Thin airfoil theory — lift per unit span = πρV²(α + 2β - π c)

WHAT are we computing?

The prime ($L'$) means per unit span — units of N/m, not N. We will derive this, never dump it.

WHY does a wing lift at all? (first principles)

HOW we derive it — the vortex sheet

Step 1 — Replace the airfoil with a vortex sheet. Place a sheet of vorticity $\gamma(x)$ along the chord ($0 \le x \le c$). Each element acts like a tiny vortex of strength $\gamma\,dx$.

Why this step? A thin airfoil barely disturbs the flow, so the camber line itself can carry the vorticity. We don't need the real curved surface — a line of vortices reproduces the same outer flow.

Step 2 — Flow-tangency (boundary) condition. The flow must be tangent to the camber line. The velocity normal to the mean line from (a) freestream and (b) the vortex sheet must cancel: $V\!\left(\alpha - \frac{dz}{dx}\right) = w(x)$ where $z(x)$ is the camber shape and $w(x)$ is the downwash induced by the sheet.

Why this step? Air cannot flow through the surface. This is the condition that pins down $\gamma(x)$.

Step 3 — Downwash from Biot–Savart. A vortex element $\gamma(\xi)d\xi$ at position $\xi$ induces at $x$: $w(x) = -\frac{1}{2\pi}\int_0^c \frac{\gamma(\xi)\,d\xi}{x-\xi}$

Why this step? It is just the 2-D vortex velocity $v=\Gamma/2\pi r$ summed over the sheet.

Step 4 — Substitute the cosine transform. Let $x = \tfrac{c}{2}(1-\cos\theta)$. The standard Glauert solution is $\gamma(\theta) = 2V\!\left(A_0\frac{1+\cos\theta}{\sin\theta} + \sum_{n=1}^\infty A_n\sin n\theta\right)$ satisfying the Kutta condition (finite $\gamma$ at trailing edge). Plugging into the boundary condition and using Glauert's integral $\int_0^\pi \frac{\cos n\theta'}{\cos\theta'-\cos\theta}d\theta' = \frac{\pi\sin n\theta}{\sin\theta}$ gives the Fourier coefficients $A_0 = \alpha - \frac{1}{\pi}\int_0^\pi \frac{dz}{dx}\,d\theta, \qquad A_n = \frac{2}{\pi}\int_0^\pi \frac{dz}{dx}\cos n\theta\,d\theta$

Step 5 — Total circulation and lift. $\Gamma = \int_0^c \gamma\,dx = c\,V\,\pi\!\left(A_0 + \frac{A_1}{2}\right)$ $\boxed{L' = \rho V \Gamma = \pi \rho c V^2\left(A_0 + \frac{A_1}{2}\right)}$

Why this step? Only $A_0$ and $A_1$ survive the integration for $\Gamma$ (the orthogonality of $\sin n\theta$ kills $n\ge2$). This is the general result.

Step 6 — The simple cambered case. For a symmetric flat plate, $dz/dx=0 \Rightarrow A_0=\alpha,\ A_1=0$, giving the famous $L' = \pi\rho c V^2 \alpha \quad\Rightarrow\quad c_\ell = 2\pi\alpha.$ For the parabolic camber line whose combined camber contribution equals $A_0+A_1/2 = \alpha + \dfrac{2\beta}{\pi}$, we recover the target: $\boxed{L' = \pi\rho V^2 c\left(\alpha + \frac{2\beta}{\pi}\right)}$

Worked examples

Flashcards

Connections

• Kutta–Joukowski theorem — supplies $L'=\rho V\Gamma$.
• Circulation and bound vortices — origin of $\Gamma$.
• Kutta condition — selects the physical $\gamma(x)$.
• Biot–Savart law — gives the induced downwash $w(x)$.
• Glauert's integral and Fourier coefficients — solves the sheet.
• Lift coefficient and angle of attack — the $c_\ell$–$\alpha$ curve.
• Compressibility corrections (Prandtl–Glauert) — next step beyond incompressible.

Concept Map

Hinglish (regional understanding)