Climb gradient calculation exercise

[MD12]

Hi,

I´m reviewing some stuff, to keep my "atpl knowledge" up to date, and just got stuck on the following exercise:

The second segment requirement for a twin engine aircraft is 2.4%. If the total drag of the aircraft at V2 is 18500lbs and each engine produces 22000lbs of thrust. The maximum permissible take-off weight will be:

a) 145800 lbs
b) 1062500 lbs
c) 91700 lbs
d) 90300 lbs

According to the book, the correct answer is A, but still I cannot get it (I know, monkey stuff).

Climb gradient=(T-D)/W

so,

Climb gradient = (44000-18500)/W

2.4/100=25500/W

From above: W=1062500 (answer b)

What am I doing wrong? Your help is really appreciated. Enjoy the rest of the weekend.

[WaterMeths]

Hi...

Try the calc again with the single engine (i.e. 22000 lbs thrust) as the 2.4 gradient is to be achieved with an engine failure.

Cheers

WM

[MD12]

Thank you! I found it out myself and was just about to post the solution. You are absolutely right.

Now it´s all clear. Tnx