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I am trying to make sure I correctly understand the moment coefficient. Both the lift and the drag coefficients vary with angle of attack and can be either positive, negative or zero. The moment coefficient pertains to the moment specifically due to the aerodynamics force (lift force on the wing mostly). The moment of a force can be calculated about any arbitrary point on the chord (or even outside of it). For a positive cambered wing, if the pole was the arbitrary point X, as the angle of attack varies (positive), the moment would vary and be negative.Is that correct? If the angle of attack goes negative, can the moment coefficient go positive?

If the pole X is chosen to be the aerodynamic center AC (25% of the chord distance from leading edge), the moment of the lift force calculated about the AC remains instead constant with varying the angle of attack.

For a positive cambered wing, I understand that there is a moment due to the lift force (mentioned above) and also another intrinsic nose-down moment (which I think it is constant for all angles of attack) solely due to the cambered shape of the wing. This intrinsic moment is nonzero when the lift is nonzero or even zero. Does the value of the moment coefficient take this intrinsic moment into account as well besides the moment due to the lift force?

Thanks. Brett

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As much as I try to explain aerodynamics without resorting to potential theory, sometimes it is indeed helpful. If you interpret the pressure distribution around a wing as the sum of

  1. a zero angle of attack distribution due to camber and
  2. the Birnbaum pressure distribution due to angle of attack,

things should become clearer.

On the left I show a cambered airfoil at zero angle of attack and on the right an uncambered airfoil at different angles of attack. The first cambered airfoil is a Joukowsky-type which has an arc section for its camber line. The one below it has an irregular camber line that is simply the center between its two surfaces. If you draw circles inside the airfoil at different chord locations, the camber line is the connection of the centers of those circles. The higher the local curvature of the camber line, the higher the local lift at that chord station.

picture one

picture two

picture three

The remarkable thing about the uncambered airfoil is that its lift distribution over chord is geometrically similar between different angles of attack. It only scales in magnitude, but keeps its general shape. Therefore, its center of pressure is at the quarter point for all angles.

The lift due to camber does not change with angle of attack and can be added to the lift of the symmetric airfoil in order to arrive at the total lift for a cambered airfoil at a nonzero angle of attack. Since the lift due to camber dominates the total lift at small angles of attack and becomes more and more insignificant as the angle of attack increases, the center of pressure will move from the point due to camber to the quarter point as the angle of attack rises.

If the lift is composed of a force acting at the quarter point plus an offset moment, this moment is only determined by the camber lift. Since that is constant over angle of attack, so is the offset moment.

This is all strictly true only in inviscid 2D flow, but gives a good approximation for real wings with enough aspect ratio and no flow separation. Viscosity will muddle the picture a little and with smaller aspect ratio the center of pressure of the symmetric airfoil will move forward, but for most airplanes the above captures reality rather well.

For a positive cambered wing, I understand that there is a moment due to the lift force (mentioned above) and also another intrinsic nose-down moment (which I think it is constant for all angles of attack) solely due to the cambered shape of the wing.

Only if you decide to let your lift force attack outside of the quarter point. Then the lift moment caused by the distance between the force vector and the quarter point will produce that moment due to the lift force. If, however, the lift attacks in the quarter point, only the intrinsic nose-down moment which is constant for all angles of attack will be left.

This intrinsic moment is nonzero when the lift is nonzero or even zero.

Yes, indeed. At the right negative angle of attack, the downforce from that angle of attack will exactly compensate the camber lift, leaving the whole airfoil at zero lift but still with the constant moment.

Does the value of the moment coefficient take this intrinsic moment into account as well besides the moment due to the lift force?

Again, only if you (against all convention) decide to define your lift force in a point different from the quarter point. As the moment coefficient is defined, it contains only the "intrinsic moment".

When separation starts (and even with a thicker boundary layer) this will gradually break down and the moment will fluctuate. But then we move outside of the linear range of aerodynamics and potential theory will no longer apply.

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  • $\begingroup$ Thank you Peter! In essence, when the total lift force is applied at the quarter point, the moment coefficient only captures the intrinsic moment due to camber. This coefficient is negative. You mention that "The lift due to camber does not change with angle of attack and can be added to the lift of the symmetric airfoil". The two figures on the left (positive camber and arcshaped camber) show different circulation distributions for zero angle. So those circulations (I guess lift comes from circulation?) would be the same no matter the angle of attack, correct? $\endgroup$ – Brett Cooper Apr 10 '18 at 17:13
  • $\begingroup$ @BrettCooper: What you write in the comment is correct. Lift is roughly circulation times chord. Circulation is another word for vortex strength. $\endgroup$ – Peter Kämpf Apr 10 '18 at 19:09
  • $\begingroup$ Hi Peter, is this constant aerodynamic moment solely due to the wing camber a pure couple moment (two equal and opposite non collinear forces), that is, a free vector that can be applied anywhere and does not depend on the reference point? $\endgroup$ – Brett Cooper Apr 14 '18 at 0:18
  • $\begingroup$ @BrettCooper: Yes, the constant moment is purely from camber, but is produced by a multitude of opposite forces (which result from the integration of pressure over a surface). There is no distinct pair of forces unless you choose to express the result of the camber-induced pressure distribution as those two forces. And yes, this moment is a free vector which does not depend on a reference point. $\endgroup$ – Peter Kämpf Jan 3 at 16:48

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