how to solve? 1

Joined: Jan 2003
Aviation Qualifications: ATPL
Posts: 279
Likes: 17
From: Europe
For the required rate of descent multiply the descent angle by the NM/minute times 100.
Taking your example...
540kt = 9NM/min
12° x 900 = 10800'/min.
But... who would fly a 12° angle
Taking your example...
540kt = 9NM/min
12° x 900 = 10800'/min.
But... who would fly a 12° angle

Joined: Jan 2003
Aviation Qualifications: ATPL
Posts: 279
Likes: 17
From: Europe
Capt. Slow
Oh, sorry for that! You are absolutely right - it is gradient, not a slope...
In that case, to find the rate required, multiply the gradient by the GS (or rather TAS).
12% x 540 = 6480'/min.
Regards
In that case, to find the rate required, multiply the gradient by the GS (or rather TAS).
12% x 540 = 6480'/min.
Regards




Well your method seems a lot simpler than mine, might adopt that for the future...
