Excel assignment
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lecture5A.pdf
ENGR 436 Transportation Engineering
Lesson 5. Horizontal Alignments (superelevation)
Announcement
• Homework 1 due today • Stay active on forums = bonus points • Homework 2 to be posted • Quiz 2 (horizontal curves – geometric properties; min. curve radius), 9/16, Wed
Review
1. Which curve has a larger D value? A. #1 B. #2 C. They have the same D.
R D
100180 ⎟ ⎠
⎞ ⎜ ⎝
⎛ =
π B
Example 2. Given a 4-degree curve with a central angle 55°25‘. PC is at station 238+45. What is the length of curve, the station of PT, the deflection angles and the chord lengths for whole stations?
PT
PI
RPC Δ/2
Δ/2 C T
E M
Δ
Assorted formulae
R D
100180 ⎟ ⎠
⎞ ⎜ ⎝
⎛ =
π 2 tan
Δ = RT
2 sin2
Δ = RC
⎟ ⎠
⎞ ⎜ ⎝
⎛ −
Δ = 1
2 secRE
⎟ ⎠
⎞ ⎜ ⎝
⎛ Δ −=
2 cos1RM
Curve length
⎟ ⎠
⎞ ⎜ ⎝
⎛ Δ= 180 π
RL
PT
PI
RPC Δ
Δ/2 C T
E M
Δ
Curve length:
Use and R
D 100180 ⎟ ⎠
⎞ ⎜ ⎝
⎛ =
π ⎟ ⎠
⎞ ⎜ ⎝
⎛ Δ= 180 π
RL
R = 18,000/π/D = 1433 ft & L = 1433*55.42*π/180 = 1385 ft Chord length = 2*sin(Δ/2)*R = 55.25 ft
Station of PT = (238+45) + L = 238*100+ 45 + 1,385 = 25229.75 feet = (252 + 30)
Station Deflection angle
Chord length (ft)
238+45 0 0 239+00 1°6’0’’ 55.25 240+00 3°6' 99.98 ... ... ... 252+00 27°6'18" 99.98 252+30 27°42'30" 30.17
Final answer:
PT
To find the chord lengths, apply the formulas to a segment of the curve:
PC Δ
R
Δ/2
PT
In this example, PC is at 238+44.75, so the distance to the next whole station is 55.25 ft. Using L = 55.25
and gives Δ1 = 2.21°
PC Δ1
R
Δ1/2
⎟ ⎠
⎞ ⎜ ⎝
⎛ Δ= 1801 π
RL
The deflection is half of this, or 1°6'18"
PT
Using the chord formula:
PC Δ
R
Δ/2
ft25.55 2
sin2 = Δ
= RC
Coming up…
• Vehicle overturning à minimum radius (Rmin) • Sight restriction à clearance distance (Ms)
How fast can a vehicle travel along a (horizontal) curve?
High-Speed Turns
What makes the vehicle follow the curve?
Radius = R
Vehicle speed u
How much centripetal force is needed?
Radius = R
Vehicle speed u, mass m
How much centripetal force is needed?
R mu
F 2
=
If no banking/slope, centripetal force comes from sideway friction.
mgfNfF ss ==
Coefficient of side friction
N = Normal force
Weight = m × g
Cross-section view
Source: Traffic Engineering, 3rd Edition, by Roess et al. 2004
Side Friction Factor
The vehicle will remain on the curve if the side friction force can provide the centripetal force:
R mu
mgfs 2
=
That means
or
gRfu s=
fg u
R 2
= s
Example 1. A curve has a radius of 800 feet; design speed is 50 mph. Is this curve safe? What design speed do you recommend for this curve?
Hint: - What side friction value to use? - 1 mph =
(,*+, -../
0,1,, 2.34562 = 1.467 ft/sec
• What are the issues?
21
• Side friction may be insufficient due to ice and snow.
• Radius can be excessive and may not work in restricted terrain or right of way (ROW).
• Superelevation
Cross-section view
α Superelevation rate = tan(α)
Normal force, N
Gravity or Weight
Side friction
A = B*
C[EF + tan (K)]
Example 2. Now that we know the curve (with radius 800 feet, side friction coefficient is 0.14, design speed is 50 mph) is NOT safe. What superelevation should be provided?
Hint 1 : 1 mph = (,*+, -../ 0,1,, 2.34562
= 1.467 ft/sec
Hint 2: A = N O
P[QRS/T5 (U)]
• In practice, superelevationrates range between 0.5% to 12%.
• The minimum value (0.5%) is used to provide proper drainage.
• For cold weathers (where icing is expected), a maximum 8% is recommended to prevent stalled vehicle from sliding towards the inside of the curve.
25
Direction of travel
fb fs
