Category: Aircraft Performance Objective Questions

250+ TOP MCQs on Air Data Computers and Answers

Aircraft Performance Multiple Choice Questions on “Air Data Computers”.

1. FSO stands for ____________
a) Full-scale output
b) Flight-scale output
c) Flight stimulation output
d) Full screen output
Answer: a
Clarification: FSO stands for full-scale output. At a specific temperature the variation in output signals as the transducer performs over a range from minimum to maximum pressure is known as full-scale output.

2. Transducers are the pressure devices used to convert signal in one form of energy to another energy form.
a) True
b) False
Answer: a
Clarification: Transducers are the pressure devices used to convert signal in one form of energy to another energy form. Pressure transducers can also be called as transmitters, senders, indicators or piezometers.

3. Air data computer system measures airspeed, mach number, temperature and height.
a) True
b) False
Answer: a
Clarification: Air data computer system measures airspeed, mach number, temperature and height. Air data computer is an important avionic component present in an aircraft that measures airspeed, mach number, temperature and height with the help of pitot-static tube.

4. Dynamic pressure is the difference between stagnation pressure and static pressure.
a) True
b) False
Answer: a
Clarification: Dynamic pressure is the difference between stagnation pressure and static pressure. In simpler way it is kinetic energy per unit volume of a fluid particle. It is represented by ‘q’. The formula for dynamic pressure is given by q=(frac{1}{2})γPM2 where q=dynamic pressure, P is pressure of the fluid, M is mach number of that medium and γ is the ratio of specific heat at constant pressure to that of specific heat at constant volume.

5. Which of the following is the correct formula for dynamic pressure?
a) q=(frac{1}{2})γPM2
b) q=(frac{1}{2})γρM2
c) q=(frac{1}{2})γPV2
d) q=(frac{1}{2})γρV2
Answer: a
Clarification: The correct formula for dynamic pressure is q=(frac{1}{2})γPM2 where q=dynamic pressure, P is pressure of the fluid, M is mach number of that medium and γ is the ratio of specific heat at constant pressure to that of specific heat at constant volume.

6. Dynamic pressure is directly proportional to the flow speed.
a) True
b) False
Answer: a
Clarification: Dynamic pressure is directly proportional to the flow speed. The formula for dynamic pressure is q=(frac{1}{2})γPM2 where q=dynamic pressure, P is pressure of the fluid, M is mach number of that medium and γ is the ratio of specific heat at constant pressure to that of specific heat at constant volume.

7. Calculate the dynamic pressure of aircraft in air when mach number and pressure are 2 and 101325N/m2.
a) 283710 N/m2
b) 541278 N/m2
c) 352769 N/m2
d) 457815 N/m2
Answer: a
Clarification: The answer is 283710. Given M=2, P=101325 N/m2 and we know γ for air is 1.4. On substituting the values in the formula q=(frac{1}{2})γPM2
We get q=(frac{1}{2})*1.4*101325*22
On solving we get q=283710 N/m2.

8. Calculate the dynamic pressure of aircraft in air when density and flow speed are 1.23 kg/m3 and 340m/s.
a) 71094 N/m2
b) 54128N/m2
c) 65487 N/m2
d) 65789 N/m2
Answer: a
Clarification: The answer is 71094 N/m2. Given speed=340m/s, ρ=1.23kg/m3. On substituting the values in the formula q=(frac{1}{2})ρV2.
We get q=(frac{1}{2})*1.23*3402
On solving we get q=71094 N/m2.

9. Find the flow speed of air when the dynamic pressure is 70516 N/m2 and density is 1.23kg/m3.
a) 338.62 m/s
b) 340 m/s
c) 313.5 m/s
d) 321.5 m/s
Answer: a
Clarification: The answer is 338.62 m/s. Given dynamic pressure=70516 N/m2 and density is 1.23kg/m3. From the formula q=(frac{1}{2})ρV2, substitute the given values.
On solving we get 70516=(frac{1}{2})*1.23*V2
V=338.62 m/s.

10. The Pitot-static system pressures are given to the air data computers to find CAS, density and temperature.
a) True
b) False
Answer: b
Clarification: The Pitot-static system pressures are given to the air data computers to find CAS (calibrated air speed), pressure height and Mach number. In addition we can find EAS (equivalent air speed) by scale-altitude correction applied to CAS (calibrated air speed).

250+ TOP MCQs on Cruise Method 3 and Answers

Aircraft Performance Multiple Choice Questions on “Cruise Method 3”.

1. Which factor must be reduced in cruise method 3?
a) Mach number
b) Angle of attack
c) Altitude
d) Temperature
Answer: b
Clarification: The factor that must be reduced in cruise method 3 (also known as constant mach number and constant altitude method) is angle of attack. Also the ratio of weight to lift coefficient and relative airspeed are kept constant i.e. W/CL ratio and u are constant throughout the method.

2. Which of the following indicates the range formula in cruise method 3?
a) R=(frac{1}{C}Big(frac{2}{rho SC_L}Big)^{frac{1}{2}}int_{W_i}^{W_f}frac{dW}{W^{frac{1}{2}}})
b) R=(Big(frac{2}{rho SC_L}Big)^{frac{1}{2}}frac{L}{D}int_{W_i}^{W_f}frac{dW}{W^{frac{1}{2}}})
c) R=(Big[frac{V_{mdi}}{C}E_{max}Big]2u_iBig{tan^{-1}Big[frac{1}{u_{i}^2}Big]+tan^{-1}Big[frac{1}{omega u_{i}^2}Big]Big})
d) R=(Big[frac{V_{mdi}}{C}E_{max}Big]2u_iBig{tan^{-1}Big[frac{1}{u_{i}^2}Big]-tan^{-1}Big[frac{1}{omega u_{i}^2}Big]Big})
Answer: d
Clarification: The correct equation of cruise method 3 is R=(Big[frac{V_{mdi}}{C}E_{max}Big]2u_iBig{tan^{-1}Big[frac{1}{u_{i}^2}Big]-tan^{-1}Big[frac{1}{omega u_{i}^2}Big]Big}) where R is range of the cruise, C is specific fuel consumption, Emax is endurance, Vmdi is relative airspeed and ω is fuel ratio.

3. Which of the following is the correct integrated range equation of cruise method 3?
a) R=(frac{1}{C}Big(frac{2W_i}{Srho}Big)^{frac{1}{2}}frac{C_L}{C_D}2Big(1-omega^{frac{-1}{2}}Big))
b) R=(frac{V}{C}int_{W_i}^{W_f}frac{dW}{D})
c) R=(Big[frac{V_{mdi}}{C}E_{max}Big]2u_iBig{tan^{-1}Big[frac{1}{u_{i}^2}Big]-tan^{-1}Big[frac{1}{omega u_{i}^2}Big]Big})
d) R=(frac{1}{C}Big(frac{2W_i}{Srho}Big)^{frac{1}{2}}frac{C_{L}^{0.5}}{C_D}2Big(1-omega^{frac{-1}{2}}Big))
Answer: c
Clarification: The correct equation of cruise method 3 is R=(frac{V}{C}int_{W_i}^{W_f}frac{dW}{D}) where R is range, V is velocity, C is specific fuel consumption, D drag and Wf, Wiare final and initial weights.

4. Cruise method 3 is also known as ____________
a) constant angle of attack and variable mach number
b) constant angle of attack and constant mach number
c) constant angle of attack and constant altitude number
d) constant mach number and constant altitude number
Answer: d
Clarification: Cruise method 3 also known as constant altitude and constant mach number method. In cruise method 3 altitude, mach number and relative airspeed are kept constant i.e. M, u and h are constant throughout the method.

5. Which factor is not a constant factor in cruise method 3?
a) Mach number
b) Angle of attack
c) Altitude
d) Relative air speed
Answer: d
Clarification: The factor that must be reduced in cruise method 3 (also known as constant angle of attack and constant altitude method) is angle of attack. Also the ratio of weight to lift coefficient and relative airspeed are kept constant i.e. W/CL ratio and u are constant throughout the method.

6. The range factor of cruise method 3 is identical to that of cruise method 1.
a) True
b) False
Answer: a
Clarification: In cruise method 3 the range factor is given by R=(Big[frac{V_{mdi}}{C}E_{max}Big]2u_iBig{tan^{-1}Big[frac{1}{u_{i}^2}Big]-tan^{-1}Big[frac{1}{omega u_{i}^2}Big]Big}) where (Big[frac{V_{mdi}}{C}E_{max}Big]) is range factor and range pf aircraft is function of the range factor of the aircraft.

7. The endurance of cruise method 3 is identical to that of cruise method 1.
a) True
b) False
Answer: b
Clarification: The endurance of cruise method 3 is not identical to that of cruise method 1. E=(Big[frac{E_{max}}{C}Big]2Big{tan^{-1}Big[frac{1}{u_{i}^2}Big]-tan^{-1}Big[frac{1}{omega u_{i}^2}Big]Big})where where V is true airspeed, C is specific fuel consumption, Emax is endurance, u is relative airspeed and ω is fuel ratio.

8. Which of the following is a reason of disadvantage of cruise method 3?
a) Increased mach number
b) Fuel saving
c) Increased time of flight
d) Increased weight of aircraft
Answer: c
Clarification: The disadvantage of cruise method 3 is that to compensate the weight of aircraft the mach number and relative airspeed are decreased and there is an increase in the time of flight thus does not making any fuel saving.

9. In which of the following cases cruise method 3 is used?
a) Commercial aircraft
b) Fighter aircraft
c) Military
d) Patrol and surveillance
Answer: d
Clarification: Cruise method 3 is used mainly for patrol and surveillance purposes as there is more endurance and time of flight and short range which is best suitable for the surveillance purpose. Rather than cruise method 3 we use cruise method 2.

10. In Cruise method 3 endurance is more concentrated than distance travelled.
a) True
b) False
Answer: b
Clarification: In cruise method 3 endurance is not much concentrated rather distance travelled is concentrated. This is the reason which makes this cruise method 3 best suitable for patrol and surveillance purposes which mainly prefer more endurance and time of flight.

250+ TOP MCQs on Take-off Performance and Answers

Aircraft Performance Multiple Choice Questions on “Take-off Performance”.

1. What is meant by minimum control speed ground?
a) The airspeed below which the rudder the will not be capable of producing a yawing moment
b) The airspeed below which the rudder the will not be capable of producing a pitching moment
c) The airspeed below which the rudder the will not be capable of producing a rolling moment
d) The airspeed below which the rudder the will not be capable of producing a dutch moment
Answer: a
Clarification: The minimum control speed ground is the airspeed below which the rudder the will not be capable of producing a yawing moment. If there is an engine failure at this moment then it is must to stop the take-off run.

2. What is lift-off speed?
a) The speed at which the aircraft generates enough power to become airborne
b) The speed at which the aircraft generates enough speed to become airborne
c) The speed at which the aircraft generates enough thrust to become airborne
d) The speed at which the aircraft generates enough lift to become airborne
Answer: d
Clarification: The lift-off speed is the speed at which the aircraft generates enough lift to become airborne. Just before reaching this speed the aircraft rotates into a nose-up altitude which is equal to the lift-off angle of attack.

3. Just before reaching the lift-off speed the aircraft is rotated into a nose-up altitude which is equal to the lift-off angle of attack.
a) True
b) False
Answer: a
Clarification: The lift-off speed is the speed at which the aircraft generates enough lift to become airborne. Just before reaching this speed the aircraft rotates into a nose-up altitude which is equal to the lift-off angle of attack.

4. What is the screen height during the take-off of an aircraft?
a) 40 ft
b) 14 ft
c) 30 ft
d) 35 ft
Answer: d
Clarification: The take-off speed for the aircraft is known to be 35 feet from the ground level. The screen height is same for all types of aircraft such as commercial, combat etc. This screen height is maintained same to avoid aircraft collision.

5. The rotation speed must allow the aircraft to rotate into lift-off attitude.
a) True
b) False
Answer: a
Clarification: The rotation speed must allow the aircraft to rotate into lift-off attitude before the lift-off speed is being achieved by the aircraft. Just before reaching lift-off speed the aircraft rotates into a nose-up altitude which is equal to the lift-off angle of attack.

6. What is meant by ground run distance?
a) The distance where the aircraft takes-off on the ground
b) The vertical distance where the aircraft takes-off on the ground
c) It is also known as the screen height
d) It is the landing distance of the aircraft
Answer: a
Clarification: The ground run distance is the distance where the aircraft takes-off on the ground. There are two parts of take-off distance. One is ground distance and the other one is the distance from which the aircraft leaves the ground. This screen height must be about 35 feet to avoid aircraft collision.

7. There are two parts of take-off distance.
a) True
b) False
Answer: a
Clarification: There are two parts of take-off distance. One is ground distance and the other one is the distance from which the aircraft leaves the ground. This screen height must be about 35 feet to avoid aircraft collision.

8. What is the use of minimum control speed airborne?
a) Maintain directional control
b) Maintain fuel ratio
c) Maintain constant thrust
d) Maintain constant power
Answer: a
Clarification: The minimum control speed airborne is used to maintain the directional control during the climb of the aircraft. The minimum control speed airborne is always greater than the minimum control speed ground.

9. What is meant by take-off safety speed?
a) Airspeed at which only the safe climb gradient is attained
b) Airspeed at which only the directional control is attained
c) Airspeed at which only the directional control is kept constant
d) Airspeed at which both the safe climb gradient and directional control is attained
Answer: d
Clarification: The take-off safety speed is the airspeed at which both the safe climb gradient and directional control is attained. This condition is attained at the time of engine failure in the airborne state.

10. What is meant by decision speed?
a) The speed at the point where there is unequal distances on the take-off run
b) The slope at the point where there is unequal distances on the take-off run
c) The slope at the point where there is equal distances on the take-off run
d) The speed at the point where there is equal distances on the take-off run
Answer: d
Clarification: The decision speed is the speed at the point where there is equal distances on the take-off run. The decision speed determines the minimum safe length of the runway from which the aircraft can take-off.

11. Transition is known as the phase where both the safe climb gradient and directional control in the climb can be attained during the failure of an engine.
a) True
b) False
Answer: a
Clarification: Transition is known as the phase where both the safe climb gradient and directional control in the climb can be attained during the failure of an engine. This is an important phase of the aircraft take-off path.

12. What is meant by airborne distance?
a) The distance between take-off point and the point at which the screen height is cleared
b) The distance between lift-off point and the point at which the screen height is not cleared
c) The distance between lift-off point and the point at which the screen height is cleared
d) The distance between take-off point and the point at which the screen height is not cleared
Answer: c
Clarification: The airborne distance is the distance between lift-off point and the point at which the screen height is cleared. The total take-off distance is the sum of both airborne distance and ground run distance.

13. The total take-off distance is the sum of both airborne distance and ground run distance.
a) True
b) False
Answer: a
Clarification: The airborne distance is the distance between lift-off point and the point at which the screen height is cleared. The total take-off distance is the sum of both airborne distance and ground run distance.

14. What is meant by minimum unstick speed?
a) The maximum speed at which the aircraft becomes airborne
b) The minimum speed at which the aircraft does not becomes airborne
c) The maximum speed at which the aircraft does not becomes airborne
d) The minimum speed at which the aircraft becomes airborne
Answer: d
Clarification: The minimum unstick speed is the minimum speed at which the aircraft becomes airborne. This occurs when the over rotation pitches up the aircraft to a geometry limited angle of attack with the tail touching the ground.

15. During the flight of the aircraft with minimum unstick speed the tail of aircraft is in contact with the runway.
a) True
b) False
Answer: a
Clarification: The minimum unstick speed is the minimum speed at which the aircraft becomes airborne. This occurs when the over rotation pitches up the aircraft to a geometry limited angle of attack with the tail touching the ground.

250+ TOP MCQs on Equations of Motion for Performance and Answers

Aircraft Performance Interview Questions and Answers for freshers on “Equations of Motion for Performance”.

1. Which of the following is a correct equation?
a) Fa+Fp+Fg=Fl
b) Fa+Fp+Fg=Fl
c) Fa+Fp+Fg=Fl
d) Fa+Fp+Fg=Fl
Answer: a
Clarification: The correct equation is Fa+Fp+Fg=Ft where Fa=aerodynamic forces, Fp=propulsive forces, Fg=gravitational forces and Fl=inertial forces. The statement says that the system of forces containing gravitational forces, aerodynamic forces and propulsive forces results in the inertial forces acting on the aircraft.

2. The system of forces acting on the aircraft are propulsive forces, aerodynamic forces, gravitational forces and these result in inertial forces.
a) True
b) False
Answer: a
Clarification: The system of forces acting on the aircraft are propulsive forces, aerodynamic forces, gravitational forces and these result in inertial forces. From Newton’s law the equation becomes Fa+Fp+Fg=Ft where Fa=aerodynamic forces, Fp=propulsive forces, Fg=gravitational forces and Fl=inertial forces.

3. What is the basic assumption taken in formulating equations of motion?
a) All the engines will operate at equal gross thrust
b) All the engines do not operate at equal gross thrust
c) All the engines will operate at equal gross drag
d) All the engines do not operate at equal gross drag
Answer: a
Clarification: The basic assumption taken in formulating equations of motion is all the engines will operate at equal gross thrust. There are other assumptions also taken in the case of convectional aircraft there are:

  • Rate of change of mass is neglected
  • Aircraft is in symmetric flight
  • Gross thrust acts in aircraft body axis
  • Total net thrust is T cosα-DM
  • Thrust component is small compared to lift component.

4. Which of the following is not the additional assumption taken in the derivation of the equations of motion?
a) Gross thrust acts in aircraft body axis
b) Thrust component is small compared to lift component
c) Aircraft is in symmetric flight
d) Rate of change of mass is neglected
Answer: c
Clarification: The aircraft is symmetrical flight during the derivation of equations of motion.The other assumptions taken are:

  • Rate of change of mass is neglected
  • Aircraft is in symmetric flight
  • Gross thrust acts in aircraft body axis
  • Total net thrust is T cosα-DM
  • Thrust component is small compared to lift component.

5. The total thrust force assumed in the derivation of equations of motion is ___________
a) FN=T cosα+DM
b) FN=T cosα-DM
c) T cosα=DM-FN
d) DM=T cosα+FN
Answer: b
Clarification: The total thrust force assumed in the derivation of equations of motion is FN=T cosα-DM where FN=net thrust force, T is thrust force, cosα is lift component and DM is drag force acting downwards.

6. In static performance the acceleration of the flight is zero.
a) True
b) False
Answer: a
Clarification: In static performance the acceleration of the flight is zero. In this case the resulting conditions are proved to be as:

  • Thrust is equal to drag and
  • Lift is equal to weight of the aircraft.

7. Excess thrust in the longitudinal equations of motion is given by ____________
a) FN+D
b) FN-D
c) D-FN
d) FN*D
Answer: b
Clarification: Excess thrust in the longitudinal equations of motion is given by FN-D where FN is net thrust force and D is drag force. This excess thrust provides the increase in potential energy and kinetic energy.

8. If the aircraft has a power-producing engine which have driven propellers then the power is to be converted to thrust to formulate the equations of motion.
a) True
b) False
Answer: a
Clarification: If the aircraft has a power-producing engine which have driven propellers then the power is to be converted to thrust to formulate the equations of motion. The conversion is given as FN=ṁ(Vi-V) where FN is net thrust force, ṁ is mass flow rate and Vi, V are velocities.

9. What is the function of excess thrust in longitudinal equations of motion?
a) Decrease the potential energy and kinetic energy
b) Increase the potential energy and kinetic energy
c) Decrease the potential energy and increase kinetic energy
d) Increase the potential energy and decrease kinetic energy
Answer: b
Clarification: The function of excess thrust in longitudinal equations of motion is to increase the potential energy and kinetic energy. Excess thrust in the longitudinal equations of motion is given by FN-D where FN is net thrust force and D is drag force.

10. The aircraft is symmetrical flight during the derivation of equations of motion.
a) True
b) False
Answer: a
Clarification: The aircraft is symmetrical flightduring the derivation of equations of motion. The other assumptions taken during the derivation of equations of motionare:

  • Rate of change of mass is neglected
  • Aircraft is in symmetric flight
  • Gross thrust acts in aircraft body axis
  • Total net thrust is T cosα-DM
  • Thrust component is small compared to lift component.

Aircraft Performance for Interviews,

250+ TOP MCQs on Cruising Performance – Effect of Alternative Fuel Flow Laws and Answers

Aircraft Performance MCQs on “Cruising Performance – Effect of Alternative Fuel Flow Laws”.

1. The range and endurance are proportional to which of the following factors?
a) fuel flow
b) net thrust
c) bypass ratio
d) airspeed
Answer: b
Clarification: The range and endurance are proportional only to net thrust. The fuel flow laws proved to be a constant idealized laws. Though fuel flow laws cannot describe all the characteristics of the engine they are proved to be a constant idealized laws.

2. Which of the following is a correct form of fuel flow?
a) C=C2θMn
b) C=C2θ(frac{1}{2})M
c) C=C2θM
d) C=C2θ(frac{1}{2})Mn
Answer: d
Clarification: The correct formula for fuel flow is C=C2θ(frac{1}{2})Mn where θ is the temperature, M is mach number and n varies from 0.2-0.6 for turbojet to turbofan. It depends on the bypass ratio of the combustion chamber.

3. Thrust produced is the functional form of engine speed and flight Mach number.
a) True
b) False
Answer: a
Clarification: Thrust produced is the functional form of engine speed and flight Mach number. The temperature decreases as the altitude increases so that during the cruise-climb situation the rotational speed increases and mach number increases.

4. The value of ‘n’ in fuel flow formula C=C2(theta^{frac{1}{2}})Mn is 0.6 for turbojet.
a) True
b) False
Answer: b
Clarification: The correct formula for fuel flow is C=C2(theta^{frac{1}{2}})Mn where θ is the temperature, M is mach number and n varies from 0.2-0.6 for turbojet to turbofan. It depends on the bypass ratio of the combustion chamber.

5. The value of ‘n’ in fuel flow formula C=C2(theta^{frac{1}{2}})Mn is 0.2 for turbofan.
a) True
b) False
Answer: b
Clarification: The correct formula for fuel flow is C=C2(theta^{frac{1}{2}})Mn where θ is the temperature, M is mach number and n varies from 0.2-0.6 for turbojet to turbofan. It depends on the bypass ratio of the combustion chamber.

6. The formula for velocity of maximum specific air range is __________
a) (big[frac{3-n}{1+n}big]^{frac{1}{4}})
b) (big[frac{3-n}{1+n}big]^{frac{1}{3}})Vmd
c) (big[frac{3-n}{1+n}big]^{frac{1}{4}})Vmd
d) (big[frac{3+n}{1+n}big]^{frac{1}{4}})Vmd
Answer: c
Clarification: The formula for velocity of maximum specific air range is (big[frac{3-n}{1+n}big]^{frac{1}{4}})Vmd where ‘n’ is the exponent and Vmd is maximum velocity. The value of ‘n’ depends on the value of bypass ratio of the engine. This implies that as the value of ‘n’ increases the bypass ratio increases.

7. The formula for velocity of maximum specific endurance is _______________
a) (big[frac{3+n}{1+n}big]^{frac{1}{4}})Vmd
b) (big[frac{2+n}{1+n}big]^{frac{1}{4}})Vmd
c) (big[frac{2+n}{2-n}big]^{frac{1}{4}})Vmd
d) (big[frac{2-n}{2+n}big]^{frac{1}{4}})Vmd
Answer: d
Clarification: The formula forvelocity of maximum specific endurance is (big[frac{2-n}{2+n}big]^{frac{1}{4}})Vmd where ‘n’ is the exponent and Vmd is maximum velocity. The value of ‘n’ depends on the value of bypass ratio of the engine. This implies that as the value of ‘n’ increases the bypass ratio increases.

8. What is the velocity of maximum specific air range of the engine when the maximum air speed is 400m/s and the exponential value is 0.2?
a) 300.84
b) 380.43
c) 440.76
d) 494.37
Answer: d
Clarification: The answer is 494.37.
Given n=0.2, Vmd=400m/s. From the formula (big[frac{3-n}{1+n}big]^{frac{1}{4}})Vmd. On substituting the values we get ([frac{3-0.2}{1+0.2}]^{frac{1}{4}})*400=494.37.

9. What is the velocity of maximum specific endurance of the engine when the maximum air speed is 400m/s and the exponential value is 0.2?
a) 350.43
b) 490.34
c) 494.37
d) 380.43
Answer: d
Clarification: The answer is 380.43.
Given n=0.2, Vmd=400m/s. From the formula (big[frac{2-n}{2+n}big]^{frac{1}{4}})Vmd. On substituting the values we get (big[frac{2-0.2}{2+0.2}]^{frac{1}{4}})*400=380.43.

10. What is the value of ‘n’ exponential in the fuel flow law for a high bypass ratio turbofan?
a) 0.4
b) 0.5
c) 0.6
d) 0.7
Answer: c
Clarification: The value of ‘n’ exponential in the fuel flow law for a high bypass ratio turbofan is 0.6. The value of ‘n’ exponential is directly proportional to the engine bypass ratio i.e. it increases when the bypass ratio increases.

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250+ TOP MCQs on Estimation of Take-off Distances and Answers

Aircraft Performance Multiple Choice Questions 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

Answer: a
Clarification: 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) FN+D-WsinγR-μrR=mV̇
b) R-L=WcosγR
c) R-L=WsinγR
d) FN-D-WsinγR-μrR=mV̇

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

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

Answer: c
Clarification: The formula for accelerating force is given by: (frac{F}{W}=Big{frac{F_N}{W}-mu_R-singamma_RBig}-frac{L}{W}Big(frac{C_D}{C_L}-mu_RBig)) 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.

4. The curly bracket in the accelerating force represents the net propulsive thrust- weight ratio.
a) True
b) False

Answer: a
Clarification: The formula for accelerating force is given by: (frac{F}{W}=Big{frac{F_N}{W}-mu_R-singamma_RBig}-frac{L}{W}Big(frac{C_D}{C_L}-mu_RBig)) 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

Answer: a
Clarification: The formula for accelerating force is given by: (frac{F}{W}=Big{frac{F_N}{W}-mu_R-singamma_RBig}-frac{L}{W}Big(frac{C_D}{C_L}-mu_RBig)) 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.

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}}})

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

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

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

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

8. In the calculations the horizontal airborne distance is assumed to be much greater than 35 feet.
a) True
b) False

Answer: a
Clarification: 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}+35Big}) 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

Answer: a
Clarification: 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

Answer: c
Clarification: 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

Answer: c
Clarification: 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.