In article ,
Don W wrote:

Alan Baker wrote:

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?
Don W.

snip


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?


Yes Alan, it does. I had just not thought of it that way. Now that
you point it out, it makes perfect sense. This is what I think you
said:

force = d (mv)/dt =
force = d (m)/dt * d(v)/dt = force = d(m)/dt * (v1-v0)

In English: force is equal to mass flow rate times the difference
in velocity before and after the propellor.

Ek= 1/2 (mv^2) -and-
Power = d(Ek)/dt = Power = d(m)/dt * (v1-v0)^2

In English: Power is the rate of change of the kinetic energy of
the airflow which is equal to the mass airflow times the square
of the difference in velocity before and after the propellor.

Is that correct? If so, it says that for fuel efficiency you
want as big a prop as you can fit turning slow. That also makes
sense because the parasitic drag on the prop goes up as the
square of the blade velocity as well.

big grin

I think something fundamental just just clicked.

Don W.

Looks correct to me, Don. Glad to have helped.

--
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."