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

We will lean on these vault ideas as we go: Kutta–Joukowski theorem, Circulation and bound vortices, Kutta condition, Biot–Savart law, and Glauert's integral and Fourier coefficients. The final payoff feeds Lift coefficient and angle of attack. The parent is Thin airfoil theory — lift per unit span = πρV²(α + 2β - π c).

Definition The ground rules (assumptions of thin-airfoil theory)

Every step below silently rests on five modelling assumptions. State them once, out loud:

Inviscid — no friction/viscosity; the air slides frictionlessly (needed for potential flow).
Incompressible — density $ρ$ is constant (low speed; if not, apply Compressibility corrections (Prandtl–Glauert)).
Irrotational outside the sheet — no spin anywhere except the bound vortices we place, so a velocity potential exists.
Two-dimensional — one wing slice, "per unit span"; no spanwise flow.
Zero thickness, small camber, small angle — the airfoil is a thin bent line, so it may be collapsed onto its camber line and all angles linearised. If any of these fails (thick wing, transonic speed, stall), the formula below no longer holds.

Step 1 — Draw the wing and name every symbol

WHAT. A wing slice is a thin curved sliver. We lay it in a picture and label the four things that will appear in the final formula: the oncoming wind, the tilt, the curve, and the length.

WHY. The contract of this page: no symbol before its picture. So before any maths, look at the sliver and meet the cast.

PICTURE.

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

Read the figure left to right:

The straight horizontal arrows are the freestream — wind of speed $V$ (metres per second) and density $ρ$ (kilograms per cubic metre) far ahead of the wing, undisturbed.
The wing slice runs from the leading edge (front, at $x = 0$ ) to the trailing edge (sharp back, at $x = c$ ). The distance $c$ is the chord — the straight length of the slice.
The wing is tilted up by a small angle $α$ , the angle of attack, measured between the chord line and the wind.
The dashed curve threading through the middle is the camber line $z (x)$ : the average of the top and bottom surfaces at each point along the chord. Its gentle bend is what "camber" means.

Step 2 — Trade the wing for a row of tiny spinners

WHAT. We erase the solid wing and drop a sheet of tiny vortices along the camber line. Each little patch $d x$ carries a spin strength $γ (x) d x$ .

WHY. A thin wing barely nudges the air, so the outer flow does not care whether the vorticity sits on a real curved skin or on the thin camber line itself. Vortices are the natural language of lift because Circulation and bound vortices tells us spin is what lifts. Working with a line of vortices is far easier than a curved solid boundary.

PICTURE.

Each small magenta circle is a vortex element: a microscopic tornado glued to the wing.
Its strength is $γ (x)$ , the vortex sheet strength — spin per unit length, units (metres/second) because it is a velocity jump across the sheet.
The total spin, added up along the whole chord, is the circulation $Γ = \int_{0}^{c} γ (x) d x .$ Here $\int_{0}^{c} \dots d x$ just means "slide $x$ from the leading edge $0$ to the trailing edge $c$ and add up every slice." $Γ$ has units m²/s.

Step 3 — The rule that pins down the spins: air cannot pass through the wing

WHAT. We demand that the total flow slides along the camber line — it never crosses it. This single sentence fixes how strong each vortex must be.

WHY. A wing is solid; air cannot tunnel through it. Two things push air across the camber line, and they must exactly cancel:

the tilted freestream, which has a small component pointing into the surface, and
the downward breeze the vortex sheet blows on itself (the downwash).

PICTURE.

The exact no-through-flow rule is that the flow's velocity component perpendicular to the surface is zero. Writing that out with the conventions above, the perpendicular freestream piece is $V sin (α - \frac{d z}{d x})$ .

With that linearisation, the flow-tangency condition is, term by term, $wind speed V (tilt, +ve nose-up α - local slope, +ve rising \frac{d z}{d x}) = downwash, +ve down w (x) .$

$\frac{d z}{d x}$ is the slope of the camber line at point $x$ — how steeply the dashed curve rises there. We use a slope because "tangent to the surface" is literally a statement about slope. On a flat plate this is zero everywhere.
$α - \frac{d z}{d x}$ is the leftover angle between wind and surface — the bit of wind still aimed into the wing, which the vortices must neutralise.
$w (x)$ is the downwash: the downward speed the vortex sheet induces at point $x$ (positive $w$ points down).

Look at the figure: the orange freestream component into the surface (up-arrow) is exactly matched by the violet downwash (down-arrow). That balance is the equation.

Step 4 — How much downwash does the sheet make? (Biot–Savart)

WHAT. We compute $w (x)$ — the breeze felt at point $x$ from every other vortex on the sheet.

WHY. Step 3 needs $w (x)$ written in terms of the unknown spins $γ$ . The Biot–Savart law gives the velocity a single 2-D vortex makes at a distance: a vortex of strength $Γ$ blows a speed $Γ/ (2 π r)$ at distance $r$ . We add up that contribution from the whole sheet.

PICTURE.

$w (x) = - \frac{1}{2 π} ⨍_{0}^{c} \frac{γ ( ξ ) d ξ}{x - ξ} .$

$ξ$ (Greek "xi") is the location of the vortex doing the pushing; $x$ is where we measure.
$x - ξ$ is the distance between them. Being downstairs in the fraction is the $1/ r$ falloff: nearer vortices shout louder, far ones whisper.
$\frac{1}{2 π}$ is the vortex-velocity constant from Biot–Savart.
The minus sign fixes the direction (a positive sheet blows down, our chosen positive $w$ ).
The slashed integral sign $⨍$ means a Cauchy principal value. Look at the fraction: when the pushing vortex sits exactly at the measuring point ( $ξ = x$ ), the denominator $x - ξ$ is zero and the fraction blows up. A single point can't push itself, so we handle this by cutting out a symmetric tiny gap around $ξ = x$ and letting it shrink: the blow-up to the left and to the right are equal and opposite, so they cancel and the sum stays finite.

In the figure the red vortex at $ξ$ blows the violet arrow felt at $x$ ; the closer they are, the bigger the arrow — that is the $1/ (x - ξ)$ in the picture. The greyed-out sliver right at $ξ = x$ is the excluded gap of the principal value.

Step 5 — A clever change of ruler: measure position by angle

WHAT. We stop measuring position with the straight ruler $x$ and start using an angle $θ$ that walks around a half-circle over the chord: $x = \frac{c}{2} (1 - cos θ), θ : 0 \to π .$

WHY. The awkward $1/ (x - ξ)$ integral becomes clean and solvable once written in this angle. It is the same trick that turns a stretched-out line into a tidy circle so that repeating patterns (sines and cosines) appear. This is exactly what Glauert's integral and Fourier coefficients is built for.

PICTURE.

At $θ = 0$ : $cos 0 = 1$ , so $x = \frac{c}{2} (1 - 1) = 0$ — the leading edge.
At $θ = π$ : $cos π = - 1$ , so $x = \frac{c}{2} (1 + 1) = c$ — the trailing edge.
Middle $θ = \frac{π}{2}$ lands at $x = c /2$ , the mid-chord.

The figure shows a point marching around the upper half-circle (the angle $θ$ ) while its shadow slides along the chord (the position $x$ ). Same point, two rulers.

Intuition Where the singular

$\frac{1 + c o s θ}{s i n θ}$ term comes from — a sketch, not faith We must invert the downwash integral of Step 4 for the spins $γ$ . That is done with a Fourier method: any tidy function on $0 \leq θ \leq π$ can be built as a sum of sine waves $sin n θ$ (this is why we changed ruler — the awkward line becomes a clean interval where sines live). So we try $γ (θ) = 2 V \sum_{n = 1}^{\infty} A_{n} sin n θ + (a special first term) .$ Two physical demands shape that special term:

Kutta condition (Kutta condition): the flow must leave the sharp trailing edge smoothly, so $γ$ must be finite (in fact zero) at $θ = π$ . Every $sin n θ$ already vanishes there — good.
Leading-edge suction: at $θ = 0$ (front) the flow wraps around the stagnation point and $γ$ is allowed to be large. We need a term that is big at $θ = 0$ but still tame at $θ = π$ . The unique such shape is $\frac{1 + cos θ}{sin θ}$ : at $θ = 0$ it blows up (front suction), and at $θ = π$ the numerator $1 + cos π = 0$ kills the blow-up (Kutta satisfied). That is why it — and not some other singular function — heads the series.

Hence the standard Glauert solution $γ (θ) = 2 V (A_{0} \frac{1 + c o s θ}{s i n θ} + \sum_{n = 1}^{\infty} A_{n} sin n θ) .$ Its coefficient $A_{0}$ multiplies the front-loaded term; the $A_{n}$ shape the rest.

$A_{0}, A_{1}, A_{2}, \dots$ are Fourier coefficients: dial numbers that mix together standard wiggle-shapes ( $sin n θ$ ) to reproduce whatever camber the wing has.

Feeding this into Steps 3–4 and using Glauert's integral $⨍_{0}^{π} \frac{c o s n θ ^{'}}{c o s θ ^{'} - c o s θ} d θ^{'} = \frac{π s i n n θ}{s i n θ}$ (again a principal value, for the same $ξ = x$ reason) peels off the dial numbers: $A_{0} = α - \frac{1}{π} \int_{0}^{π} \frac{d z}{d x} d θ, A_{n} = \frac{2}{π} \int_{0}^{π} \frac{d z}{d x} cos n θ d θ .$

$A_{0}$ carries the tilt $α$ plus the average slope of the camber.
$A_{n}$ ( $n \geq 1$ ) carry the shape of the camber, weighted by $cos n θ$ .

Step 6 — Add up the spins: only two dials survive

WHAT. We integrate $γ$ to get the total circulation $Γ$ , and find that almost every dial cancels — only $A_{0}$ and $A_{1}$ remain.

WHY. When we sum $sin n θ$ over the full sweep $0 \to π$ , the wiggles for $n \geq 2$ have equal positive and negative humps that cancel exactly (this is orthogonality — different sine waves don't overlap on average). Only the flat-average pieces from $A_{0}$ and $A_{1}$ leave a residue.

PICTURE.

$Γ = \int_{0}^{c} γ d x = c V π (A_{0} + \frac{A _{1}}{2}) .$

$c V π$ is the size factor from converting the integral back to the straight ruler.
$A_{0} + \frac{A _{1}}{2}$ is the only surviving combination: tilt-and-average ( $A_{0}$ ) plus half of the first camber wiggle ( $A_{1}$ ).

The figure shows the tall $A_{0}, A_{1}$ bars surviving and the $A_{2}, A_{3}, \dots$ bars greyed out ("cancelled"). Multiply by $ρ V$ (Kutta–Joukowski, Step 2): $L^{'} = ρ V Γ = π ρ c V^{2} (A_{0} + \frac{A _{1}}{2})$

This is the fully general thin-airfoil lift.

Step 7 — The flat plate (degenerate case: no camber) and defining $c_{ℓ}$

WHAT. Switch off camber: the camber line is dead straight, so $\frac{d z}{d x} = 0$ everywhere. We also introduce a tidy dimensionless bookkeeping number, the lift coefficient $c_{ℓ}$ .

WHY. Always test a formula on the simplest input. A flat tilted plate is the baseline every cambered wing is measured against. And to compare wings of different size and speed on one graph we strip out $ρ$ , $V$ and $c$ into a pure number.

PICTURE.

With $\frac{d z}{d x} = 0$ : both camber integrals vanish, so $A_{0} = α$ and $A_{1} = 0$ . Then $L^{'} = π ρ c V^{2} α, c_{ℓ} = \frac{π ρ c V ^{2} α}{\frac{1}{2} ρ V ^{2} c} = 2 π α .$

No camber ⇒ no extra lift at zero tilt; the line passes through the origin.
The slope $\frac{d c _{ℓ}}{d α} = 2 π$ — this slope will never change no matter the camber.

The figure's $c_{ℓ}$ – $α$ line goes straight through $(0, 0)$ : tilt is the only source of lift.

Step 8 — Adding camber: what $z (x)$ and $β$ actually are

WHAT. Turn camber back on with a concrete, standard camber shape and finally define the camber parameter $β$ on it.

WHY. So far $β$ has been a promise. Here it earns its meaning: it is a number that measures how strongly the camber line bends, read straight off the shape of $z (x)$ .

PICTURE.

Now push this camber slope through the coefficient integrals of Step 5. Using the change of ruler $x = \frac{c}{2} (1 - cos θ)$ , the slope becomes $\frac{d z}{d x} (θ) = \frac{4 z _{m a x}}{c} (1 - \frac{2 x}{c}) = β (1 - (1 - cos θ)) = β cos θ .$

We substituted $\frac{2 x}{c} = 1 - cos θ$ , so $1 - \frac{2 x}{c} = cos θ$ .
The end-slope constant $\frac{4 z _{m a x}}{c}$ is exactly $β$ , so the whole slope is $β cos θ$ — a single clean cosine.

Feed $\frac{d z}{d x} = β cos θ$ into the coefficient formulas, using $\int_{0}^{π} cos θ d θ = 0$ and $\int_{0}^{π} cos θ cos θ d θ = \frac{π}{2}$ : $A_{0} = α - \frac{1}{π} \int_{0}^{π} β cos θ d θ = α - 0 = α,$ $A_{1} = \frac{2}{π} \int_{0}^{π} β cos θ cos θ d θ = \frac{2}{π} \cdot β \cdot \frac{π}{2} = β .$

$A_{0} = α$ : the average camber slope is zero (the front and back slopes are equal and opposite), so only tilt survives here.
$A_{1} = β$ : the first cosine mode of the camber is exactly $β$ .

Therefore the surviving combination of Step 6 is $A_{0} + \frac{A _{1}}{2} = α + \frac{β}{2} .$

Substitute into the boxed general lift of Step 6: $L^{'} = π ρ V^{2} c (α + \frac{β}{2}), c_{ℓ} = \frac{L ^{'}}{\frac{1}{2} ρ V ^{2} c} = 2 π α + π β .$

$α$ = tilt term, $\frac{β}{2}$ = camber term. They add; camber acts like extra tilt built into the wing.
On the graph, camber slides the whole line left by a constant $π β$ — the slope stays glued at $2 π$ .

Now every case, each visible on the figure:

Positive lift ( $α + \frac{β}{2} > 0$ ): the usual flying wing.
Zero lift ( $α + \frac{β}{2} = 0$ ): the zero-lift angle $α_{L = 0} = - \frac{β}{2}$ . A cambered wing ( $β > 0$ ) reaches this at a negative tilt — it still lifts when pointed slightly down.
Negative lift ( $α + \frac{β}{2} < 0$ ): tilt so far below $α_{L = 0}$ that even the camber cannot save it — the wing pushes down.

Recall Note on the parent's form

The parent note wrote the camber term as $\frac{2 β}{π}$ , which corresponds to a different definition of the camber parameter (a different normalisation of the mean-line slope). With the clean geometric definition $β = 4 z_{m a x} / c$ used here, the honest result of the integrals is $A_{0} + A_{1} /2 = α + β /2$ . The physics — camber shifts the line, never its $2 π$ slope — is identical; only the label on the knob differs.

The one-picture summary

WHAT. One figure that chains all eight steps: wing → vortex sheet → tangency + downwash → angle ruler → surviving dials → $Γ$ → lift → the two sign cases.

Trace the arrows: the solid wing becomes a sheet of spins; the no-through-flow rule plus Biot–Savart fix those spins; the angle ruler cleans up the integral; summing kills all dials but $A_{0}$ and $A_{1}$ ; Kutta–Joukowski turns their sum into $L^{'} = π ρ V^{2} c (α + \frac{β}{2}) .$

Recall Feynman: the whole walk in plain words

We start with a bent card in the wind. We can't easily do maths on a bent solid card, so we pretend it's a row of tiny spinning tops laid along its curve. The one rule the tops must obey: the air has to slide along the card, never through it. That rule says how much each top has to spin — and figuring out how hard one top blows on another is just the tiny-tornado rule (Biot–Savart). The straight ruler makes the sums ugly, so we re-label positions by an angle that walks around a half-circle; suddenly the mess turns into tidy sine waves. When we add up all the spinning to get the total swirl $Γ$ , every wiggle cancels except the first two — the tilt $α$ and one lump of camber. Multiply the total swirl by how fast and how heavy the air is, and out pops the lift. Tilt and curve are the two ways to make the card scoop harder; curve just pre-tilts the card, so a curved wing still lifts even when you point it slightly downward.

Recall Check yourself

Where does $Γ$ come from physically? ::: The Kutta condition forces smooth flow off the sharp trailing edge, which demands a definite bound circulation. Why only $A_{0}$ and $A_{1}$ in the lift? ::: Summing $sin n θ$ over $0 \to π$ cancels all $n \geq 2$ by orthogonality. What is $c_{ℓ}$ ? ::: The dimensionless lift coefficient $c_{ℓ} = L^{'} / (\frac{1}{2} ρ V^{2} c)$ . What does camber do to the $c_{ℓ}$ – $α$ line? ::: Shifts it left by a constant; the slope stays $2 π$ . Zero-lift angle for camber $β$ (with $β = 4 z_{m a x} / c$ )? ::: $α_{L = 0} = - β /2$ . What five assumptions underlie the whole theory? ::: Inviscid, incompressible, irrotational, two-dimensional, thin/small-angle.

Connections

Kutta–Joukowski theorem — turns $Γ$ into $L^{'}$ (Step 2, Step 6).
Circulation and bound vortices — why spin means lift (Step 2).
Kutta condition — picks the finite-trailing-edge spin (Step 5).
Biot–Savart law — the downwash integral (Step 4).
Glauert's integral and Fourier coefficients — the angle ruler and dials (Step 5–6).
Lift coefficient and angle of attack — where $c_{ℓ} = 2 π α + π β$ lives (Step 7–8).
Compressibility corrections (Prandtl–Glauert) — what changes when the air compresses.