250+ TOP MCQs on Landing Gear – Tire Sizing and Answers

Aircraft Design Multiple Choice Questions on “Landing Gear – Tire Sizing”.

1. Following diagram represents _________

a) wheel diameter for braking
b) lift curve
c) drag curve slope
d) drag polar

Answer: a
Clarification: Above figure is used for initial and statistical estimation of required wheel rim diameter. This estimation is based on kinetic energy absorption at the time of braking. Lift curve represents relationship between lift and AOA.

2. A general aviation aircraft has impact Weight on wheel about 100 unit. Find the required diameter.
a) 7.53 unit
b) 9.753 unit
c) 10.98 unit
d) 11 unit

Answer: a
Clarification: Given, Weight on wheel W = 100 unit.
For, general aviation aircraft Diameter D = 1.51*W0.349 = 1.51*100^0.349 = 7.53 unit.

3. Only dynamic loading should be used for determining the nose wheel tire sizing.
a) True
b) False

Answer: b
Clarification: Tires selection is done by finding the smallest tire which can carry the calculated maximum loads. Both static and dynamic loading should be used to evaluate size of nose wheel tire.

4. Evaluate width of the wheel if weight carried by it is 200 unit. Consider general aviation aircraft.
a) 3.73 unit
b) 9 unit
c) 7 unit
d) 9/7 unit

Answer: a
Clarification: Given, Weight W = 200 unit
Width of general aviation is given by,
Width w = 0.715*W^0.312 = 0.715*200^0.312 = 3.73 unit.

5. A twin business engine is designed in such way that the load of 250 unit is carried out by the tire. Find the appropriate ratio of the diameter to the width of main wheel.
a) 3.8
b) 2.13
c) 5.6
d) 9.8

Answer: a
Clarification: Given, twin business jet, Load W = 250 unit.
Ratio of diameter to width D / w = 2.299*W^0.091 = 2.299*250^0.091 = 3.8.

6. A bomber aircraft has following data. Weight on wheel is 80 unit. Calculate the width of main wheel.
a) 0.8546
b) 3.45
c) 6.78
d) 0.2345

Answer: a
Clarification: For bomber, weight W = 80 unit.
Width of main wheel d = 0.1043*W^0.48 = 0.1043*80^0.48 = 0.8546 unit.

7. Jet fighter aircraft is designed to operate from rough unpaved runway. Find the diameter of wheel if wheel needs to carry weight of 200 unit.
a) 10.23 unit
b) 23.1
c) 34.2 unit
d) 1.023 unit

Answer: a
Clarification: Given, rough unpaved runway, Weight on wheels W = 200 unit.
Diameter d = 1.59*W^0.302 = 1.59*200^0.302 = 7.87 unit
Now, actual diameter D = d + 0.3*d = 7.87 + 0.3*7.87 = 10.23 unit.

8. An aircraft has weight of 300N at the time of landing. Calculate braking K.E. if stall speed is 10 m/s.
a) 1.529 KJ
b) 156 KJ
c) 23 KJ
d) 23.45 KJ

Answer: a
Clarification: Given, Weight W = 300N, stall speed v = 10 m/s
Braking K.E. = 0.5*W*v² / g = 0.5*300*10² / 9.81 = 1.529 KJ.

9. An aircraft is designed to have landing weight of 2500N and stall speed V of 9.8m/s. How much kinetic energy should be absorbed by wheel? Consider number of wheels with brake is 2.
a) 6125J
b) 5.98 KJ
c) 2100
d) 235 KJ

Answer: a
Clarification: Given, landing weight W = 2500N, stall speed v = 9.8m/s. Number of wheels n = 2.
Kinetic energy K.E. = 0.5*W*v²/g*n = 0.5*2500*9.8²/9.81*2 = 6125 J.

10. Determine weight carried by wheel if inflation pressure is 100 psi and footprint area is 80 inch².
a) 8000 lb force
b) 900
c) 200 lb of force
d) 120

Answer: a
Clarification: Given, inflation pressure P = 100 psi, area a = 80 in².
Weight W = P*a = 100*80 = 8000 lb of force.

11. If tire pressure is 25 psi then, determine minimum dynamic hydroplaning speed.
a) 45 knots
b) 25 psi
c) 369 knot
d) 25 knots

Answer: a
Clarification: Given, tire pressure p = 25 psi
Minimum dynamic hydroplaning speed V can be estimated by using,
V = 9*p^0.5 = 9*25^0.5 = 9*5 = 45 knots.