FAA Flight Instructor Airplane (FIA)Flight Maneuvers and EndorsementsHard
A commercial pilot applicant enters a steep turn at a 60° bank angle and maintains a constant altitude. What load factor is the airplane experiencing during this turn?
- A3.0 G
- B1.0 G
- C2.0 G
- D1.4 G
Show answer & explanationAnswer & explanation
Correct answer: C. 2.0 G
Load factor in a coordinated level turn equals 1/cos(bank angle). At 60° bank, cos(60°) = 0.5, so load factor = 1 / 0.5 = 2.0 G. This doubling of apparent weight also raises the stall speed by approximately 41% at this bank angle.
Why the other options are wrong
- A. 3.0 G would require a bank angle of about 70.5°, steeper than given.
- B. 1.0 G would only occur in unaccelerated, wings-level flight.
- D. 1.4 G corresponds to a 45° bank turn (1/cos45° ≈ 1.41), not 60°.
Load Factor in Level Turns
The load factor imposed during a coordinated, constant-altitude turn is calculated as 1 divided by the cosine of the bank angle.
- 45° bank ≈ 1.41 G
- 60° bank = 2.0 G
- Stall speed increases as the square root of the load factor
Memory trick: 'Sixty means double' — 60° bank always yields 2 G's.