Camber changes the circulation required to satisfy flow tangency even when the geometric angle of attack is zero. Thin-airfoil theory captures that effect by expanding the vortex-sheet solution in trigonometric modes.

Use the chord mapping

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

with

$$0\le\theta\le\pi.$$

Let the camber-line slope, evaluated at the corresponding chord position, be

$$\frac{dz_c}{dx}=z_c'(\theta).$$

A vortex-sheet distribution satisfying the Kutta condition can be written as

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

The factor

$$\frac{1+\cos\theta}{\sin\theta}$$

produces the same leading-edge behavior as the symmetric-airfoil solution while still vanishing at the trailing edge. The sine modes also vanish at

$$\theta=\pi,$$

so the Kutta condition is built into the representation.

Coefficients from camber-line slope

Substituting the trigonometric representation into the thin-airfoil tangency equation and projecting onto the orthogonal cosine modes gives

$$\boxed{ A_0 =\alpha-rac1\pi \int_0^\pi z_c'(\theta),d\theta },$$

and, for

$$n\ge1,$$

$$\boxed{ A_n =\frac2\pi \int_0^\pi z_c'(\theta)\cos(n\theta),d\theta }.$$

The airfoil geometry therefore enters through trigonometric projections of the camber-line slope.

The angle of attack appears only in $A_0$. Changing $\alpha$ shifts the overall circulation without changing the geometry-dependent higher coefficients.

Total circulation and lift

The total positive sheet-strength integral is

$$G=\int_0^c\gamma(x),dx.$$

Using

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

we get

$$G =U_\infty c \int_0^\pi \left[ A_0(1+\cos\theta) +\sum_{n=1}^{\infty}A_n\sin(n\theta)\sin\theta \right]d\theta.$$

Orthogonality eliminates every sine mode except $n=1$. Since

$$\int_0^\pi(1+\cos\theta)d\theta=\pi,$$

and

$$\int_0^\pi\sin^2\theta,d\theta=\frac\pi2,$$

we obtain

$$\boxed{ G=\pi U_\infty c\left(A_0+\frac{A_1}{2}\right) }.$$

The section lift coefficient is therefore

$$C_L=\frac{2G}{U_\infty c},$$

so

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

Substituting the coefficient definitions gives

$$\boxed{C_L=2\pi(\alpha-\alpha_{L=0})},$$

where the zero-lift angle of attack is

$$\boxed{ \alpha_{L=0} =\frac1\pi \int_0^\pi z_c'(\theta)(1-\cos\theta),d\theta }.$$

Thus camber shifts the angle at which lift vanishes but does not change the ideal thin-airfoil lift-curve slope:

$$\boxed{\frac{dC_L}{d\alpha}=2\pi\ \text{per radian}}.$$

A symmetric airfoil has

$$z_c'=0,$$

so

$$\alpha_{L=0}=0$$

and the familiar result

$$C_L=2\pi\alpha$$

is recovered.

Worked camber example

Suppose a simple camber-line slope is represented in the angular coordinate by

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

with

$$m=0.080.$$

Then

$$A_0 =\alpha-rac{m}{\pi}\int_0^\pi\cos\theta,d\theta =\alpha,$$

because the cosine integral over $0$ to $\pi$ is zero.

For the first mode,

$$A_1 =\frac{2m}{\pi}\int_0^\pi\cos^2\theta,d\theta =\frac{2m}{\pi}\frac\pi2 =m.$$

All higher coefficients vanish by orthogonality.

Therefore

$$C_L =2\pi\left(\alpha+\frac m2\right).$$

The zero-lift angle is

$$\boxed{\alpha_{L=0}=-\frac m2=-0.040,\mathrm{rad}}.$$

Converting to degrees,

$$\alpha_{L=0} =-0.040\frac{180}{\pi} \approx\boxed{-2.29^\circ}.$$

At geometric angle of attack

$$\alpha=0,$$

the airfoil still has

$$C_L=2\pi(0.040) \approx\boxed{0.251}.$$

Camber has therefore shifted the lift curve horizontally: the same ideal slope remains, but zero lift occurs at a negative angle.

What camber changes—and what it does not

Within thin-airfoil theory:

• camber changes the vortex-sheet distribution;
• camber changes the zero-lift angle;
• camber can produce lift at zero geometric angle of attack;
• the ideal two-dimensional lift-curve slope remains $2\pi$ per radian.

The detailed pressure distribution and pitching moment also depend on the higher trigonometric coefficients. Those moment effects require additional aerodynamic-moment concepts beyond the lift relation developed here.