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

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

- Runway friction
- Wheel spin-up
- Side-wind loads.

2. Which of the following are the correct equations for take-off run?

a) F_{N}+D-Wsinγ_{R}-μ_{r}R=mV̇

b) R-L=Wcosγ_{R}

c) R-L=Wsinγ_{R}

d) F_{N}-D-Wsinγ_{R}-μ_{r}R=mV̇

View Answer

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

- F
_{N}-D-Wsinγ_{R}-μ_{r}R=mV̇ - R+L=Wcosγ
_{R}

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

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, μ

_{R}is runway coefficient of the rolling friction, γ

_{R}is runway slope, L is lift, C

_{D}and C

_{L}are the coefficients of drag and lift.

4. The curly bracket in the accelerating force represents the net propulsive thrust- weight ratio.

a) True

b) False

View Answer

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, μ

_{R}is runway coefficient of the rolling friction, γ

_{R}is runway slope, L is lift, C

_{D}and C

_{L}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

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, μ

_{R}is runway coefficient of the rolling friction, γ

_{R}is runway slope, L is lift, C

_{D}and C

_{L}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.

6. The formula for ground run is given by the formula __________

a) S_{G}=\(\frac{V_{LOF}^2}{2g(A+BV^2)_{V_{LOF}}}\)

b) S_{G}=\(\frac{V_{LOF}^2}{2g(A+BV^2)_{0.7V_{LOF}}}\)

c) S_{G}=\(\frac{V_{LOF}^2}{2g(A-BV^2)_{0.7V_{LOF}}}\)

d) S_{G}=\(\frac{V_{LOF}^2}{2g(A-BV^2)_{V_{LOF}}}\)

View Answer

Explanation: The formula for ground run is given by the formula S

_{G}=\(\frac{V_{LOF}^2}{2g(A-BV^2)_{0.7V_{LOF}}}\) where

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

7. The airborne distance is given by __________

a) S_{A}=\(\frac{W}{(F_N-D)_{av}}\Big\{\frac{V_{2}^2+V_{LOF}^2}{2g}+35\Big\}\)

b) S_{A}=\(\frac{W}{(F_N+D)_{av}}\Big\{\frac{V_{2}^2+V_{LOF}^2}{2g}+35\Big\}\)

c) S_{A}=\(\frac{W}{(F_N+D)_{av}}\Big\{\frac{V_{2}^2-V_{LOF}^2}{2g}+35\Big\}\)

d) S_{A}=\(\frac{W}{(F_N-D)_{av}}\Big\{\frac{V_{2}^2-V_{LOF}^2}{2g}+35\Big\}\)

View Answer

Explanation: The airborne distance is given by S

_{A}=\(\frac{W}{(F_N-D)_{av}}\Big\{\frac{V_{2}^2-V_{LOF}^2}{2g}+35\Big\}\) where

- W is weight
- D is drag
- G is acceleration due to gravity
- V is velocity
- F
_{N}is normal force.

8. In the calculations the horizontal airborne distance is assumed to be much greater than 35 feet.

a) True

b) False

View Answer

Explanation: In the calculations the horizontal airborne distance is assumed to be much greater than 35 feet. The airborne distance is given by S

_{A}=\(\frac{W}{(F_N-D)_{av}}\Big\{\frac{V_{2}^2-V_{LOF}^2}{2g}+35\Big\}\) where

- W is weight
- D is drag
- G is acceleration due to gravity
- V is velocity
- F
_{N}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

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.

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

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

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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