prop rpm question

At 1000 rpm or so, my airplane will taxi and get up to, what, 15-20
kts? But at double the rpm it will fly at 80-90 kts, though it would
take a long time to take off. Surely double the rpm produces more
than double the propellor thrust...or does it? Anyway, it seems very
nonlinear, that is, double the rpm and I get much more than double the
performance. Why is that?

"Bob Fry" wrote in message
...
At 1000 rpm or so, my airplane will taxi and get up to, what, 15-20
kts? But at double the rpm it will fly at 80-90 kts, though it would
take a long time to take off. Surely double the rpm produces more
than double the propellor thrust...or does it? Anyway, it seems very
nonlinear, that is, double the rpm and I get much more than double the
performance. Why is that?

The laws of physics (certainly those relating to mechanics - velocity,
acceleration, thrust, drag and all that) are rarely linear. On the ground
your aircraft is probably not in an optimum attitude for drag reduction, so
it's probably not fair to compare it with an aircraft in the sky. And will
your aircraft fly straight and level at 1000rpm? If so, how fast?

The main thing dictating how your aircraft performs is drag. "Normal" drag
increases with the square of the speed you fly at - so if you double the
speed, you roughly quadruple the drag (hence everything has a terminal
velocity when falling to earth - as you get faster, the drag increases
faster than your speed increases, and you stop accelerating once drag equals
the acceleration caused by gravity). Remember also that at low speeds you
have induced drag, which is high at low speeds but vanishes as you get
faster.

The moral of the story, though, is that if you think something ought to
behave in a linear manner, you're probably mistaken.

Cheers,

D.

Uh, thanks for trying...I guess:

DC "Bob Fry" wrote in message
DC ...
At 1000 rpm or so, my airplane will taxi and get up to, what,
15-20 kts? But at double the rpm it will fly at 80-90 kts,
though it would take a long time to take off. Surely double
the rpm produces more than double the propellor thrust...or
does it? Anyway, it seems very nonlinear, that is, double the
rpm and I get much more than double the performance. Why is
that?

DC The laws of physics (certainly those relating to mechanics -
DC velocity, acceleration, thrust, drag and all that) are rarely
DC linear.

Eh? F=ma and many others are.

DC On the ground your aircraft is probably not in an
DC optimum attitude for drag reduction, so it's probably not fair
DC to compare it with an aircraft in the sky. And will your
DC aircraft fly straight and level at 1000rpm? If so, how fast?

Kee-rist. Drag, and attitude (angle of attack, really) have little to
do with the explanation I was looking for. No, of course it won't fly
straight and level at 1000 rpm. That's nearly full idle landing rpm.

DC The main thing dictating how your aircraft performs is
DC drag. "Normal" drag increases with the square of the speed you
DC fly at

I think you mean parasitic drag.

DC - so if you double the speed, you roughly quadruple the
DC drag (hence everything has a terminal velocity when falling to
DC earth - as you get faster, the drag increases faster than your
DC speed increases, and you stop accelerating once drag equals
DC the acceleration caused by gravity). Remember also that at low
DC speeds you have induced drag, which is high at low speeds but
DC vanishes as you get faster.

Induced drag--drag caused by the wing producing lift at a vector not
perpendicular to flight--never vanishes unless lift vanishes.

Anyway I'll restate the question, plus post to r.a.'s garbage heap,
r.a.piloting.

At 1000 rpm the prop produces some amount of thrust (lift), call it
T[1000]. This thrust is only enough to move the plane in a moderate
taxi.

At double that rpm, 2000 rpm, the prop produces another amount of
thrust, call it T[2000]. Now I'm not positive, but it sure seems that

T[2000] T[1000]

Why, if rpm only doubles, does thrust (seem to) much more than double?

"Bob Fry" wrote in message
...
Uh, thanks for trying...I guess:
snip
Anyway I'll restate the question, plus post to r.a.'s garbage heap,
r.a.piloting.

At 1000 rpm the prop produces some amount of thrust (lift), call it
T[1000]. This thrust is only enough to move the plane in a moderate
taxi.

At double that rpm, 2000 rpm, the prop produces another amount of
thrust, call it T[2000]. Now I'm not positive, but it sure seems that

T[2000] T[1000]

Why, if rpm only doubles, does thrust (seem to) much more than double?

The short way around is to look at your engine's power chart and evaluate
how many excess HP it is developing at 1000 rpm vs 2,000 rpm.

If we assume the plane in question is a C-152, the engine is making very
little power at 1,000 rpm. I'd guess 15 hp, of which at least 5 hp is spent
in friction inside the engine, leaving 10 hp for thrust. At 2,000 rpm, the
engine is probably making 60 hp, of which 10 is spent on internal friction.
Therefore, you have 50 hp for thrust.

Another way to look at it is that your prop has an advance rate. Let's say
it the advance rate is 4 feet per revolution. At 1,000 rpm, and no drag on
the airplane (rolling or aerodynamic), the airplane would have a terminal
velocity of 4,000 fpm, or about 48 mph. Of course, there is rolling and
aerodynamic drag, and there is prop drag too, so the engine can only drag
the plane along at, say, 30 mph, assuming a flat smooth runway.

At 2,000 rpm, with no drag, the terminal velocity would be 8,000 fpm, or
about 85 mph. Of course, there is still aerodynamic and prop drag, but there
is no rolling resistance, so you get more bang for your RPM buck. Of course,
it helps that your engine is delivering 60 hp, as opposed to 15 hp when it
turns 1,000 rpm.

Yet another way to look at it is that when your prop spins at 2x the speed,
it requires 4x the power to turn it... KE=1/2MV^2..

And yes, all of this stuff is ideal world, no prop efficiency losses, etc...

KB

KB

I wish to leave the engine out of the discussion, but let's
continue...

"KB" == Kyle Boatright writes:

KB If we assume the plane in question is a C-152,

Close enough, it's an Aircoupe with a C90.

But let's look just at the prop. Why does a prop produce so much more
thrust, much more than double, when it's turned at only twice the
rate?

KB Another way to look at it is that your prop has an advance
KB rate. Let's say it the advance rate is 4 feet per
KB revolution.

Yep, 48" pitch.

KB At 1,000 rpm, and no drag on the airplane (rolling
KB or aerodynamic), the airplane would have a terminal velocity
KB of 4,000 fpm, or about 48 mph. Of course, there is rolling and
KB aerodynamic drag, and there is prop drag too, so the engine
KB can only drag the plane along at, say, 30 mph, assuming a flat
KB smooth runway.

KB At 2,000 rpm, with no drag, the terminal velocity would be
KB 8,000 fpm, or about 85 mph.

Hmmmm...so prop thrust is indeed only twice at double the
rpm?...ideally speaking of course.

The idealized (no viscosity etc.) math seems to say that it is linear,
but intuitive feel says not.

On Mon, 16 Jan 2006 21:15:49 -0800, Bob Fry
wrote:

At 1000 rpm or so, my airplane will taxi and get up to, what, 15-20
kts? But at double the rpm it will fly at 80-90 kts, though it would
take a long time to take off. Surely double the rpm produces more
than double the propellor thrust...or does it? Anyway, it seems very
nonlinear, that is, double the rpm and I get much more than double the
performance. Why is that?

http://www.allstar.fiu.edu/aero/Propulsion2.htm

I'm not an engineer, nor do I play one on TV. I do find it interesting
that the speed or velocity of the aircraft is a factor in figuring
thrust in both the propeller and the jet propulsion formulas.

Perhaps someone else can explain in it terms that you and I can
understand.

TC

This won't be a complete ( or maybe even correct answer) but ...

The prop is an airfoil. Here *lift* will be *thrust*. The thrust force
generated will be

T = C A d/2 V^2. V is the speed of the airfoil (prop). So the force is
nonlinear, it goes as the square.
For the rest of the parameters: C is coeff of lift (depends on angle of
attack), A is airfoil area, d= air density.

The prop will have induced and parasitic drag. The relative wind angle of
attack on the prop foil will depend on the aircraft speed. At low speed it
is at high angle of attack. The induced drag is large and the engine can't
get up to full rpm. As the aircraft speed increases, the AOA decreases, the
induced drag decreases and the can rev up to higher rpm (as constnt
throttle). There will be an AOA where the drag is minimum, This is where the
prop is most efficient. It is a narrow range, because, as you go faster,
parasitic drag on the prop kicks in. Constant speed props are used to adjust
the pitch to remian efficient over a wider raneg of airspeeds.

Kershner states that the maximum thrust force occurs when the plane is
standing still (at a fixed throttle setting, I guess), and decreases as you
go faster. I do not understand this. Is it beacese AOA is largest? I am
trying to see how this relates to power. Power would be force*distance/time
or force*velocity. Maybe the thrust decreases slowly with airspeed, but the
power still goes up as you go faster.

This is just a hand waving argument. Please, anyone who knows more, feel
free to correct this picture.

Dave






"Bob Fry" wrote in message
...
I wish to leave the engine out of the discussion, but let's
continue...

"KB" == Kyle Boatright writes:

KB If we assume the plane in question is a C-152,

Close enough, it's an Aircoupe with a C90.

But let's look just at the prop. Why does a prop produce so much more
thrust, much more than double, when it's turned at only twice the
rate?

KB Another way to look at it is that your prop has an advance
KB rate. Let's say it the advance rate is 4 feet per
KB revolution.

Yep, 48" pitch.

KB At 1,000 rpm, and no drag on the airplane (rolling
KB or aerodynamic), the airplane would have a terminal velocity
KB of 4,000 fpm, or about 48 mph. Of course, there is rolling and
KB aerodynamic drag, and there is prop drag too, so the engine
KB can only drag the plane along at, say, 30 mph, assuming a flat
KB smooth runway.

KB At 2,000 rpm, with no drag, the terminal velocity would be
KB 8,000 fpm, or about 85 mph.

Hmmmm...so prop thrust is indeed only twice at double the
rpm?...ideally speaking of course.

The idealized (no viscosity etc.) math seems to say that it is linear,
but intuitive feel says not.

"Bob Fry" wrote in message
...
I wish to leave the engine out of the discussion, but let's
continue...

"KB" == Kyle Boatright writes:

KB If we assume the plane in question is a C-152,

Close enough, it's an Aircoupe with a C90.

But let's look just at the prop. Why does a prop produce so much more
thrust, much more than double, when it's turned at only twice the
rate?

KB Another way to look at it is that your prop has an advance
KB rate. Let's say it the advance rate is 4 feet per
KB revolution.

Yep, 48" pitch.

KB At 1,000 rpm, and no drag on the airplane (rolling
KB or aerodynamic), the airplane would have a terminal velocity
KB of 4,000 fpm, or about 48 mph. Of course, there is rolling and
KB aerodynamic drag, and there is prop drag too, so the engine
KB can only drag the plane along at, say, 30 mph, assuming a flat
KB smooth runway.

KB At 2,000 rpm, with no drag, the terminal velocity would be
KB 8,000 fpm, or about 85 mph.

Hmmmm...so prop thrust is indeed only twice at double the
rpm?...ideally speaking of course.

The thrust is probably 4x, like the engine's power. The advance rate makes
the prop want to pull the airplane at 2x the speed, but again, 2x the speed
requires 4x the power...



The idealized (no viscosity etc.) math seems to say that it is linear,
but intuitive feel says not.

: The thrust is probably 4x, like the engine's power. The advance rate makes
: the prop want to pull the airplane at 2x the speed, but again, 2x the speed
: requires 4x the power...

Check your equations. Drag force goes with the square of the velocity.
Drag power (force*velocity) therefore goes with the *cube* of velocity. So, if that's
the governing equation, 2x the speed requires 8x the power.

If you look at the operator handbooks for identical airframes with different
engine options (e.g. PA28-140/150/160/180/235, PA24-180/250/260/400), you'll see that
they almost exactly follow this cubic (i.e. cube-root) equation.

-Cory


--

************************************************** ***********************
* Cory Papenfuss *
* Electrical Engineering candidate Ph.D. graduate student *
* Virginia Polytechnic Institute and State University *
************************************************** ***********************

Bob Fry wrote:

Uh, thanks for trying...I guess:

DC "Bob Fry" wrote in message
DC ...
At 1000 rpm or so, my airplane will taxi and get up to, what,
15-20 kts? But at double the rpm it will fly at 80-90 kts,
though it would take a long time to take off. Surely double
the rpm produces more than double the propellor thrust...or
does it? Anyway, it seems very nonlinear, that is, double the
rpm and I get much more than double the performance. Why is
that?

DC The laws of physics (certainly those relating to mechanics -
DC velocity, acceleration, thrust, drag and all that) are rarely
DC linear.

Eh? F=ma and many others are.

DC On the ground your aircraft is probably not in an
DC optimum attitude for drag reduction, so it's probably not fair
DC to compare it with an aircraft in the sky. And will your
DC aircraft fly straight and level at 1000rpm? If so, how fast?

Kee-rist. Drag, and attitude (angle of attack, really) have little to
do with the explanation I was looking for. No, of course it won't fly
straight and level at 1000 rpm. That's nearly full idle landing rpm.

DC The main thing dictating how your aircraft performs is
DC drag. "Normal" drag increases with the square of the speed you
DC fly at

I think you mean parasitic drag.

DC - so if you double the speed, you roughly quadruple the
DC drag (hence everything has a terminal velocity when falling to
DC earth - as you get faster, the drag increases faster than your
DC speed increases, and you stop accelerating once drag equals
DC the acceleration caused by gravity). Remember also that at low
DC speeds you have induced drag, which is high at low speeds but
DC vanishes as you get faster.

Induced drag--drag caused by the wing producing lift at a vector not
perpendicular to flight--never vanishes unless lift vanishes.

Anyway I'll restate the question, plus post to r.a.'s garbage heap,
r.a.piloting.

At 1000 rpm the prop produces some amount of thrust (lift), call it
T[1000]. This thrust is only enough to move the plane in a moderate
taxi.

At double that rpm, 2000 rpm, the prop produces another amount of
thrust, call it T[2000]. Now I'm not positive, but it sure seems that

T[2000] T[1000]

Why, if rpm only doubles, does thrust (seem to) much more than double?

Because prop thrust increases as the square of the RPM and thus you get
four times more thrust at twice the RPM. There are other factors that
come into play as well such as advance ratio, but I assume you can Google...

Matt