Pitching Moment of an Airplane

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I think that, choosing ##CG## as the reference pole, these are the five moments that need to add to zero:
  1. The force ##W## produces no moment (zero lever arm): ##\tau_W = 0 ##
  2. The lift force ##L_{wings}##, applied at ##AC_{wing}##, produces a moment given by ##L_{wings} ## times the distance between ##CG## and ##AC_{wing}##. This moment produces CCW rotation.
  3. The ##L_{tail}##, applied at ##AC_{tail}##, produces a torque given by ##L_{wings}## times the distance between ##CG## and ##AC_{tail}##. This moment produces CW rotation.
  4. The intrinsic aerodynamic moment ##M_a## solely due to the wing. This is a CW moment.
  5. The intrinsic aerodynamic moment ##M_{tail}## solely due to the horizontal stabilizer CW. This is a CW moment.
 

CWatters

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Perhaps have a look at the Wikipedia definition of AC...

https://en.m.wikipedia.org/wiki/Aerodynamic_center

It's the point about which the pitching moment doesn't vary with small changes in AOA. Im not an expert but there are two aspects of stability that need to be got right...

Static stability. This is what you are talking about in your post #26. Eg The moments sum to zero.

Dynamic stability: This determines what happens when the aircraft is disturbed, eg what happens if turbulence lifts the nose? Does the aircraft stay nose high? Return to level flight? Diverge (eg pitch up more and more)? Oscillate? Normally you want it to return to level flight without overshooting/oscillating. I've never looked at the maths but my understanding is that the maths is made simpler if you take moments about the AC rather than the GoG or some other point, because the pitching moment doesn't change with AOA.

I'm afraid you are getting to the limits of my knowledge at this point.

PS regarding #26...

The down force on the tail produces a nose up torque. If that's CW then the pitching moment of the wing (nose down) must be CCW. So 4. appears to be wrong.
 

CWatters

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Ok I'm a bit wrong with my description of static and dynamic stability. See this YouTube video...

 
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Thanks CWatters! I truly appreciate your help.

My lengthy discussion is due to the fact that I am trying to build a model airplane and I am trying to figure out he wing placement (hence the lift which is applied to the aerodynamic center AC located at 25% of the chord) must be placed related to the total ##CG## and to the rear horizontal stabilizer. For most airplanes, stability is achieved when the CG is in front of the AC which is in front of the AC of the rear stabilizer wing...

Have you designed a RC plane before? If so, how did you determine those distances?

thanks.
 

CWatters

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I cheated. Mk 1 eye ball used to get the basic dimensions, the basis that if it looks right it probably is!

CoG set somewhere between 25% and 33% initially and small adjustments made later - perhaps by moving the battery pack backwards or forwards. Typically forward for improved stability on scale models and training aircraft, back for aerobatic and speed models.

I flew fast electric powered competition gliders some years ago and for those we would set them up with a slightly rearward CoG as that allowed us to reduce the down load on the tail. That in turn meant the wing had to produce less lift and hence less drag so they went faster. I understand commercial jets use a similar slightly rearward cog trick to reduce fuel consumption. If you move it back too far they get unstable and tricky to fly.

We would also make trial flights adjusting the elevator trim until it flew as required. Then we would change the rigging angle of the tail plane until no elevator trim was required, again to reduce drag. Our tail planes used thin virtually symmetrical sections.

Another issue for us was adverse yaw. We did away with the rudder to save weight and used differential aileron movement (more up aileron than down) to control yaw in turns. If you didn't set this up right you lost speed in turns.

I flew in several international competitions but never placed higher than about 13th out of 50-70 competitors. This is back in the 1980/90s. My main problem was making them light enough. I always seemed to need more resin than necessary to wet out the glass/Kevlar. The rules set a minimum wing loading so a heavy aircraft was also bigger and slower.
 
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Thanks CWatters. That is awesome and impressive. I am trying to make a RC model high-lifting airplane (that can carry a lot of weight, given a certain take off speed, etc.). I see how I can determine the minimum take off speed given a certain wing areas, and motor power. I wonder how quickly the plane will reach that speed. That will determine the minimum takeoff distance (runway). Do you have any insight in that?

Just to make sure, you mention that the CG of the entire airplane should be placed at distance, from the leading edge of the wing, that is 25%-33% the mean chord line (starting from the trailing edge of the wing 25-33$ of the chord backward).

Thanks and happy Easter!!!
 
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I flew fast electric powered competition gliders some years ago
Wow, that sounds like fun. But you don't say if the gliders were remote control models, or full size with you on board as pilot.
 

CWatters

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They were F5B class RC models. I also flew full size gliders for awhile but not long enough to go cross country, all local flying.

There are quite a few videos of F5B models on YouTube. An example...

 
Last edited:

CWatters

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Thanks CWatters. That is awesome and impressive. I am trying to make a RC model high-lifting airplane (that can carry a lot of weight, given a certain take off speed, etc.). I see how I can determine the minimum take off speed given a certain wing areas, and motor power. I wonder how quickly the plane will reach that speed. That will determine the minimum takeoff distance (runway). Do you have any insight in that?
For that sort of aircraft you want a thick wing section with quite a bit of camber and possibly flaps. It's tricky working out how fast it will accelerate to that speed. More power the better. A fine pitch prop will also accelerate faster but don't go too fine or the top speed will be too slow.

Just to make sure, you mention that the CG of the entire airplane should be placed at distance, from the leading edge of the wing, that is 25%-33% the mean chord line (starting from the trailing edge of the wing 25-33$ of the chord backward).

Thanks and happy Easter!!!
Yes 25-33% of the chord back from the leading edge.
 

FactChecker

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I guess the aerodynamic moment is due to the lift force, always applied at the CP.
This is wrong and I don't know if it is clearly corrected later. There is a torque on the wing that has nothing to do with the location of the wing versus the CP. In creating lift, airflow is diverted down. That produces a twisting torque on the wing which is different from the leverage arm that the lift is applied to. Both must be accounted for.
 

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