Visual walkthrough — Finite wing theory — induced drag, Prandtl's lifting line

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Step 1 — Why a real wing must shed vortices

WHAT. We start with the one object we are allowed to assume from the parent: a wing carries circulation, written $\Gamma$ (the Greek capital "gamma"). Think of $\Gamma$ as how strongly the air is set spinning around the wing — big $\Gamma$ = strong spin = lots of lift. On a finite wing this spin is not the same everywhere along the span: strong in the middle, zero at the tips (there is no wing past the tip to hold any spin).

WHY. A spinning "tube" of air (a vortex filament) obeys a hard rule from Helmholtz: a vortex line cannot simply stop inside the fluid. So if the bound spin $\Gamma$ weakens as we move outward, the "lost" spin cannot vanish — it must peel off and trail behind the wing.

PICTURE. Below: the span runs left–right. The blue curve is $\Gamma (y)$, largest at the centre, dropping to zero at each tip. Wherever $\Gamma$ is dropping, little orange trailing vortices peel off downstream.

The strength of trailing vorticity shed per unit span is exactly $-\frac{d\Gamma }{dy}$. The minus sign says: where $\Gamma$ falls, positive trailing vorticity appears.

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Step 2 — One trailing filament's kick: Biot–Savart

WHAT. Pick a single trailing vortex that peels off at station $\eta$ (Greek "eta" — just another spanwise position, the source of the vortex). We ask: how much downward air-speed does it create back at some other station $y$ (the observer)?

WHY Biot–Savart, and why the $4\pi$. The Biot–Savart law is the tool that converts "a line of spin of strength $\Gamma$ at distance $r$" into "the speed it induces at a point." For a full infinite straight vortex line it gives speed $\frac{\Gamma }{2\pi r}$. But a trailing filament is semi-infinite: it starts at the wing and runs to infinity downstream only — that is exactly half an infinite line. Half the line, half the speed: $\frac{\Gamma }{2\pi r}\stackrel{\text{half}}{\to }\frac{\Gamma }{4\pi r}.$

PICTURE. The filament starts at $\eta$ and shoots to the right (downstream) to infinity. At the observation point $y$ on the lifting line, the induced velocity points down. The perpendicular distance is $r=|y-\eta |$.

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Step 3 — Add up all filaments: the downwash integral

WHAT. The real trailing sheet is a continuous row of these filaments, one at every $\eta$ from tip to tip. To get the total downwash $w(y)$ at station $y$, we add up all their little kicks — that is what the integral sign $\int$ means: "sum of infinitely many slices."

WHY integrate. Each filament at $\eta$ gave $dw$. There is a filament at every $\eta$. Summing (integrating) over all of them is the only honest way to get the combined effect.

PICTURE. Many orange filaments across the span; the black arrow $w(y)$ at the observer is the total downward push, the pile-up of all the little $dw$ arrows.

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Step 4 — Downwash tilts the flow: the induced angle

WHAT. The wing flies forward at speed $V_{\infty }$ (the "$\infty$" means far upstream, undisturbed). Locally it also feels the downwash $w$ pushing air down. Add these two as arrows: the air the wing actually "feels" is tilted downward by a small angle we call ${\alpha }_i$, the induced angle ($\alpha$ is Greek "alpha" = angle; subscript $i$ = induced).

WHY a tangent, and why we can drop it. The two velocities form a right triangle: forward leg $V_{\infty }$, downward leg $w$. The angle between the resultant and the forward direction has $\tan {\alpha }_i=\frac{w}{V_{\infty }}$ — because tangent = opposite over adjacent, and here "opposite" is the downward $w$, "adjacent" is the forward $V_{\infty }$. Since $w\ll V_{\infty }$ (downwash is tiny compared to flight speed), the angle is small and $\tan {\alpha }_i\approx {\alpha }_i$ (small-angle rule). That is why we may write: ${\alpha }_i(y)=\frac{w(y)}{V_{\infty }}.$

PICTURE. The velocity triangle, and — crucially — the lift arrow. Lift is always perpendicular to the flow the wing actually feels. Tilt the felt flow down by ${\alpha }_i$, and the whole lift arrow tilts back by ${\alpha }_i$. Its small rearward shadow is the induced drag.

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Step 5 — Close the loop: the lifting-line equation

WHAT. We now have two independent facts about the same local lift, and we set them equal.

1. Aerodynamics side (Thin airfoil theory): the strip's lift coefficient is $c_{\ell }=2\pi ({\alpha }_{\text{eff}}-{\alpha }_{L=0})$, where the effective angle is what the airfoil truly sees after downwash steals ${\alpha }_i$: ${\alpha }_{\text{eff}}=\alpha -{\alpha }_i$.
2. Circulation side (Kutta–Joukowski theorem): the same strip has $c_{\ell }=\frac{2\Gamma }{V_{\infty }c(y)}$, where $c(y)$ is the local chord (front-to-back width).

WHY. Both must describe one physical strip, so they are equal. Substituting the downwash integral for ${\alpha }_i$ turns this into a single equation whose only unknown is the shape $\Gamma (y)$.

PICTURE. A single spanwise strip, its geometric angle $\alpha$, the stolen ${\alpha }_i$, and the effective angle that remains — the "accounting" of angles at one strip.

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Step 6 — The trick that solves it: a Fourier sine series

WHAT. We change coordinate from $y$ to an angle $\theta$ by $y=-\frac{b}{2}\cos \theta$. As $\theta$ sweeps $0\to \pi$, the tip-value $y$ sweeps $-b/2\to +b/2$. Then we guess $\Gamma$ as a sum of sine waves: $\Gamma (\theta )=2b{V}_{\infty }{\sum }_{n=1}^{\infty }{A}_n\sin (n\theta ).$

WHY sines, why this $\theta$. At the tips ($\theta =0$ and $\theta =\pi$) every $\sin (n\theta )=0$ automatically — so this guess cannot violate the "zero spin at the tips" rule, no matter what the coefficients $A_n$ are. Sines are also orthogonal (Step 7), which will make lift and drag collapse to clean formulas.

PICTURE. The first few building blocks: $\sin \theta$ is one clean hump (the elliptic shape), $\sin 2\theta$ is a wiggle that is positive on one side and negative on the other, $\sin 3\theta$ wigglier still. Real $\Gamma$ = a weighted stack of these.

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Step 7 — Harvest lift and drag: only $A_1$ lifts, everyone drags

WHAT. Feed the series into the total-lift integral $L=\rho V_{\infty }\int \Gamma dy$ and the induced-drag integral $D_i=\rho V_{\infty }\int \Gamma {\alpha }_{i}dy$. Two clean results fall out: $C_L=\pi AR{A}_1,C_{D,i}=\pi AR{\sum }_{n=1}^{\infty }n{A}_n^2.$

WHY only $A_1$ in lift. Lift's integral contains ${\int }_0^{\pi }\sin (n\theta )\sin \theta d\theta$. By orthogonality this is zero for every $n\ne 1$ — the wiggly harmonics cancel themselves out over the span, contributing exactly nothing to lift. But the drag integral pairs each harmonic with itself, giving $n{A}_n^2\ge 0$ — every harmonic pays.

PICTURE. Two bar charts: lift bars are zero except $A_1$; drag bars are non-zero for all $n$. "Lift loves A-one; drag pays for the rest."

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Step 8 — The best-possible wing (and edge cases)

WHAT. Minimise drag. Since $\delta \ge 0$ and every extra harmonic only adds drag while adding no lift, the minimum is reached by killing them all: $A_2=A_3=\cdots =0$, leaving pure $\Gamma \propto \sin \theta$ — an elliptical lift distribution. Then: $\boxed{C_{D,i}=\frac{C_L^2}{\pi AR}},C_{D,i}=\frac{C_L^2}{\pi eAR}\text{in general.}$

WHY it is the floor, and the edge cases:

• Elliptic ($\delta =0$, $e=1$): downwash is constant across the whole span — no wasted swirl anywhere. Minimum drag.
• Any non-elliptic ($\delta >0$, $e<1$): strictly more drag for the same lift.
• $AR\to \infty$ (the 2D limit): $C_{D,i}\to 0$ — we recover the infinite wing's zero drag (d'Alembert's paradox). The finite-wing lift slope $a=\frac{2\pi }{1+2\pi /(\pi AR)}\to 2\pi$, the 2D value.
• $C_L=0$ (no lift): $C_{D,i}=0$ — induced drag is purely drag-due-to-lift; carry no lift, pay no induced penalty.

PICTURE. Downwash across the span: flat blue line for elliptic, dipping/humping orange for non-elliptic (more downwash near tips = wasted energy).

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The one-picture summary

The whole chain in one frame: varying $\Gamma$ → shed trailing sheet (Helmholtz) → downwash (Biot–Savart, $4\pi$) → tilted lift → induced drag; solved by a sine series where only $A_1$ lifts and the elliptic case sets the floor $C_L^2/(\pi AR)$.

Recall Feynman retelling — the whole walk in plain words

A wing spins the air to make lift, and it spins hardest in the middle, fading to nothing at the tips. That fading spin has to go somewhere, so it peels off the back edge as trailing whirlpools (Helmholtz's rule: a whirlpool can't just end mid-air). Those whirlpools blow air gently downward right where the wing is flying — that's downwash. Because the wing now feels its own wind tilted slightly down, the lift arrow (always square to the felt wind) tips backward a little, and that backward lean is a brand-new drag — even with zero friction. To predict the exact shape of the spin we write it as a stack of sine-wave humps; the tips come out zero for free, and a beautiful cancellation means only the single clean hump makes lift while every extra wiggle only makes drag. So the cleverest wing uses just the clean hump — an oval-shaped lift — giving constant, minimal downwash and the smallest drag physically possible: $C_L^2/(\pi AR)$. Long and lean wins.

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Connections

• Finite wing theory — induced drag, Prandtl's lifting line — the parent this page unpacks.
• Biot–Savart law · Helmholtz vortex theorems · Kutta–Joukowski theorem · Thin airfoil theory — the four tools used step by step.
• d'Alembert's paradox · Aspect ratio & wing design · Oswald efficiency factor — where the limits and results connect.