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vicleecy
18th Jan 2007, 00:58
Goodday all crew,

Anyone has a 737-800 flight manual in hand telling the climb gradient lost of a 737-800 during a turn?

many thanks!

Victor

A37575
18th Jan 2007, 01:09
Don't specifically know about the NG but the 737Classics and Jurassics flight manuals shows a loss of 0.6% during the duration of the turn at 15 degrees angle of bank one engine inop.

john_tullamarine
18th Jan 2007, 07:10
Depends on how you do the sums but figure somewhere in the range 0.5-0.7 percent loss in the turn.

bubbers44
18th Jan 2007, 20:44
So if normal single engine rate of climb was 600 fpm you would climb at 596.4 fpm in a 15 degree bank. Not worthy of discussion in my mind.

wondering
19th Jan 2007, 11:56
As in the original question, itīs a gradient loss and not a percent loss.

mutt
19th Jan 2007, 12:15
Bubbers44....

What GRADIENT is a twin engined aircraft required to achieve during takeoff. Now subtract 0.6% from that GRADIENT.....

Suddenly it becomes worthy of discussion :)

Mutt

bubbers44
19th Jan 2007, 13:01
Thanks Mutt, Missed the gradient part. I just did a typical max wt. climb limited example where the 2.4 climb gradient is reduced .6 % and the roc decreases from 360 fpm to 270 fpm or 25%. At TGU in Honduras we took an enormous prohibitive penalty for taking off to the south because of terrain and an immediate 180 required after take off. We routinely took off on a 6000 ft strip with a 10 knot tail wind because of that in a 757. I believe the required bank angle was 25 degrees too.

mbga9pgf
19th Jan 2007, 16:46
Thanks Mutt, Missed the gradient part. I just did a typical max wt. climb limited example where the 2.4 climb gradient is reduced .6 % and the roc decreases from 360 fpm to 270 fpm or 25%. At TGU in Honduras we took an enormous prohibitive penalty for taking off to the south because of terrain and an immediate 180 required after take off. We routinely took off on a 6000 ft strip with a 10 knot tail wind because of that in a 757. I believe the required bank angle was 25 degrees too.


Isnt the answer Cos(15 degrees)?

After all, all you are doing is angling the lift vector round to give you force in the direction of the turn?

Dont see how this could vary from a/C to A/C apart from changes in induced drag (surely the same as you are climbing/turning at same speed, just reducing your lift vector by Cos(15 degrees)?

:} :bored: