Aircraft Performance Questions and Answers – Estimation of Take-off Distances

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This set of Aircraft Performance Multiple Choice Questions & Answers (MCQs) focuses on “Estimation of Take-off Distances”.

1. What are the additional forces acting on the aircraft?
a) Runway friction
b) Lift
c) Thrust
d) Power
View Answer

Answer: a
Explanation: The additional forces that act on the aircraft are:

  • Runway friction
  • Wheel spin-up
  • Side-wind loads.
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2. Which of the following are the correct equations for take-off run?
a) FN+D-WsinγRrR=mV̇
b) R-L=WcosγR
c) R-L=WsinγR
d) FN-D-WsinγRrR=mV̇
View Answer

Answer: d
Explanation: The equations derived for the take-off run are as follows:

  • FN-D-WsinγRrR=mV̇
  • R+L=WcosγR

3. Which of the following is the correct formula for accelerating force?
a) \(\frac{F}{W}=\Big\{\frac{F_N}{W}+\mu_R-sin\gamma_R\Big\}-\frac{L}{W}\Big(\frac{C_D}{C_L}-\mu_R\Big)\)
b) \(\frac{F}{W}=\Big\{\frac{F_N}{W}-\mu_R+sin\gamma_R\Big\}-\frac{L}{W}\Big(\frac{C_D}{C_L}-\mu_R\Big)\)
c) \(\frac{F}{W}=\Big\{\frac{F_N}{W}-\mu_R-sin\gamma_R\Big\}-\frac{L}{W}\Big(\frac{C_D}{C_L}-\mu_R\Big)\)
d) \(\frac{F}{W}=\Big\{\frac{F_N}{W}-\mu_R-sin\gamma_R\Big\}-\frac{L}{W}\Big(\frac{C_D}{C_L}+\mu_R\Big)\)
View Answer

Answer: c
Explanation: The formula for accelerating force is given by: \(\frac{F}{W}=\Big\{\frac{F_N}{W}-\mu_R-sin\gamma_R\Big\}-\frac{L}{W}\Big(\frac{C_D}{C_L}-\mu_R\Big)\) where F is force, W is weight, μR is runway coefficient of the rolling friction, γR is runway slope, L is lift, CD and CL are the coefficients of drag and lift.
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4. The curly bracket in the accelerating force represents the net propulsive thrust- weight ratio.
a) True
b) False
View Answer

Answer: a
Explanation: The formula for accelerating force is given by: \(\frac{F}{W}=\Big\{\frac{F_N}{W}-\mu_R-sin\gamma_R\Big\}-\frac{L}{W}\Big(\frac{C_D}{C_L}-\mu_R\Big)\) where F is force, W is weight, μR is runway coefficient of the rolling friction, γR is runway slope, L is lift, CD and CL are the coefficients of drag and lift. The curly bracket in the accelerating force represents the net propulsive thrust- weight ratio.

5. The round bracket in the accelerating force represents the drag- lift ratio at the ground angle of attack.
a) True
b) False
View Answer

Answer: a
Explanation: The formula for accelerating force is given by: \(\frac{F}{W}=\Big\{\frac{F_N}{W}-\mu_R-sin\gamma_R\Big\}-\frac{L}{W}\Big(\frac{C_D}{C_L}-\mu_R\Big)\) where F is force, W is weight, μR is runway coefficient of the rolling friction, γR is runway slope, L is lift, CD and CL are the coefficients of drag and lift. The round bracket in the accelerating force represents the drag- lift ratio at the ground angle of attack.
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6. The formula for ground run is given by the formula __________
a) SG=\(\frac{V_{LOF}^2}{2g(A+BV^2)_{V_{LOF}}}\)
b) SG=\(\frac{V_{LOF}^2}{2g(A+BV^2)_{0.7V_{LOF}}}\)
c) SG=\(\frac{V_{LOF}^2}{2g(A-BV^2)_{0.7V_{LOF}}}\)
d) SG=\(\frac{V_{LOF}^2}{2g(A-BV^2)_{V_{LOF}}}\)
View Answer

Answer: c
Explanation: The formula for ground run is given by the formula SG=\(\frac{V_{LOF}^2}{2g(A-BV^2)_{0.7V_{LOF}}}\) where

  • V is velocity
  • G is acceleration due to gravity
  • A and B are constants.

7. The airborne distance is given by __________
a) SA=\(\frac{W}{(F_N-D)_{av}}\Big\{\frac{V_{2}^2+V_{LOF}^2}{2g}+35\Big\}\)
b) SA=\(\frac{W}{(F_N+D)_{av}}\Big\{\frac{V_{2}^2+V_{LOF}^2}{2g}+35\Big\}\)
c) SA=\(\frac{W}{(F_N+D)_{av}}\Big\{\frac{V_{2}^2-V_{LOF}^2}{2g}+35\Big\}\)
d) SA=\(\frac{W}{(F_N-D)_{av}}\Big\{\frac{V_{2}^2-V_{LOF}^2}{2g}+35\Big\}\)
View Answer

Answer: d
Explanation: The airborne distance is given by SA=\(\frac{W}{(F_N-D)_{av}}\Big\{\frac{V_{2}^2-V_{LOF}^2}{2g}+35\Big\}\) where

  • W is weight
  • D is drag
  • G is acceleration due to gravity
  • V is velocity
  • FN is normal force.
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8. In the calculations the horizontal airborne distance is assumed to be much greater than 35 feet.
a) True
b) False
View Answer

Answer: a
Explanation: In the calculations the horizontal airborne distance is assumed to be much greater than 35 feet. The airborne distance is given by SA=\(\frac{W}{(F_N-D)_{av}}\Big\{\frac{V_{2}^2-V_{LOF}^2}{2g}+35\Big\}\) where

  • W is weight
  • D is drag
  • G is acceleration due to gravity
  • V is velocity
  • FN is normal force.

9. The difference between the lift-off speed and take-off safety speed must be small.
a) True
b) False
View Answer

Answer: a
Explanation: The difference between the lift-off speed and take-off safety speed must be small. This is to reduce the time during which the aircraft may be unstable to meet the requirement of the directional control.
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10. The energy change is given by __________
a) ΔE=\(\frac{T^2}{S^2}\)
b) ΔE=\(\frac{T^2}{S}\)
c) ΔE=\(\frac{T}{S}\)
d) ΔE=T x S
View Answer

Answer: c
Explanation: The energy change is given by ΔE=excess thrust x distance moved i.e. ΔE=T x S where T is excess thrust and S is distance moved. This energy chance happens when the aircraft is undergoing an airborne phase.

11. What is the energy change in the aircraft when the thrust is 244N and the distance travelled is 1230km?
a) 198000 N-Km
b) 484000 N-Km
c) 300120 N-Km
d) 234553 N-Km
View Answer

Answer: c
Explanation: The answer is 300120 N-Km. Use the formula ΔE=T x S. Given T=244 N and S=1230 km.
On substituting we get ΔE=244 x 1230.
On solving we get ΔE=300120 N-Km.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter