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In regards to recent question about neutral point, as being defined as center of lifts from wing and tail, could one also define a "neutral point" for down forces CG AND down force (up elevator) from tail. Lift force is now only from wing

Notice if the down force center is behind the lift center, the nose pitches up. Would the center of this rotation be at the wing, at the center of all moments (between Clift and Cdown), or somewhere else?

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Disclaimer: I am no expert and if you get a better answer from one of the other contributors here, you should listen to them, not me.

From what I can tell, it looks like you're asking whether you can define a point that includes all downward forces on the airplane, including center of gravity and tail downforce. As I understand it, yes you can, and this point is center of gravity, which includes all the associated downward forces.

The airplane then would pitch about the center of gravity, as it usually does.

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  • $\begingroup$ Also don't forget about the thrust vector. If the thrust vector is below the center of gravity as is the case with high wing airplanes, thrust produces an "up" force or a torque of positive pitch. Pitch is determined by the relation between the moments of up forces (against gravity) and down forces (CG). $\endgroup$
    – DLH
    Oct 10, 2018 at 20:43
  • $\begingroup$ Thank you so much! Now I feel so much better about the theory. Including all down forces works. Now my tricycle gear can rotate in peace when I pull back on the stick! $\endgroup$
    – Robert DiGiovanni
    Oct 10, 2018 at 22:33
  • $\begingroup$ Centre of gravity is, as the name suggests, the action point of gravity alone. The downforce—if it exists, because there might as well be upforce—of the tail acts way behind centre of gravity (or ahead for a canard), so the action point of this sum is not the center of gravity. But that point is not particularly useful anyway. $\endgroup$
    – Jan Hudec
    Oct 11, 2018 at 21:11
  • $\begingroup$ Yes, up force will factor into lift torques (to create a new centre of pressure), down force creates a new centre of down. Notice "tail" consists of elevator AND stabilizer (both aerodynamic). Small elevator deflection will rotate nose until stabilizer, now deflected into airstream, stops it. New pitch angle is trim! Meanwhile back at the wing, the center of wing lift is now (adjusted) directly over CG and of equal force, and plane is in straight and level flight. Next up, rolls! $\endgroup$
    – Robert DiGiovanni
    Oct 12, 2018 at 9:07
  • $\begingroup$ @RobertDiGiovanni, tail does NOT consist of elevator AND stabiliser, it is a stabiliser with movable portion called the elevator. You can't talk about their lifts separately, because they are a single airfoil that changes camber. $\endgroup$
    – Jan Hudec
    Oct 12, 2018 at 11:25
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Note: this answer was made to a version of the answer that was improperly edited and made little sense.

Forces do not converge at points.

You can take a set of forces—any set of forces—with different action points and determine the action point of their sum, which is the point in which all their moments cancel.

Since the force is sum of the constituent forces and the moment (around arbitrary fixed point) is the sum of the constituent moments, you can simply find the moment arm that satisfies the two conditions and that is the action point¹.

Now you can do this with any set of forces, but it may not be useful. In case of aircraft, it is useful to sum the lift forces—their action point is called centre of pressure. And compare it to gravity, action point of which is the centre of gravity.

Summing all down forces is much less useful though, because the set of down forces changes with angle of attack and elevator deflection. As the angle of attack increases, at some point the lift force of the horizontal stabiliser changes from downward to upward!

Note that this point is not the neutral point. The neutral point is the point around which the total moment of lift forces does not change with angle of attack. That however, does not mean that total moment is zero around this point. Quite contrary; for statically stable aircraft, the moment around neutral point is pitching up, and the center of gravity is ahead of the neutral point to balance.

¹ In the general form, the equations are $\vec F=\sum\vec F_i$ and $\vec F\times\vec r=\sum\vec F_i\times\vec r_i$. In the linear case where all arms are along one line (e.g. longitudinal) and all the forces are in the same parallel direction (e.g. vertical), you can solve that as $r = \frac{\sum F_ir_i}{\sum F_i}$, but there is no division operator corresponding to cross-product for the vector case.

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  • $\begingroup$ Appreciate the response. Someone edited the question title. My focus is understanding points of rotation in an aerogravimetric environment, which I picture as a yin and yang of CG and aerodynamic forces around a point. This is why summing up and down forces (Including gravity on center of mass) may create a point of rotation not at the CG. My theory, putting it to the test in a public forum. I value the opinions found here. $\endgroup$
    – Robert DiGiovanni
    Oct 11, 2018 at 21:36
  • $\begingroup$ You do well to sum the lift forces as centre of pressure. Why not sum the down forces as centre of down (includes CG). Think of it this way: plane straight, level flight. Send 20 passengers to rear. CG moves back, nose up. Deflect elevator producing 20 passengers of down force. Centre of down moves back, nose goes up. It works! In both cases tail will drop until up force from H stab (opposing elevator) brings forces back into balance. $\endgroup$
    – Robert DiGiovanni
    Oct 12, 2018 at 1:15
  • $\begingroup$ @RobertDiGiovanni, you can describe the situation using any point of rotation, you can just get different linear motion of that point. However, standard law of motion state that acceleration of the centre of gravity equals sum of forces divided by mass and rotation of the centre of gravity equals sum of moments divided by moment of inertia. If you pick any other point, it will make additional movement due to the rotation, complicating your equations. $\endgroup$
    – Jan Hudec
    Oct 12, 2018 at 11:24
  • $\begingroup$ Not any point. It is a dance around total lift, and total down force, out of balance. That is the point of rotation in reality. Once back in balance, rotation stops. Notice this is not a contradiction of CG theory, only adding to it. Try it. 2 see saws, one up, one down. Find their centers. How will it turn if out of balance (not vertically aligned)? Thanks and good day! $\endgroup$
    – Robert DiGiovanni
    Oct 12, 2018 at 13:59
  • $\begingroup$ @RobertDiGiovanni, See, a see-saw is an object with a fixed pivot point. A fixed pivot point will absorb any force needed for it to stay put. A see-saw still rotates around its centre of mass according to the sum of moments and its centre of mass moves according to the sum of forces—but this includes the forces through the pivot, which is why it is easier to analyse it using the moments around the pivot, because at that coordinate system, the forces through pivot add no moments. $\endgroup$
    – Jan Hudec
    Oct 12, 2018 at 19:38

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