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

This page drills the formula from the parent note through every kind of input it can face — positive and negative angles, positive and negative camber, zero camber, zero angle, the degenerate flat plate, a stall-warning limit, a real-world word problem, and a sneaky exam twist. Before any numbers, we build a map of all the cases so you never meet a scenario we didn't show.

Everything rests on the two boxed results from the parent, restated in plain words:

Definition Dynamic pressure — the "push" of moving air

Before Example 5 uses it, meet the quantity $q = \frac{1}{2} ρ V^{2}$ , called dynamic pressure. In plain words: a stream of air moving at speed $V$ carries kinetic energy $\frac{1}{2} m v^{2}$ per unit mass; $q$ is that energy density, and it is exactly the extra pressure the air can press onto a surface it slams into. Its units are Pa = N/m². Multiplying $q$ by an area gives a force. For a wing section, the reference area per metre of span is just the chord $c$ (a $c \times 1$ strip), so $q c$ has units (N/m²)(m) = N/m — a force per span. The lift coefficient $c_{ℓ}$ is defined as the dimensionless ratio that turns this reference push into the actual lift: $c_{ℓ} \equiv \frac{L ^{'}}{\frac{1}{2} ρ V ^{2} c} ⟹ L^{'} = q \frac{1}{2} ρ V^{2} c c_{ℓ} .$ This is not a new physics formula — it is the definition of $c_{ℓ}$ rearranged. We will use it in Example 5 to go from a $c_{ℓ}$ back to a force.

The scenario matrix

Every problem this formula can pose falls into one of these cells. The worked examples below are tagged with the cell they cover, and together they hit all of them.

Cell	Case class	What is special	Example
O	Flat plate, zero angle	$β = 0, α = 0 \Rightarrow L^{'} = 0$ (the trivial boundary)	Ex 0
A	Flat plate, positive $α$	$β = 0$ : only the tilt lifts	Ex 1
B	Cambered wing, positive $α$	Both knobs add	Ex 2
C	Negative angle of attack	Sign flips — lift can go down	Ex 3
D	Zero-lift angle (degenerate $L^{'} = 0$ )	Terms cancel exactly (flat plate ⇒ $α = 0$ ; cambered ⇒ shifted)	Ex 4
E	Zero angle, pure camber	$α = 0$ : lift from curve alone	Ex 5
E−	Negative camber ( $β < 0$ )	Curve reversed — shifts the line the other way	Ex 5b
F	Limiting / stall-warning behaviour	Formula grows without bound; where physics quits	Ex 6
G	Real-world word problem	Total lift, mass it can hold	Ex 7
H	Exam twist: solve backwards	Given $L^{'}$ , find the required $α$	Ex 8

The sign story ties cells O–E− together — worth a picture before we compute.

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

Figure s01 — Master sign-map: the $c_{ℓ}$ – $α$ lines for a flat plate (blue) and a cambered wing (green). Alt text: two parallel straight lines of slope $2 π$ ; the blue line through the origin, the green line shifted up-left by $4 β$ , with markers for Cell O (origin), Cell D (green zero-crossing at negative angle), and Cell E (green value on the vertical axis).

Read the figure carefully — it is the master map for cells O through E−:

The horizontal axis is the angle of attack $α$ in degrees; the vertical axis is the lift coefficient $c_{ℓ}$ .
The blue line is the flat plate, $c_{ℓ} = 2 π α$ : it passes exactly through the origin, so at $α = 0$ it gives $c_{ℓ} = 0$ (that is Cell O, the trivial boundary).
The green line is a positively cambered wing ( $β = 0.05$ ), $c_{ℓ} = 2 π α + 4 β$ : the same slope $2 π$ but slid up and to the left by the constant $4 β$ .
The red dot where the green line crosses zero is the cambered zero-lift angle (Cell D), sitting at a small negative $α$ .
The orange square on the vertical axis is the green line's value at $α = 0$ — pure-camber lift with no tilt (Cell E).
Everything left of a line's zero-crossing lies below the axis: negative (downward) lift (Cell C). Notice both lines are parallel: camber shifts, it never tilts. A negative camber ( $β < 0$ , Cell E−) would slide the line the opposite way, down and to the right.

Worked examples

Baseline flow for most examples: $ρ = 1.225$ kg/m³, $V = 50$ m/s, $c = 1.5$ m. Then the constant lump $π ρ V^{2} c = π (1.225) (5 0^{2}) (1.5) = 14432 N/m$ appears again and again — call it $K = 14432$ N/m, so $L^{'} = K (α + \frac{2 β}{π})$ .

Cell O — flat plate, zero angle (the trivial boundary)

Worked example Example 0 — nothing tilted, nothing curved

Flat plate ( $β = 0$ ) sitting exactly level in the wind ( $α = 0$ ). Find $L^{'}$ .

Forecast: any lift at all?

Both knobs are off: $α = 0$ and $β = 0$ , so the bracket $(α + \frac{2 β}{π}) = 0$ . Why this step? This is the boundary where both lift sources vanish — the reference "do nothing" state every other case is measured against.
$L^{'} = K \times 0 = 0 N/m$ . Why this step? Multiplying the constant $K$ by a zero bracket completes the formula and confirms the physical expectation — we carry the multiplication through explicitly so the zero is derived, not just asserted.

Verify: on the figure this is the origin, where the blue line crosses zero ✓. A flat plate aligned with the flow produces no net turning of the air, hence no lift — consistent with symmetry (nothing distinguishes up from down) ✓.

Cell A — flat plate, positive angle

Cell B — cambered wing, positive angle

Worked example Example 2 — both knobs on

Same flow, now $β = 0.05$ rad, still $α = 5^{\circ}$ .

Forecast: more or less than Ex 1, and by roughly how much?

Convert: $α = 5 \times π /180 = 0.08727$ rad. Why this step? Same radian rule as Ex 1 — the linearisation $sin α \approx α$ demands radians, so we convert before any arithmetic.
Camber term: $\frac{2 β}{π} = \frac{2 ( 0.05 )}{π} = 0.03183$ . Why this step? Camber adds an effective extra angle, not a slope change.
Sum inside brackets: $0.08727 + 0.03183 = 0.11910$ . Why this step? We add the tilt and camber contributions first because the formula multiplies $K$ by their sum; combining them now keeps the final multiplication a single clean step.
$L^{'} = 14432 \times 0.11910 = 1719 N/m$ . Why this step? Multiplying the benchmark constant $K$ by the combined bracket produces the total lift for the cambered wing — the number we compare against the flat plate.

Verify: ratio to flat plate $= 1719/1259 = 1.37$ , a $\sim 37%$ gain — matches the parent's " $\sim 36%$ " ✓. The gain is a shift, not a steeper slope (Cell A and B share the same $2 π$ slope).

Cell C — negative angle of attack

Worked example Example 3 — pointing the nose down

Flat plate ( $β = 0$ ), $α = - 3^{\circ}$ . Find $L^{'}$ and its direction.

Forecast: what sign will the lift have?

$α = - 3 \times π /180 = - 0.05236$ rad. Why this step? Negative tilt is a genuine negative number in the formula, and it must still be in radians for the linearisation to hold.
$L^{'} = K α = 14432 \times (- 0.05236) = - 755.6 N/m$ . Why this step? Nothing special — the linear formula just runs into negative $α$ .
The minus sign means the aerodynamic force points downward (a "down-force"). Why this step? We interpret the negative numeric result physically — reading the sign is the whole point of this case, since it tells us the wing pushes down, not up.

Verify: magnitude equals Ex 1 scaled by $3/5$ : $1259 \times (3/5) = 755$ N/m ✓. On the figure, this point sits left of the origin on the flat-plate (blue) line, below the axis ✓.

Cell D — zero-lift angle (exact cancellation)

Worked example Example 4 — the balance point (both the flat-plate and cambered case)

(a) Flat plate ( $β = 0$ ): at what $α$ is $L^{'} = 0$ ? (b) Cambered wing $β = 0.05$ rad: at what $α$ is $L^{'} = 0$ ?

Forecast: for each, positive, zero, or negative angle?

Set $L^{'} = 0$ : since $K \neq = 0$ , we need $α + \frac{2 β}{π} = 0$ . Why this step? A product is zero only when the bracket is zero.
(a) Flat plate: with $β = 0$ the bracket is just $α$ , so $α_{L = 0} = 0$ . Why this step? A symmetric flat plate has no built-in camber offset, so its zero-lift line passes through the origin — consistent with Cell O.
(b) Cambered: $α_{L = 0} = - \frac{2 β}{π} = - \frac{2 ( 0.05 )}{π} = - 0.03183$ rad $= - 1.82 4^{\circ}$ . Why this step? We solve the zero-bracket condition for $α$ and then convert to degrees, because the answer to a "what angle" question is most useful quoted in degrees even though it was derived in radians.

Verify: (a) plug $α = 0, β = 0$ : bracket $= 0$ ⇒ $L^{'} = 0$ ✓ (matches Cell O). (b) plug back — $α + 2 β / π = - 0.03183 + 0.03183 = 0$ ✓. So the zero-lift boundary is $α = 0$ for a flat plate and shifts negative by exactly the camber amount — one consistent rule, two cases. On the figure, (a) is the blue line's origin crossing and (b) is the red dot on the green line.

Cell E — zero angle, pure camber

Worked example Example 5 — lift with no tilt

Cambered wing $β = 0.05$ rad, $α = 0$ . Find $c_{ℓ}$ and $L^{'}$ .

Forecast: can a wing lift while sitting level in the wind?

$c_{ℓ} = 2 π (0) + 4 β = 4 (0.05) = 0.2$ . Why this step? With $α = 0$ only the camber term survives in $c_{ℓ} = 2 π α + 4 β$ .
$L^{'} = \frac{1}{2} ρ V^{2} c c_{ℓ} = 0.5 (1.225) (2500) (1.5) (0.2) = 459.375 N/m$ . Why this step? We use the definition of $c_{ℓ}$ from the callout above, rearranged to $L^{'} = q c c_{ℓ}$ with $q = \frac{1}{2} ρ V^{2}$ the dynamic pressure — this converts the dimensionless lift number into an actual force per span.
Cross-check with the direct form: $L^{'} = K \cdot \frac{2 β}{π} = 14432 (0.03183) = 459.4$ N/m ✓. Why this step? Recomputing the same lift by the other route (direct $K$ -formula instead of the $c_{ℓ}$ definition) is a self-consistency test — agreement confirms we made no slip.

Verify: two independent routes give the same $\approx 459$ N/m ✓. Yes — camber alone lifts at zero tilt; that is exactly why real wings are curved. See Lift coefficient and angle of attack for the full curve.

Cell E− — negative camber

Worked example Example 5b — reflex camber (the mirror of Cell E)

A tail surface with negative camber $β = - 0.05$ rad, sitting level ( $α = 0$ ). Find $c_{ℓ}$ and $L^{'}$ .

Forecast: if positive camber lifted up, what does negative camber do?

$c_{ℓ} = 2 π (0) + 4 β = 4 (- 0.05) = - 0.2$ . Why this step? $β$ enters linearly, so a sign flip on the curve flips the sign of its lift contribution.
$L^{'} = \frac{1}{2} ρ V^{2} c c_{ℓ} = 0.5 (1.225) (2500) (1.5) (- 0.2) = - 459.375 N/m$ . Why this step? Same $c_{ℓ}$ -definition as Ex 5; the negative $c_{ℓ}$ carries straight through to a negative (downward) force.

Verify: exact mirror of Ex 5: same magnitude, opposite sign ✓. On the figure this would be the green line reflected through the origin — slid down and to the right, giving a positive zero-lift angle. Reflex/negative camber is used on tailless aircraft to trim (push down at the rear), showing the formula covers the full $β$ sign range.

Cell F — limiting behaviour / where the theory quits

Worked example Example 6 — pushing

$α$ toward stall Flat plate. The formula says $c_{ℓ} = 2 π α$ forever. Compute $c_{ℓ}$ at $α = 5^{\circ}, 1 5^{\circ}, 2 5^{\circ}$ and mark where physics disagrees.

Forecast: does lift really keep climbing linearly to $9 0^{\circ}$ ?

$c_{ℓ} (5^{\circ}) = 2 π (0.08727) = 0.548$ . Why this step? $5^{\circ}$ is a small angle where the linear theory is trustworthy — this is our reliable anchor point below stall.
$c_{ℓ} (1 5^{\circ}) = 2 π (0.26180) = 1.645$ . Why this step? $1 5^{\circ}$ sits right at the edge of the real stall region — the theory here already predicts more lift ( $1.645$ ) than a real airfoil can hold ( $\approx 1.5$ ), so this number is the first one we should distrust.
$c_{ℓ} (2 5^{\circ}) = 2 π (0.43633) = 2.742$ . Why this step? Well past stall the linear theory has no upper bound — it predicts ever-rising lift, which is physically impossible.
But real airfoils stall near $c_{ℓ} \approx 1.5$ (around $12 - 1 5^{\circ}$ ): flow separates, the Kutta condition assumption breaks, and measured lift collapses. Why this step? The derivation assumed small angles and attached flow; beyond stall those assumptions fail.

Verify: the numbers grow linearly (ratio $0.548 : 1.645 : 2.742 = 1 : 3 : 5$ , matching $5 : 15 : 25$ ) ✓. The limit to remember: trust the straight line only for small $α$ ; the physical ceiling is stall, not the formula.

Figure s02 — Cell F stall warning: the straight linear-theory line (blue) climbs without limit, while the real airfoil (red dashed) peaks near $c_{ℓ} \approx 1.5$ and drops at stall. Alt text: a rising straight blue line and a red dashed curve that follows it up to about $1 3^{\circ}$ , then bends over and falls; the shaded gap after stall marks where the formula over-predicts lift.

The figure shows the straight theory line rocketing past the real (dashed) curve, which peaks and drops at stall — the shaded region is where the formula lies.

Cell G — real-world word problem

Worked example Example 7 — how much can this wing hold up?

A light aircraft wing: span $b = 10$ m, chord $c = 1.5$ m, cruising at $V = 50$ m/s in air of $ρ = 1.225$ kg/m³, angle of attack $α = 4^{\circ}$ , camber $β = 0.04$ rad. Find total lift and the mass it supports ( $g = 9.81$ m/s²).

Forecast: enough to lift a car? a person? a plane?

Convert: $α = 4 \times π /180 = 0.06981$ rad. Why this step? As in every example, the linearised formula needs the angle in radians before we can add it to the camber term.
Camber term $\frac{2 β}{π} = \frac{2 ( 0.04 )}{π} = 0.02546$ . Why this step? $β$ is already given in radians (a slope, dimensionless), so no conversion — but we must form $2 β / π$ to match the units of $α$ before adding.
Per-span lift $L^{'} = K (α + 2 β / π) = 14432 (0.06981 + 0.02546) = 14432 (0.09527) = 1375 N/m$ . Why this step? Combining the two knobs and multiplying by $K$ gives the force on each metre of span — the per-strip lift we will replicate across the span.
Total lift over the span: $L = L^{'} \times b = 1375 \times 10 = 13750$ N. Why this step? $L^{'}$ is force per metre; multiplying by the span $b = 10$ m adds up all ten $1$ -metre strips (strip theory) to get the whole-wing force.
Supported mass $m = L / g = 13 750/9.81 = 1402 kg$ . Why this step? In level flight lift balances weight $m g$ , so dividing the lift force by $g$ inverts $L = m g$ to reveal the mass the wing can hold aloft.

Verify: units — (N/m)(m) = N ✓, N ÷ (m/s²) = kg ✓. About 1.4 tonnes — a plausible small aircraft ✓. (This uses strip theory; true 3-D wings lose some lift to tip vortices — see Circulation and bound vortices.)

Cell H — exam twist: solve it backwards

Worked example Example 8 — find the required angle

Baseline flow ( $K = 14432$ N/m), flat plate ( $β = 0$ ). The designer needs exactly $L^{'} = 2000$ N/m. What angle of attack (in degrees) is required?

Forecast: bigger or smaller than the $5^{\circ}$ that gave 1259 N/m?

Rearrange $L^{'} = K α$ for $α$ : $α = L^{'} / K = 2000/14 432 = 0.13858$ rad. Why this step? We invert the formula because the unknown is now the angle, not the force — dividing the target lift by the constant $K$ isolates $α$ .
Convert: $α = 0.13858 \times 180/ π = 7.9 4^{\circ}$ . Why this step? Answers about tilt are quoted in degrees, so we convert the radian result back at the very end for a human-readable design angle.

Verify: plug $α = 0.13858$ rad back: $L^{'} = 14432 \times 0.13858 = 2000$ N/m ✓. Larger than $5^{\circ}$ (as expected, since we want more lift) and still comfortably below stall from Cell F ✓.

Recall Rapid self-test across all cells

Ex 0 flat plate at zero angle ::: $0$ N/m Ex 1 flat-plate lift at 5° ::: $1259$ N/m Ex 2 cambered lift ::: $1719$ N/m ( $\sim 37%$ more) Ex 3 lift at $α = - 3^{\circ}$ ::: $- 756$ N/m (down-force) Ex 4a flat-plate zero-lift angle ::: $0^{\circ}$ Ex 4b cambered zero-lift angle for $β = 0.05$ ::: $- 1.8 2^{\circ}$ Ex 5 pure-camber lift at $α = 0$ ::: $459$ N/m, $c_{ℓ} = 0.2$ Ex 5b negative camber ::: $- 459$ N/m, $c_{ℓ} = - 0.2$ Ex 6 where the line lies ::: beyond stall ( $\sim 12$ – $1 5^{\circ}$ ) Ex 7 mass supported ::: $\approx 1402$ kg Ex 8 angle for 2000 N/m ::: $7.9 4^{\circ}$

Connections

Kutta–Joukowski theorem — the $L^{'} = ρ V Γ$ these numbers ultimately come from.
Lift coefficient and angle of attack — the $c_{ℓ}$ – $α$ line that Cells C–E live on.
Kutta condition — the assumption that fails in Cell F (stall).
Circulation and bound vortices — why 3-D span multiplication (Cell G) is only approximate.
Compressibility corrections (Prandtl–Glauert) — how these numbers change at high $V$ .