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martiben81
27th Nov 2006, 00:28
Why does a derated thrust takeoff allow for a higher takeoff weight? I see the relation of a derated takeoff lowering Vmcg. Does that then allow for a higher weight when the runway is the limiting factor in the takeoff weight? Any pointers in the right direction would be appreciated. Thanks

Jet Man
27th Nov 2006, 01:16
Using a de-rate will not increase the performance limited take-off weight it will decrease it. Yes Vmcg will be lower but it will take more time/runway to achieve this lower speed.

Sqwak7700
27th Nov 2006, 03:16
Does that then allow for a higher weight when the runway is the limiting factor in the takeoff weight?

You are almost there. It will allow a higher weight if runway length is not a factor (ie, have a really long runway).

Lets say that you are limited by your Vmcg for take-off. You can decrease Vmcg by reducing the TO thrust (less thrust, less rudder needed to fight it). But, in doing so, it will take you more runway length to get off the ground.

Hopefully that makes sense.

Oasis
27th Nov 2006, 07:11
A contaminated runway may also cause you to be runway length limited, the slippery runway is effectively shortened.

idg
28th Nov 2006, 11:15
There have been plenty of discussion threads about this very topic in Tech Log so a quick search there should bring results.

Not sure of an unlimited runway length, but I certainly know of a situation where the runway is short and higher MTOW can be achieved with a derate. In my experience a short bodied TriStar(-500) which has a very high Vmcg because of the reduced rudder arm, could gain a 20T weight advantage out of a (relatively) short field by derating thrust. This permitted a much lower V1 and therefore allowed the a/c to reach it quicker (even allowing for higher weight and reduced thrust) and therefore was able to stop within the runway length available, rather than rushing down the runway to achieve a V1 (higher than Vmcg) and roar off the runway with no useful load whatsoever!

Zeke
29th Nov 2006, 00:10
Not sure of an unlimited runway length, but I certainly know of a situation where the runway is short and higher MTOW can be achieved with a derate.

Correct, the reduction in the Vmca sometimes generates a takeoff performance benefit (higher MTOW) off a short runway.

V1 is the maximum speed at which it is still possible to reject the takeoff and stop the aircraft within the runway limits with V1 > Vmcg, and with the ASDA being often the most constraining limitation on a short runway, lower Vmcg, one can lower V1.

A lower Vmcg will lower the ASDR (lower V1) for a given takeoff mass, and lead to better takeoff performance when the MTOW without derate is ASDA/Vmcg limited.

flyboy007
9th Dec 2006, 07:15
IDG, you are correct.

As odd as it sounds, we sometimes 'pretend' we have less powerful engines (eg derate, but it gets around the max derate rule) to allow us to lift more off shorter runways. Typically it comes into effect on the L1011-500 for field lengths around about the 7000'-8000' point.

Shorter fields than this and we use full thrust, and accept the vmcg limit, and longer fields full thrust when we are less likely to be vmcg limited.

It comes down the the fact that it's a stopping problem, not a going problem.

Weird, but true.