Aspect ratio — effect on induced drag: Exercises

Before we start, let us re-earn every symbol so a reader who arrives cold can follow line one.

Definition The symbols we will use — nothing assumed

$b$ = span, the straight tip-to-tip distance of the wing (a length, in metres).
$S$ = planform area, the wing's shadow area seen from directly above (in $m^{2}$ ).
$c$ = chord, the front-to-back width of the wing at one spanwise station (a length). For a rectangular wing $S = b c$ .
$A R = b^{2} / S$ = aspect ratio, a pure number ("how many chords fit along the span"). Big = long and skinny.
$C_{L}$ = lift coefficient, a dimensionless "how hard is this wing working to lift" number. Larger at slow speed.
$e$ = span efficiency, a number $\leq 1$ measuring how close the lift's spanwise shape is to the ideal ellipse. $e = 1$ is perfect.
$C_{D, i}$ = induced-drag coefficient, the dimensionless drag that exists only because the wing makes lift.
$α_{i} = C_{L} / (π A R)$ = induced angle, the tiny angle (in radians) by which the tip vortices tilt the local airflow downward.
$q = \frac{1}{2} ρ V^{2}$ = dynamic pressure, the "push" of the moving air; $ρ$ is air density, $V$ is flight speed.

Figure — Aspect ratio — effect on induced drag

The figure above is our master reference: the black bar is a wing seen from the front, the orange curls are the tip vortices, and the red angle $α_{i}$ is the downward tilt of the flow. Every problem is a variation on reading this picture correctly.

Level 1 — Recognition

Here you only need to pick the right formula and plug in. No traps beyond arithmetic.

Problem 1.1

A rectangular wing has span $b = 10 m$ and constant chord $c = 2 m$ . Find its planform area $S$ and its aspect ratio $A R$ .

Recall Solution 1.1

WHAT: We want $S$ then $A R$ . WHY these: area comes first because $A R = b^{2} / S$ needs it. A rectangle's area is length $\times$ width: $S = b c = 10 \times 2 = 20 m^{2} .$ Now the aspect ratio: $A R = \frac{b ^{2}}{S} = \frac{1 0 ^{2}}{20} = \frac{100}{20} = 5.$ Check: for constant chord $A R = b / c = 10/2 = 5$ — same answer. Good.

Problem 1.2

State, in words and symbol, what happens to $C_{D, i}$ if you keep $C_{L}$ and $e$ fixed but triple the aspect ratio.

Recall Solution 1.2

WHAT: track the dependence. In $C_{D, i} = C_{L}^{2} / (π e A R)$ , with $C_{L}, e$ held fixed, $C_{D, i} \propto 1/ A R$ . Tripling $A R$ multiplies the denominator by 3, so: $C_{D, i}^{new} = \frac{1}{3} C_{D, i}^{old} .$ Induced drag drops to one third. This is exactly the "long thin wing wins" story from Aspect ratio — effect on induced drag.

Problem 1.3

An elliptical-loaded wing has $A R = 8$ and flies at $C_{L} = 0.6$ . Compute $C_{D, i}$ .

Recall Solution 1.3

WHY $e = 1$ : "elliptical loading" is the ideal case, so $e = 1$ and we may use the short formula. $C_{D, i} = \frac{C _{L}^{2}}{π A R} = \frac{0. 6 ^{2}}{π \times 8} = \frac{0.36}{25.13} = 0.01432.$

Level 2 — Application

Now you must rearrange the formula or chain two facts together.

Problem 2.1

A glider wing has $A R = 25$ , $e = 0.95$ , and must produce $C_{L} = 1.0$ in a slow thermalling turn. Find $C_{D, i}$ .

Recall Solution 2.1

Full formula because $e \neq = 1$ : $C_{D, i} = \frac{C _{L}^{2}}{π e A R} = \frac{1. 0 ^{2}}{π \times 0.95 \times 25} = \frac{1}{74.61} = 0.01340.$ Notice how tiny it is despite the high $C_{L} = 1.0$ : the enormous $A R = 25$ is doing the work. That is why sailplanes look the way they do.

Problem 2.2

A wing produces $C_{D, i} = 0.020$ at $C_{L} = 0.8$ with $e = 1$ . What is its aspect ratio?

Recall Solution 2.2

WHAT: solve for $A R$ . Start from $C_{D, i} = C_{L}^{2} / (π A R)$ and isolate $A R$ : $A R = \frac{C _{L}^{2}}{π C _{D, i}} = \frac{0. 8 ^{2}}{π \times 0.020} = \frac{0.64}{0.06283} = 10.19.$ Sanity check: plug $A R = 10.19$ back: $0.64/ (π \cdot 10.19) = 0.020$ . ✓

Problem 2.3

Two wings fly at the same $C_{L} = 0.7$ and same $e$ . Wing A has $A R = 6$ , wing B has $A R = 15$ . By what factor is B's induced drag smaller?

Recall Solution 2.3

With $C_{L}, e$ fixed, $C_{D, i} \propto 1/ A R$ , so the ratio is just the inverse ratio of the aspect ratios: $\frac{C _{D, i, B}}{C _{D, i, A}} = \frac{A R _{A}}{A R _{B}} = \frac{6}{15} = 0.40.$ B's induced drag is $0.40 \times$ A's — i.e. 60% smaller.

Level 3 — Analysis

Here you weigh induced drag against its rival, parasite drag — see Drag polar and Parasite drag.

Problem 3.1

The total drag coefficient is $C_{D} = C_{D, 0} + C_{D, i}$ where $C_{D, 0} = 0.015$ is the (roughly constant) parasite part. A wing has $A R = 9$ , $e = 0.85$ . At $C_{L} = 1.1$ , is induced or parasite drag larger, and what is total $C_{D}$ ?

Recall Solution 3.1

Induced part: $C_{D, i} = \frac{1. 1 ^{2}}{π \times 0.85 \times 9} = \frac{1.21}{24.03} = 0.05035.$ Compare: $C_{D, i} = 0.05035$ vs $C_{D, 0} = 0.015$ . Induced drag is larger — over 3× the parasite drag — because $C_{L} = 1.1$ is high (slow flight). Total: $C_{D} = 0.015 + 0.05035 = 0.06535.$

Problem 3.2

Same wing as 3.1 but now in fast cruise at $C_{L} = 0.3$ . Recompute $C_{D, i}$ and the total. Which term now dominates?

Recall Solution 3.2

$C_{D, i} = \frac{0. 3 ^{2}}{π \times 0.85 \times 9} = \frac{0.09}{24.03} = 0.003745.$ $C_{D} = 0.015 + 0.003745 = 0.018745.$ Now parasite drag dominates ( $0.015$ vs $0.00375$ ). The lesson: induced drag is a low-speed / high- $C_{L}$ problem; parasite drag is a high-speed problem. This is the whole reason gliders (slow, high $C_{L}$ ) crave high $A R$ while fighters (fast, low $C_{L}$ ) do not — see Glide ratio & L/D max.

The figure shows how the two drag pieces cross over as $C_{L}$ grows: the flat gray parasite line and the rising orange $C_{L}^{2}$ induced curve.

Level 4 — Synthesis

Combine the formula with real dimensional quantities: forces, span, area, speed.

Problem 4.1

An aircraft weighs $W = 60, 000 N$ , wing area $S = 30 m^{2}$ , span $b = 20 m$ , $e = 0.9$ . It flies straight and level (so lift $L = W$ ) at density $ρ = 1.0 kg/m^{3}$ and speed $V = 60 m/s$ . Find $A R$ , $C_{L}$ , and the actual induced-drag force $D_{i}$ in newtons.

Recall Solution 4.1

Step 1 — aspect ratio. $A R = b^{2} / S = 2 0^{2} /30 = 400/30 = 13.33$ . Step 2 — dynamic pressure. $q = \frac{1}{2} ρ V^{2} = 0.5 \times 1.0 \times 6 0^{2} = 0.5 \times 3600 = 1800 Pa$ . Step 3 — lift coefficient. Level flight means $L = W$ , so $C_{L} = \frac{L}{q S} = \frac{60000}{1800 \times 30} = \frac{60000}{54000} = 1.111.$ Step 4 — induced-drag coefficient. $C_{D, i} = \frac{C _{L}^{2}}{π e A R} = \frac{1.11 1 ^{2}}{π \times 0.9 \times 13.33} = \frac{1.235}{37.70} = 0.03276.$ Step 5 — turn the coefficient back into a force by reversing $C_{D, i} = D_{i} / (q S)$ : $D_{i} = C_{D, i} q S = 0.03276 \times 1800 \times 30 = 1769 N .$ So of the plane's total drag, about $1770 N$ is purely the price of making lift.

Problem 4.2

For the same aircraft, a designer proposes stretching the span to $b = 26 m$ (keeping $S = 30 m^{2}$ and everything else fixed). What is the new induced-drag force, and how much power (at $V = 60 m/s$ ) does the change save? (Power $= D_{i} V$ .)

Recall Solution 4.2

New aspect ratio: $A R^{'} = 2 6^{2} /30 = 676/30 = 22.53$ . $C_{L}$ is unchanged (same $W, q, S$ ) at $1.111$ , so $C_{D, i}^{'} = \frac{1.11 1 ^{2}}{π \times 0.9 \times 22.53} = \frac{1.235}{63.71} = 0.01938.$ New force: $D_{i}^{'} = 0.01938 \times 1800 \times 30 = 1047 N .$ Power saved: $Δ P = (D_{i} - D_{i}^{'}) V = (1769 - 1047) \times 60 = 722 \times 60 = 43, 320 W \approx 43.3 kW .$ A longer, thinner wing here saves over 43 kW of engine power — real fuel money. (In practice you'd then check the extra wetted area and structural weight; see the next level.)

Level 5 — Mastery

Full reasoning, optimisation, and edge cases. No hand-holding in the problem statements.

Problem 5.1 — The induced/parasite optimum

Total drag coefficient is $C_{D} = C_{D, 0} + \frac{C _{L}^{2}}{π e A R}$ . For a fixed weight and speed (so $C_{L}$ is fixed), induced drag falls with $A R$ but parasite drag rises because more span means more wetted area; model this as $C_{D, 0} = k A R$ with $k = 0.0016$ . Take $C_{L} = 0.9$ , $e = 0.9$ . Find the $A R$ that minimises total $C_{D}$ , and the minimum $C_{D}$ .

Recall Solution 5.1

WHAT: minimise $C_{D} (A R) = k A R + \frac{C _{L}^{2}}{π e} \cdot \frac{1}{A R}$ over $A R$ . WHY calculus: the function has one term growing with $A R$ and one shrinking; the minimum is where their slopes cancel, i.e. where the derivative is zero. Write $C = \frac{C _{L}^{2}}{π e} = \frac{0.81}{π \times 0.9} = \frac{0.81}{2.827} = 0.2865$ . So $C_{D} = k A R + C / A R$ . Differentiate and set to zero: $\frac{d C _{D}}{d A R} = k - \frac{C}{A R ^{2}} = 0 \Rightarrow A R^{⋆} = \frac{C}{k} = \frac{0.2865}{0.0016} = 179.1 = 13.38.$ Minimum total drag (note at the optimum the two terms are equal, each $= k A R^{⋆}$ ): $C_{D}^{⋆} = k A R^{⋆} + \frac{C}{A R ^{⋆}} = 2 k A R^{⋆} = 2 \times 0.0016 \times 13.38 = 0.04283.$ Interpretation: the sweet spot is where induced drag equals the extra parasite drag it costs to shrink it. Push $A R$ higher and parasite/weight penalties win — exactly the "higher AR isn't always better" caveat, now made quantitative.

Problem 5.2 — Winglet as an effective- $A R$ boost

A wing has geometric $A R = 8$ , $e = 0.80$ , flying at $C_{L} = 0.9$ . A winglet raises the effective span efficiency to $e^{'} = 0.92$ without changing $b$ or $S$ . Because $e$ and $A R$ enter only as the product $e A R$ , we can quote an effective aspect ratio $A R_{eff} = (e^{'} / e) A R$ at the old $e$ . Find (a) the drag reduction as a percentage, and (b) $A R_{eff}$ .

Recall Solution 5.2

(a) Since $C_{D, i} \propto 1/ e$ with everything else fixed: $\frac{C _{D, i}^{'}}{C _{D, i}} = \frac{e}{e ^{'}} = \frac{0.80}{0.92} = 0.8696.$ That is a reduction of $1 - 0.8696 = 0.1304$ , i.e. 13.0% less induced drag. (b) The product $e A R$ went from $0.80 \times 8 = 6.40$ to $0.92 \times 8 = 7.36$ . Expressed as an effective aspect ratio at the original efficiency $e = 0.80$ : $A R_{eff} = \frac{e ^{'} A R}{e} = \frac{0.92 \times 8}{0.80} = 9.2.$ The winglet makes an $A R = 8$ wing behave, drag-wise, like an $A R = 9.2$ one — without physically growing the span.

Problem 5.3 — The degenerate cases

Explain, using the formula and the physics, what $C_{D, i}$ does in each limit: (a) $C_{L} \to 0$ (no lift); (b) $A R \to \infty$ (infinite wing); (c) $e \to 0$ (pathologically bad lift shape). Confirm each is consistent with the parent note.

Recall Solution 5.3

(a) $C_{L} \to 0$ : numerator $C_{L}^{2} \to 0$ , so $C_{D, i} \to 0$ . Physics: no lift means no pressure difference, no tip leakage, no vortices — nothing to tilt. Matches "no lift ⇒ no induced drag." (b) $A R \to \infty$ : denominator $\to \infty$ , so $C_{D, i} \to 0$ . Physics: an infinitely long wing has tips infinitely far away — the 2-D case where induced drag is exactly zero (see Lifting-line theory (Prandtl)). This is why induced drag is a purely finite-wing, 3-D effect. (c) $e \to 0$ : denominator $\to 0$ , so $C_{D, i} \to \infty$ . Physics: $e = 0$ is an impossibly bad, spiky lift distribution that dumps all its energy into concentrated tip vortices. Real wings never get near it ( $e \approx 0.7$ – $1.0$ ), but the limit correctly warns that a poor spanwise shape is punished hard. The best possible case, $e = 1$ , is the Elliptical lift distribution.

Recall One-line self-test before you close the page

$C_{D, i}$ formula ::: $C_{D, i} = C_{L}^{2} / (π e A R)$ Convert coefficient to force ::: $D_{i} = C_{D, i} q S$ with $q = \frac{1}{2} ρ V^{2}$ Optimum $A R$ when $C_{D, 0} = k A R$ ::: $A R^{⋆} = C_{L}^{2} / (π e k)$ , and there the two drags are equal Three limits that give zero or infinite $C_{D, i}$ ::: $C_{L} \to 0$ (zero), $A R \to \infty$ (zero), $e \to 0$ (infinite)

Connections

Aspect ratio — effect on induced drag — the parent formula every problem uses
Drag polar — the $C_{D} = C_{D, 0} + C_{L}^{2} / (π e A R)$ split behind Levels 3–5
Parasite drag — the rival term in the optimisation
Wingtip devices (winglets) — the effective- $A R$ idea in Problem 5.2
Elliptical lift distribution — the $e = 1$ ideal
Lifting-line theory (Prandtl) — why $A R \to \infty$ kills induced drag
Glide ratio & L/D max — where the drag balance pays off