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Thin-aerofoil theory & the lift curve

Why is the lift curve a straight line, and why is its slope almost always close to $2\pi$ per radian? Thin-aerofoil theory answers both with a beautiful piece of classical mathematics. This is the most maths-heavy article — take it slowly.

The idea: replace the wing with a sheet of vorticity

A thin, slightly cambered section barely disturbs the flow, so we model it as its camber line alone, replaced by a continuous sheet of tiny vortices of strength $\gamma(x)$ per unit length. We then demand the one physical condition: the flow must be tangent to the camber line everywhere (nothing flows through the wing). That single requirement determines $\gamma(x)$ , and from it the circulation and lift.

A clever change of variable

Map the chord $0 \le x \le c$ onto an angle $0 \le \theta \le \pi$ :

x = \frac{c}{2}\,(1 - \cos\theta)

(1)

The leading edge is $\theta = 0$ , the trailing edge $\theta = \pi$ . The vortex strength is written as a Fourier series tailored to satisfy the flow-tangency (and Kutta) conditions:

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

(2)

The leading $A_0$ term carries the singular behaviour at the nose; the sine series vanishes at the trailing edge, automatically enforcing the Kutta condition. Imposing flow tangency fixes every coefficient as an integral of the camber-line slope $dz/dx$ :

A_0 = \alpha - \frac{1}{\pi}\int_0^{\pi}\frac{dz}{dx}\,d\theta

(3)

A_n = \frac{2}{\pi}\int_0^{\pi}\frac{dz}{dx}\cos(n\theta)\,d\theta

(4)

$\alpha$: angle of attack [rad]
$dz/dx$: local slope of the camber line [—]

The payoff: lift

Total circulation is the integral of the sheet, $\Gamma = \int_0^{c}\gamma\,dx$ . Carrying it through and using Kutta–Joukowski, only the first two coefficients survive in the lift:

c_l = \pi\,(2A_0 + A_1) = 2\pi\!\left[\, \alpha + \frac{1}{\pi}\int_0^{\pi}\frac{dz}{dx}(\cos\theta - 1)\,d\theta \right]

(5)

Collect the camber integral into a single constant, the zero-lift angle, to get the standard linear form:

c_l = 2\pi\,(\alpha - \alpha_{L=0}), \qquad \alpha_{L=0} = -\frac{1}{\pi}\int_0^{\pi}\frac{dz}{dx}(\cos\theta - 1)\,d\theta

(6)

The moment and the aerodynamic centre

The same analysis gives the pitching moment about the quarter-chord:

c_{m,c/4} = -\frac{\pi}{4}\,(A_1 - A_2)

(7)

Notice it contains no $\alpha$ . The moment about the quarter-chord is independent of angle of attack: the quarter-chord is the aerodynamic centre, the natural pivot about which lift changes produce no extra twisting. For a symmetric section $A_1 = A_2 = 0$ , so the moment is zero.

How reality bends the line

Fig. 1. The lift curve is straight and steep, then a real wing stalls near

15^\circ

. The inviscid model (dashed) keeps climbing — it has no mechanism to stall.

Thin-aerofoil theory is inviscid and assumes a thin section, so two corrections appear in practice:

Thickness nudges the inviscid slope slightly above $2\pi$ . A common estimate is $a_0 \approx 2\pi\,(1 + 0.77\,t/c)$ , giving $\approx 6.86$ /rad for a 12% section — which is what our panel solver reports.
Viscosity pulls the real slope back below $2\pi$ (typically $\approx 0.10$ /deg $\approx 5.7$ /rad) and, above $\sim 15^\circ$ , causes the flow to separate and the lift to collapse — stall. Our inviscid model cannot capture this, so its line rises forever.