FAA Commercial Pilot Airplane (CAX)Aerodynamics and PerformanceHard
An airplane's unaccelerated (1G) stall speed is 58 KCAS. If the pilot enters a coordinated turn with a 60° bank angle, what is the approximate accelerated stall speed?
- A116 KCAS
- B82 KCAS
- C71 KCAS
- D58 KCAS
Show answer & explanationAnswer & explanation
Correct answer: B. 82 KCAS
At 60° bank, load factor = 1/cos(60°) = 2.0. Accelerated stall speed = unaccelerated stall speed × √(load factor) = 58 × √2 = 58 × 1.414 ≈ 82 KCAS.
Why the other options are wrong
- A. This doubles the stall speed, overestimating the actual increase (which follows a square root, not linear, relationship).
- C. This underestimates the effect of the 2.0 load factor on stall speed.
- D. This is the unaccelerated stall speed, not accounting for the increased load factor.
Accelerated Stall Speed
Stall speed increases with the square root of load factor: Vs(accelerated) = Vs(1G) × √n, where n is the load factor for the given bank angle.
- Formula: Vs2 = Vs1 × √n
- 60° bank produces load factor of 2.0
- Stall speed increases, but not linearly with load factor
- Steep turns significantly raise the risk of an accelerated stall
Memory trick: Square root the G to find the new stall degree!