# How to justify that M_AC=0 for a symmetric airfoil?

1. I learnt that for a symmetric airfoil, the pitch moment around the aerodynamic center should be null:

$$M_{AC}=0$$

$$\hspace{11pt}$$ How could I intuitively explain this statement?

1. I also learnt that for an airfoil with positive curvature, $$M_{AC}<0$$ and for an airfoil with negative curvature, $$M_{AC}>0$$. I'm new at my Flight Stability course, and I don't know how to achieve these conclusions. I do know the definition of aerodynamic center, but it isn't enough for me to make such observations.

How could I deduce these statements?

• This answer should help. Let me know if it doesn't. – Peter Kämpf Oct 8 '16 at 15:13

2. Since the airfoil is symmetrical, the moment around the a.c., $M_{a.c.}$ for an angle of attack $\alpha$ will be of equal magnitude and opposite sign at angle of attack $-\alpha$: $M_{a.c.}(\alpha)=-M_{a.c.}(-\alpha)$
The two statements can only hold if $M_{a.c.} = 0$
• After thinking about the question again, I should say that this answer as it is, can be considered self-contradictory. Since $M_{a.c.}$, by definition, doesn't depend on $\alpha$, it will always have the same value regardless of the signal of $\alpha$. Therefore the statement "$M_{a.c.}$ for an angle of attack $\alpha$ will be of equal magnitude and opposite sign at angle of attack $-\alpha$" is wrong. (To be continued) – Élio Pereira Oct 7 '19 at 12:07
• (Continuation) "Opposite sign" should be replaced by "same sign'', and no important conclusion can be taken from this. However, in reality $M_{a.c.}$ is defined for small angles, where the linearisation of the dynamic equations may be performed. The truth is that $M_{a.c.}$ doesn't exist in exact terms. But if "$a.c.$" in "$M_{a.c.}$"' were substituted by some point in the chord of the airfoil near the "$a.c.$" point that would be obtained in a linear world, then the answer can be considered right. – Élio Pereira Oct 7 '19 at 12:07