Conformal mapping and the Joukowski airfoil

Why it matters

Conformal mapping was the first method that gave the exact potential flow, and hence the exact pressure distribution and lift, around a realistic airfoil shape. The Joukowski airfoil shows clearly where lift comes from (circulation fixed by a sharp trailing edge), why thickness slightly raises the lift slope, and why camber shifts the zero-lift angle — results that thin airfoil theory later reproduces approximately.

Key ideas

Complex potential. In 2-D incompressible irrotational flow, φ and ψ both satisfy Laplace's equation and obey the Cauchy–Riemann relations, so w(z) = φ + iψ is an analytic function of z = x + iy. Its derivative is the complex velocity dw/dz = u − iv. Elementary flows become simple functions: uniform flow V∞z, source (Λ/2π) ln z, vortex (iΓ/2π) ln z (clockwise positive), doublet κ/(2πz).

Conformal mapping. An analytic function z = f(ζ) maps the ζ-plane to the z-plane, preserving angles between curves wherever dz/dζ ≠ 0. Because an analytic function of an analytic function is analytic, a potential flow in the ζ-plane becomes a potential flow in the z-plane:

• streamlines map to streamlines, so a circle that is a streamline in ζ becomes the body contour in z;
• the complex potential is carried over unchanged, w(z) = w(ζ(z));
• velocities scale by the mapping derivative, (u − iv)_z = (u − iv)_ζ / (dz/dζ);
• the circulation round corresponding curves, and therefore the lift, is the same in both planes when the mapping tends to z ≈ ζ far away.

Joukowski transformation. z = ζ + b²/ζ, with b a real constant. Far from the origin z ≈ ζ, so the free stream is unchanged. The derivative dz/dζ = 1 − b²/ζ² vanishes at ζ = ±b — the two critical points, where angles are not preserved. A circle that passes through ζ = +b produces a sharp trailing edge there (the smooth circle is folded through 180°).

Shapes produced:

• Circle of radius b centred at the origin → flat plate from z = −2b to +2b (chord 4b).
• Circle of radius a > b centred at the origin → ellipse with semi-axes a + b²/a and a − b²/a.
• Circle through ζ = b, centre shifted slightly along the negative real axis (centre −εb, radius a = b(1 + ε)) → symmetric Joukowski airfoil, chord ≈ 4b, thickness ratio t/c ≈ 1.3ε for small ε.
• Centre also shifted upward → cambered Joukowski airfoil; the angle β that the line from the centre to ζ = b makes below the real axis sets the camber (zero-lift angle α_L=0 = −β).

Lift. In the circle plane the flow is uniform flow at angle α + doublet + vortex. The Kutta condition requires the point ζ = b (the trailing edge) to be a stagnation point on the circle, which fixes Γ = 4πaV∞ sin(α + β). The Kutta–Joukowski theorem then gives L′ = 4πρ∞V∞²a sin(α + β). With c ≈ 4b, c_l ≈ 2π(a/b) sin(α + β): lift slope slightly above 2π because a/b = 1 + ε > 1 (thickness), and lift at α = 0 because of β (camber).

Limitations. The trailing edge is a cusp (zero angle), which is structurally impractical; the shape family is restricted; and the theory is inviscid — no drag, no stall. Modern design uses panel methods and viscous codes, but the Joukowski airfoil remains the standard test case for those codes.

Formulas

• w = φ + iψ, dw/dz = u − iv — complex potential (m²/s) and complex velocity (m/s).
• z = ζ + b²/ζ — Joukowski mapping; b (m).
• dz/dζ = 1 − b²/ζ² — mapping derivative; zero at ζ = ±b.
• (u − iv)_z = (u − iv)_ζ / (dz/dζ) — velocity transformation.
• Ellipse from circle radius a at origin: semi-axes a + b²/a and a − b²/a.
• Γ = 4π·a·V∞·sin(α + β) — Kutta condition at ζ = b; a circle radius (m).
• L′ = ρ∞V∞Γ = 4πρ∞V∞²·a·sin(α + β) (N/m).
• c_l ≈ 2π(a/b)·sin(α + β), using c ≈ 4b; α_L=0 = −β.
• t/c ≈ 1.3ε — symmetric Joukowski airfoil, small ε, with a = b(1 + ε).

Worked examples

Example 1 (standard). With b = 1 m, a circle of radius a = 1.25 m centred at the origin is mapped by z = ζ + b²/ζ. Find the body shape, its chord and thickness ratio. What would a circle of radius 1 m give?

1. Semi-major axis a + b²/a = 1.25 + 0.8 = 2.05 m; semi-minor axis a − b²/a = 1.25 − 0.8 = 0.45 m.
2. Ellipse of chord 2 × 2.05 = 4.10 m and thickness 2 × 0.45 = 0.90 m; t/c = 0.90/4.10 = 0.220.
3. With a = b = 1 m the semi-minor axis is zero: a flat plate from −2 m to +2 m, chord 4b = 4 m.

Answer: an ellipse 4.10 m × 0.90 m (t/c ≈ 22 %); a = b gives a flat plate of chord 4 m.

Example 2 (GATE level). A symmetric Joukowski airfoil is generated with b = 1 m from a circle centred at ζ = −0.1 m of radius a = 1.1 m. It is placed in air (ρ = 1.225 kg/m³) at V∞ = 30 m/s and α = 6°. Find the exact chord, the circulation, the lift per unit span and c_l, and compare with 2πα.

1. Trailing edge: ζ = +1 → z = 1 + 1 = 2 m. Leading edge: ζ = −0.1 − 1.1 = −1.2 → z = −1.2 + 1/(−1.2) = −2.033 m. Chord c = 4.033 m (≈ 4b).
2. Symmetric, so β = 0: Γ = 4πaV∞ sin α = 4π × 1.1 × 30 × sin 6° = 414.7 × 0.1045 = 43.35 m²/s.
3. L′ = ρ∞V∞Γ = 1.225 × 30 × 43.35 = 1593 N/m.
4. c_l = 2Γ/(V∞c) = 2 × 43.35/(30 × 4.033) = 0.717.
5. Thin-plate value: 2πα = 2π × 0.1047 = 0.658. The thick section gives about 9 % more lift at this α (a/b = 1.1).

Answer: c = 4.03 m, Γ = 43.3 m²/s, L′ ≈ 1.59 kN/m, c_l ≈ 0.72 versus 0.66 from 2πα. (This section is about 12 % thick; the first-order estimate 1.3ε = 13 % slightly over-predicts.)

Common mistakes

• Writing the Joukowski map as z = ζ + 1/ζ and then using circles of arbitrary radius without noting that b = 1 has been assumed.
• Forgetting that the circle must pass through ζ = +b to give a sharp trailing edge; otherwise you get an ellipse-like rounded body with no Kutta point.
• Applying the Kutta condition at the leading edge, or at z = 2b in the circle plane — it is applied at the critical point ζ = b on the circle.
• Using c_l = 2πΓ/(V∞c); the correct relation is c_l = 2Γ/(V∞c).
• Expecting conformal mapping to predict drag or stall.

For GATE AE

Expect: shapes produced by the Joukowski map for a given circle (flat plate, ellipse, airfoil), chord and thickness of the image, critical points of the mapping, circulation from the Kutta condition, and lift from Γ = 4πaV∞ sin(α + β). Conceptual MCQs test what conformal mapping preserves (angles, circulation, the Laplace equation) and what it does not (velocity magnitudes, which scale with |dz/dζ|).

Quick check

1. Where is the Joukowski mapping z = ζ + b²/ζ not conformal?
2. A circle of radius 0.5 m at the origin with b = 0.5 m maps to what?
3. What value of Γ does the Kutta condition give for a circle of radius a at angle of attack α (symmetric case)?
4. Why is the Joukowski trailing edge a cusp?

Answers: 1. At ζ = ±b, where dz/dζ = 0. 2. A flat plate of chord 4b = 2 m. 3. Γ = 4πaV∞ sin α. 4. The circle passes through the critical point ζ = b, where angles are doubled, so the smooth 180° contour becomes a 360° fold — a zero-angle edge.