In article ,
Don W wrote:

Can someone explain to me why 300HP applied to a large rotor
at ~700 RPM is enough to lift a 2000lb helicopter straight up,
but the same 300HP applied to a smaller diameter propellor
at ~2600 RPM can not even come close to allowing a 2000 LB
airplane to climb vertically?

This is really bugging me. BTW, does anyone have any idea
what the thrust produced by the propellor of the hypothetical
300 HP (say LYC-IO540 powered) airplane would be? Obviously
the thrust produced by the 300HP helicopter exceeds 2000 LBs.

TIA,

Don W.

It's got everything to do with the amount of air they move and the
difference in efficiency between moving a little air at high speed or a
lot of air at lower speed.

Let's look at this qualitatively.

In order to lift an object by moving air, you need to create enough
force. Force is equal to a change in momentum with respect to time. That
is, you can think of force as being equal to changing the momentum of a
constant mass at a constant rate of acceleration (F = ma), *or* you can
think of it as applying a constant speed change to a flow of mass (F =
m/s * v). But as long as multiplying the two together gives you the same
total force, it does matter from a momentum perspective.

But! From and energy and power perspective it matters a lot.

Kinetic energy is proportional to the mass being moved but also
proportional to the *square* of the speed you move it at.

So if you go from rotor moving x mass per second at y speed to a
propellor moving x/2 mass per second at 2y speed, then your power goes
down by half from the change in mass, but *up* by four from the change
in speed. IOW, move half the mass to achieve the same force and you need
to use twice the power.

Does that help?

--
Alan Baker
Vancouver, British Columbia
"If you raise the ceiling 4 feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect
if you sit in the bottom of that cupboard."