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Old 23rd September 2009 | 09:08
  #9 (permalink)  
Capt Pit Bull
 
Joined: Aug 1999
Posts: 1,050
Likes: 4
From: England
The problem here is that the folks that asked the OP what the 'power' was probably were not using the term in its formal sense anyway.

As far as answering the question.... well, since the velocity of the aircraft is variable, the power being developed at different points during the take off run is also variable.

If you want an answer for 'the' (singular) power I guess you could give an average value reasonably easily:

Kinetic Energy = 1/2 M V^2 (all in S.I. approximations)
= 1/2 x 400,000 x 93^2 = 1,730,000,000 (1.73 Gigajoules)

Power = Energy / Time = 1,730,000,000 / 45 = approx 38.5 Megawatts = approx 51,600 horsepower.

Meerkats says "Simples!"

Power = (F * X) / t = (164, 064 * 6100) / 45 = 22, 239,787 ft-lbs / sec
1 HP = 550 ft-lbs / sec so this is
40, 436 HP or

Power = (F * X) / t = (199, 611 * 4993) / 37 = 26,936,695 ft-lbs / sec
1 HP = 550 ft-lbs / sec so this is
48,976 HP – a pretty large variation!
I wouldn't sweat it

If you were an astrophysicist anything within 3 orders of magnituade would be bang on

pb
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