07a Takeoff Performance

7/27/2019 07a Takeoff Performance

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TP, Page 1

FLIGHT

OPERATIONS

ENGINEERING

Takeoff Performance



TP, Page 2

Takeoff Performance

• Field length requirements

• Tire speed requirements

• Brake energy requirements

• Climb requirements

• Obstacle requirements



TP, Page 3

Field Length Considerations

• Ground acceleration capability

• Takeoff speeds

• Accelerate - Go considerations

• Accelerate - Stop considerations

• Field length calculation and considerations



TP, Page 4

Ground Acceleration Capability

FSlope

Σ Forces = Mass * Acceleration

Friction Drag Thrust



TP, Page 5

Ground Roll Acceleration Equation

Σ Forces = Mass * Acceleration

Thrust - Drag - Friction - Fslope = Mass * Acceleration

T - D - µ ( W - L ) - Wsin φ = aWg

small angles sin φ = φ in radians

T - D - µ ( W - L ) - W φ = aW

g


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Rearranging

T - D - µ ( W - L ) - W φ = aW

g

a = [ T - D - µ ( W - L ) - W φ ]

g

W

Combine Drag and Lift term into one since both are a

function of dynamic pressure

a = [ T - µ W - (D - µ L ) - W φ ]g

W




• Factors effecting airplane acceleration capability

– Thrust - rating, altitude, temperature, bleeds

– µ, rolling friction, function of the airplane’s gear – CD, CL airplane configuration, flap setting, CDL items

– q, dynamic pressure, the faster the airspeed the worsethe acceleration

– Weight, less weight results in better acceleration

– Hand calculations assume weight is constant

– Computer programs take into account fuel burn

a = [ T - µ W - ( CD - µ CL ) q S - W φ ]g

W



Workbook Problem

Do problem 9 in the performance workbook

• Determine the airplane’s all engine groundacceleration capability at 0 ktas and 150 ktas for thefollowing conditions.



Sample Acceleration Calculation

Acceleration - ft/s2

Speed - kts

0.01.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

0 20 40 60 80 100 120 140 160

Weight = 240000 lb SLOPE= 0Sea level, Std. day WIND= 0 µ 0.0165

NO ENG = 2 (CD-µ CL)= 0.08ALL ENGINE

V - GS V - TAS DYNAMIC - q F - THRUST F - SLOPE µW (CD- µCL)qS ACCEL - ft/s/s ACCEL - kt/s

0 0 0.00 35532 0 3960 0 9.0 5.3

20 20 1.35 34653 0 3960 211 8.7 5.240 40 5.42 33775 0 3960 845 8.4 5.0

60 60 12.19 32896 0 3960 1902 8.0 4.8

80 80 21.67 32017 0 3960 3382 7.6 4.5

100 100 33.86 31139 0 3960 5284 7.1 4.2

120 120 48.75 30260 0 3960 7609 6.6 3.9

140 140 66.36 29381 0 3960 10357 6.0 3.5

150 150 76.18 28942 0 3960 11889 5.6 3.3



Ground Distance Calculation



Ground Distance Calculation

• To determine the field length required we need to

determine how much runway it took to accelerate frombrake release, ground speed = 0, to some other speed

– Rotation speed

– Engine failure speed



Fundamental Time, Distance Relationship

∆ s

∆ t

∆ V

∆ t

∆ V

a

Velocity, V = or ∆ s = V ∆ t

Acceleration, a = or ∆ t =

Substitute and solve for distance:

V ∆ V

as =




Integrating you obtain:

Where the beginning velocity is brake release, zero groundspeed and the end velocity is the final ground speed

aVdV

S

g Final V

0∫=




If acceleration, a, were a constant with velocity, thenthe integral would be easy.

Is acceleration constant with velocity during theground run?

a

VdV Sg Final

V

0

∫=




• No, acceleration, a, varies as a function of theairplanes airspeed.

• As the airplane’s speed increases the acceleration, a,reduces because of the thrust decay and the increasein q, dynamic pressure.

• Both these are functions of true airspeed, therefore wewill change the integral to terms of true airspeed.

( )a

dV V V S W

T Final V

W V

−

∫=



Evaluating the Integral

Methods to evaluate integral:

• “Step integration” method – Break up the integral into many small steps and assume

the acceleration is constant over the step

• “Average” acceleration – Calculate an average acceleration and assume theacceleration is constant over velocity range

– Note: Same equations as the step integration only you are

using one large step

( )

a

dV V V S W

T Final V

W V

-∫=



Step Integration Method

An integral can be broken up into many small integrals

the sum of which is equal to the original integral

( ) ( )

( ) ( )a

dV V V

a

dV V V

adV V V

adV V V S

W

T Final V

1T Final V

W

1T Final V

2T Final V

W 2V

1V

W 1V

W V

−

∫+−

∫

+−

∫+−∫=

−

−

−K

K




• Assume acceleration is constant and equal to the

average of the acceleration at the initial stepvelocity and the final step velocity

• Evaluate the integral for one small step as an

example, step 1 - 2

( ) 2V

1V 2ave1

W

2ave1

2

2ave1

W

2V

1V21 a

VV

a2 V a

dV V V S

−−−

− −=−

∫=




Rearranging

−−

−=

−−−−

−

2ave1

W 1

2ave1

2

1

2ave1

W 2

2ave1

2

221 a

V V

a2

V

a

V V

a2

V S

−−

−=

−−−−

−

2ave1

W 1

2ave1

W 2

2ave1

2

1

2ave1

2

221 a

V V

a

V V

a2

V

a2

V S

−−

−=

−−

−

2ave1

W 1W 2

2ave1

2

1

2

221 a

W V V V

a2

V V S



TP, Page 20


Rearranging

As the old college professor said - “recognizing”

( )( ) ( )

−−

−+=

−−

−

21ave

W 12

21ave

121221 a

V V V

a2

V V V V S

( ) ( ) ( )

−−

−+=

−−

−

21ave

W 12

21ave

121221 a

V V V

a

V V

2

V V S



TP, Page 21


( ) ( )

−−

−=

−−

−−

21ave

W 12

21ave

1221ave21 a

V V V

a

V V V S

( )( )21ave

12W 21ave21 a

V V V V S−

−

−−−=

( )

21ave

21W 21ave

21 a

V V V S

−

−−

−

∆−=



TP, Page 22


• General expression for calculating distance of an

individual step

• Total distance is the sum of the steps

( )

aveStep

StepW aveStep

Step a

V V V S

∆−=

( ) ( )1Final Final Step2Final 1Final Step

21Step1W StepStepTotal

SS

SSSS

−−−−−

−−

++

+= Σ=

LL

L



TP, Page 23

Weight = 240000 lb SLOPE= 0

Sea level, Std. day WIND= 0

NO ENG = 2ALL ENGINE

V - GS V - TAS ACCEL- ft/s/s ACCEL - kt/s S – Step - ft Sum S

0 0 9.0 5.3 0

20 20 8.7 5.2 64 64

40 40 8.4 5.0 199 264

60 60 8.0 4.8 346 610

80 80 7.6 4.5 510 1120

100 100 7.1 4.2 697 1817

120 120 6.6 3.9 917 2734

140 140 6.0 3.5 1183 3917

150 150 5.6 3.3 713 4630

Example of Distance Calculation

Total distance from brake release to 150knots based on 20 knot steps = 4630 feet

Note: 1 knot step = 4635 feet



TP, Page 24

Average Acceleration Method

• Average acceleration is just using the step

integration method with a single step

• The trick is in determining the appropriate averageacceleration to give an adequate approximation of

the distance



TP, Page 25


• Example: Compute the average acceleration byusing the acceleration at 0 and 150 kts

– From earlier calculations

– a0 = 9.0 ft/s2

– a150 = 5.6 ft/ s2

• Notice the distance does not match the stepintegration answer

( )ft 43906878.1

2

6.50.9

150075S

2

1500=∗

+

−=

−



TP, Page 26


• Looking at the acceleration chart, the accelerationfall off is more proportional to V2

• Why?


Speed - kts

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

0 20 40 60 80 100 120 140 160



TP, Page 27

Average Acceleration Calculation

• Try calculating average acceleration at average of V2.

• From earlier chart a106.1

= 6.9 ft/s2

4.0

4.5

5.0

5.5

6.0

6.5

7.0

7.5

8.0

8.5

9.0

0 5000 10000 15000 20000 25000 30000


V2 - kts2

( ) 1.1062

0150V 22

accel avefor =+=



TP, Page 28


• Example: Compute the average acceleration by

using the acceleration at 106.1 kts. – a106.1 = 6.9 ft/s2

• Notice the distance does match the step integrationanswer of 4630 ft. quite well

( ) ft 46446878.19.6

150075S2

1500=∗−=

−

Summary of Ground Acceleration



TP, Page 29

Summary of Ground AccelerationCalculation Methods

• Step integration on velocity is used in the computer

programs to calculate the ground run with allengines operating

• Forces are a function of speed

• Current computer programs take credit for fuel burnoff during ground run

• Average acceleration method is quick easy way to

determine the effect of various parameters on thetakeoff ground run



TP, Page 30

Items that Affect Ground Run

• Slope, how does slope effect ground run

– Uphill slope, worse acceleration, longer distance – Downhill slope better acceleration, shorter distance

• Wind, how does wind effect ground run

– At a given true airspeed the acceleration is the same

– Effect of wind is to change the ∆V the airplaneaccelerates through

( )

aveStep

StepW veStepa

Step

a

V V V S

∆−

=



TP, Page 31

Effect of Wind

Effect of head windis to reduce ∆VEffect of tail wind isto increase ∆V

Weight = 240000 lb SLOPE = 0Sea level, Std. day WIND = 20 µ 0.0165

NO ENG = 2 (CD-µCL)= 0.08ALL ENGINE

V - GS V - TAS DYNAMIC - q F - THRUST F - SLOPE µW (CD-µCL)qS ACCEL - ft/s/s ACCEL - kt/s

0 20 1.35 34653 0 3960 211 8.7 5.2

20 40 5.42 33775 0 3960 845 8.4 5.0

40 60 12.19 32896 0 3960 1902 8.0 4.8

60 80 21.67 32017 0 3960 3382 7.6 4.5

80 100 33.86 31139 0 3960 5284 7.1 4.2

100 120 48.75 30260 0 3960 7609 6.6 3.9

120 140 66.36 29381 0 3960 10357 6.0 3.5

130 150 76.18 28942 0 3960 11889 5.6 3.3

V - GS V - TAS ACCEL - ft/s/s ACCEL - kt/s S-Step-ft

0 20 8.7 5.2

20 40 8.4 5.0 66

40 60 8.0 4.8 208

60 80 7.6 4.5 364

80 100 7.1 4.2 542

100 120 6.6 3.9 750

120 140 6.0 3.5 1001

130 150 5.6 3.3 614

Sum S

0

66

274

639

1181

1931

2932

3546



TP, Page 32

Effect of Flap

• Increased takeoff flap will typically reduce theground acceleration capability due to increased drag

But

• Acceleration will be to a lower speed (VR,V2)

• Overall effect will be shorter distance

( )

aveStep

StepW aveStep

Step a

V V

S

∆−

=

V



TP, Page 33

Effect of Thrust

• Direct relationship – More thrust - shorter distance

– Less thrust - more distance

For example,

Same conditions as earlier only thrust has been

reduced by 10%.

Result 14% increase indistance required to

accelerate from 0 –150 for this example.

( )

aveStep

StepW aveStep

Step a

V V S

∆−

=

V

V - GS V - TAS ACCEL - ft/s/s ACCEL - kt/s S-Step-ft

0 8.0 4.8

20 7.8 4.6 72

40 7.5 4.4 223

60 7.2 4.2 389

80 6.7 4.0 574100 6.3 3.7 788

120 5.8 3.4 1042

140 5.2 3.1 1356

Sum S

0

72

295

684

12582046

3088

4444

150 4.9 2.9 824 5268

0

20

40

60

80100

120

140

150



TP, Page 34

Engine Out Ground Acceleration

• What is different in the calculation of the distance

required to accelerate following an engine failure – Failed engine’s thrust spins down as a function

of time, not airspeed

– Pilot inputs rudder to steer the airplane – Additional drag



TP, Page 35

Spindown Characteristics

Thrust

TO Thrust

Time From the Event, Sec

Fuel Cut

Throttle ChopSpindown

Factor

0

0.1

0.2

0.3

0.40.5

0.6

0.7

0.8

0.9

1

0 2 4 6 8 10 12



TP, Page 36

Spindown Characteristics

• Fuel cut used for the continued takeoff following an

engine failure – Lowest thrust

• Throttle chop is used for the AFM emergency stop

calculation from an event just prior to V1

– Higher thrust conservative for stop calculation

– Note: older airplanes used fuel cut following

engine failure just prior to V1



TP, Page 37

Calculation of Engine Out Accel/Distance

• Same equation used to calculate engine out acceleration

• Engine thrust now changing rapidly with time. Typically astep integration based on time is required. This becomesan iterative process.

a = [ T - µ W - ( CD - µ CL ) q S - W φ ]g

W



TP, Page 38

Workbook Problem

• Do problem 10 in the performance workbook

• Determine the airplane’s engine inoperativeground acceleration capability at 150 ktasfor the following conditions.



TP, Page 39

Engine Out Acceleration

Event Time V - GS V - TAS E1-FC E2 ACCEL - ft/s/s ACCEL - kt/s S-Step-ft Sum S

Eng Fail 0 150 150.0 1 1 5.4 3.2

1 153.3 153.3 0.22 1 2.3 1.4 256 256

2 155.6 155.6 0.08 1 1.7 1.0 261 517

3 157.3 157.3 0.038 1 1.5 0.9 264 781

4 158.8 158.8 0.018 1 1.4 0.8267 1047

4.9 160.0 160.0 0.01 1 1.3 0.8 242 1290

Speed - kts


0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

0 20 40 60 80 100 120 140 160 180



TP, Page 40

Summary

• Looked at ground calculation for both all engine and

engine out acceleration between two speeds – All engine is between brake release and engine

failure or rotation speed

– Engine inoperative is between engine failure and rotation speed

• Primary method of calculation is a step integration

– All engine is step integration based on speed – Engine inoperative is based on time



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07a Takeoff Performance