Summary: Assuming constant angle of attack $\alpha$, speed $V$, and lift $L$ applied at the center of pressure CP (whose position is $x_{CP}$), the distance $(x_{CP} - x_O)$ between the moment reference point $O$ (whose position is $x_O$) and $CP$ determines the sign (positive, negative or zero) of the pitching moment as long as $L\neq 0$:

$$M_O = (x_{CP} - x_O)L$$

However, the moment $M_O$ depends on $\alpha$, i.e. $M_O=M_O(\alpha)$, since changing $\alpha$ while keeping the reference point $O$ position constant, varies the magnitude $M_O$ since both lift $L$ and the position $x_{CP}$ vary.

Interestingly, when $O=AC$ and $x_O=x_{AC}$, the moment does not change with changing $\alpha$:

$$ M_{O}(\alpha) = M_{AC} = (x_{CP} - x_{AC})L = constant$$

We can move lift $L$ from its application point $x_{CP}$ to point $x_{AC}$ while adding a constant free pitching $M_{AC}$ whose magnitude is $(x_{CP}-x_{AC})L$. The moment $M_{AC}=0$ for symmetric wings and $M_{AC}\neq0$ for cambered wings.

In the case of a cambered wing, the equation $$ M_{AC} = (x_{CP} - x_{AC})L$$ predicts that $$M_{AC}=0$$ when lift $L=0$. However, we know that a cambered wing has a a constant nonzero moment $M_{AC}$ for any $\alpha$ even when $L=0$ at the zero-lift $\alpha_0$. How do we transform the equation for $M_{AC}$ so it becomes equal to the sum of two terms, one solely due to camber and one solely due to lift: $$ M_{O}(\alpha) = M_{AC} = (x_{CP} - x_{AC})L= M_{camber}+M_{lift}$$

where the term $M_{camber}\neq 0$ for any $\alpha$?

  • $\begingroup$ The center of pressure moves to infinity as lift approaches zero for a cambered airfoil. No need to reduce the moment to zero just because lift disappears. $\endgroup$ Commented Jan 2, 2019 at 22:41

3 Answers 3


Interestingly, when $O=AC$ and $x_O=x_{AC}$, the moment does not change with changing $α$

Yes, but only if your lift is also applied at the AC! That's the whole point of the aerodynamic center. You replace the pressure distribution with a lift force (actually a resultant force, but let's neglect drag here) and a moment. If you place the lift force at the center of pressure, then the resultant moment about that point is zero. Then, the moment about any other point $O$ is only "created" by the resultant force. So in that case indeed: $$M_O = (x_{CP} - x_O)L$$

If you now place the lift at your AC and also O=AC, then:

$$M_O = M_{AC}$$

This $M_{AC}$ is independent of $\alpha$. If you now move to any different O, you get: $$M_O = M_{AC} + L (x_O - x_{AC})$$

The first term $M_{AC}$ is what you are calling "$M_{camber}$" and the second term is your "$M_{lift}$".

  • $\begingroup$ I understand your explanation. On my book (I can attach page) the moment $M_{AC}$ due to lift $L$ about the aerodynamic center $AC$ is constant even when $L$ is applied at the $CP$ and even without the vector trick of transferring lift $L$ so it acts at $AC$ with an added constant couple moment. If $L$ acts at $AC$, the lever arm is zero but the free couple moment is nonzero.When $\alpha$ increases, $L$ increases and the lever arm $(x_{CP} - X_{AC})$ decreases.So I agree that $$M_O=M_{AC}$$ But, I think the decomposition into the 2 mentioned moment terms is only possible if $L$ acts at $AC$. $\endgroup$ Commented Jan 2, 2019 at 18:06

In the case of a cambered wing, the equation $M_{AC} = (x_{CP} - x_{AC})\cdot L$ predicts that $M_{AC}=0$ when lift $L=0$.

No, it doesn't.

It only predicts that either $(x_{CP} - x_{AC})$ approaches infinity or that lift decreases to zero. Both conditions will satisfy $M_{AC} = (x_{CP} - x_{AC})\cdot L$, but only the first one will satisfy $M_{AC} = const.$ as well.

plot from XFLR5

Plot from XFLR5 (own work). The overlaid line shows the center of pressure on a wing with washout at small angle of attack where the inner wing creates positive lift while the wingtips create negative lift. At approx. 75% of span the sign of the local lift force changes (= local lift is zero) and the center of pressure switches from negative infinity to positive infinity (apologies for the line not really going to infinity because it has been calculated only at discrete points along the span, but I hope the plot gets the point across).

Why would this happen? At $L = 0$ there is both positive and negative lift along the wing chord. An airfoil with positive camber will show negative lift in the forward part of the airfoil and positive lift in the rear part. Even if their sum is zero, there is enough local lift to create a sizeable pitching moment. Below you see the XFOIL results for a NACA 4409 airfoil. Note that lift is effectively zero while the pitching moment remains unchanged.

pressure distribution of NACA 4409 at -4.25°

Blue is upper side pressure while red is lower side pressure relative to static pressure. The suction peak near the nose has traveled to the bottom side due to the low angle of attack while the rear part of the airfoil, which is less affected by AoA changes, still shows positive lift from camber. For illustration, the same thing with arrows indicating local pressure:

pressure vectors on NACA 4409 at -4.25°

  • $\begingroup$ So as the net lift $L\rightarrow 0$ the location of $x_{CP}\rightarrow \infty$: the force decreases but the lever arm increases keeping $M_{AC}$ constant. The nose down moment is caused by a force couple: the lift closer to the LE is directed down and the other lift force is directed up. I didn't know the positive lift was at the wing inner region while negative lift at the outer area. The force couple would also be predicted for an infinite span wing, right? What causes the lift in the front to b downward? The higher pressure at the top surface relative to the wing's lower surface. But why? $\endgroup$ Commented Jan 3, 2019 at 3:26
  • 1
    $\begingroup$ @BrettCooper: Remember that zero-lift angle of attack is negative: The negative lift near the leading edge comes from the negative angle of attack while the positive lift near the trailing edge is from camber. The changing lift of the sample wing in the plot comes from the change in incidence over span (washout). $\endgroup$ Commented Jan 3, 2019 at 6:50

Thank you, Daniel and Peter. Just to wrap things up on this topic, the two-term moment decomposition for $M_O$ is mathematically possible when:

a) $L$ is moved to act at $AC$ instead of $CP$.

b) the moment $M_0$ generated by $L$ acting at $AC$ is calculated about a moment reference point $O\neq AC$

If a) and b) are satisfied, the pitching moment $M_O$ becomes:$$M_O= M_AC+L(x_O-x_{AC})$$ which is the sum of the two moment terms $M_AC$ and $L(x_O-x_{AC})$

However, if $L$ is not moved to $AC$, the moment expression calculated for $L$ acting at $CP$ about an arbitrary point $O$ is given by $$M_O= (x_{CP} - x_{O}) L$$

and this last expression cannot be converted into a sum of two moment terms (one solely due to camber and one solely due to lift).

  • $\begingroup$ The moment reference point is the quarter chord point / neutral point / aerodynamic center (AC), so if lift is applied at AC, it creates no moment. To faithfully reproduce reality (where lift would be summed up at some variable location along the chord), you need to add a moment (offset moment - you should know this link!). When lift is shifted to the AC, this moment is constant over angle of attack. Please let me know what is unclear in the linked answer! $\endgroup$ Commented Jan 3, 2019 at 16:40
  • $\begingroup$ Thanks. I will study that past thread and post. When you say that there is no moment, you are referring to the lift generated moment when the moment is calculated about $AC$. When $L$ is applied at $AC$, the moment about an arbitrary point is $$M_{O}=M_{AC} +(x_{O}-x_{AC})L$$ If $O=AC$, $x_O=x_{AC}$ and $$M_{AC}=M_{AC} +0$$ since the lever arm $(x_{AC}-x_{AC})=0$ $\endgroup$ Commented Jan 4, 2019 at 15:16
  • $\begingroup$ What you say is correct, but your symbology is confusing. Why don't you call the constant moment from camber $M_C$ and avoid calling two different things $M_{AC}$ ? Then you can write $$M_0 = M_C + (x_0 - x_{AC})\cdot L$$ and $$M_{AC} = M_C+0$$ This would probably look less confusing. $\endgroup$ Commented Jan 4, 2019 at 17:55
  • $\begingroup$ Great. Thanks. I would think that, experimentally, in a wind tunnel setup, the best way to empirically measure $M_C$ would be to support the wing from the point $AC$, i.e. use $AC$ as the fixed pivotal point about which to support the wing. Once we find the AoA where $L=0$, the moment experienced by the wing is only $M_{AC}$...Still, not sure how to measure $M_{AC}$ empirically anyway... $\endgroup$ Commented Jan 4, 2019 at 19:36

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