Aviation

AVN1107 AIRCRAFT OPERATIONS

LECTURER: KELLI RUSSELL

Lecture 4 - Semester 2, 2020 1

TOPICS

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IN-FLIGHT PERFORMANCE

INTRODUCTION TO LOADING

13TH AUGUST

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WEEK 3 TUTORIAL

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DOES THE OBSTACLE INFRINGE THE MINIMUM TAKE- OFF GRADIENT OF 1:20?

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300m 20m

(300 ÷ 20) + 30 = 45m

45

Obstacle is within the approach and take-off area.

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300 m

300 ÷ 20 = 15m Object is 20m tallObstacle infringes by 5

meters ∴ 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 needs to be reduced by 100 m

New TODA will be 800 – 100 = 700 m

OBSTACLE INFRINGEMENT AND CALCULATION OF NEW TODA

WHAT IS THE PRESSURE & DENSITY ALTITUDES FOR AERODROME ‘A’ PRESSURE ALTITUDE (AERODROME ‘A’)

Pressure Altitude = [(ISA) – (Actual Pressure)] x (30) + (Aerodrome Elevation)

= [(1013) – (1002)] x 30 + 1,500

= 1,830 feet

ISA TEMPERATURE @ 1830’ = 15 – (1.8 X 2) = 11.4°C

ISA DEVIATION = ACTUAL OAT – ISA TEMPERATURE

= 18 – 11.4 = 6.6 (ROUND UP TO 7)

DENSITY HEIGHT (AERODROME ‘A’)

Density Height = Pressure Height + ISA DEVIATION x 120

= 1830 + (7 x 120) = 2,670 feet

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WHAT IS THE PRESSURE & DENSITY ALTITUDES FOR AERODROME ‘B’ PRESSURE ALTITUDE (AERODROME ‘B’)

Pressure Altitude = [(ISA) – (Actual Pressure)] x (30) + (Aerodrome Elevation)

= [(1013) – (1010)] x 30 + 2,840

= 2,930 feet

ISA TEMPERATURE @ 2930’ = 15 – (2.9 X 2) = 9.2°C

ISA DEVIATION = ACTUAL OAT – ISA TEMPERATURE

= 16 – 9 = 7

DENSITY HEIGHT (AERODROME ‘B’)

Density Height = Pressure Height + ISA DEVIATION x 120

= 2,930 + (7 x 120) = 3,770 feet

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WHAT IS THE HEADWIND & CROSSWIND COMPONENT FOR AERODROME ‘A’ Wind: 220°M/5kts

Runway Heading: Assume runway 18 for headwind. i.e. 180°M

Headwind speed = wind speed x cos (angular difference between runway heading and wind)

= 5 x cos (220 – 180)

= 3.83

∴ HWC = 3.8 KTS Crosswind speed = wind speed x sin (angular difference)

= 5 x sin (220 – 180)

∴ X-wind = 3.2 KTS

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WHAT IS THE HEADWIND & CROSSWIND COMPONENT FOR AERODROME ‘B’ Wind: 250°M/10kts

Runway Heading: Assume runway 28 for headwind. i.e. 280°M

Headwind speed = wind speed x cos (angular difference between runway heading and wind)

= 10 x cos (280 – 250)

= 8.66

∴ HWC = 8.7 KTS Crosswind speed = wind speed x sin (angular difference)

= 10 x sin (280 – 250)

∴ X-wind = 5 KTS

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CAN YOU TAKE-OFF FROM AERODROME ‘A’ AT 1050 KGS? IF NOT, WHAT IS THE MTOW?

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775 No 775 kgs

CAN YOU LAND AT AERODROME ‘A’ AT 1050 KGS

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Yes, you can land at 1050kg. Not limited by climb & LDR only 540 m

CAN YOU LAND AT AERODROME ‘B’ AT 1050 KGS?

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Yes, you can land at 1050 kgs Notice for light style a/c that the weight doesn’t make a lot of difference (why?)

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Dyson-Holland, 2015

Exercise 1: TODA = 1200m. Runway 09/27. One obstacle (30m tall) is located 480m east of the runway end and 45m north of the centreline. Will this obstacle affect TODA?

60m

45m

45m

(480/20) + 30m = 54m

Yes, the obstacle is within the area

480

?

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Exercise 1: TODA = 1200m. Runway 09/27. One obstacle (30m tall) is located 480m east of the runway end and 45m north of the centreline. Will this obstacle affect TODA?

24m (480/20)

The obstacle infringes by 6 meters and thus TODA needs to be reduced by 120m

IN-FLIGHT PERFORMANCE

In-flight performance refers to the performance of aircraft during their climb, cruise and descent phases of a flight

The most important phase in-flight performance considers is the cruise phase

“In-flight performance is generally about finding the optimum height, speed and power settings to obtain the fastest cruise using the least fuel for a particular flight” (Dyson-Holland, 2015, p. 85)

Range in this context can be defined as a distance per unit of fuel

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RANGE

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Distance travelled per unit of fuel burned in either ANM or GNM

Specific Range

(SR)

Distance is measured relative to the air (i.e., nil wind) Found from the TAS and the fuel consumption for a particular weight and power setting For best specific air range we require the best ratio of TAS to fuel flow and, since fuel flow (FF) for an engine-propeller combination depends upon power, best TAS/power ratio

Specific Air

Range (SAR)

Distance is measured relative to the ground (i.e., taking into account effects of the wind) Found from the GS and the fuel consumption for a particular weight and power setting

Specific Ground Range (SGR)

FACTORS AFFECTING IN- FLIGHT PERFORMANCE

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IASMAX L/D or IASMNM DAirspeed

Dependent on the engine speed, the mixture setting and Density Altitude Refer AFM/POH for correct power setting

Fuel Consumption

Headwind – lower in-flight ground speed Tailwind – increase in-flight ground speed

Wind

A heavier weight will result in a shallower climb. Overloading aircraft can lead to overheating during the climb-out phases, additional wear and tear on engine parts, increased fuel consumption, slower cruising speeds and reduced range.

Weight

BEST SPECIFIC AIR RANGE • The best TAS/power ratio occurs at a speed where a tangent drawn from the origin (i.e. 0 KTAS) of the

power-required curve, just touches the power curve. At any other speed, the ratio is less.

• Air nautical miles per gallon = TAS ÷ fuel flow

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BEST SPECIFIC GROUND RANGE (SGR) • Taking into account the effect of the wind.

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Best ground range speed is higher in a headwind

Robson, 2018

Best ground range speed is lower in a tailwind

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Power required curve ≠ Total drag curve

IAS min drag remains the same regardless of density altitude but TAS will change

IAS min drag is higher than IAS min power

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Dyson-Holland, 2015

“In general, most light aeroplanes are designed so that during cruise, with recommended power settings, the propeller will be at or close to its maximum efficiency” (Dyson-Holland, 2015, p. 87)

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Dyson-Holland, 2015 “For most small aeroplanes, the weight changes are relatively small and so often ignored. [Thus] The in-flight performance data is specified for maximum take-off weight only” (Dyson- Holland, 2015, p. 91)

SPECIFIC RANGE AS A FUNCTION OF WEIGHT

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MAXIMUM RATE-OF-CLIMB DATA

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Dyson-Holland, 2015

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LOADING

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The Centre of Gravity (CG) must be within its limit all the time

“The CG is a point at which the aircraft would balance if it were suspended at that point” (FAA, 2008, p. 9-2)

Too far forward CG results in nose of aeroplane being too heavy Too far rearward CG results in nose of aeroplane being too light (tail heavy)

Calculating the position of the CG (Robson, 2009, p. 322)

Calculate individual moments Calculate total weight and total moment Calculate CG arm = total moment/total weight

LOADING

• Principles of weight and balance computation (FAA, 2008)

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DATUM POINT

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http://www.rfc.ca/data/pdf/Pilot%20Resources/C172%20Weight%20and %20Balance%20Worksheet.pdf

Refer Pilot Operating Handbook (POH)

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LOADING

Determine weight (Commercial operations) (CAAP 215-1)

Crew

• Actual weight including the weight of any crew baggage; or,

• Standard weight • Male crew = 85kgs

• Female crew = 65 kgs

Pax

• Actual weight including all personal items and carry-on baggage for six seats or less.

• Standard weight specified in CAAP 235-1

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Robson, 2009

LOADING

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Normal category vs. utility category (Robson, 2009)

Normal category Certified for non-aerobatic manoeuvres (normally limited to stalls and 60°AoB Typical load factor: +3.8g ~ -1.5g

Utility category Certified for limited aerobatic manoeuvres (e.g., Spin if certified) Typical load factor: +4.5g ~ -1.8g

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BEW – 800kgs CG – 2.49m Pilot – 82kgs Pax 1 – 85kgs Pax 2 – 65kgs Baggage – 42kgs Fuel – 150litres* Fuel burn – 70litres *Litres ->kg – x 0.72

800 2.49 1992 167 384.1 65 211.25 42 163.38

1074 2.56 2750.73 108 284.04 1182 2.57 3034.77 50.4 118.944

1131.6 2.57 2915.826

Massey University, n.d.

LOADING – ADJUST THE CG

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Removing weight and finding the new CG

Adding weight finding the new CG

Moving weight

REMOVING WEIGHT AND FINDING THE NEW CG

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TOTAL WEIGHT x ARM = TOTAL MOMENT - ITEM WEIGHT x ARM = - MOMENT

NEW TOTAL WEIGHT NEW TOTAL MOMENT

NEW TOTAL MOMENT = NEW CG POSITION NEW TOTAL WEIGHT

EXAMPLE – REMOVING WEIGHT & FINDING NEW CG

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• CG limit: 120 ~ 130 • Current CG: 125 • Weight to be removed: 50kgs from a baggage area 1 (150) to

achieve the MCTOW of 4000kgs • Will the new CG be within the limit?

ADDING WEIGHT AND FINDING THE NEW CG

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TOTAL WEIGHT x ARM = TOTAL MOMENT

+ ITEM WEIGHT x ARM = + MOMENT

NEW TOTAL WEIGHT NEW TOTAL MOMENT

NEW TOTAL MOMENT = NEW CG POSITION NEW TOTAL WEIGHT

EXAMPLE – ADDING WEIGHT & FINDING NEW CG

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• CG limit: 120 ~ 130 • Current CG: 125 • Current Wt: 4,000kgs • Can you load additional weight to bring a/c’s MCTOW of 4,050kgs

without exceeding the CG limit?

MOVING WEIGHT

When weight is moved from one station to another the total weight remains the same but the CG point will change.

WEIGHT MOVED (w) = CG CHANGE (d)

TOTAL WEIGHT (W) DISTANCE MOVED (D)

OR

W x d = w x D

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EXAMPLE OF MOVING WEIGHT

• CG limit: 120 ~ 130 • Current CG: 132 • Current Wt: 4,000kgs • How much weight to be moved from a baggage area 2 (180) to a baggage area 1

(50) so that CG is at its aft limit?

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LOADING QUESTIONS

1. The mass (also commonly known as take-off weight) of the aeroplane is 5,700 kg and the CG is computed to be 325cm aft of the datum. The aft CG limit is 290cm. Calculate the amount of freight that has to be moved from the rear freight compartment (395cm aft of the datum) to the forward baggage area (55cm aft of the datum) to move the CG to the aft limit.

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2. The mass (also commonly known as all up weight) of a jump plane is 2,500kg with current CG at 90cm aft of the datum. Two skydivers (88kg each) are dropped from 150cm aft of datum. What is the new CG immediately after the drop?

LOADING QUESTIONS

3. The mass (also commonly known as take-off weight) of the aeroplane is 2,300 kgs and the CG is computed to be 310cm aft of the datum. The aft CG limit is 295cm. Calculate the amount of freight that has to be moved from the rear freight compartment (405cm aft of the datum) to the forward baggage area (155cm aft of the datum) to move the CG to the aft limit

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