"Don W" wrote in message
. net...
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.

I am not a helicopter guy, so please don't expect my to carry this thread
very far; but I'll try at the most basic level.

Lift as generated by throwing air downward in order to maintain the
altitude, or the constant rate of ascent or descent, of an object is based
upon a momentum equation--rather than an energy equation. Therefore,
throwing twice as much air downward half as fast will support the same
weight; but will require about half as much energy per unit time, or about
one half the horsepower. Remember that horsepower is a measure of energy,
or work, per unit of time.

The helicopter is thus supported on the downwash from its rotor, producing a
vertical thrust at least equal to its weight (actually more when hovering)
and literally glides forward in response to tilt.

I stopped for a moment to dress in my flame retarding coveralls in
anticipation of the response to my use of the word "glide"; however that is
what it does. It even recovers a little efficiency, compared to its
hovering condition, by virtue of continuously transitioning onto new and
undisturbed air. At its most "efficient" speed, a helicopter might be more
than half as efficient as a really atrocious airplane.

OTOH, in the case of an airplane propeller, we need to make the energy
equation work--while the wings deal with the momentum equation. We can
choose a wingspan appropriate for the intended weight and cruising speed and
a wing area to meet our stall speed requirements, determine the expected
drag in cruise, choose a propeller disk area and number of blades
appropriate for reasonable efficiency in cruise, and match the result to an
engine, and possibly a PRSU since the propeller disk area determines the
diameter and the maximum RPM. Finally, determine that the available power
can supply sufficient thrust for take-off and climb. Traditionally, small
airplanes produce a maximum static thrust on the order of one fifth of their
weight when tied in place, and much less in cruise. The propeller, of
course, constantly transitions into new and undisturbed air and its
efficiency improves from zero at the start of the take-off roll to an
acceptable figure in cruise.

One size does not fit all.

BTW, a 700 rpm rotor is pretty small and may be really inefficient--even by
helicopter standards!

I hope this helps.

Peter