FAA Private Pilot Airplane (PAR)Aerodynamics and Aircraft SystemsHard
An airplane has a maneuvering speed (Va) of 120 KIAS at its maximum gross weight of 2,400 pounds. If the airplane is flown today at a reduced weight of 1,900 pounds, approximately what is the new maneuvering speed?
- A120 KIAS
- B113 KIAS
- C95 KIAS
- D107 KIAS
Show answer & explanationAnswer & explanation
Correct answer: D. 107 KIAS
Maneuvering speed varies with the square root of the weight ratio: Va(new) = Va(old) × √(W(new)/W(old)) = 120 × √(1,900/2,400) = 120 × √0.792 = 120 × 0.890 ≈ 107 KIAS. As weight decreases, Va decreases because the wing stalls at a lower airspeed before structural limit loads can be exceeded.
Why the other options are wrong
- A. Va does not stay the same when weight decreases; it must be reduced at lighter weights.
- B. This value overestimates Va; it does not correctly apply the square-root relationship.
- C. This value is too low; it does not match the square-root weight ratio calculation.
Maneuvering Speed (Va) and Weight
The maximum speed at which full or abrupt control deflection can be applied without exceeding the airplane's structural load limits; Va decreases as weight decreases.
- Va(new) = Va(old) × √(W(new)/W(old))
- Lower weight means the wing stalls at a lower speed, reducing the speed at which loads become critical
- Flying above the reduced Va at light weight risks structural damage during abrupt maneuvers or turbulence
Memory trick: Lighter plane, lighter Va — the wing gives up sooner.