Thin-airfoil theory does more than predict the lift-curve slope. It also predicts a special moment reference: the quarter-chord point.

For a cambered thin airfoil, the vortex-sheet distribution is

$$\gamma(\theta) =2U_\infty\left[ A_0\frac{1+\cos\theta}{\sin\theta} +\sum_{n=1}^{\infty}A_n\sin(n\theta) \right],$$

with the chord mapping

$$\boxed{x=\frac c2(1-\cos\theta)},$$

and

$$dx=\frac c2\sin\theta,d\theta.$$

The section lift coefficient is

$$\boxed{c_l=2\pi\left(A_0+\frac{A_1}{2}\right)}.$$

The angle of attack appears only in $A_0$; the coefficients $A_1,A_2,\ldots$ are determined by camber-line geometry.

Moment about the leading edge

The thin-airfoil pressure difference gives the elemental lift

$$dL'=\rho U_\infty\gamma(x),dx.$$

With positive pitching moment defined nose up, an upward force acting aft of the leading edge contributes a nose-down moment:

$$dM'_{LE}=-x,dL'.$$

Therefore

$$\boxed{M'{LE} =-\rho U\infty\int_0^cx\gamma(x),dx}.$$

To evaluate the integral, substitute the angular coordinate:

$$\int_0^cx\gamma(x),dx =\frac{U_\infty c^2}{2} \int_0^\pi (1-\cos\theta) \left[ A_0(1+\cos\theta) +\sum_{n=1}^{\infty}A_n\sin(n\theta)\sin\theta \right]d\theta.$$

For the $A_0$ term,

$$\int_0^\pi(1-\cos\theta)(1+\cos\theta)d\theta =\int_0^\pi\sin^2\theta,d\theta =\frac\pi2.$$

For the sine modes, orthogonality gives zero for all but $n=1$ and $n=2$:

$$\int_0^\pi(1-\cos\theta)\sin\theta\sin\theta,d\theta =\frac\pi2,$$

and

$$\int_0^\pi(1-\cos\theta)\sin(2\theta)\sin\theta,d\theta =-\frac\pi4.$$

Therefore

$$\int_0^cx\gamma(x),dx =\frac{\pi U_\infty c^2}{4} \left(A_0+A_1-\frac{A_2}{2}\right).$$

Using

$$q_\infty=\frac12\rho U_\infty^2$$

and

$$c_{m,LE}=\frac{M'{LE}}{q\infty c^2},$$

we obtain

$$\boxed{ c_{m,LE} =-\frac\pi2 \left(A_0+A_1-\frac{A_2}{2}\right) }.$$

Unlike the lift coefficient, the leading-edge moment changes with angle of attack because it contains $A_0$.

Shift the moment to quarter chord

Moving the moment reference from the leading edge to

$$x=\frac c4$$

gives

$$c_{m,c/4}=c_{m,LE}+\frac14c_l.$$

Substitute the thin-airfoil formulas:

$$c_{m,c/4} =-\frac\pi2 \left(A_0+A_1-\frac{A_2}{2}\right) +\frac14\left[2\pi\left(A_0+\frac{A_1}{2}\right)\right].$$

The $A_0$ terms cancel exactly. After collecting the remaining terms,

$$\boxed{ c_{m,c/4}=\frac\pi4(A_2-A_1) }.$$

This is the key result.

Why quarter chord is the aerodynamic center

For a fixed airfoil shape, $A_1$ and $A_2$ depend on camber geometry but not on angle of attack. Therefore

$$\boxed{\frac{dc_{m,c/4}}{d\alpha}=0}.$$

But the aerodynamic center is defined as the point where the pitching moment is independent of angle of attack. Hence thin-airfoil theory predicts

$$\boxed{x_{ac}=\frac c4}.$$

This result is independent of camber shape within the assumptions of the theory.

The quarter-chord moment itself need not be zero. It is constant with angle of attack, not necessarily zero.

Symmetric airfoil

For a symmetric thin airfoil,

$$A_1=A_2=0.$$

Therefore

$$\boxed{c_{m,c/4}=0}.$$

When lift is nonzero, the center of pressure also lies at quarter chord because

$$\frac{x_{cp}}c =\frac14-\frac{c_{m,c/4}}{c_l} =\frac14.$$

The coincidence of center of pressure and aerodynamic center is a special consequence of symmetry, not the general definition of either point.

Cambered-airfoil example

Consider the simple camber-line slope

$$z_c'(\theta)=m\cos\theta$$

with

$$m=0.080.$$

The thin-airfoil coefficient projection gives

$$A_1=m=0.080,$$

$$A_2=0.$$

Thus

$$c_{m,c/4} =\frac\pi4(0-0.080) \approx\boxed{-0.0628}.$$

The negative sign means a nose-down moment under the stated convention.

This value does not depend on $\alpha$ within thin-airfoil theory. At the same time,

$$c_l=2\pi(\alpha-\alpha_{L=0})$$

changes linearly with angle of attack.

The center of pressure therefore moves according to

$$\boxed{ \frac{x_{cp}}c =\frac14-\frac{c_{m,c/4}}{c_l} },$$

while the aerodynamic center remains fixed at

$$c/4.$$

As $c_l$ approaches zero, the center of pressure can move arbitrarily far away even though the finite quarter-chord pitching moment remains perfectly well defined.

Model scope

The exact quarter-chord result belongs to classical two-dimensional incompressible thin-airfoil theory. Real aerodynamic centers can shift because of finite thickness, viscosity, separation, compressibility, and three-dimensional wing effects.

The theoretical result remains foundational because it explains why quarter chord is such a common moment reference: the leading angle-of-attack-dependent contribution to lift produces no change in pitching moment about that point.