Just to make sure I'm on the right line
Let’s say you had a climbing CAS of 90kts with a 15kt headwind, where your pressure height is 4,000’ and temperature is 12C, how many feet per minute (FPM) do you have to achieve to ensure obstacle clearance using 6% gradient?
Using a flight computer, enter PH 4,000, Temp 12 and CAS 90.
TAS = 86kts
Now deduct the headwind:
86 - 15 = 71kts ground speed (GS)
Using the rule of thumb GS / FPM = Climb gradient (in %)
6 x 71 = 426FPM
To check the answer is correct (assuming slight error in rule of thumb), let’s work it the long way.
71kts = 71 x 6076 (feet in NM) / 60 (minutes is hour) = 7190 FPM (horizontally)
7190 x 0.06 (gradient) = 431FPM
Copy copy?