I've been struggling a bit to find a satisfying explanation for why the component of net fluid force along a 2D wing's chord cannot cause a torque. At the moment (no pun intended), I'm only considering a symmetrical wing, not an airfoil. Say, a rectangle.

I have two explanations:

  1. The cross-product formula gives zero torque, since the component of net fluid force along the wing's chord is parallel to the moment arm, and so the sine of the angle between them is zero.

  2. The component of net fluid force along the wing's chord cannot generate a torque; it can only translate the wing, with a push or a pull.

  3. The component of net fluid force along the wing's chord cannot generate a torque because such a force is generated from fluid pressure around the wing that cancels out, and so only the perpendicular component of net fluid force is associated with a pressure imbalance around the wing, which can cause a torque, namely if the aerodynamic center is not at the center of mass. I feel this is not quite right ...

The above explanations are unsatisfactory to me, for some reason.

What point am I missing that could help me feel more confident about the idea? Preferably, something that's physically intuitive.

Should I instead not overthink it, and simply treat the cross-product formula as a definition of the torque force?


1 Answer 1


The airfoil has no curvature, so the pressure distribution, when the angle of attack is zero, on the upper and lower are the same. Bernoulli's equation for incompressible flow (or the compressible equivalent) gives an inverse relation between velocity and pressure. Since the air has to travel along the same length and shape on both sides, the velocity distribution is the same and therefore so is the pressure distribution. Pressure cancels on both sides and there is no aerodynamic force or torque. For this situation, the torque at the aerodynamic center (which never depends on the angle of attack, by definition) is zero.

When we increase the angle of attack, the pressure distributions are no longer the same (generating lift and drag), but the torque at the aerodynamic center has not changed (because it doesn't depend on the angle of attack).

The total torque on the wing is just the torque of the airfoils integrated along the wing. If we integrate the torque at the aerodynamic center along the wing, we get zero torque. And zero torque at one point means zero torque at all points, for a rigid body.

  • $\begingroup$ Hi Steve, I'm not confident in this answer, but your pressure-differential argument is one that I'm interested in. I've put it into my own words and thoughts, in my edit to my question -- feel free to edit your answer or leave a comment. Thanks so much. $\endgroup$
    – user59350
    Commented Jul 11, 2021 at 19:48

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