Olendirk
3rd Apr 2004, 06:24
Hey guys!
with a ground speed of 540 kt and 12% slope what is my descend rate?
Thanks!
OD
WX Man
3rd Apr 2004, 08:55
Groundspeed = 540kt = 9nm/min = 54720fpm
12% of 54720 = 6566fpm
You must be a tug pilot or a parachute drop pilot.
Keith.Williams.
3rd Apr 2004, 14:34
You can cut the arithmetic a little by using
ROC = % x 1.0133 x TAS
It is actually the same method as used by WX Man, but some of the stages have been combined into the 1.0133.
The more 3s you put onto the end of the 1.0133, the more accurate will be your answer.
Jinkster
4th Apr 2004, 11:24
My method:
Convert percentage to degrees = (12/ 100) x 60 = 7.2deg
deg x 100 x groundspeed divide by 60
7.2 x 100 x 540 / 60 = 6480FPM
:ok: