Elliptic lift distribution and effect of aspect ratio

Why it matters

The elliptic lift distribution is the benchmark for every wing: it gives the least induced drag for a given lift and span, uniform downwash and simple closed-form results. Comparing a real wing with it — through the span efficiency e and the aspect ratio — is how designers judge wing planforms, and the aspect-ratio formulas let you convert wind-tunnel data from one wing to another.

Key ideas

The distribution. The circulation (and hence lift per unit span, L′ = ρ∞V∞Γ) varies along the span as an ellipse: Γ(y) = Γ₀·√(1 − (2y/b)²), maximum Γ₀ at the root and zero at the tips. In the Fourier series of lifting-line theory it is the single term A₁ sin θ.

Uniform downwash. Substituting into the lifting-line downwash integral gives a constant downwash w = Γ₀/(2b) across the whole span. The induced angle is therefore the same at every station: α_i = C_L/(πAR).

Minimum induced drag. With all higher Fourier coefficients zero, δ = 0 and e = 1, so C_D,i = C_L²/(πAR) — the smallest possible value for a planar wing of given span and lift. Any other distribution has e < 1.

How to get it. An untwisted wing with an elliptic planform (chord ∝ √(1 − (2y/b)²)) and the same airfoil all along the span produces an elliptic lift distribution at every α (the Spitfire is the famous example). Elliptic planforms are expensive to build; a straight-tapered wing with taper ratio around 0.3–0.45 comes within about 1 % of e = 1, and twist can tailor the distribution further at the design C_L.

Effect of aspect ratio.

• Lift slope: a = a₀/(1 + a₀/(πAR)) (a₀, a per radian). As AR → ∞, a → a₀; low-AR wings have much lower slopes and so need more α for the same C_L.
• Induced drag falls as 1/AR at fixed C_L; in dimensional form at fixed lift, D_i = L²/(q∞πb²) — what matters is span, not area.
• Two elliptic wings of different AR at the same C_L differ only in α_i, so their geometric angles are related by α₂ = α₁ + (C_L/π)(1/AR₂ − 1/AR₁) and their drag coefficients by C_D2 = C_D1 + (C_L²/π)(1/AR₂ − 1/AR₁) (profile drag unchanged). This is Prandtl's classic check against experiment.

Practical trade-offs. High AR (gliders, airliners, long-endurance UAVs) gives low induced drag but heavier, more flexible wings, lower roll rate and less fuel volume per unit span. Low AR (fighters) gives a stiff, compact wing that tolerates high load factors and supersonic flight. Winglets give part of the benefit of extra span without increasing it.

Stall behaviour. With an elliptic distribution, c_l is the same at every station, so the whole span tends to stall at once — undesirable for control. Washout or a tapered planform with tip-root section changes is used to make the root stall first.

Formulas

• Γ(y) = Γ₀·√(1 − (2y/b)²) — Γ₀ root circulation (m²/s), b span (m).
• L = ρ∞V∞Γ₀·(πb/4) ⇒ Γ₀ = 4L/(ρ∞V∞πb).
• w = Γ₀/(2b) (m/s) — uniform downwash.
• α_i = w/V∞ = C_L/(πAR) (rad).
• C_D,i = C_L²/(πAR); general wing C_D,i = C_L²/(πeAR).
• D_i = L²/(q∞πeb²) (N).
• a = a₀/(1 + a₀/(πAR)) — per radian; for a non-elliptic wing replace a₀/(πAR) by a₀(1 + τ)/(πAR).
• C_L = a(α − α_L=0).
• α₂ = α₁ + (C_L/π)(1/AR₂ − 1/AR₁) (rad) and C_D2 = C_D1 + (C_L²/π)(1/AR₂ − 1/AR₁) — same C_L, elliptic wings.

Worked examples

Example 1 (standard). An elliptic wing of span 12 m and area 18 m² carries 12 kN at 60 m/s at sea level (ρ = 1.225 kg/m³). Find C_L, Γ₀, the downwash, α_i, C_D,i and D_i.

1. AR = b²/S = 144/18 = 8; q∞ = 0.5 × 1.225 × 60² = 2205 Pa.
2. C_L = L/(q∞S) = 12 000/(2205 × 18) = 0.302.
3. Γ₀ = 4L/(ρ∞V∞πb) = 48 000/(1.225 × 60 × π × 12) = 17.3 m²/s.
4. w = Γ₀/(2b) = 17.32/24 = 0.722 m/s.
5. α_i = w/V∞ = 0.0120 rad = 0.69°. Check: C_L/(πAR) = 0.302/25.13 = 0.0120 ✓.
6. C_D,i = C_L²/(πAR) = 0.0914/25.13 = 0.00364; D_i = q∞S·C_D,i = 2205 × 18 × 0.00364 = 144 N (= L·α_i ✓).

Answer: C_L = 0.302, Γ₀ = 17.3 m²/s, w = 0.72 m/s, α_i = 0.69°, C_D,i = 0.0036, D_i ≈ 144 N.

Example 2 (GATE level). An airfoil has a₀ = 0.105 per degree and α_L=0 = −2°. It is used on an untwisted elliptic wing of AR = 6. Find C_L and C_D,i at α = 6°, and the angle an AR = 10 elliptic wing would need for the same C_L.

1. a₀ = 0.105 × 57.30 = 6.016 per rad.
2. a = a₀/(1 + a₀/(πAR)) = 6.016/(1 + 6.016/18.85) = 6.016/1.319 = 4.561 per rad = 0.0796 per degree.
3. C_L = a(α − α_L=0) = 0.0796 × 8 = 0.637.
4. C_D,i = C_L²/(πAR) = 0.4055/18.85 = 0.0215 (α_i = C_L/(πAR) = 0.0338 rad = 1.94°).
5. AR = 10, same C_L: α₂ = α₁ + (C_L/π)(1/AR₂ − 1/AR₁) = 6° + (0.637/π)(0.1 − 0.1667) rad = 6° − 0.01352 rad = 6° − 0.77° = 5.23°.

Answer: C_L ≈ 0.637, C_D,i ≈ 0.0215 at AR = 6; the AR = 10 wing needs α ≈ 5.2°.

Common mistakes

• Calling a distribution "elliptic" because the planform is elliptic while the wing is twisted — twist changes the lift distribution.
• Using a₀ per degree inside a₀/(πAR); the formula needs radians (or convert π to 180 consistently).
• Forgetting that at fixed lift, induced drag depends on span, not on AR as such: adding chord does not help.
• Assuming e = 1 for any real aircraft; typical span efficiencies are 0.85–0.98, and the aircraft Oswald factor is lower still.
• Getting the sign of the AR conversion wrong — a higher AR needs a smaller α for the same C_L.

For GATE AE

Common items: finite-wing lift slope from a₀ and AR; C_L at a given α with α_L=0; induced angle and induced drag of an elliptic wing; Γ₀ from total lift; uniform downwash; and conversion of α or C_D between two aspect ratios. Learn the formulas in radians and convert only at the end.

Quick check

1. Elliptic wing, AR = 8, C_L = 0.5: find C_D,i.
2. What is the downwash distribution of an elliptic wing?
3. a₀ = 2π per rad, AR = 8: find a.
4. Which planform gives an elliptic lift distribution without twist?

Answers: 1. 0.25/(8π) = 0.00995. 2. Uniform, w = Γ₀/(2b). 3. a = 2π/(1 + 0.25) = 5.03 per rad. 4. An elliptic planform with the same section along the span.