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I was wondering if there is a way to determine the pivot point when an aircraft changes pitch, assuming a standard slightly nose heavy and back elevator trimmed Cessna 172. Will it pivot around the CG, the Center of lift, or somewhere in between.

Logically, I would think it would be "from where it is held up", but some writers stress CG.

Can you help me understand?

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3 Answers 3

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By the very definition, a free-flying object can only pivot around its CG.

Using anything else as a pivot (e.g. Aerodynamic Center) involves a combination of momentum and translation (linear movement) of the CG and will quickly get you into mathematical problems once you start combining the forces acting on the object. Make no mistake, an object only rotates around its CG.

If this is just a mental visualisation exercise, you can 'imagine' the pivot point to be at the center of pressure (i.e. the famous plane suspended by strings at the wings), and as the plane is already flying at a few hundred knots in +x, the little +y acceleration error will not make much of a difference as a mental exercise, but the error is there mathematically.

Later edit:if you still have problems visualising it, place your aircraft in outerspace where there is no air or gravity.Put some big thrusters on the wing and smaller thrusters on the tail. CG (center of mass) is very much forward of the wing. Now fire the thrusters.

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  • $\begingroup$ I think this is an oversimplification. Consider a car on a rail. This will not rotate around the CG. In other words, when there's a direct relation between pitch and vertical speed, the object no longer necessarily rotates around the CG. (Of course, it's much easier to consider rotations around the CG and translation of the CG separately, but the superposition of rotation and translation may give an apparent rotation around a different point) $\endgroup$
    – Sanchises
    Commented Aug 31, 2018 at 7:55
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    $\begingroup$ @Sanchises a car on a rail is constrained, while a plane is not. I agree that this is a first order simplification, but your counterexample does not really help, imho. $\endgroup$
    – Federico
    Commented Aug 31, 2018 at 8:31
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    $\begingroup$ But that only applies if it was a free floating object. So that would be the case for an airplane doing a "bunt".. a ballistic arc. But the airplane is "resting" on something, the center point of the lifting force resisting gravity. You still have a center of mass trying to rotate the overall mass about the center of lift, opposed by the tail downforce. It's the C of G that is the application point of the nose down pitching moment. The pivot axis that force is acting on has to be aft of that, at the center of lift. $\endgroup$
    – John K
    Commented Aug 31, 2018 at 11:17
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    $\begingroup$ @JohnK as I say in my answer, that depends on your frame of reference. you can define one that rotates around the CoG, or around the CoL; neither is wrong, they will simply lead to different equations. CoG equations tend to be simpler. $\endgroup$
    – Federico
    Commented Aug 31, 2018 at 12:02
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    $\begingroup$ @robertdigiovanni People standing don't rotate by their CG when they fall over. Oh, but they do. That center of gravity is falling and moving forward from momentum. But the rotational aspect is still around their CoG. $\endgroup$
    – TomMcW
    Commented Aug 31, 2018 at 18:33
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It depends on your frame of reference

You have not specified a frame of reference, so I will look at the two most common choices.

  1. Body-fixed frame of reference

This is the frame that moves with the aircraft. By definition, any rotation happens w.r.t. the origin point of such frame. By standard practice, the origin is the center of gravity (the equations are much simpler). Thus in this case Radu094's answer applies.

  1. Earth-fixed frame of reference

In this frame, not only you are pitching, but you are also moving forward (in the case of fixed wing aircraft; if you are in a hovering helicopter, then it is still the Center of Gravity).

This forward motion completely changes the position of the "pivot point", that will end up being outside the aircraft.

Assuming a vertical load $n$, and a forward velocity $V$, your aircraft will pull up around a point placed $R$ meters above the aircraft when the aircraft started pulling up:

$$R = \frac{V^2}{g \cdot (n -1)}$$

enter image description here

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  • $\begingroup$ Consider a rectangular wooden board that you hold at one end. Place in a stream of moving water and hold at one end. It will align downstream. Now let go and watch it twirl merrily downstream, AROUND ITS CENTER. Got it? $\endgroup$
    – Robert DiGiovanni
    Commented Aug 31, 2018 at 13:03
  • $\begingroup$ @RobertDiGiovanni you are not holding the aircraft though. your example has the same problem of Sanchises' car. also, the "around its center" is in the body fixed reference frame, because in the earth-fixed one the center is moving, thus you say that you agree with my answer. $\endgroup$
    – Federico
    Commented Aug 31, 2018 at 13:07
  • $\begingroup$ @RobertDiGiovanni also, if you want to focus on the "person falling forward" analogy or similar, then you want to focus on the earth-fixed reference system, i.e. the second part of my answer. $\endgroup$
    – Federico
    Commented Aug 31, 2018 at 13:49
  • $\begingroup$ Standard practice? This is what is being studied. The important distinction here is any object resisting gravity is NOT weightless and there for is held up at a point we know as center of lift. Your second example shows G force being created by a turn. I accept it. Fortunately for us, CG and CL are pretty close in aircraft, but this may make further study of interest. $\endgroup$
    – Robert DiGiovanni
    Commented Aug 31, 2018 at 16:24
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    $\begingroup$ @RobertDiGiovanni I would appreciate it if you would stop typing words in all-caps. It comes across as condescending, like you think you need to slow down in order for us to understand what you're saying. $\endgroup$
    – Sophie Swett
    Commented Sep 1, 2018 at 4:49
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The pivot point is whatever you choose! However, choosing the centre of gravity makes any analysis simplest.

A motion of a free moving body (e.g. an airplane) can be described as combination of motion of an arbitrary reference point and rotation around that point. As far as describing the path taken by the body, any reference point will do.

However, for dynamic analysis, that is understanding the relation between the forces acting on the body and its movement, using centre of mass (a.k.a. centre of gravity) is the only reasonable option. That is because both the Newton's Second Law of Motion,

$$ a = \frac{F}{m}, $$

and its rotational version,

$$ \alpha = \frac{\tau}{I} $$

($a$ is acceleration, $F$ is force, $m$ is mass, $\alpha$ is angular acceleration, $\tau$ is torque and $I$ is moment of inertia), only work in these simple forms for the centre of mass (you can verify that by doing the integration over the mass of the body).

Alternatively in some cases you might want to chose the pivot point so that it is not accelerating, but depending on the path taken, that may not be possible. For example when the pilot pulls on the control column, an downward force will be generated at the tail. Its moment will cause upward angular acceleration in pitch, but it is still an unbalanced force, so it will cause a downward acceleration of the centre of gravity first. This combination means a non-accelerating pivot lies somewhere far ahead of the plane.

But as the attitude and path change, the angle of attack will increase and the wings will build up an upward force that eventually balances, and exceeds, the downward force at the tail and the plane will accelerate upward. At which point, a non-accelerating pivot lies somewhere far above the plane (see the diagram in Frederico's answer). So in case of an airplane, looking for a non-accelerated pivot is futile. It rarely works in situations other than where the pivot actually has a solid support, so it can't accelerate.

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  • $\begingroup$ Ahhh now I get what you are saying. Now it make sense. Thanks. $\endgroup$
    – John K
    Commented Sep 1, 2018 at 1:03
  • $\begingroup$ It is important to realize, accepting the center of gravity model, that the upforces in the wing and H stab generated by deflecting the tail down are instantaneous with with the tail deflection. The plane will not sink, then rise, it pitches up and climbs immediately, putting more down pressure on the tail with vertical motion. $\endgroup$
    – Robert DiGiovanni
    Commented Sep 3, 2018 at 4:35
  • $\begingroup$ @RobertDiGiovanni, no, the force on the h-stab is instantaneous with the tail deflection, but the force on the wing is not, because that is a function of the angle of attack and the force on the tail needs some time to act on the plane to change its attitude and velocity. So the plane might actually sink a tiny bit first (deflecting the elevator takes a bit of time too, and alpha will start building up when the torque does, so it might not actually sink, but there is a distinctive lag between the forces). $\endgroup$
    – Jan Hudec
    Commented Sep 3, 2018 at 11:06
  • $\begingroup$ Agree there, this is why I like to test to see how much. Interesting to look at conservation of energy in this system. An aircraft in flight is in dynamic equilibrium, as opposed to being static. Does adding elevator input energy to the system, or is it altering its orientation in the wind field only? $\endgroup$
    – Robert DiGiovanni
    Commented Sep 3, 2018 at 15:38
  • $\begingroup$ @RobertDiGiovanni, deflecting the elevator does not itself add power, but it creates drag, so it changes the distribution of energy. $\endgroup$
    – Jan Hudec
    Commented Sep 4, 2018 at 7:24

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