FAA Commercial Pilot Airplane (CAX)Navigation and Flight PlanningHard
An airplane experiences an engine failure at 6,500 feet AGL. The best-glide configuration normally yields a 10:1 glide ratio in still air, but a 10-knot headwind component is expected to reduce the effective glide ratio to 8:1. Approximately how many nautical miles can the airplane glide under these conditions?
- A8.6 NM
- B12.8 NM
- C6.4 NM
- D10.7 NM
Show answer & explanationAnswer & explanation
Correct answer: A. 8.6 NM
Horizontal glide distance = altitude × glide ratio = 6,500 ft × 8 = 52,000 ft. Converting to nautical miles: 52,000 ÷ 6,076 ≈ 8.56 NM, which rounds to approximately 8.6 NM.
Why the other options are wrong
- B. This overestimates the distance beyond what either ratio supports.
- C. This corresponds to using the still-air 10:1 ratio incorrectly divided, not matching the headwind-adjusted figure.
- D. This value would result from using the still-air 10:1 ratio, ignoring the headwind reduction.
Glide Distance Calculation
Glide distance over the ground is found by multiplying altitude above the landing point by the effective glide ratio, adjusted for wind, then converting units as needed.
- Glide ratio expresses horizontal distance traveled per unit of altitude lost
- Headwinds reduce, and tailwinds increase, the effective glide ratio over the ground
- 1 nautical mile = 6,076 feet for unit conversion
Memory trick: Wind steals your glide, so ratio shrinks