stage. The remainder of this chapter will focus only on non-overlapping phases. The reader is referred to the Highway Capacity Manual [Transportation Research Board 2016] for details on determining critical lane groups for overlapping phases.

In addition, the sum of the flow ratios for the critical lane groups can be used to calculate a suitable cycle length, which will be discussed in the next section. This is given by

=

 =    

 1

n

c cii

v Y

s (7.7)

where

Yc = sum of flow ratios for critical lane groups, (v/s)ci = flow ratio for critical lane group i, and

n = number of critical lane groups.

The total lost time for the cycle will also be used in the calculation of cycle length. In determining the total lost time for the cycle, the general rule is to apply the lost time for a critical lane group when its movements are initiated (the start of its green interval). The total cycle lost time is given as

( ) =

= 1

n

L ci i

L t (7.8)

where

L = total lost time for cycle in seconds,

(tL)ci = total lost time for critical lane group i in seconds, and n = number of critical lane groups.

EXAMPLE 7.3 CALCULATE SUM OF FLOW RATIOS

Calculate the sum of the flow ratios for the critical lane groups for the three-timing-stage timing plan determined in Example 7.1 given the saturation flow rates in Table 7.1.

7.4 Development of a Traffic Signal Phasing and Timing Plan 265

Table 7.1 Saturation Flow Rates for Three-Timing-Stage Design at Intersection of Maple Street and Vine Street

Timing Stage 1 Timing Stage 2 Timing Stage 3

EB L: 1750 veh/h

WB L: 1750 veh/h

EB T/R: 3400 veh/h

WB T/R: 3400 veh/h

SB L: 450 veh/h

NB L: 475 veh/h

SB T/R: 1800 veh/h

NB T/R: 1800 veh/h

SOLUTION

Note that the saturation flow rates are relatively low for the SB and NB left (L) turns because they are permitted only, and the opposing through and right-turn (T/R) vehicles limit the number of usable gaps for these vehicles. The saturation flow rates for the WB and EB through and right-turn (T/R) movements account for both through lanes. The flow ratios will now be calculated, with the critical lane group for each timing stage indicated with a check mark in Table 7.2. As indicated in the table, the critical lane group for timing stages 1, 2, and 3, respectively, are the EB left turn, the WB through and right turn, and the NB through and right turn.

Table 7.2 Flow Ratios and Critical Lane Groups for Three-Timing-Stage Design at Intersection of Maple Street and Vine Street

Timing Stage 1 Timing Stage 2 Timing Stage 3

EB L: 300

0.171 1750

= ✓

WB L: 250

0.143 1750

=

EB T/R: 1100

0.324 3400

=

WB T/R: 1150

0.338 3400

= ✓

SB L: 70

0.156 450

=

NB L: 90

0.189 475

=

SB T/R: 370

0.206 1800

=

NB T/R:

390 0.217

1800 =

✓

The sum of the flow ratios for the critical lane groups for this timing plan will be needed for the next section. Since this timing plan does not include any overlapping phases, this value is simply the sum of the highest lane group v/s ratios for the three timing stages, as follows:

=

 =    

= + + =

 1

0.171 0.338 0.217 0.726

n

c cii

v Y

s

Assuming 2 seconds of start-up lost time and 2 seconds of clearance lost time (1 second of yellow time plus 1 second of all-red time), for each critical lane group, gives a lost time of 4 s/timing stage. The total lost time for the cycle is then 12 seconds (3 timing stages × 4 s/ timing stage).

266 Chapter 7 Traffic Control and Analysis at Signalized Intersections

EXAMPLE 7.4 CALCULATE SUM OF FLOW RATIOS AND TOTAL LOST TIME

Suppose it is necessary to run the NB and SB movements in a split-phase configuration (with timing stage 3 for SB movements and a new timing stage 4 for NB movements). Calculate the sum of the flow ratios for the critical lane groups and total cycle lost time for this situation, assuming that the EB and WB movement phasing remains the same. Table 7.3 summarizes the calculation of the flow ratios and the identification of the critical lane groups.

SOLUTION

The sum of the flow ratios for the critical lane groups for this phasing plan is

=

  = + + + =   

1

0.171 0.338 0.206 0.217 0.932 n

cii

v s

The total lost time for the cycle is 16 seconds (4 timing stages × 4 s/timing stage).

Table 7.3 Flow Ratios and Critical Lane Groups for Four-Timing-Stage Design (Split Phasing for N-S Movements) at Intersection of Maple Street and Vine Street

Timing Stage 1 Timing Stage 2 Timing Stage 3 Timing Stage 4

EB L: 300

0.171 1750

= ✓ EB T/R: 1100

0.324 3400

= SB L: 70

0.040 1750

= NB L: =

WB L: 250

0.143 1750

= WB T/R: 1150

0.338 3400

= ✓ SB T/R: =370 0.206

1800 ✓ NB T/R: =390

0.217 1800

✓

7.4.5 Calculate Cycle Length In practice, cycle lengths are generally kept as short as possible, typically between 60 and 75 seconds. However, complex intersections with five or more timing stages can have cycle lengths of 120 seconds or more. The minimum cycle length necessary for the lane group volumes and phasing plan of an intersection is given by