FAA Private Pilot Helicopter (PRH)Performance and LimitationsMedium
A pilot enters a level turn at a 60° angle of bank. Compared to straight-and-level flight, this maneuver will require the rotor system to produce approximately:
- A1.4 times the normal lift, increasing power required
- BHalf the normal lift, decreasing power required
- CThe same amount of lift, only redirected
- D2 times the normal lift, significantly increasing power required
Show answer & explanationAnswer & explanation
Correct answer: D. 2 times the normal lift, significantly increasing power required
Load factor in a level turn equals 1/cos(bank angle). At 60°, cos 60° = 0.5, so load factor = 1/0.5 = 2. The rotor system must produce twice the aircraft's weight in lift, which substantially increases power required and can push the helicopter toward its power and blade-stall limits.
Why the other options are wrong
- A. 1.4G corresponds to a 45° bank, not 60°.
- B. Lift requirement increases in a turn, it never decreases.
- C. Lift must increase in magnitude, not just change direction, to maintain altitude in a bank.
Load Factor in Turns
Load factor (G) in a level turn equals 1 divided by the cosine of the bank angle; steeper banks require exponentially more lift and power.
- 45° bank ≈ 1.4G
- 60° bank = 2G
- Increased load factor raises power required and stall/retreating blade stall risk
Memory trick: Steeper bank = heavier 'invisible passenger' the rotor must lift