Motor Torque and Roll Moment Resulting

[Peter Germann]

Is motor torque equal to roll moment?
Turning the prop at 9'400 static RPM while logging 519 Watts motor power (IN, I assume) I have found the torque resulting being 0.396 NM or 56 ozin (at 75% motor efficiency shaft power is assumed to be 389 W). IF motor torque is equal to the roll moment resulting, this would require 55 Grams (2 oz) to compensate with tip weight (1.5 m or 59 in span)
 
As this seems to be quite a bit, I did a rather pragmantic test trying to "measure" the static run roll moment: With the airplane restrained on its back and the wing balanced, it took approx. 25 Gr (1 oz) only to compensate the roll moment at the same RPM.

Am I missing a point?

Peter Germann

[Igor Burger]

Yes, you are missing something - aerodynamic forces 

... no I am not kidding, imagine prop closed in tube and front of prop are only tubes and after also ... so air can go in and out only straight, you will have turbine without measurable moment :- ))) ... and that is how the wing and tail cance most of prop moment ... CW turning prop will make CCW monet on nose, because it rotatest air CW, but that air hits the wing at som angle, wing stops its movement and it will make some CW moment :- ))

[Peter Germann]

Thank you, Igor.
If I understand correctly, this means that the spiral airflow mass hitting the wing (and othe surfaces) compensates part of the motor torque. If so, the moment I have measured is (more or less) equal to motor torque minus the opposite torque resulting from the spiral airflow.
Consequently, my rather simple experiment suggests that I should add 16 Grams (0.56 oz) to the tip in order to compensate the remaining in-flight motor torque and keep the wing level in horizontal flight. As tip weight will not help (torquewise) when flying overhead I may want to displace the vertical c.g. instead.

Peter G.

[Igor Burger]

I wrote it in some other thread, if you displace vertical CG it will also not help (so much), because it will be too little overhead (centrifugal force minus gravity) ... if you do it on tip weight, it will work just opposite in reversed flight, the proper solution is flaps warp compensating that difference ... because its effect is "aproximately" constant as we fly "aproximately" constant speed

however there is still some minor difference because of torque and speed change which can be compensated by vertical CG position, but I would say it has to be UNDER the LO in level flight, but all will change over head, so I think it will be better to keep it in line. ... just thoughts, I never tried to balance anything by vertical CG position. My target is to use hig rpm low diameter prop which has smaller torque then large high pitch props giving the same power and that makes it easier to trim. So far works well.

[Peter Germann]

so I think it will be better to keep it in line.

As I am, probably perhaps not very rational, a bit reluctant about twisting flaps, this probably then comes down to a pair (=no yaw) of roll trim tabs to take care of the remaining and to have the v.c.g. in center.
Is it possible to quantify the remaining (after spiral flow effect) torque for 13 x 5.5 2-bl at 9’500 as opposed to 11 x 5.5 3-bl at 11’500?

[Brett Buck]

Thank you, Igor.
If I understand correctly, this means that the spiral airflow mass hitting the wing (and othe surfaces) compensates part of the motor torque. If so, the moment I have measured is (more or less) equal to motor torque minus the opposite torque resulting from the spiral airflow.
Consequently, my rather simple experiment suggests that I should add 16 Grams (0.56 oz) to the tip in order to compensate the remaining in-flight motor torque and keep the wing level in horizontal flight. As tip weight will not help (torquewise) when flying overhead I may want to displace the vertical c.g. instead.

   The torque is far less in flight.

     Brett

[Peter Germann]

   The torque is far less in flight.

When doing the static torque test it took, at flight RPM, 25 Grams (0.88 oz) of weight on the tip to keep the wing level. This indicates a torque of 25 Gr tip weight x 75 cm half wing span  or 1’875 Grcm (26.06 ozin). Motor In Power logged at this time was 519 Watts.
In level flight, Motor In Power then was recorded to be 340 Watts or 0.66 x static power. Is it correct that this translates to in-flight torque being 0.66 x less? Resulting level flight torque then would have been 1’237 Grcm (17.17 ozin).

[Igor Burger]

When doing the static torque test it took, at flight RPM, 25 Grams (0.88 oz) of weight on the tip to keep the wing level. This indicates a torque of 25 Gr tip weight x 75 cm half wing span  or 1’875 Grcm (26.06 ozin). Motor In Power logged at this time was 519 Watts.
In level flight, Motor In Power then was recorded to be 340 Watts or 0.66 x static power. Is it correct that this translates to in-flight torque being 0.66 x less? Resulting level flight torque then would have been 1’237 Grcm (17.17 ozin).

Yes it is correct if it is constant rpm setup. It also means that if you ballance model perfectly level by those 25 * 0.66 = 16.5g , you will have 2 * 16.5 = 33g too much in tip in reversed flight, and it is terrible misalighment if I can feel difference of 1g or even 0.5g if it is flight after flight.

[Peter Germann]

Yes it is correct if it is constant rpm setup. It also means that if you ballance model perfectly level by those 25 * 0.66 = 16.5g , you will have 2 * 16.5 = 33g too much in tip in reversed flight, and it is terrible misalighment if I can feel difference of 1g or even 0.5g if it is flight after flight.

Yes, which is why I thought it would be better to install a pair of roll trim tabs compensating motor torque.

[Brett Buck]

Yes, which is why I thought it would be better to install a pair of roll trim tabs compensating motor torque.

   Or just tweak it until the wings are level upright and inverted. That ends up taking out all the torque and misalignments no matter what the source, at least in the static sense.
 
   Brett

[Howard Rush]

When I made a test stand to measure engine torque, I mounted a propeller on a motor not attached to the stand, blew on the test stand as an engine would, and positioned a vane on the stand to cancel the torque from the swirly air.  Then I could measure the engine torque alone.

Aerodynamic rolling moment from the swirly air on a wing will be different in flight than on a test stand. 

[Peter Germann]

When I made a test stand to measure engine torque, I mounted a propeller on a motor not attached to the stand, blew on the test stand as an engine would, and positioned a vane on the stand to cancel the torque from the swirly air.  Then I could measure the engine torque alone. Aerodynamic rolling moment from the swirly air on a wing will be different in flight than on a test stand. 

Quite ingenious, Howard...
Learning from all of the above inputs I will add a pair of roll trim tabs for roll moments compensation and I will adjust the v.c.g. to be at the wing's centerline.
Thank you all for helping me along, I appreciate your efforts very much.
kind regards, Peter Germann

[Air Ministry .]

tests at Supermarine found tourque effect equivilant to propellor static ( twisting effect ) at same airspeed . 400 mph stationary prop , roll moment = airflow induced roll on prop .

theres a thing .   

[Douglas Ames]

For you guys running CW props with the outbd. wing hanging low -
What if you transfered (some) tip weight to an inboard tipweight box of equal distance on the inbd. wing?
Wouldn't that raise the outbd. wing and shift the lateral CG (dynamic) to help level the wings?
No, I didn't say "remove" tip weight, that would reduce line tension, just transfer a portion of it as a trim aid.

On a conventional CCW set-up a portion of the tip weight acts as a damper to engine torque, correct?
...along with generating cent. force and counter-balancing line weight. Thoughts?

Not intending to hijack the thread...

[phil c]

Quite ingenious, Howard...
Learning from all of the above inputs I will add a pair of roll trim tabs for roll moments compensation .....
kind regards, Peter Germann

"Brett: Or just tweak it until the wings are level upright and inverted."

Tweaking the flaps a very small bit is better than adding trim tabs.  Not only the work and finishing involved, but trim tabs add another discontinuity in the wing that is at an angle of attack relative to the wind.  Gusts and uneven airflow can cause them to stall or generate variable drag.  Tweaking the flaps spreads the discontinuity over most of the wing so it has less concentrated effect.

[Peter Germann]

[Tweaking the flaps spreads the discontinuity over most of the wing so it has less concentrated effect.
[/quote]
Good point, tThank you, I will be twisting...
Peter G.

[Air Ministry .]

This all depends on many things . TRACTIVE EFFORT being paramont .

a bit like this .
depending on prop pitch , air density , gravitational effort , and other things .

Like a sail on a boat , in broard terms , even .

If turbulant air is where its at , then things start happening .

[Wolfgang Nieuwkamp]

Hello Peter,

the tip weight should be the weight of one line, plus the difference in weight between the inner and outboard wing.
The purpose is to bring the weight just under the lift vector.

The motor torque, and the torque caused by a deflected rudder above the symmetry axis, can be compensated aerodynamically, and/or by lowering the lead out exits (for tractor props).

Regards,

Wolfgang

[Howard Rush]

the tip weight should be the weight of one line, plus the difference in weight between the inner and outboard wing.

This is another persistent stunt myth.  It's more complicated than that when you consider maneuvering.  I think it's best to find the right amount experimentally.

[RandySmith]

This is another persistent stunt myth.  It's more complicated than that when you consider maneuvering.  I think it's best to find the right amount experimentally.

Right, a few of the other things that will effect tip weight is wing asymmetry , flap area , flap span , difference in flap span , many people use a wider tip on the outboard flap, this will change tip weight , also wing aspect ratio will effect  tip weight.

Randy

[Howard Rush]

Randy is correct.