Post on 14-Apr-2018
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TP, Page 1
FLIGHT
OPERATIONS
ENGINEERING
Takeoff Performance
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TP, Page 2
Takeoff Performance
• Field length requirements
• Tire speed requirements
• Brake energy requirements
• Climb requirements
• Obstacle requirements
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TP, Page 3
Field Length Considerations
• Ground acceleration capability
• Takeoff speeds
• Accelerate - Go considerations
• Accelerate - Stop considerations
• Field length calculation and considerations
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TP, Page 4
Ground Acceleration Capability
FSlope
Σ Forces = Mass * Acceleration
Friction Drag Thrust
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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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Ground Roll Acceleration Equation
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
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Ground Roll Acceleration Equation
• 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
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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.
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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
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Ground Distance Calculation
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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
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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 =
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Fundamental Time, Distance Relationship
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∫=
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Fundamental Time, Distance Relationship
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
∫=
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Fundamental Time, Distance Relationship
• 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
−
∫=
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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
-∫=
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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
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Step Integration Method
• 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
−−−
− −=−
∫=
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Step Integration Method
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
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TP, Page 20
Step Integration Method
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
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TP, Page 21
Step Integration Method
( ) ( )
−−
−=
−−
−−
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
−
−−
−
∆−=
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TP, Page 22
Step Integration Method
• 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
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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
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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
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TP, Page 25
Average Acceleration Method
• 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=∗
+
−=
−
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TP, Page 26
Average Acceleration Method
• Looking at the acceleration chart, the accelerationfall off is more proportional to V2
• Why?
Acceleration - ft/s2
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
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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
Acceleration - ft/s2
V2 - kts2
( ) 1.1062
0150V 22
accel avefor =+=
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Average Acceleration Method
• 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
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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
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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
∆−
=
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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
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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
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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
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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
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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
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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
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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
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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.
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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
Acceleration - ft/s2
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
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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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