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Old February 5th 07, 07:35 PM posted to rec.aviation.piloting
Orval Fairbairn
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Default Increasing power required with altitude.. what's a good plain english explanation?

In article ,
"Danny Deger" wrote:

"xerj" wrote in message
...
No, same IAS, same drag, same thrust, same power requirement from the
engine to generate the thrust. The statement that power is drag time
velocity is incorrect. That is the point where the error is made.


All of the definitions of power that I have seen have been along the lines
of P = T * V, or something that equates to that.

For instance:-

"The formula for Thrust Horsepower (THP) is:
THP = D x V"

from http://selair.selkirk.bc.ca/aerodyna...nce/Page4.html.

That is wrong?


You can certainly define a term called Thrust Horse Power as thrust x
velocity. And this link definition of Brake Horse Power is correct (torque
times RPM). But there is no reason to think these terms are equal in an
aircraft. A great deal of the power out of the engine (all of the power if
in steady state level flight) goes into the air and not the airframe. It is
my understanding that for a given thrust at a given IAS (actually Equivelant
Air Speed, EAS, is the better term), the engine power requirement is
basically the same for different altitudes. I wish I had a good aircraft
performance handbook to confirm this.



That is incorrect! A classic problem in sophomore aero engineering is to
determine the maximum altitude at which an aircraft will fly,
simplifying the problem by assuming turbosupercharging to allow constant
power and discounting compressibility effects, given its stall IAS and
lift/drag curves.

At very high altitudes a plane will fly very fast at low IAS (min porew
required speed/alpha.

The power = speed*thrust is valid and is a basic tenet of aero
engineering.