Prandtl's lifting-line theory and induced drag

Why it matters

Prandtl's lifting-line theory was the first method to predict how a finite wing's lift and drag differ from those of its airfoil sections. It explains why the lift slope falls as aspect ratio decreases, gives the induced drag that dominates cruise and climb drag at low speed, and identifies the elliptic lift distribution as the ideal. It is still used for preliminary wing sizing.

Key ideas

From one horseshoe to many. A single horseshoe vortex gives infinite downwash at the tips. Prandtl instead superposed an infinite number of horseshoe vortices of different spans, all with their bound parts on one line (the lifting line, at the quarter chord). The result is a bound circulation Γ(y) that varies along the span, falling to zero at the tips, and a continuous trailing vortex sheet whose strength at each station is −dΓ/dy. Wherever the circulation changes, vorticity is shed (Helmholtz).

Downwash and induced angle. By Biot–Savart, the sheet induces a downwash at each point of the lifting line, w(y₀) = (1/4π) ∫ (dΓ/dy)/(y₀ − y) dy over −b/2 to b/2. The local relative wind is tilted down by the induced angle α_i = w/V∞, so each section sees an effective angle of attack α_eff = α − α_i.

Fundamental equation. Each section is assumed to behave like a 2-D airfoil at its effective angle: c_l = a₀(α_eff − α_L=0) and, from Kutta–Joukowski, c_l = 2Γ/(V∞c). Combining gives an integro-differential equation for Γ(y): α(y₀) = 2Γ(y₀)/(a₀V∞c(y₀)) + α_L=0(y₀) + (1/4πV∞) ∫ (dΓ/dy)/(y₀ − y) dy. Geometry (chord c, twist α, section α_L=0) goes in; Γ(y) comes out.

Forces. Lift is the sum of section lifts, L = ρ∞V∞ ∫ Γ dy. Because the local lift is perpendicular to the local relative wind, it is tilted back by α_i and has a component along V∞: the induced drag D_i = ρ∞ ∫ w Γ dy. Physically, it is the rate at which the wing puts kinetic energy into the trailing vortex system.

Fourier solution. With y = −(b/2) cos θ, write Γ(θ) = 2bV∞ Σ Aₙ sin nθ. Then

• C_L = πA₁AR — only the first coefficient carries lift;
• C_D,i = πAR Σ nAₙ² = C_L²/(πAR)·(1 + δ), with δ = Σ_{n≥2} n(Aₙ/A₁)² ≥ 0;
• span efficiency e = 1/(1 + δ) ≤ 1; e = 1 only for the elliptic distribution (all Aₙ = 0 for n ≥ 2), which gives the minimum induced drag for a given lift and span. For a symmetric wing only odd n appear.

Lift slope of a finite wing. For an elliptic wing, a = a₀/(1 + a₀/(πAR)); for other planforms a factor (1 + τ) multiplies the second term, with τ a small planform correction (about 0.05–0.25, read from a chart in your textbook).

Assumptions and limits. Incompressible, inviscid, attached flow; straight, unswept wing of moderately high aspect ratio (AR ≳ 4); small angles; trailing wake flat and aligned with V∞ (no roll-up). It fails for low-AR, delta or highly swept wings (use vortex-lattice or lifting-surface methods) and near stall.

Formulas

• w(y₀) = (1/4π) ∫ (dΓ/dy)/(y₀ − y) dy (m/s); α_i = w/V∞ (rad).
• α_eff = α − α_i.
• L = ρ∞V∞ ∫ Γ(y) dy (N); D_i = ρ∞ ∫ w(y)Γ(y) dy (N).
• Γ(θ) = 2bV∞ Σ Aₙ sin nθ, y = −(b/2) cos θ.
• C_L = πA₁AR; C_D,i = C_L²/(πeAR), e = 1/(1 + δ), δ = Σ_{n≥2} n(Aₙ/A₁)².
• D_i = L²/(q∞πeb²) — the same result in dimensional form, using AR = b²/S; q∞ = ½ρ∞V∞² (Pa).
• a = a₀/[1 + (a₀/πAR)(1 + τ)] — finite-wing lift slope per radian; τ = 0 for elliptic.

Worked examples

Example 1 (standard). An aircraft weighing 50 kN flies level at 70 m/s at sea level (ρ = 1.225 kg/m³). Its wing has span 15 m, area 30 m² and span efficiency e = 0.9. Find C_L, C_D,i and the induced drag.

1. q∞ = 0.5 × 1.225 × 70² = 3001 Pa; AR = b²/S = 225/30 = 7.5.
2. C_L = L/(q∞S) = 50 000/(3001 × 30) = 0.555.
3. C_D,i = C_L²/(πeAR) = 0.3084/(π × 0.9 × 7.5) = 0.3084/21.21 = 0.0145.
4. D_i = q∞S·C_D,i = 3001 × 30 × 0.0145 = 1309 N. Check: L²/(q∞πeb²) = 2.5 × 10⁹/(3001 × π × 0.9 × 225) = 1309 N ✓.

Answer: C_L = 0.555, C_D,i = 0.0145, D_i ≈ 1.31 kN.

Example 2 (GATE level). A lifting-line solution for an untwisted wing of AR = 8 gives Fourier coefficients A₁ = 0.020, A₃ = 0.002 (others negligible). Find C_L, δ, e and C_D,i, and check C_D,i directly from the coefficients.

1. C_L = πA₁AR = π × 0.020 × 8 = 0.503.
2. δ = 3(A₃/A₁)² = 3 × (0.1)² = 0.030.
3. e = 1/(1 + δ) = 1/1.03 = 0.971.
4. C_D,i = C_L²(1 + δ)/(πAR) = 0.2527 × 1.03/(25.13) = 0.01035.
5. Direct: C_D,i = πAR(A₁² + 3A₃²) = 25.13 × (0.0004 + 0.000012) = 0.01035 ✓.

Answer: C_L ≈ 0.503, δ = 0.03, e ≈ 0.971, C_D,i ≈ 0.0104.

Common mistakes

• Writing D_i = L²/(πeb²ρV²) — the dynamic pressure is ½ρV², so the correct form is D_i = 2L²/(πeρV²b²).
• Using the wing's geometric α instead of α_eff = α − α_i to enter the section lift curve.
• Thinking a high aspect ratio alone removes induced drag; C_D,i falls as 1/AR but is zero only with zero lift.
• Believing induced drag is independent of speed. At fixed lift, D_i ∝ 1/V², so it dominates at low speed.
• Confusing span efficiency e (lifting-line, induced drag only) with the Oswald factor, which also absorbs lift-dependent viscous drag of the whole aircraft.
• Applying lifting-line theory to low-AR or highly swept wings.

For GATE AE

Expect: C_D,i or D_i from C_L (or weight), AR and e; finite-wing lift slope from a₀ and AR; induced angle α_i = C_L/(πAR) for an elliptic wing; C_L, δ and e from given Fourier coefficients; and MCQs on assumptions and on how D_i varies with speed, span and density at fixed lift. Practise switching between the coefficient form and the dimensional form L²/(q∞πeb²).

Quick check

1. C_L = 0.8, AR = 10, e = 0.9: find C_D,i.
2. How does D_i change if speed doubles at constant lift?
3. Which Fourier coefficient alone determines C_L?
4. For an elliptic wing with a₀ = 2π per rad and AR = 6, find the lift slope.

Answers: 1. 0.64/(π × 10 × 0.9) = 0.0226. 2. It falls to one quarter. 3. A₁. 4. a = 2π/(1 + 2/6) = 4.71 per rad.