Finite wing theory — induced drag, Prandtl's lifting line: Worked examples

Before anything, let us re-earn the three symbols we will lean on hardest, in plain words:

The scenario matrix

Every problem this topic can throw at you is one (or a blend) of these cells. Each row is a "case class"; the last column names the worked example that nails it.

#	Case class	What makes it tricky	Example
1	Ideal / elliptic ( $δ = 0$ )	The clean formula $C_{D, i} = C_{L}^{2} / (π A R)$	A
2	Non-elliptic ( $δ > 0$ , extra $A_{n}$ )	Must build $δ$ from harmonics	B
3	Limiting: $A R \to \infty$	Recover the 2D infinite wing	C
4	Degenerate: $C_{L} = 0$	Does drag vanish? Sign of $C_{L}^{2}$	C
5	Lift-slope / effective-AoA	The $a = a_{0} / (1 + a_{0} / π A R)$ machinery	D
6	Real-world word problem	Extract $C_{L}$ from weight, speed, area	E
7	Efficiency factor $e < 1$	Convert $δ \leftrightarrow e$ , quote a $%$	F
8	Exam twist: fixed area, vary span	Area constant but $A R$ moves — sign of the effect	G
9	Downwash geometry / sign	Where does air go up vs down; sign of $α_{i}$	H (figure)

We now walk cell by cell. Guess before you read the steps — that is where the learning lives.

Example A — Cell 1: the ideal elliptic wing

Worked example Statement

An elliptically-loaded wing has $A R = 8$ and flies at $C_{L} = 0.5$ . Find its induced drag coefficient.

Forecast: Will $C_{D, i}$ be bigger or smaller than $C_{L}$ itself? Guess a rough number in your head now.

Step 1 — pick the formula. Elliptic loading means only $A_{1} \neq = 0$ , so $δ = \sum_{n \geq 2} n (A_{n} / A_{1})^{2} = 0$ . $C_{D, i} = \frac{C _{L}^{2}}{π A R} .$ Why this step? $δ = 0$ collapses the general $\frac{C _{L}^{2}}{π A R} (1 + δ)$ to its cleanest form — this is the floor, the least drag physically possible for that lift and span.

Step 2 — plug numbers. $C_{D, i} = \frac{0. 5 ^{2}}{π \cdot 8} = \frac{0.25}{25.133} = 0.00995.$ Why this step? Just arithmetic — but notice how tiny it is compared to $C_{L} = 0.5$ . Making lift is cheap on a long wing.

Verify: Units — everything is dimensionless, good. Sanity — $C_{D, i} ≪ C_{L}$ , exactly what we expect for an efficient wing (a lift-to-induced-drag ratio of $\approx 50$ ).

Example B — Cell 2: a non-elliptic wing

Worked example Statement

A wing carries $A_{2} / A_{1} = 0.08$ and $A_{3} / A_{1} = 0.04$ (all higher $A_{n} = 0$ ), with $A R = 6$ and $C_{L} = 0.6$ . Find $C_{D, i}$ .

Forecast: Will this beat or lose to an elliptic wing of the same $A R$ and $C_{L}$ ? By roughly what $%$ ?

Step 1 — build $δ$ . $δ = \sum_{n \geq 2} n (\frac{A _{n}}{A _{1}})^{2} = 2 (0.08)^{2} + 3 (0.04)^{2} = 0.0128 + 0.0048 = 0.0176.$ Why this step? $δ$ is literally "how much do the non-lifting harmonics cost". Each $A_{n}$ ( $n \geq 2$ ) adds drag but no lift (orthogonality kills them in the lift integral).

Step 2 — the drag. $C_{D, i} = \frac{C _{L}^{2}}{π A R} (1 + δ) = \frac{0.36}{π \cdot 6} (1.0176) = 0.01910 \cdot 1.0176 = 0.01944.$ Why this step? The $(1 + δ)$ multiplier converts the elliptic floor into the real value.

Verify: The elliptic floor would be $0.36/ (π \cdot 6) = 0.01910$ . Our answer is $0.01944$ , i.e. $1.76%$ higher — matching $δ$ exactly. Consistent. ✓

Example C — Cells 3 & 4: the limiting and degenerate cases

Worked example Statement

(i) What happens to $C_{D, i}$ as $A R \to \infty$ at fixed $C_{L}$ ? (ii) What is $C_{D, i}$ when $C_{L} = 0$ (a symmetric wing at zero lift)?

Forecast: One of these gives exactly zero. Both? Guess which.

Step 1 — the $A R \to \infty$ limit. $lim_{A R \to \infty} \frac{C _{L}^{2}}{π A R} = 0.$ Why this step? An infinitely long wing has tips "at infinity" — the leaking-around-the-tip effect is diluted to nothing. We recover the 2D infinite wing with no induced drag, exactly d'Alembert's paradox. The finite theory contains the infinite one as a limit — a good sanity check on the whole framework.

Step 2 — the $C_{L} = 0$ case. $C_{D, i} = \frac{0 ^{2}}{π A R} = 0.$ Why this step? Induced drag is drag-due-to-lift. No lift ⇒ no downwash ⇒ no rearward tilt ⇒ no induced drag. Note the square: $C_{L}^{2}$ means the sign of $C_{L}$ never matters — a wing lifting downward ( $C_{L} < 0$ , e.g. an inverted race-car wing) pays the same induced-drag price as one lifting up.

Verify: For $C_{L} = - 0.5$ , $A R = 8$ : $C_{D, i} = (- 0.5)^{2} / (π \cdot 8) = 0.00995$ — identical to Example A. Down-force costs the same as up-force. ✓

Example D — Cell 5: the finite-wing lift slope

Worked example Statement

An elliptic wing has $A R = 10$ and 2D lift slope $a_{0} = 2 π$ . (i) What is its 3D lift slope $a$ ? (ii) If you want $C_{L} = 0.7$ and the zero-lift angle is $α_{L = 0} = - 2^{\circ}$ , what geometric angle of attack do you set?

Forecast: Will the 3D slope be above or below $2 π = 6.283$ ? And will the required angle be bigger or smaller than the 2D wing would need?

Step 1 — 3D slope. $a = \frac{a _{0}}{1 + \frac{a _{0}}{π A R}} = \frac{2 π}{1 + \frac{2 π}{10 π}} = \frac{2 π}{1 + 0.2} = \frac{6.2832}{1.2} = 5.236 /rad .$ Why this step? Part of the geometric angle is "eaten" by the induced angle $α_{i}$ , so each degree of $α$ buys less lift — the slope drops below $2 π$ . This is the finite-wing softening.

Step 2 — required angle. With $C_{L} = a (α - α_{L = 0})$ , $α - α_{L = 0} = \frac{C _{L}}{a} = \frac{0.7}{5.236} = 0.13369 rad = 7.66 0^{\circ} .$ So $α = α_{L = 0} + 7.66 0^{\circ} = - 2^{\circ} + 7.66 0^{\circ} = 5.66 0^{\circ}$ . Why this step? We invert the lift-slope relation to go from "lift I want" back to "angle I must set". The $- 2^{\circ}$ shift accounts for the airfoil's built-in camber.

Verify: A 2D wing ( $a_{0} = 2 π$ ) would need only $0.7/6.2832 = 0.1114$ rad $= 6.38 3^{\circ}$ above $α_{L = 0}$ , i.e. $α = 4.38 3^{\circ}$ . The finite wing needs more angle ( $5.66 0^{\circ} > 4.38 3^{\circ}$ ) — exactly the "softer, needs more incidence" story. ✓

Example E — Cell 6: a real-world word problem

Worked example Statement

A small aircraft weighs $W = 12, 000 N$ , has wing area $S = 16 m^{2}$ , span $b = 12 m$ , and cruises at $V_{\infty} = 50 m/s$ in air of density $ρ = 1.225 kg/m^{3}$ . Assume elliptic loading. Find (i) $C_{L}$ , (ii) $A R$ , (iii) the induced drag force $D_{i}$ in newtons.

Forecast: Will $D_{i}$ be tens, hundreds, or thousands of newtons?

Step 1 — $C_{L}$ from force balance. In steady level flight, lift = weight: $C_{L} = \frac{W}{\frac{1}{2} ρ V _{\infty}^{2} S} = \frac{12000}{0.5 \cdot 1.225 \cdot 5 0 ^{2} \cdot 16} = \frac{12000}{24500} = 0.4898.$ Why this step? The dimensionless $C_{L}$ is defined so that $L = \frac{1}{2} ρ V^{2} S C_{L}$ ; we run that backwards from the known lift (=weight).

Step 2 — aspect ratio. $A R = \frac{b ^{2}}{S} = \frac{1 2 ^{2}}{16} = \frac{144}{16} = 9.$ Why this step? Straight from the definition — span squared over area.

Step 3 — induced drag coefficient, then force. $C_{D, i} = \frac{C _{L}^{2}}{π A R} = \frac{0.489 8 ^{2}}{π \cdot 9} = \frac{0.23994}{28.274} = 0.008486.$ $D_{i} = \frac{1}{2} ρ V_{\infty}^{2} S C_{D, i} = 0.5 \cdot 1.225 \cdot 2500 \cdot 16 \cdot 0.008486 = 207.9 N .$ Why this step? We convert the dimensionless drag back into a real force using the same dynamic-pressure factor $\frac{1}{2} ρ V^{2} S$ .

Verify: Units of $\frac{1}{2} ρ V^{2} S$ : $kg/m^{3} \cdot (m/s)^{2} \cdot m^{2} = kg\cdotm/s^{2} = N$ ✓. Sanity: $D_{i} \approx 208 N$ is under $2%$ of the $12, 000 N$ weight — plausible for induced drag alone. ✓

Example F — Cell 7: efficiency factor round-trip

Worked example Statement

A wing is measured to have Oswald efficiency $e = 0.90$ at $A R = 7$ , $C_{L} = 0.55$ . (i) What $δ$ does this imply? (ii) What is $C_{D, i}$ ? (iii) How much drag, in $%$ , is wasted versus a perfect elliptic wing?

Forecast: $10%$ inefficiency in $e$ — does that mean exactly $10%$ extra drag?

Step 1 — $δ$ from $e$ . Since $e = \frac{1}{1 + δ}$ , $δ = \frac{1}{e} - 1 = \frac{1}{0.90} - 1 = 1.1111 - 1 = 0.1111.$ Why this step? $e$ and $δ$ are two languages for the same idea; $e$ is what wind-tunnel people quote, $δ$ is what the Fourier series produces.

Step 2 — the drag. $C_{D, i} = \frac{C _{L}^{2}}{π e A R} = \frac{0.5 5 ^{2}}{π \cdot 0.90 \cdot 7} = \frac{0.3025}{19.792} = 0.015285.$ Why this step? The $e$ in the denominator is the compact way to carry the penalty.

Step 3 — percentage waste. The elliptic floor is $C_{L}^{2} / (π A R) = 0.3025/ (π \cdot 7) = 0.013756$ . Extra fraction: $\frac{0.015285 - 0.013756}{0.013756} = 0.1111 = 11.11%.$ Why this step? This is exactly $δ$ again — confirming $δ$ is the fractional drag penalty.

Verify: $e = 0.90$ gives $11.11%$ extra, not $10%$ — because the penalty is $1/ e - 1$ , not $1 - e$ . A common trap. ✓

Example G — Cell 8: the exam twist (fixed area, grow the span)

Worked example Statement

A wing has area $S = 20 m^{2}$ held fixed, and you stretch the span from $b = 10 m$ to $b = 14 m$ (making it thinner in chord). At a fixed $C_{L} = 0.6$ and elliptic loading, what happens to $C_{D, i}$ ?

Forecast: Area didn't change — so does the induced drag stay the same, rise, or fall?

Step 1 — write $C_{D, i}$ in span form. $C_{D, i} = \frac{C _{L}^{2}}{π A R} = \frac{C _{L}^{2}}{π} \cdot \frac{S}{b ^{2}} .$ Why this step? Substituting $A R = b^{2} / S$ exposes the true dependence: span squared in the denominator, area in the numerator. Span, not area, is king.

Step 2 — the two cases. $A R_{1} = \frac{1 0 ^{2}}{20} = 5, C_{D, i, 1} = \frac{0.36}{π \cdot 5} = 0.022918.$ $A R_{2} = \frac{1 4 ^{2}}{20} = 9.8, C_{D, i, 2} = \frac{0.36}{π \cdot 9.8} = 0.011693.$ Why this step? Same $S$ , but the longer span raises $A R$ from $5$ to $9.8$ , so drag falls.

Step 3 — the ratio. $\frac{C _{D, i, 2}}{C _{D, i, 1}} = \frac{5}{9.8} = 0.5102,$ a drop of nearly half. Why this step? It confirms the "long & lean" mantra from the parent note: at fixed area, stretching span alone slashes induced drag.

Verify: Area truly unchanged (both $20 m^{2}$ ), yet drag halved — the exam-trap answer "area fixed ⇒ drag fixed" is wrong. The controlling quantity is $b^{2}$ . ✓

Example H — Cell 9: downwash geometry and its sign (with figure)

Worked example Statement

An elliptic wing produces $C_{L} = 0.6$ at $A R = 8$ and $V_{\infty} = 60 m/s$ . For an elliptic wing the downwash $w$ is uniform across the span. Find (i) the induced angle $α_{i}$ in degrees, (ii) confirm geometrically that the lift tilts rearward, and (iii) state the sign of $w$ and where the flow goes up instead.

Forecast: Will $α_{i}$ be a fraction of a degree, or several degrees?

Step 1 — induced angle from $C_{L}$ and $A R$ . For an elliptic wing $α_{i} = \frac{C _{L}}{π A R}$ (uniform). $α_{i} = \frac{0.6}{π \cdot 8} = 0.023873 rad = 1.36 8^{\circ} .$ Why this step? For elliptic loading the downwash is constant, so $α_{i} = w / V_{\infty}$ is one number for the whole wing. It equals $C_{L} / (π A R)$ — the same quantity that appears in the lift-slope formula, which is no coincidence.

Step 2 — read the geometry (see figure). The wing meets $V_{\infty}$ (horizontal) plus a downward $w$ . Their sum, the local relative wind, points down by $α_{i}$ . Lift is $⊥$ to that local wind, so lift itself is rotated back by $α_{i}$ from true-vertical. Its horizontal component is the induced drag: $C_{D, i} = C_{L} α_{i}$ . Why this step? This is the physical origin of induced drag — geometry, not friction. The red arrow in the figure is the rearward tilt.

Step 3 — the check on that geometric relation. $C_{L} α_{i} = 0.6 \cdot 0.023873 = 0.014324,$ which must equal $C_{D, i} = \frac{C _{L}^{2}}{π A R} = \frac{0.36}{π \cdot 8} = 0.014324$ . Identical. ✓ Why this step? It proves $C_{D, i} = C_{L} α_{i}$ is not a separate rule — it's the same thing seen through the tilted-lift triangle.

Step 4 — sign and where air goes up. Behind the wing and along the span, $w$ points down (negative, downwash) — that is the sign we used. But outboard of the tips, the trailing tip vortex circulates the other way, so just beyond each wingtip the air is pushed up (upwash). This is why birds fly in a V: the trailing bird sits in a neighbour's tip-upwash and saves energy. Why this step? Covers the full sign picture — downwash inside the span, upwash outside — which the parent note only hinted at.

Verify: $α_{i} = 1.3 7^{\circ}$ is a small angle, so $tan α_{i} \approx α_{i}$ is safe, and $C_{D, i} = C_{L} α_{i}$ matched the closed form to the last digit. ✓

Recap

Recall Which formula for which cell?

Elliptic ideal ::: $C_{D, i} = C_{L}^{2} / (π A R)$ with $δ = 0$ Non-elliptic ::: multiply by $(1 + δ)$ , $δ = \sum_{n \geq 2} n (A_{n} / A_{1})^{2}$ $A R \to \infty$ or $C_{L} = 0$ ::: $C_{D, i} \to 0$ 3D lift slope ::: $a = a_{0} / (1 + a_{0} / π A R)$ Force from coefficient ::: multiply by $\frac{1}{2} ρ V^{2} S$ $e$ from $δ$ ::: $e = 1/ (1 + δ)$ , penalty $= 1/ e - 1$ Fixed area, grow span ::: $C_{D, i} \propto S / b^{2}$ , span wins Induced angle ::: $α_{i} = C_{L} / (π A R)$ , and $C_{D, i} = C_{L} α_{i}$