FAA Airline Transport Pilot (ATM)Aircraft Performance and Weight and BalanceHard
An airplane's 1G, wings-level stall speed in the landing configuration is 130 KIAS. During a coupled ILS approach, the autopilot briefly commands a 60° bank angle turn to intercept the localizer. Assuming a level, coordinated turn, what is the approximate new stall speed in this turn?
- A150 KIAS
- B184 KIAS
- C260 KIAS
- D130 KIAS (unchanged)
Show answer & explanationAnswer & explanation
Correct answer: B. 184 KIAS
In a level coordinated turn at 60° bank, load factor n = 1/cos(60°) = 1/0.5 = 2.0. New stall speed = Vs × √n = 130 × √2 ≈ 130 × 1.414 ≈ 183.8 KIAS, which rounds to approximately 184 KIAS.
Why the other options are wrong
- A. 150 KIAS underestimates the load factor effect at 60° bank.
- C. 260 KIAS would require a load factor of 4, not 2.
- D. Stall speed always increases with bank angle in a level turn due to added load factor.
Load Factor in Turns
In a level, coordinated turn, load factor n = 1/cos(bank angle); stall speed increases by a factor of √n compared to level, wings-level flight.
- At 45° bank, n ≈ 1.41, stall speed increases ~19%.
- At 60° bank, n = 2.0, stall speed increases by √2 (~41%).
- At 30° bank, n ≈ 1.15, a smaller stall speed increase.
- Steep turns near approach speeds can reduce stall margin significantly.
Memory trick: Steeper bank, heavier 'G', higher stall speed — 60° doubles your load.