"xerj" wrote in message
...
Gettin' a bit confused here. (nothing new in that)

In the big sprawling thread I started down below, there's been a couple of
themes that have come up.

One is that I am pretty sure that for the same IAS (not TAS) at a higher
altitude, more power is required. However, one contributor to the thread
has stated that this is not the case:-

"Power is net force time velocity. Thrust equals drag, net force is
zero.
The energy change of the airframe overtime is zero. All energy from the
engine is going into the air. The power to move air to make the same
thrust
is the same regardless of velocity. Same IAS, same engine power
requirement. Look at some aircraft performance charts."

I'd always understood that power = thrust x velocity, hence the deduction
that it requires more power to go the same IAS at a higher alt. At the
same IAS the drag and hence the thrust is the same. Plug that into the
equation and you get the power required, which is more because TAS is
higher at altitude.

As for aircraft performance charts, they're for the most part in TAS, not
IAS.

I found an aircraft performance chart and I stand corrected. At 75% power
the aircraft flies 140 TAS at sea level and 150 TAS at 8K feet. This is 140
IAS at sea level and 124 IAS as 8K feet. I am very surpprised, because I
always thought you would get the same IAS for the same power.

So for the same power, the IAS is less at altitude. I do know that the
calculations from engine power to thrust and power effect to the overall
system is more of an art than I science. I do stand by my statement that
for the purpose of using the fact that force times velocity is power , it is
not correct to say trust times velocity is power. The net force on the
airframe is zero, thrust equals drag so net force is zero. All of the
energy is going into the air, not the airframe. But it looks like due to
the higher velocity of the airstream at altitude (higher TAS) the power
requirement to move the air for a given IAS is greater. Keep in mind the
energy/power from the engine is going into moving the air, not accelerating
the airframe.

As an example of the problem of using thrust time velocity as power,
calculate the power being generated by engines doing a runup on the runway
before takeoff. Velocity is zero, thus "power" is zero. The equation
brakes down.

Live and learn though. IAS does drop with altitude for the same power out
of the engine. This effect may be one reason jets are inherently faster at
altitude. While thrust of a jet does drop with altitude, this drop is not
as bad on aircraft speed as the drop in power in a piston or turbo prop.

Thanks xerj. Even on old engineer can learn a few new things :-)

Danny Deger




However, the same author as the snippet above says:-

"The statement that power is drag time velocity is
incorrect."

Is it? I've seen that formula mentioned in almost every text on power that
I've seen.

Is there something I'm missing?

Not trying to be a PITA, just seeking clarification of something I was
sure was right. And I know that operationally TAS is much more important
than IAS except for, say, stall speed, best glide and the like. So it's a
largely an academic question, I realise. It was (sort of) started as a way
of finding a plain language non-mathematical explanation for the question
"why does the same IAS require more power at altitude?". I haven't found
that plain language explanation yet, but now I'm getting conflicting
answers as to the very definition of power.

Can someone clear it up?

TIA!