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

Symbols reused here (all built in the parent): $L^{'}$ = lift per unit span (N/m), $Γ$ = circulation, $α$ = angle of attack in radians, $β$ = camber slope parameter, $c$ = chord, $V$ = freestream speed, $ρ$ = density, $c_{ℓ}$ = section lift coefficient, $γ (x)$ = vortex-sheet strength, $A_{0}, A_{1}, \dots$ = Glauert Fourier coefficients.

True or false — justify

Camber makes the lift-curve steeper (bigger

d c_{ℓ} / d α

False. The slope is fixed at

2 π

per radian; camber adds the constant

4 β

c_{ℓ} = 2 π α + 4 β

, sliding the whole line left, never tilting it.

A symmetric (uncambered) airfoil produces zero lift at

α = 0

True. With

d z / d x = 0

we get

A_{0} = α

A_{1} = 0

, so

c_{ℓ} = 2 π α

, which is

0

when

α = 0

— no camber means no lift without tilt.

A cambered airfoil produces zero lift at

α = 0

False. Its zero-lift angle is

α_{L = 0} = - 2 β / π < 0

, so at

α = 0

it still lifts; you must tilt it slightly down to kill the lift.

Doubling the freestream speed

V

doubles the lift per unit span.

False.

L^{'} = ρ V Γ

but

Γ

itself is proportional to

V

(see

Γ = π c V (A_{0} + A_{1} /2)

), so

L^{'} \propto V^{2}

— doubling

V

quadruples

L^{'}

The lift depends on the full curved surface of the airfoil, so we cannot replace it with a straight line of vortices.

False. Thin airfoil theory places the vortex sheet on the mean camber line (nearly straight for a thin wing); it reproduces the same outer flow because the airfoil barely disturbs the stream.

All the Fourier coefficients

A_{0}, A_{1}, A_{2}, A_{3}, \dots

contribute to the lift.

False. Only

A_{0}

and

A_{1}

survive the circulation integral; the

sin n θ

terms for

n \geq 2

integrate to zero by orthogonality, so higher coefficients affect the moment, not the lift.

Kutta–Joukowski

L^{'} = ρ V Γ

is an approximation valid only for thin airfoils.

False. Kutta–Joukowski is exact for any 2-D body in inviscid incompressible flow with circulation

Γ

; thin-airfoil theory is only the approximate way we estimate

Γ

If the circulation

Γ

is zero, the section still produces some lift from the tilt of the plate.

False.

L^{'} = ρ V Γ

, so

Γ = 0

gives exactly zero lift regardless of

α

— no circulation, no lift, full stop.

The Kutta condition is an extra physical assumption we impose, not something the equations force.

True. Inviscid theory alone allows infinitely many circulations; the Kutta condition (finite velocity leaving the sharp trailing edge) selects the single physical one, mimicking what viscosity actually does.

Spot the error

"Because we tilt the wing by

α = 5^{\circ}

, we plug

α = 5

into

c_{ℓ} = 2 π α

Error: units. The linearisation used

sin α \approx α

in radians; you must convert

5^{\circ} = 0.0873

rad first, otherwise the lift is inflated by a factor of

\approx 57

"Camber contributes

\frac{2 β}{π}

to the lift, and since

β

is an angle we should also express it in degrees like

α

Error: consistency.

β

is a slope/angle parameter entering the same linearised algebra, so it too must be the radian-equivalent quantity; mixing degrees for one and radians for the other corrupts the sum

α + 2 β / π

"Air over the top travels a longer path and must meet the bottom air at the trailing edge, so it speeds up — that's the lift."

Error: the equal-transit-time myth. Top and bottom particles do not arrive together; the real cause is the bound circulation set by the Kutta condition, which genuinely makes the upper flow faster.

"The vortex sheet strength is

γ (θ) = 2 V (A_{0} \frac{1 + c o s θ}{s i n θ} + \sum A_{n} sin n θ)

, and at the trailing edge this blows up."

Error. At the trailing edge

θ = π

\frac{1 + c o s θ}{s i n θ} \to 0

(both numerator and

sin θ \to 0

, ratio

\to 0

), so

γ

stays finite there — that is precisely how the form encodes the Kutta condition.

"Since

L^{'} = ρ V Γ

has

ρ

in it, in a vacuum the wing would still lift a little from its own spin."

Error.

ρ = 0

in vacuum gives

L^{'} = 0

; there is no fluid to carry momentum downward, so no lift no matter how the "sheet" spins.

L^{'}

is the total lift of the whole wing."

Error: the prime.

L^{'}

is lift per unit span (N/m). Total lift needs

L^{'} \times

span (times a correction for finite-wing 3-D effects, which this 2-D theory ignores).

"To find

Γ

we integrate the downwash

w (x)

along the chord."

Error.

Γ = \int_{0}^{c} γ d x

integrates the vortex-sheet strength

γ

, not the downwash

w

;

w

appears only in the flow-tangency condition used to find

γ

Why questions

Why does the derivation put the vortices on the camber line instead of the actual airfoil surface?

A thin airfoil barely perturbs the flow, so to first order the mean line and the surface are indistinguishable to the outer stream; using the line makes the integral 1-D and analytically solvable.

Why do only

A_{0}

and

A_{1}

appear in the lift, while

A_{2}, A_{3}, \dots

vanish?

The circulation integral

\int_{0}^{π} γ d θ

meets the orthogonality of

sin n θ

over

[0, π]

; every

n \geq 2

term integrates to zero, leaving only the

A_{0}

and

A_{1}

contributions.

Why must the angle of attack be small for this theory?

We linearised

sin α \approx α

and treated the sheet as flat; large angles break both approximations and, physically, cause flow separation (stall) that inviscid theory cannot capture.

Why is the lift-curve slope universally

2 π

regardless of camber shape?

The slope comes only from the

A_{0} = α - (camber integral)

term's

α

-dependence; camber changes the constant part of

A_{0}

and

A_{1}

but not how they respond to

α

, so

d c_{ℓ} / d α = 2 π

always.

Why does camber let a wing lift at zero angle of attack?

Curvature deflects air downward even with no tilt, adding the constant

4 β

c_{ℓ}

; the aircraft can then cruise level while still generating upward force.

Why do we need the Kutta condition at all — isn't the flow uniquely determined?

Pure inviscid theory admits a whole family of solutions differing by circulation; the Kutta condition (smooth departure at the sharp edge) picks the one real viscosity would enforce, fixing

Γ

uniquely.

Why does the Biot–Savart law enter the derivation?

We need the downwash each vortex element induces at other chord points to enforce flow-tangency; Biot–Savart (here the 2-D

v = Γ/2 π r

) is exactly the tool that gives an induced velocity from a vorticity distribution.

Why is

c_{ℓ}

dimensionless while

L^{'}

has units?

c_{ℓ} = L^{'} / (\frac{1}{2} ρ V^{2} c)

divides the lift by dynamic pressure times chord, cancelling all units so the same

c_{ℓ}

describes geometrically similar wings at any scale or speed.

Edge cases

What is

L^{'}

for a symmetric flat plate at

α = 0

Zero:

A_{0} = 0

A_{1} = 0

, so

Γ = 0

and

L^{'} = ρ V \cdot 0 = 0

— a flat plate aligned with the flow makes no lift.

What is the lift at the zero-lift angle

α = - 2 β / π

for a cambered wing?

Exactly zero — that angle is defined as the one where the

α

term cancels the camber term, so

c_{ℓ} = 0

even though the wing is curved.

What happens to the formula at

ρ = 0

(vacuum)?

L^{'} = π ρ V^{2} c (\dots) = 0

; with no air there is nothing to push down, so lift vanishes — the formula correctly degenerates.

What happens as the chord

c \to 0

L^{'} \to 0

linearly in

c

; an infinitely thin-chord "wing" has no surface to turn air and produces no lift per span.

If we make

α

negative but

β

large enough that

α + 2 β / π > 0

, does the wing still lift up?

Yes — the sum determines the sign of

Γ

; a strongly cambered wing can lift upward even at a modest negative geometric angle of attack.

At the trailing edge (

θ = π

) what is the vortex-sheet strength

γ

It is finite (in fact tends to zero); the Glauert form was chosen precisely so

γ

does not diverge there, enforcing the Kutta condition.

What does the theory predict just before stall (large

α

It keeps predicting a linear

c_{ℓ} = 2 π α + 4 β

with no upper limit — a known failure, since real flow separates and lift drops; the inviscid model has no mechanism for stall.

If two airfoils have the same

α + 2 β / π

but different camber shapes, do they have the same lift?

Yes, the same

L^{'}

, because lift depends only on

A_{0} + A_{1} /2

; the different shapes differ in pitching moment (from higher

A_{n}

), not in lift.

Recall One-line summary of the traps

Slope is always $2 π$ (camber only shifts); always radians; lift is $ρ V Γ$ so $\propto V^{2}$ and vanishes with $ρ$ or $Γ$ ; only $A_{0}, A_{1}$ lift; the Kutta condition is an imposed physical choice, not an equation output.

Connections

Parent: lift per unit span — the results these traps probe.
Kutta–Joukowski theorem — why $L^{'} = ρ V Γ$ is exact, not approximate.
Kutta condition — the imposed choice several traps hinge on.
Circulation and bound vortices — why $Γ = 0$ means zero lift.
Biot–Savart law — the downwash tool.
Glauert's integral and Fourier coefficients — why only $A_{0}, A_{1}$ survive.
Lift coefficient and angle of attack — the fixed $2 π$ slope.
Compressibility corrections (Prandtl–Glauert) — what breaks when $V$ is not small.