Lift — Kutta-Joukowski theorem L = ρV∞Γ: Worked examples

This page is the drill hall for the parent topic. We take the single formula $L^{'} = ρ V_{\infty} Γ$ and push it through every case it can throw at you — positive and negative circulation, zero inputs, degenerate flows, limiting behaviour, a real-world word problem, and an exam twist. By the end you should never meet a scenario you have not already seen worked.

The scenario matrix

Every problem this theorem produces lands in one of these cells. Each worked example is tagged with the cell it covers.

#	Cell class	What is being tested	Worked in
A	Forward, $Γ > 0$	plain plug-in, upward lift	Ex 1
B	Sign flip, $Γ < 0$	negative/downward lift, direction	Ex 2
C	Zero input	$Γ = 0$ or $V_{\infty} = 0$ → zero lift (d'Alembert)	Ex 3
D	$Γ$ from geometry	thin-airfoil $Γ = π V_{\infty} c α$	Ex 4
E	Limiting/scaling	how $L^{'}$ scales when $V_{\infty}$ doubles	Ex 5
F	Real-world word problem	total lift over a span, back-solve $Γ$	Ex 6
G	Non-airfoil source ( $Γ$ from spin)	Magnus, sideways force	Ex 7
H	Exam twist / vector direction	$L^{'} = ρ V_{\infty} \times Γ$ , is lift vertical?	Ex 8

Constants used throughout: sea-level air $ρ = 1.23 kg/m^{3}$ unless stated.

Cell A — Forward flight, positive circulation

Cell B — Negative circulation, downward force

Worked example Example 2 — flip the swirl, flip the lift

Same air and speed as Ex 1 ( $ρ = 1.23$ , $V_{\infty} = 40 m/s$ ), but now the flow swirls counter-clockwise: $Γ = - 25 m^{2} / s$ . What is $L^{'}$ and which way does it point?

Forecast: the number should be the same size as Ex 1 — but which sign?

Step 1. Same theorem, $L^{'} = ρ V_{\infty} Γ$ . Why this step? The theorem is linear in $Γ$ ; flipping the sign of $Γ$ flips the sign of $L^{'}$ and nothing else.

Step 2. $L^{'} = 1.23 \times 40 \times (- 25) = - 1230 N/m$ . Why this step? A negative result means the force points opposite to the "up" of the positive- $Γ$ convention — i.e. downward here (this is exactly the case of an upside-down wing, or an inverted race-car wing producing downforce).

Step 3. Report magnitude and direction: $∣ L^{'} ∣ = 1230 N/m$ , pointing down.

Verify: $L^{'} (Γ) = - L^{'} (- Γ)$ , so Ex 1 and Ex 2 have equal magnitude, opposite sign ✓. Race-car front wings really do use this to press tyres onto the track.

Cell C — Zero inputs and d'Alembert's paradox

Worked example Example 3 — the two ways to get zero lift

(a) A perfectly streamlined shape in ideal flow generates no circulation: $Γ = 0$ . (b) A wing sitting still in a wind tunnel that is switched off: $V_{\infty} = 0$ . Find $L^{'}$ in each case.

Forecast: both give zero — but the reasons are different. Can you say why before reading on?

Step 1 (case a). $L^{'} = ρ V_{\infty} \cdot 0 = 0$ . Why this step? With no swirl there is no top/bottom speed difference, so Bernoulli gives no pressure difference. This is exactly d'Alembert's paradox: ideal flow with no circulation produces neither drag nor lift.

Step 2 (case b). $L^{'} = ρ \cdot 0 \cdot Γ = 0$ . Why this step? No oncoming momentum to deflect. Circulation alone (a stationary vortex) transports nothing past the wing, so there is no net force from freestream deflection. You must fly for the swirl to become lift.

Step 3. Both cases: $L^{'} = 0$ .

Verify: The formula is a product of three factors; if any factor is $0$ , the product is $0$ ✓. Physically distinct causes, identical zero result.

Cell D — Circulation forced by geometry (thin airfoil)

Worked example Example 4 — you're given the shape, not

$Γ$ A thin airfoil has chord $c = 1.5 m$ , flies at $V_{\infty} = 55 m/s$ , angle of attack $α = 4^{\circ}$ , air $ρ = 1.23$ . Find $Γ$ , then $L^{'}$ .

Forecast: here $Γ$ is not handed to you — the Kutta condition fixes it. Guess whether $L^{'}$ lands near a thousand or several thousand N/m.

Step 1. Convert the angle: $α = 4^{\circ} = 4 \times \frac{π}{180} = 0.0698 rad$ . Why this step? The thin-airfoil formula $Γ = π V_{\infty} c α$ is built from small-angle calculus, so $α$ must be in radians (degrees would be wrong by a factor $180/ π \approx 57$ ).

Step 2. $Γ = π V_{\infty} c α = π \times 55 \times 1.5 \times 0.0698 = 18.09 m^{2} / s$ . Why this step? The Kutta condition (smooth flow off the sharp trailing edge) picks exactly this $Γ$ — no more, no less.

Step 3. $L^{'} = ρ V_{\infty} Γ = 1.23 \times 55 \times 18.09 = 1224 N/m$ .

Verify: Cross-check via the lift coefficient. Thin-airfoil says $c_{ℓ} = 2 π α = 2 π (0.0698) = 0.4386$ . Then $L^{'} = \frac{1}{2} ρ V_{\infty}^{2} c c_{ℓ} = 0.5 \times 1.23 \times 5 5^{2} \times 1.5 \times 0.4386 = 1224 N/m$ ✓ — the two roads meet.

Cell E — Scaling and limiting behaviour

Figure — Lift — Kutta-Joukowski theorem L = ρV∞Γ

Worked example Example 5 — what happens when you double the speed?

A wing gives $L^{'} = 1224 N/m$ at $V_{\infty} = 55 m/s$ (from Ex 4, same airfoil and $α$ ). You now fly at $V_{\infty} = 110 m/s$ . Predict the new $L^{'}$ without re-plugging every number.

Forecast: the K–J theorem is linear in $V_{\infty}$ — so is the answer just $2 \times$ ? Look at the figure before deciding.

Step 1. Note that for a fixed shape and angle, $Γ = π V_{\infty} c α \propto V_{\infty}$ . Why this step? The circulation itself grows with speed (faster flow swirls harder off the same trailing edge). This is the "hidden second $V_{\infty}$ " the parent note warned about.

Step 2. So $L^{'} = ρ V_{\infty} Γ \propto V_{\infty} \times V_{\infty} = V_{\infty}^{2}$ . Why this step? One factor of $V_{\infty}$ is explicit, one is buried in $Γ$ . Multiplying, $L^{'}$ scales as the square of speed — even though the theorem is linear in each factor separately.

Step 3. Double the speed → $L^{'}$ grows by $2^{2} = 4$ : new $L^{'} = 4 \times 1224 = 4896 N/m$ .

Verify: Direct compute. New $Γ = π \times 110 \times 1.5 \times 0.0698 = 36.17 m^{2} / s$ ; $L^{'} = 1.23 \times 110 \times 36.17 = 4894 N/m$ (rounding) ✓. The red $V_{\infty}^{2}$ curve in the figure is what you feel on takeoff: a small speed gain buys a big lift gain.

Cell F — Real-world word problem (back-solve $Γ$ )

Worked example Example 6 — how much swirl holds a light plane up?

A small aircraft has mass $m = 1200 kg$ , wingspan $b = 11 m$ , cruises at $V_{\infty} = 60 m/s$ in air $ρ = 1.23$ . Assuming uniform lift along the span, what circulation $Γ$ does each metre of wing carry?

Forecast: total lift must equal weight. Estimate the weight in newtons first ( $\approx$ 12 000 N), then guess $Γ$ .

Step 1. Weight to support: $W = m g = 1200 \times 9.81 = 11772 N$ . Why this step? In steady level flight, total lift $L = W$ . This anchors the whole problem in a real force.

Step 2. Lift per unit span needed: $L^{'} = \frac{L}{b} = \frac{11 772}{11} = 1070.2 N/m$ . Why this step? K–J gives $L^{'}$ (N/m), so we must convert total lift to per-span first — the classic "per unit span" trap.

Step 3. Invert the theorem for $Γ$ : $Γ = \frac{L ^{'}}{ρ V _{\infty}} = \frac{1070.2}{1.23 \times 60} = 14.50 m^{2} / s$ . Why this step? We know $L^{'}$ , $ρ$ , $V_{\infty}$ ; the only unknown is $Γ$ , so solve algebraically.

Verify: Rebuild forward: $L^{'} = 1.23 \times 60 \times 14.50 = 1070 N/m$ ; $\times b = 1070 \times 11 = 11770 N \approx W$ ✓. About $14.5 m^{2} / s$ of swirl per metre keeps the plane airborne.

Cell G — Circulation from spin (Magnus effect)

Worked example Example 7 — the curving football

A football of no aerofoil shape at all is kicked at $V_{\infty} = 25 m/s$ with heavy spin that drags air into a circulation $Γ = 8 m^{2} / s$ around it. Find the sideways force per unit length.

Forecast: the ball is round, not wing-shaped — does K–J still apply? (It does.)

Step 1. Apply the same theorem: $F^{'} = ρ V_{\infty} Γ$ . Why this step? K–J cares only about $ρ$ , $V_{\infty}$ , $Γ$ — not whether the swirl comes from shape or spin. This is the Magnus effect, a direct cousin of wing lift.

Step 2. $F^{'} = 1.23 \times 25 \times 8 = 246 N/m$ . Why this step? Same product structure; only the source of $Γ$ differs.

Step 3. Direction: perpendicular to the flight path, on the side where spin and airflow agree — that is the "bend" of a curved free kick.

Verify: Units → N/m ✓. Halving the spin ( $Γ \to 4$ ) would halve the curve to $123 N/m$ ✓, matching everyday intuition that more spin = more bend.

Cell H — Exam twist: is the lift really vertical?

Worked example Example 8 — direction, not magnitude

A wing flies with freestream $V_{\infty}$ pointing horizontally at $30 m/s$ , circulation $Γ = 20 m^{2} / s$ , air $ρ = 1.23$ . The exam asks: (i) magnitude of $L^{'}$ , (ii) is the lift straight up, and (iii) what if the freestream came in tilted $1 0^{\circ}$ nose-down?

Forecast: we all "know" lift is vertical — but is it defined relative to the ground or the wind?

Step 1. Magnitude: $L^{'} = ρ V_{\infty} Γ = 1.23 \times 30 \times 20 = 738 N/m$ . Why this step? The scalar form gives magnitude regardless of orientation.

Step 2. Direction from the vector form $L^{'} = ρ V_{\infty} \times Γ$ . Why this step? The cross product means lift is always perpendicular to $V_{\infty}$ , not to the ground.

Step 3. (ii) With horizontal freestream, "perpendicular to the wind" is straight up, so lift is vertical — true here by coincidence of geometry. (iii) If the freestream tilts $1 0^{\circ}$ nose-down, the lift tilts with it to stay perpendicular: it points $1 0^{\circ}$ away from vertical. The magnitude $738 N/m$ is unchanged.

Verify: The cross-product magnitude $∣ ρ V_{\infty} \times Γ ∣ = ρ V_{\infty} Γ$ (they are perpendicular in 2-D), $= 738 N/m$ ✓. See the figure: rotating $V_{\infty}$ rotates the lift arrow by the same angle, always $9 0^{\circ}$ apart.

Active Recall

Recall If

$Γ$ flips sign, what happens to $L^{'}$ ? Question ::: A wing gives $+ 1230 N/m$ with clockwise $Γ$ . What does counter-clockwise $- Γ$ give? Answer ::: $- 1230 N/m$ — same magnitude, downward (downforce). $L^{'}$ is linear in $Γ$ .

Recall Two ways to get zero lift?

Question ::: Name the two independent inputs that each force $L^{'} = 0$ . Answer ::: $Γ = 0$ (no swirl — d'Alembert's paradox) or $V_{\infty} = 0$ (no flight, no momentum to deflect).

Recall Why does doubling speed quadruple lift for a fixed wing?

Question ::: The theorem is linear in $V_{\infty}$ ; why is observed lift $\propto V_{\infty}^{2}$ ? Answer ::: Because $Γ = π V_{\infty} c α$ also grows with $V_{\infty}$ , so $L^{'} = ρ V_{\infty} Γ \propto V_{\infty}^{2}$ .

Recall Back-solving

$Γ$ from level flight? Question ::: Total lift equals weight $W$ , span $b$ . Formula for $Γ$ ? Answer ::: $Γ = \frac{W}{ρ V _{\infty} b}$ (from $L^{'} = W / b = ρ V_{\infty} Γ$ ).