Propellors vs Rotors

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.

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."

"Alan Baker" wrote in message
...
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."

Good explanation.

This is why I've been trying to get glider tug guys interested in gearing a
small engine to a large slow prop.

Bill Daniels

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.

Alan:
Does this mean that a helicopter in hover is continuously pushing
down a mass of air equal to the weight of the aircraft? The corollary
is that a rotary wing or fixed wing aircraft in unaccelerated flight is
displacing a mass of air equal to its weight, i.e., it is not flying
because of low pressure above the wing but because of the upward force
on the airfoil from the displacement of air downward. True?
Russell Thorstenberg
Houston, Texas

In article .com,
wrote:

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.

Alan:
Does this mean that a helicopter in hover is continuously pushing
down a mass of air equal to the weight of the aircraft?

No. It means that it is continuously changing the velocity of a mass
flow of air.

Let's say the helicopter has a mass of 1000 kg. To hover, it requires a
force of 9800 Newtons be exerted upward on it. Therefore, the rotors
must exert a force of 9800 Newtons downward on the air. If we assume to
speed at which it makes the air flow -- say 50 meters per second -- then
we can calculate the mass flow involved.

F = M(ass)/s * v; or M/s = F/v = 9800/50 = 196 kg/s.

So if the helicopter is moving the air downward at 50 meters per second,
then the amount of mass to which it must impart that velocity is 196
kg/s. If it only moves the air downward at 10 meters per second, then it
must be moving 980 kg/s at that speed.

Every one knows "F = ma", but few realize that "F = m/s * v" is equally
valid. The same force that will accelerate a fixed mass at a give rate
can also change a mass flow's velocity by a fixed amount.

The corollary
is that a rotary wing or fixed wing aircraft in unaccelerated flight is
displacing a mass of air equal to its weight, i.e., it is not flying
because of low pressure above the wing but because of the upward force
on the airfoil from the displacement of air downward. True?

It's not an either/or situation. Both are true. It flies because of the
low pressure and the low pressure (among other things) directs air
downward.

Russell Thorstenberg
Houston, Texas

--
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."

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.

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."

Or, as a very good aero guy told me

"It is better to annoy a lot of air a little than to annoy a little air
a lot."

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?

Think of the airplane as a boat. Take a 30 foot boat and put in 300 hp
motor and a 6" propellor. It will not move the boat much. But put on a 2
foot prop and low rpm and you will move the boat quite nicely. Air is like
water.

Colin

"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