FAA Commercial Pilot Airplane (CAX)Aerodynamics and PerformanceHard
During a dive recovery, a pilot pulls out along a circular arc at a constant velocity of 200 feet per second and a radius of 2,000 feet. Using the pull-up load factor formula n = 1 + V²/(gR), where g = 32.2 ft/sec², what approximate load factor does the pilot experience?
- A1.6G
- B2.0G
- C2.6G
- D1.2G
Show answer & explanationAnswer & explanation
Correct answer: A. 1.6G
Using n = 1 + V²/(gR): n = 1 + (200²)/(32.2 × 2,000) = 1 + 40,000/64,400 = 1 + 0.621 = 1.62G, which rounds to approximately 1.6G. This formula shows that load factor in a pull-up increases with the square of velocity and decreases as the radius of the arc increases.
Why the other options are wrong
- B. Overestimates the result, possibly from a math error in the division.
- C. Too high; does not match the correct arithmetic using the given values.
- D. Underestimates the velocity-squared term in the numerator.
Pull-Up Load Factor
In a vertical pull-up maneuver, load factor is calculated as n = 1 + V²/(gR), where V is velocity, g is gravitational acceleration, and R is the radius of the pull-up arc.
- Formula: n = 1 + V²/(gR)
- Load factor increases with velocity squared
- Load factor decreases as pull-up radius increases (gentler arc)
Memory trick: Velocity squared divides by gravity times radius, plus one for gravity's prize.