FAA Airline Transport Pilot (ATM)Aircraft Performance and Weight and BalanceEasy
An airplane's clean, wings-level stall speed is 120 knots. During an instrument approach, the pilot enters a steady 60-degree bank turn while maintaining altitude. What is the approximate stall speed in this turn?
- A170 knots
- B240 knots
- C120 knots, since bank angle does not affect stall speed
- D139 knots
Show answer & explanationAnswer & explanation
Correct answer: A. 170 knots
In a level turn, load factor n = 1/cos(bank angle). At 60 degrees, cos(60°) = 0.5, so n = 2. Stall speed increases by the square root of the load factor: new stall speed = 120 × √2 ≈ 120 × 1.41 ≈ 170 knots.
Why the other options are wrong
- B. This overstates the increase; it would correspond to a load factor of 4, not 2.
- C. Bank angle increases load factor, which does raise stall speed.
- D. This value does not match the √2 multiplier for a 2G load factor.
Load Factor Effect on Stall Speed
Stall speed increases with the square root of the load factor experienced in a maneuver, such as a banked turn: Vs(new) = Vs(1G) × √n.
- Load factor n = 1/cos(bank angle) in a level turn
- 60-degree bank produces a load factor of 2
- Stall speed increases by √n, not by n directly
- At 60° bank, stall speed increases by about 41%
Memory trick: Steep bank, steep rise: 60 degrees doubles your G's and grows stall speed by 41%