Propellors vs Rotors

----------------------much snipping-------------

If I'm not mistaken, Hughes once built some kinda
giant tip-thrust powered test-freak that had a rotor speed of about 16
rpm. I've seen the videos, but I can't recall the name.


Fairchild-Hiller once built something like that, supposedly with ram jets in
the tips, and Hughes may have as well; although I am fairly sure that the
rotor speed was within the normal range for helicopters.

There was also a propane fueled pulse jet powered single blade helicopter of
the strap-onto-the-pilot variety several years ago. ( I even had a "blurb"
about it--and a related APU for gliders--which may still be buried among
other books and catalogs.) Although I don't know if it ever flew out of
tether...

Peter

P.S.: As to the original post: the pilot shops at most airports have
passable texts to introduce the theory of helicopters, and a lot of
accomplished helicopter pilots and mechanics used to hang around on
"rec.aviation.rotorcraft: so that lurking over there could pay dividends...

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."

wright1902glider wrote:

Now here's a simple 19th Century way to prove the point. Attach a
length of yarn to the "lower" surface of a slow-flying plane. Watch the
yarn and see what direction it takes in flight. Is it straight back? Or
is it down? If air is indeed being forced downward by the wing,
shouldn't we be able to see the results in the yarn?

Harry


Thank you, Harry!

"Attach a length of yarn to the "lower" surface of a slow-flying plane.
Watch the yarn"

Is this why they invented wing walking? How do I do that on a low wing
monoplane?

Colin

This pushes and pulls on the rotor-blade control horns,
constantly changing the pitch of the blade as it flys around in a
circle. If you tilt the dinner plates forward, the blade flys at a
lower AOA in the front 1/2 of the rotor disk than it does at the back
1/2. Since its producing more lift in the back 1/2 than in the front
1/2, the blade flies higher in back. Stay with me here. As the blade
flies higher, its coning angle relative to the rotor head increases to
a greater angle than it does in the forward 1/2 of the rotor disk..
Therefore, its line of thrust relative to the fuselage is not vertical,
but is actually inclined forward.

Except the swash plate is not inclined forward in forward
flight. It's inclined to the right in forward flight, if the rotor
turns counterclockwise as seen from above, as in most American
'copters.
The rotor is a gyroscope, and trying to tilt one edge of it
will result instead in tilting the edge 90 degrees away in the
direction of rotation, like any other gyro. The rotor blades reach
their maximum pitch on the left side of the machine, and the blades
reach their maximum flap at the rear, tilting the disc forward. The
advancing blade on the right has minimum pitch and the front of the
disc is lowest.This is a happy coincidence, since we also need a lot
more AOA on the retreating blade to partly make up for its much lower
airspeed through the air compared to the advancing blade on the right.
Assymetrical lift has to be dealt with or the machine will roll over as
soon as it move forward, so the retreating blade's higher pitched AOA,
the blade's small downward flap approaching the retreating side and
rising flap approaching the advancing side also contributes to AOA
changes, and the lead/lag hinges on many rotors allow the blades to
accellerate on the retrating side and decellerate on the advancing side
at and therefore reduce some of the airspeed difference. Symmetrical
airfoils are used to minimize vibration caused by center of pressure
changes with AOA changes.
Helicopters are a lot more complex than they seem.

Dan

The 300 hp in the helicopter is moving it's wing fast enough to lift the
Helo. The
fixed wing engine is also moving it's wing fast enough to lift the aircraft.
The Helicopter itself need not move forward, so the lift appears vertical,
but the wing is indeed climbing at a shallow angle, just like the fixed
wing.

Al


"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.

ELIPPSE wrote:
Well, the way I hears it, the low pressure on the wing is produced by
the Coanda effect as air travels up and over the wing, not because of a
change of velocity.The velocity change hypothesis has been disproven
over and over! The mass of air displaced downward is the result (read:
reaction) to the air having clung-to and been moved around the surface
by the Coanda effect. Lift is caused by a pressure difference, not the
downwash. There is no such force as suction, and downwash doesn't
generate lift.


http://jef.raskincenter.org/publishe...da_effect.html

scroll down to OTHER PARADOXES

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.

Peter Dohm wrote:

"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.

This makes sense to me now.

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.

Nit picking here, but the propellor is actually doing a lot of work
even when the aircraft is not moving. It just does not translate
into useful work on the airplane. You are thinking of the work as
thrust * velocity (airplane) which is correct from the viewpoint of
the airplane, but not the system. The prop is moving plenty of air,
dust, leaves etc. although that does not help you get where you are
going.

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

Yes it does. Thanks.

Don W.

Hi Colin,

I cant remember what the max rotor rpm on the R22 Beta is although
I have a few hours in them. I just make sure that both guages
stay in the green arc and that little light and the annoying horn
don't come on ;-).

Don W.

COLIN LAMB wrote:

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


The Schweizer 300 (formerly Hughes 269C) has a rotor rpm power on of 442 to
471 rpm. Power off range is 390 to 504 rpm. Esceed these limits and you
are quite likely to break something.

The Schweizer has 190 hp and gross weight is 2050 lbs.

Colin