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5th Jul 2007, 18:32

#3 (permalink)

DBate

Join Date: Jan 2003

Location: Europe

Posts: 270

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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

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