FAA Aircraft Dispatcher (ADX)Navigation and ATCHard
A published SID requires a minimum climb gradient of 290 feet per nautical mile to 5,000 feet MSL. If the aircraft maintains a groundspeed of 140 knots during the climb, what rate of climb (in feet per minute) is required to comply with this gradient?
- A677 fpm
- B406 fpm
- C507 fpm
- D812 fpm
Show answer & explanationAnswer & explanation
Correct answer: A. 677 fpm
Rate of climb (fpm) = (climb gradient in ft/NM × groundspeed in knots) ÷ 60. Calculation: (290 × 140) ÷ 60 = 40,600 ÷ 60 = 676.7 fpm, which rounds to approximately 677 fpm.
Why the other options are wrong
- B. 406 fpm would result from using a groundspeed of about 84 knots, not 140.
- C. 507 fpm underestimates the required climb rate for this gradient and groundspeed.
- D. 812 fpm overstates the requirement; it would correspond to a gradient closer to 348 ft/NM.
Climb Gradient to Rate of Climb Conversion
Published climb gradients (ft/NM) must be converted to a rate of climb (fpm) based on groundspeed to verify aircraft performance compliance with departure procedures.
- Formula: ROC (fpm) = (gradient ft/NM × groundspeed kt) ÷ 60
- Standard minimum obstacle clearance gradient is 200 ft/NM
- Higher published gradients override the 200 ft/NM standard when obstacles require it
- Groundspeed, not airspeed, is used in the calculation
Memory trick: 'Gradient times speed, divide by sixty' — the standard performance formula.