Aspect ratio — effect on induced drag: Worked examples

This page is a drill hall. The parent note Aspect ratio — effect on induced drag built the one formula that runs this whole topic:

$C_{D, i} = \frac{C _{L}^{2}}{π e A R}$

Here we don't re-derive it — we stress-test it. We ask: what happens for every kind of input you could be handed? Big span, zero span, zero lift, a table of numbers, a real airplane, a sneaky exam twist. By the end you should never meet a case you haven't already seen.

Before anything, let's make sure every letter is a picture, not a squiggle.

Definition The five letters, each as a picture

$C_{L}$ — the lift coefficient. Think of it as a dial the pilot turns: pull the nose up, $C_{L}$ goes up. It has no units. $C_{L} = 0$ means "wings producing zero lift" (knife-edge or gliding at exactly zero angle).
$A R$ — the aspect ratio, $b^{2} / S$ , span squared over wing area. Picture "how many chord-lengths fit along the wing." Glider: long and skinny, $A R \sim 30$ . Fighter: stubby, $A R \sim 3$ .
$e$ — the span efficiency, a number between $0$ and $1$ . It answers "how close is your lift shape to the perfect elliptical one?" $e = 1$ is perfect; a plain rectangle is $\approx 0.8$ .
$π$ — just $3.14159 \dots$ , the constant that falls out of the elliptical integral in Lifting-line theory (Prandtl).
$C_{D, i}$ — the induced-drag coefficient, the answer we compute. Bigger = more drag born purely from making lift on a finite wing.

The scenario matrix

Here is every kind of case this topic can throw at you. Each worked example below is tagged with the cell it fills.

Cell	What varies / degenerate input	Question it answers	Example
A. Forward compute	ordinary $C_{L}, e, A R$	plug in, get $C_{D, i}$	Ex 1
B. $1/ A R$ scaling	change $A R$ only	how does span pay off?	Ex 2
C. $C_{L}^{2}$ scaling	change $C_{L}$ only	why does slow flight hurt?	Ex 3
D. $e$ effect	non-elliptical shape	cost of a bad lift shape	Ex 4
E. Degenerate: $C_{L} = 0$	zero lift	is there any induced drag?	Ex 5
F. Limit: $A R \to \infty$	infinite (2-D) wing	does induced drag vanish?	Ex 6
G. Backward solve	given $C_{D, i}$ , find $A R$	design/sizing problem	Ex 7
H. Real-world word problem	full drag polar, actual forces	fuel/thrust in Newtons	Ex 8
I. Exam twist	ratio trap, both inputs move	resist "cancel" mistakes	Ex 9

We'll hit all nine.

Example 1 — Cell A: the plain forward computation

Worked example Statement

A wing with elliptical loading ( $e = 1$ ) flies at $C_{L} = 0.8$ with aspect ratio $A R = 6$ . Find $C_{D, i}$ .

Forecast: Guess the order of magnitude first. Lift coefficients are around $1$ ; $π \cdot 6 \approx 19$ . So $C_{D, i}$ should be roughly $0.64/19 \approx 0.03$ — a few percent. Hold that guess.

Step 1 — Square the lift. $C_{L}^{2} = 0. 8^{2} = 0.64$ Why this step? The formula's numerator is $C_{L}^{2}$ , not $C_{L}$ — lift is the "troublemaker" and it enters squared. Squaring first keeps the arithmetic clean.

Step 2 — Build the denominator. $π e A R = π \cdot 1 \cdot 6 = 18.85$ Why this step? With $e = 1$ the shape is ideal, so nothing extra shrinks the denominator; it's just $π A R$ .

Step 3 — Divide. $C_{D, i} = \frac{0.64}{18.85} = 0.0340$ Why this step? This is the definition, executed. Numerator (harm) over denominator (span working for us).

Verify: Matches our forecast of " $\approx 0.03$ ." Units: all inputs are dimensionless, so $C_{D, i}$ is dimensionless — correct for a coefficient. ✓

Example 2 — Cell B: doubling the span (the $1/ A R$ payoff)

Worked example Statement

Same wing as Ex 1 ( $C_{L} = 0.8$ , $e = 1$ ) but now stretched to $A R = 12$ . Compare with the $A R = 6$ answer.

Forecast: We only changed $A R$ , and it sits alone on the bottom. Doubling the bottom should halve the answer. Guess $0.0170$ .

Step 1 — Keep the numerator frozen. $C_{L}^{2} = 0.64 (unchanged — we did not touch C_{L})$ Why this step? Isolating the variable that changed makes the scaling visible. $C_{L}$ and $e$ are the same, so only the $A R$ in the denominator moves.

Step 2 — Recompute the denominator. $π \cdot 1 \cdot 12 = 37.70$ Why this step? Doubling $A R$ from $6$ to $12$ exactly doubles $π A R$ .

Step 3 — Divide, then take the ratio. $C_{D, i} = \frac{0.64}{37.70} = 0.0170, \frac{0.0170}{0.0340} = \frac{1}{2}$ Why this step? The ratio is the point: because $C_{D, i} \propto 1/ A R$ , twice the aspect ratio = half the induced drag. Look at the blue vs yellow bars in the figure — the second is exactly half-height.

Verify: $0.0170 \times 2 = 0.0340$ recovers the $A R = 6$ value. Consistent. ✓

Example 3 — Cell C: slowing down (the $C_{L}^{2}$ explosion)

Worked example Statement

A wing with $A R = 10$ , $e = 1$ flies in cruise at $C_{L} = 0.4$ and in landing at $C_{L} = 1.2$ . Compare $C_{D, i}$ .

Forecast: $C_{L}$ triples ( $0.4 \to 1.2$ ). Because it enters squared, guess $3^{2} = 9$ times more drag.

Step 1 — Cruise case. $C_{D, i} = \frac{0. 4 ^{2}}{π \cdot 10} = \frac{0.16}{31.42} = 0.00509$ Why this step? Establish the small-lift baseline so we can measure the growth.

Step 2 — Landing case. $C_{D, i} = \frac{1. 2 ^{2}}{π \cdot 10} = \frac{1.44}{31.42} = 0.0458$ Why this step? Same wing, same $π A R$ ; only the numerator swelled from $0.16$ to $1.44$ — a factor of $9$ .

Step 3 — Read the ratio. $\frac{0.0458}{0.00509} = 9.0$ Why this step? Confirms the square law. The parabola in the figure shows why: near the origin it's flat, but at high $C_{L}$ it shoots up steeply.

Verify: $(1.2/0.4)^{2} = 3^{2} = 9$ , matching the drag ratio exactly. This is why slow, high-lift flight (takeoff/landing) is dominated by induced drag — see Glide ratio & L/D max. ✓

Example 4 — Cell D: the price of a non-elliptical shape

Worked example Statement

Two wings, both $A R = 8$ , both at $C_{L} = 0.7$ . Wing X is elliptical ( $e = 1$ ); Wing R is a plain rectangle ( $e = 0.8$ ). By what percent is the rectangle worse?

Forecast: $e$ sits on the bottom. Dropping it from $1.0$ to $0.8$ divides by a smaller number, so drag rises. $1/0.8 = 1.25$ , so guess +25%.

Step 1 — Elliptical wing. $C_{D, i} = \frac{0.49}{π \cdot 1 \cdot 8} = \frac{0.49}{25.13} = 0.0195$ Why this step? $e = 1$ is the benchmark — the least drag physically possible for this $C_{L}$ and $A R$ (the Elliptical lift distribution minimum).

Step 2 — Rectangular wing. $C_{D, i} = \frac{0.49}{π \cdot 0.8 \cdot 8} = \frac{0.49}{20.11} = 0.0244$ Why this step? The rectangle's lift piles up more toward the tips, making stronger tip vortices; $e = 0.8$ encodes that penalty.

Step 3 — Percent increase. $\frac{0.0244 - 0.0195}{0.0195} = 0.25 = 25%$ Why this step? Turns two numbers into the design-relevant statement: a poor lift shape costs a quarter more induced drag. This is why designers use taper and washout to fake an ellipse.

Verify: $0.0195/0.8 = 0.0244$ exactly — dividing by $e = 0.8$ is multiplying by $1.25$ . ✓

Example 5 — Cell E: the degenerate case $C_{L} = 0$

Worked example Statement

A wing ( $A R = 7$ , $e = 0.85$ ) is flown at exactly zero lift (gliding at the zero-lift angle, or knife-edge). What is its induced drag?

Forecast: No lift means nothing to tilt backward. Guess exactly zero.

Step 1 — Plug $C_{L} = 0$ into the formula. $C_{D, i} = \frac{0 ^{2}}{π \cdot 0.85 \cdot 7} = \frac{0}{18.69} = 0$ Why this step? Because the numerator is $C_{L}^{2}$ , a zero there kills the whole thing regardless of how big $A R$ or $e$ are.

Step 2 — Sanity-check the physics. Why this step? Zero lift means equal pressure top and bottom near the tips ⇒ nothing leaks around ⇒ no trailing vortices ⇒ no downwash ⇒ lift (which is zero anyway) tilts back by nothing. The chain from the parent note simply collapses.

Verify: $C_{D, i} = 0$ . Note the wing still has Parasite drag — skin friction and form drag don't need lift. Induced drag is the only term that vanishes at zero lift. This directly refutes the mistake "induced drag depends on shape only." ✓

Example 6 — Cell F: the limit $A R \to \infty$ (the infinite 2-D wing)

Worked example Statement

Take a fixed lift ( $C_{L} = 1.0$ , $e = 1$ ) and let the span grow without bound: $A R = 10, 100, 1000, \to \infty$ . What does $C_{D, i}$ approach?

Forecast: $A R$ is on the bottom; pushing it to infinity should drive the fraction to zero.

Step 1 — Tabulate the trend. $A R = 10 : \frac{1}{π \cdot 10} = 0.0318, A R = 100 : 0.00318, A R = 1000 : 0.000318$ Why this step? Each 10× increase in $A R$ divides the drag by 10 — a clean $1/ A R$ decay you can see marching toward zero.

Step 2 — Take the formal limit. $lim_{A R \to \infty} \frac{C _{L}^{2}}{π e A R} = 0$ Why this step? An infinite wing has no tips. No tips ⇒ no leakage ⇒ no vortices ⇒ the flow is truly 2-D. Induced drag is a purely 3-D, tip-born effect, so in 2-D it must be exactly zero — and the algebra agrees.

Verify: The curve $1/ (π A R)$ hugs the axis but never crosses it — induced drag approaches zero, matching "2-D airfoil has no induced drag." ✓

Example 7 — Cell G: solving backward for the wing you need

Worked example Statement

A designer wants induced drag no worse than $C_{D, i} = 0.012$ while flying at $C_{L} = 0.9$ with an expected $e = 0.9$ . What is the minimum aspect ratio?

Forecast: We're inverting the formula. Since drag falls as $A R$ rises, there's a smallest $A R$ that just meets the target. Expect a mid-teens number.

Step 1 — Rearrange the formula for $A R$ . $C_{D, i} = \frac{C _{L}^{2}}{π e A R} ⟹ A R = \frac{C _{L}^{2}}{π e C _{D, i}}$ Why this step? $A R$ was buried in the denominator; multiply both sides by $A R$ and divide by $C_{D, i}$ to lift it out. Now it's an explicit design equation.

Step 2 — Plug in. $A R = \frac{0. 9 ^{2}}{π \cdot 0.9 \cdot 0.012} = \frac{0.81}{0.03393} = 23.87$ Why this step? Direct substitution of the three known numbers.

Step 3 — Interpret "minimum." Why this step? $A R = 23.9$ hits the target exactly. Any larger $A R$ gives less drag (better), so the minimum acceptable aspect ratio is $\approx 24$ — glider territory.

Verify: Feed $A R = 23.87$ back into the forward formula: $0.81/ (π \cdot 0.9 \cdot 23.87) = 0.012$ . Round-trip closes. ✓

Example 8 — Cell H: a real-world word problem (Newtons, not just coefficients)

Worked example Statement

A drone cruises at $V = 30 m/s$ , air density $ρ = 1.2 kg/m^{3}$ , wing area $S = 2.0 m^{2}$ , span $b = 6.0 m$ , $e = 0.9$ . Its weight is $W = 250 N$ and in steady level flight lift equals weight. Find the induced-drag force $D_{i}$ in Newtons.

Forecast: Induced drag is usually a few percent of weight in cruise. Guess a handful of Newtons.

Step 1 — Dynamic pressure $q$ . $q = \frac{1}{2} ρ V^{2} = \frac{1}{2} (1.2) (3 0^{2}) = 540 Pa$ Why this step? Coefficients turn into forces through $q S$ ; we need $q$ first. Units: $kg\cdotm^{- 3} \cdot m^{2} s^{- 2} = Pa$ . ✓

Step 2 — Lift coefficient from level flight. $L = W = 250 N \Rightarrow C_{L} = \frac{L}{q S} = \frac{250}{540 \cdot 2.0} = 0.2315$ Why this step? We're not given $C_{L}$ ; steady flight ( $L = W$ ) hands it to us. This is the classic word-problem link between forces and coefficients.

Step 3 — Aspect ratio. $A R = \frac{b ^{2}}{S} = \frac{6. 0 ^{2}}{2.0} = 18$ Why this step? The formula needs $A R$ , and we're given raw geometry ( $b$ and $S$ ) instead — so build it.

Step 4 — Induced-drag coefficient. $C_{D, i} = \frac{C _{L}^{2}}{π e A R} = \frac{0.231 5 ^{2}}{π \cdot 0.9 \cdot 18} = \frac{0.05360}{50.89} = 0.001053$ Why this step? Now every ingredient is in hand; this is the forward compute (Cell A) applied to real numbers.

Step 5 — Convert back to a force. $D_{i} = C_{D, i} q S = 0.001053 \cdot 540 \cdot 2.0 = 1.14 N$ Why this step? The question asked for a force in Newtons, so we undo the coefficient with the same $q S$ . Units: dimensionless $\times$ Pa $\times m^{2} = N$ . ✓

Verify: $D_{i} = 1.14 N$ is $1.14/250 = 0.46%$ of weight — small, exactly as expected for a high- $A R$ ( $= 18$ ) cruise drone. This tiny induced drag is precisely why long-span drones loiter so efficiently. ✓

Example 9 — Cell I: the exam twist (don't let terms "cancel")

Worked example Statement

Trap question: "A wing's aspect ratio is doubled and its lift coefficient is doubled. Since one is on top and one on the bottom, the induced drag is unchanged, right?"

Forecast: Suspicious — $C_{L}$ enters squared but $A R$ enters to the first power. They should not cancel. Guess: drag actually doubles.

Step 1 — Write the before and after symbolically. $C_{D, i}^{old} = \frac{C _{L}^{2}}{π e A R}, C_{D, i}^{new} = \frac{( 2 C _{L} ) ^{2}}{π e ( 2 A R )}$ Why this step? Substituting $2 C_{L}$ and $2 A R$ keeps the algebra honest instead of eyeballing "top vs bottom."

Step 2 — Simplify the ratio. $\frac{C _{D, i}^{new}}{C _{D, i}^{old}} = \frac{4 C _{L}^{2}}{2 A R} \cdot \frac{A R}{C _{L}^{2}} = \frac{4}{2} = 2$ Why this step? The $4$ comes from squaring the doubled $C_{L}$ ; the $2$ from the doubled $A R$ . They give $4/2 = 2$ , not $1$ . The square law wins.

Step 3 — State the corrected answer. Why this step? Induced drag doubles, it does not stay the same. The trap relies on forgetting that $C_{L}$ is squared.

Verify (concrete numbers): Start $C_{L} = 0.5, A R = 8, e = 1$ : $C_{D, i} = 0.25/ (8 π) = 0.009947$ . Now $C_{L} = 1.0, A R = 16$ : $C_{D, i} = 1.0/ (16 π) = 0.019894$ . Ratio $= 2.000$ . ✓

Recall Self-test: which cell is each of these?

"Wing at zero lift" ::: Cell E (degenerate $C_{L} = 0$ → $C_{D, i} = 0$ ) "Given a target drag, find the span" ::: Cell G (backward solve) " $C_{L}$ triples, drag ×9" ::: Cell C ( $C_{L}^{2}$ scaling) "Infinite/2-D wing" ::: Cell F ( $A R \to \infty$ , drag → 0) "Rectangle vs ellipse" ::: Cell D (span-efficiency $e$ effect)

Connections

Aspect ratio — effect on induced drag — the parent formula every example uses
Lifting-line theory (Prandtl) — where $C_{D, i} = C_{L}^{2} / (π A R)$ is born
Trailing vortices & downwash — the physics behind Cell E and F
Elliptical lift distribution — the $e = 1$ benchmark in Cell D
Drag polar — Example 8's forces live inside $C_{D} = C_{D, 0} + C_{L}^{2} / (π e A R)$
Parasite drag — the term that does not vanish at zero lift (Cell E)
Wingtip devices (winglets) — a way to raise effective $A R$ / $e$ in Cell G
Glide ratio & L/D max — why the $C_{L}^{2}$ growth of Cell C matters at low speed