Biot-Savart law, horseshoe vortex and downwash

Why it matters

A finite wing cannot be treated as an airfoil: the pressure difference spills round the tips and leaves a pair of trailing vortices behind it. These vortices induce a downward velocity — downwash — over the wing and the tail, which tilts the lift backwards (induced drag) and changes the tail's angle of attack. The Biot–Savart law is the tool that turns a vortex system into induced velocities, and the horseshoe vortex is the simplest model of a lifting wing.

Key ideas

Biot–Savart law. A vortex filament of strength Γ induces at a point P a velocity contribution from each element dl: dV = (Γ/4π)·(dl × r)/|r|³, where r is the vector from the element to P. The induced velocity is perpendicular to both the filament and r, and its sense follows the right-hand rule (thumb along the filament direction, fingers give the swirl). The law is purely kinematic — it follows from the definition of vorticity in incompressible flow, exactly as in magnetism.

Straight segment. Integrating along a straight segment gives a velocity of magnitude V = (Γ/4πh)(cos θ₁ − cos θ₂), where h is the perpendicular distance from P to the line of the filament and θ₁, θ₂ are the angles between the filament direction and the lines drawn from its start and its end to P. Special cases:

• infinite filament (θ₁ = 0, θ₂ = π): V = Γ/(2πh) — the 2-D point vortex;
• semi-infinite filament, P opposite its end (θ₁ = π/2, θ₂ = π): V = Γ/(4πh) — half the infinite value.

Helmholtz's theorems (inviscid flow): a vortex filament has constant strength along its length and cannot end in the fluid. So the bound vortex that gives the wing its lift (L′ = ρ∞V∞Γ) cannot simply stop at the wing tips; it turns downstream as two trailing vortices and, in principle, closes through the starting vortex far behind.

Horseshoe vortex. The simplest finite-wing model: a bound vortex of constant strength Γ along the span b (normally placed on the quarter-chord line), and two semi-infinite trailing vortices from the tips to infinity downstream. Physical tip vortices are a rolled-up version of the vortex sheet that a real wing sheds along its whole span.

Downwash. The trailing vortices induce a downward velocity w between them (and upwash outboard of the tips). The bound vortex induces nothing on itself but contributes at points ahead of or behind the wing.

• On the bound vortex, at spanwise station y: w(y) = (Γ/4π)[1/(b/2 + y) + 1/(b/2 − y)] (downward), equal to Γ/(πb) at mid-span and infinite at the tips. This unrealistic tip singularity is why Prandtl replaced the single horseshoe by a continuous distribution (lifting-line theory, next topic).
• Far downstream on the centre line, the trailing vortices look infinite: w = 2Γ/(πb) — twice the value at the wing.

Consequences. The local flow at the wing is tilted down by the induced angle α_i ≈ w/V∞, so the effective angle of attack is α_eff = α − α_i; the local lift, perpendicular to the local flow, is tilted back by α_i and has a component along V∞ — the induced drag D_i = L·α_i. At the tail, the downwash angle ε reduces the tail's angle of attack and must be accounted for in trim and stability.

Formulas

• dV = (Γ/4π)·(dl × r)/|r|³ — Biot–Savart; Γ in m²/s, lengths in m, V in m/s.
• V = (Γ/4πh)·(cos θ₁ − cos θ₂) — straight segment.
• V = Γ/(2πh) — infinite filament; V = Γ/(4πh) — semi-infinite filament, point abeam its end.
• w(y) = (Γ/4π)·[1/(b/2 + y) + 1/(b/2 − y)] — downwash on the bound vortex of a horseshoe of span b.
• w(0) = Γ/(πb) — at mid-span on the bound vortex; w = 2Γ/(πb) far downstream on the centre line.
• α_i = w/V∞ (rad); D_i = L·α_i — small angles.
• L = ρ∞V∞Γb — total lift of a horseshoe vortex (N).

Worked examples

Example 1 (standard). A wing of span 10 m is represented by a horseshoe vortex of strength Γ = 30 m²/s in a 50 m/s stream. Find the downwash on the bound vortex at mid-span and at y = 2.5 m, and the induced angle at mid-span.

1. Mid-span: w(0) = Γ/(πb) = 30/(π × 10) = 0.955 m/s.
2. y = 2.5 m: w = (30/4π)[1/(5 + 2.5) + 1/(5 − 2.5)] = 2.387 × (0.1333 + 0.4) = 1.273 m/s.
3. α_i(0) = w/V∞ = 0.955/50 = 0.0191 rad = 1.09°.

Answer: w = 0.955 m/s at mid-span and 1.27 m/s at y = 2.5 m; α_i ≈ 1.09° at mid-span. The downwash grows towards the tips — an artefact of the single horseshoe.

Example 2 (GATE level). For the same horseshoe vortex (b = 10 m, Γ = 30 m²/s), find the downwash at a point on the centre line 5 m behind the bound vortex, in the plane of the vortex, and compare with the far-downstream value.

1. Bound vortex: perpendicular distance h = 5 m; ends at ±5 m, so θ₁ = 45°, θ₂ = 135° and cos θ₁ − cos θ₂ = √2. w_bound = (30/(4π × 5)) × 1.414 = 0.477 × 1.414 = 0.675 m/s.
2. Each trailing vortex runs from its tip (x = 0) to x = +∞; P is 5 m downstream of its start and h = 5 m from its line. The start-to-P line makes θ₁ = 45° with the filament direction and the far end gives θ₂ = 180°, so cos θ₁ − cos θ₂ = 0.7071 + 1 = 1.7071. w_trail = (30/(4π × 5)) × 1.7071 = 0.815 m/s each.
3. All three contributions are downward: w = 0.675 + 2 × 0.815 = 2.31 m/s.
4. Far downstream: w∞ = 2Γ/(πb) = 60/(10π) = 1.91 m/s.

Answer: w ≈ 2.31 m/s at 5 m behind the wing, versus 1.91 m/s far downstream and 0.955 m/s at the wing. Close behind the wing the bound vortex adds to the downwash, which is why tail downwash must be computed, not guessed.

Common mistakes

• Using Γ/(2πh) (infinite filament) for a trailing vortex seen from the wing; at the bound vortex each trailing vortex is semi-infinite and gives Γ/(4πh).
• Forgetting that the bound vortex induces no velocity on itself but does at other points.
• Adding contributions as scalars without checking each one's direction with the right-hand rule.
• Using h as the distance to the end of the filament rather than the perpendicular distance to its line.
• Treating downwash as uniform for a single horseshoe; it is uniform only for an elliptic lift distribution (next topic).

For GATE AE

Expect: induced velocity from straight, semi-infinite and finite segments; downwash at the wing or at the tail from a horseshoe vortex; induced angle and induced drag from w; and MCQs on Helmholtz's theorems and the structure of the horseshoe model. Draw the vortex system, mark directions, then add vector contributions.

Quick check

1. Velocity induced by an infinite vortex filament of Γ = 5 m²/s at 2 m?
2. Velocity induced at 2 m abeam the end of a semi-infinite filament of Γ = 20 m²/s?
3. Horseshoe: Γ = 30 m²/s, b = 10 m — downwash at the wing centre?
4. Why must the bound vortex turn downstream at the tips?

Answers: 1. 5/(4π) = 0.398 m/s. 2. 20/(8π) = 0.796 m/s. 3. 0.955 m/s. 4. Helmholtz: a vortex filament cannot end in the fluid.