A spar that must lift itself stops the span at a quarter of the weight

A bending moment never chooses a span took the two celebrated answers to the least-drag wing with a structural limit — Prandtl’s bell of 1933, held to the elliptic wing’s lift and the moment of inertia of its lift, and the wing held to its root bending moment — and showed that neither problem chooses a span. Hold the lift and the moment and let the span grow, and the least drag keeps falling, the tips loading downward to carry the moment back to the root. The two famous spans, $\sqrt{3/2}$ and 4/3 of the elliptic wing’s, are where the least-drag loading’s tips stop lifting, and nothing in either problem says a tip may not push down.

That essay ended on what does stop a span in an aeroplane: the wing itself. A longer wing needs a heavier spar to carry the same lift the same distance, and the spar’s weight has to be lifted as well, so a longer wing must make more lift, and lift costs induced drag. The question is whether this, rather than a rule about downloads, picks a span and a loading — and which.

The spar in Glauert’s series #

Write the circulation as Glauert does, $\Gamma = 2bU\sum A_n\sin n\theta$ with $y = \tfrac{b}{2}\cos\theta$. The lift is $\pi q b^2 A_1$, the induced drag $\pi q b^2\sum nA_n^2$. A spar stressed to its limit at every station, of uniform depth $t$, has flanges whose area at each station is the bending moment there over $t$, and integrating by parts shows that their volume is half the lift’s moment of inertia about the centre line:

$$\int_0^{b/2} M\,dy = \tfrac12\int_0^{b/2}\ell\,y^2\,dy = \frac{\pi}{64}\,q\,b^4 A_1(1 + \varepsilon), \qquad \varepsilon = \frac{A_3}{A_1}.$$

Only the first and third harmonics appear; every other one integrates to zero against $y^2$. So the whole problem lives on two numbers, the span and the ratio $\varepsilon$: $\varepsilon = 0$ is the ellipse, $\varepsilon = -1/3$ is Prandtl’s bell, and anything more negative pushes down at the tips. The lift is $L = \pi q b^2 A_1$ and the induced drag in terms of it is

$$D_i = \frac{L^2\,(1 + 3\varepsilon^2)}{\pi q b^2}.$$

The spar weighs the flange volume times a material constant over the depth. That leaves one choice to make: what happens to the depth as the span changes.

The area held. An aeroplane’s wing area is usually fixed by its landing speed — the lift at the slowest permitted speed — and not by its span. Lengthen such a wing and its chord shrinks as $1/b$, its thickness with it, and the spar’s depth with that. The spar’s weight then goes as $b^5 A_1(1 + \varepsilon)$, and as a share of the lift it carries, as $\sigma = x^3(1 + \varepsilon)$ with $x$ the span in units of a length set by the material, the thickness ratio and the wing loading.

The chord held. A sailplane’s chord is set by other things — the Reynolds number, the space for the spar and the controls — and lengthening its wing adds area at the same chord. The spar’s depth is then fixed, its weight goes as $b^4$, and $\sigma = x^2(1 + \varepsilon)$. The new area also weighs something and drags through the air; those are charged separately below.

Either way the lift the wing must make is the payload plus the spar, $L = W_0/(1 - \sigma)$, and the quantity to minimise over span and loading together is $D_i \propto (1 + 3\varepsilon^2)/\big(x^2(1-\sigma)^2\big)$.

A quarter, whatever the loading #

Fix the loading and choose the span. With the area held, the drag is least where $x(1 - \sigma)$ is greatest, and $\tfrac{d}{dx}\big[x - x^4(1 + \varepsilon)\big] = 0$ gives $x^3(1 + \varepsilon) = 1/4$:

$$\sigma^* = \tfrac14.$$

At the best span the spar is a quarter of everything the wing lifts — the payload, the rest of the structure and itself — and that is true for every loading, the ellipse included, and for any material. With the chord held the same step gives $\sigma^* = 1/3$. The fractions are exact in the model and depend only on how fast the spar’s weight grows with span: if it grows as $b^n$ the best span spends $1/(n+1)$ of the lift on the spar. A lighter material, a deeper section or a lower wing loading changes the best span — all three move the length $x$ is measured in — and leaves the share alone.

That is how the structure stops the span, and it stops it without any rule about the loading. Lengthening the wing beyond that point costs more in the extra lift the extra spar needs than it saves by spreading the wake.

What a quarter would mean in an aeroplane #

A quarter is a great deal of spar. A transport aeroplane’s whole wing structure is about a tenth of its take-off weight and its spar a fraction of that; a sailplane’s wings, the most span-hungry structure flown, are about a third of its loaded weight, skins and ribs included. No aeroplane spends a quarter of its weight on bending material, and the reason is not that their designers miscounted. The weight-optimal span answers one question — what spreading the wake is worth against lifting the spar — and every real wing is stopped short of it by something the question leaves out: the profile drag of its own area, which the two drags a wing pays set against the induced drag at every speed; the cost of making it; the width of the gate it must park at; a flutter speed that a long, light wing pushes down. The quarter is a ceiling on how far span is worth buying, not a description of what is bought.

It is still the right reference, because it says which way every other cost pushes. All of them stop the span short of where the spar alone would, so in all of them the induced drag is a larger share of the total than at the weight-optimal span, and the case for shedding tip load — the case below — is weaker, not stronger. The span is the whole story of the induced drag only when the span is free; when something else has fixed it, the loading has less to buy.

The best loading, with the area held #

Now let the loading choose as well. At the best span for each $\varepsilon$, with $\sigma = 1/4$, the drag is $(1 + 3\varepsilon^2)\,(1 + \varepsilon)^{2/3}$ times the best ellipse’s, and setting its derivative to zero leaves a quadratic,

$$12\varepsilon^2 + 9\varepsilon + 1 = 0, \qquad \varepsilon^* = \frac{\sqrt{33} - 9}{24} = -0.1356.$$

The best wing sheds a little load from its tips, less than a fifth of the way from the ellipse towards the zero-slope tip at which the bell stops. Its span is $(1 + \varepsilon^*)^{-1/3} = 1.0498$ times the best ellipse’s, and its drag 0.9575 times. Five per cent more span, four per cent less drag — modest numbers, on the flat bottom of a minimum that the optimum that does not matter would recognise, and on a transport aeroplane four per cent of the induced drag is a percentage point of fuel. The tips still lift.

The other root of the quadratic, $\varepsilon = -(9 + \sqrt{33})/24 = -0.614$, is a maximum, and beyond it the drag falls again without limit as $\varepsilon \to -1$: tips that push down hard take the flange volume towards zero, the span runs off, and the problem held to a fixed moment, found unbounded, is still unbounded. With the area held the wing finds a genuine local optimum on the lifting side, separated from the runaway by a hump, so the rule against a download is not needed to find it — only to keep a search from wandering past the hump.

With the chord held, Prandtl’s bell #

Hold the chord, so that the spar’s depth does not change, and the best span carries a spar of a third of the lift. At that share the same derivative gives $9\varepsilon^2 + 6\varepsilon + 1 = 0$, which is $(3\varepsilon + 1)^2 = 0$: a double root at $\varepsilon = -1/3$. The drag at the best span falls monotonically all the way to the bell and on past it; there is no hump, and with downloads forbidden the best wing is exactly the bell, at $\sqrt{3/2}$ of the best ellipse’s span and $8/9$ of its drag.

Those are Prandtl’s numbers from a different question, and the reason is short. At the best span both the elliptic wing and the bell carry a spar of a third of the lift, so both carry the same lift for the same payload, and since the depth is the same their flange volumes — their moments of inertia of lift — are the same. Prandtl’s problem, minimum drag at a given lift and a given moment of inertia, is the weight problem with the spar’s depth fixed; his constraint was the spar’s weight all along, and his span, $\sqrt{3/2}$, is the span at which a wing whose spar must lift itself stops paying for more. The bell is not an answer to an artificial question. It is the answer for a wing whose spar is as deep at the longer span as at the shorter. The same trade decided the box wing whose lighter spar turned it into a biplane: once the spar’s weight is on the bill, the least-drag shape is the one that carries its bending most cheaply, not the one with the least drag for a given span. The loading nobody used was nobody’s choice because nobody held the depth.

The double root is worth a sentence of its own, because it says how the bell sits in the design space. A double root means the drag’s slope with respect to the loading vanishes at the bell and does not change sign there: approaching from the ellipse, the drag keeps falling all the way to the bell with a slope that dies away as it arrives. So for a wing whose spar keeps its depth, every loading between the ellipse and the bell is a step in the right direction, the last steps buy almost nothing, and the bell is where the slope finally reaches zero rather than a sharp minimum. A designer who stops short — a third harmonic of −0.25 instead of −0.33 — gives up very little. The Horten brothers’ flying wings of the 1940s and the bell-loaded Prandtl-D wings flown at NASA’s Armstrong centre in the 2010s used loadings near the bell for a reason that has nothing to do with drag, the favourable yaw an outboard washout produces; the weight argument says they were not paying for it in drag either.

What a wing’s own area costs #

With the chord held, a longer wing also has more area, so more skin and ribs to lift, and more surface to drag through the air. Both are charged here in the simplest way: skin weight in proportion to the span, and a profile drag in proportion to the span, measured by the share of the total drag it is at the optimum.

The answer moves smoothly from the bell back towards the ellipse. A wing with no profile drag at all is a bell; one whose profile drag is a tenth of its total sheds three-quarters as much tip load as the bell, gains fifteen per cent of span and saves nine per cent of drag; one whose profile drag is half its total is nearly elliptic and gains three per cent of span. The mechanism is that every extra metre of span now costs drag directly, through its area, as well as through the weight of its spar, and an elliptic loading is the one that gets the most from a given span.

That answers the question of whether a sailplane and a light aeroplane want different loadings, with a qualification. The flight condition decides, not the type of aircraft. For a wing with a parabolic polar, the profile drag is exactly half the total at the best lift-to-drag ratio; it is less when flying slower and more when flying faster. A sailplane circling in a thermal flies slowly, with induced drag dominant, and the weight-optimal wing for that condition is near the bell; the same sailplane running between thermals at twice that speed, with profile drag dominant, wants a wing near the ellipse. A light aeroplane cruising far above its best-glide speed sits further still towards the ellipse. A designer optimising for one condition gets one loading; the others pay for it.

The convention, and what the model leaves out #

The induced drag throughout is the price of having ends, computed in the Trefftz plane. Every drag is the total at a fixed payload and dynamic pressure, in units of the best elliptic wing’s under the same costs, so that the comparison is between best wings and not between a best wing and an arbitrary one. The spar is a fully stressed flange pair of uniform depth along the span, sized by bending alone; the “area held” case scales its depth with the chord, the “chord held” case holds it. The loading is restricted to the two harmonics the spar’s weight can see; any other harmonic only adds induced drag, so including it would not change an optimum.

The structure is idealised in the ways that matter most at the tips. A real spar is not stressed to its limit everywhere — its outer part is sized by stiffness, by gust loads, by the minimum gauge of a skin that can be built — so the outboard wing is heavier than the flange volume says, which favours shedding tip load further. The spar carries no torsion and no shear, and the wing’s own weight is not spread along the span to relieve the bending, as it is on a real wing whose fuel sits in the outer panels; that relief lowers the flange volume of any loading and favours a longer span. The wake is flat and the induced drag is Trefftz’s, so the curvature of a heavily loaded wing in flight is ignored. And the rule against downloads is a convention the model needs when the area is held only past the hump at −0.614, and when the chord is held, at the bell itself.

Who connected weight to span #

Prandtl’s 1933 paper posed the bell problem as a minimum of drag at given lift and moment of inertia, motivated by structural weight, and its constraint is the flange volume of a uniform fully stressed spar, as integrating the moment by parts shows. Robert T. Jones returned to the problem at NACA in 1950 with the root moment and then with the weight of a wing that has to carry its own spar, and found the bell’s neighbourhood again. The relation between a structure’s weight and the span it should have is older than either: it is the reasoning behind every sailplane designer’s trade between a longer wing and a heavier one, and the quarter and the third here are the simplest forms of it, in which the spar’s weight grows as a single power of the span.

How it was checked #

With the area held the numerical optimum — a scan over the loading ratio refined by golden section, with the span chosen by golden section at each — matches $(\sqrt{33} - 9)/24$ to five parts in a billion, and both it and the best ellipse carry a spar of a quarter of the lift to a part in a billion; the span and drag ratios match $(1 + \varepsilon)^{-1/3}$ and $(1 + 3\varepsilon^2)(1+\varepsilon)^{2/3}$ to the same order. With the chord held it lands on $\varepsilon = -1/3$ to two parts in a million — a double root, so located less sharply — and on $\sqrt{3/2}$ and 8/9, the drag to rounding, with spars of a third. The drag at the best span for loadings that push down falls from 1.10 at −0.5 to 0.50 at −0.95 of the ellipse’s, which is the unbounded fall the rule against downloads is there to stop. With a profile drag charged the best ratio moves monotonically towards zero. And the model refuses a wing of no span, a negative span and a spar held some way it does not know.

Still open: the spar that stiffness sizes #

The spar here is sized by strength alone, so its outboard flanges thin to nothing where the moment does. A real outer wing is sized by stiffness — enough torsional and bending rigidity to keep flutter beyond the flight envelope and the tip’s deflection within limits — and its weight there does not fall with the moment. The next calculation gives the spar a stiffness floor: a minimum flange area along the span, scaled from the flutter speed of the shake that is not resonance, and charges its weight as lift. It asks how much that floor moves the best loading towards the bell, whether the quarter at the best span survives once part of the spar’s weight no longer depends on the loading, and at what floor the outer wing becomes cheaper to delete than to carry.