FAA Airline Transport Pilot (ATM)Transport Aerodynamics and High-Altitude OperationsMedium
A transport aircraft has a wings-level 1g stall speed of 120 KIAS. During a steady 45° banked turn, what is the approximate new stall speed?
- A143 KIAS
- B127 KIAS
- C120 KIAS
- D170 KIAS
Show answer & explanationAnswer & explanation
Correct answer: A. 143 KIAS
Load factor in a 45° bank = 1/cos(45°) = 1.414. Stall speed increases with the square root of load factor: 120 × √1.414 = 120 × 1.189 ≈ 143 KIAS.
Why the other options are wrong
- B. This underestimates the load factor's effect (would correspond to about 30° bank).
- C. This ignores the load factor increase from banking.
- D. This overstates the effect, closer to what a 60° bank (n=2, √2=1.41... actually 60° gives 170) would produce.
Bank Angle Effect on Stall Speed
Stall speed increases with the square root of load factor (n), and load factor increases as 1/cos(bank angle) in a level turn.
- 45° bank: n=1.41, Vs increases ~19%
- 60° bank: n=2, Vs increases ~41%
- Steep turns significantly raise stall speed margin needs
Memory trick: Steeper bank, heavier feel, higher stall speed