A stiffness floor keeps the span and gives the tips their load back

A spar that must lift itself charged a wing’s spar to its lift. A spar stressed to its limit at every station weighs in proportion to the volume of its flanges, which is half the lift’s moment of inertia about the centre line; only the first and third of Glauert’s harmonics reach that integral, so the whole problem lived on the span and one ratio, $\varepsilon = A_3/A_1$. The best span came out where the spar weighs exactly a quarter of everything the wing lifts, for every loading and every material, and the best loading shed a little load from the tips: $\varepsilon^* = (\sqrt{33} - 9)/24 = -0.136$, a sixth of the way from the ellipse to Prandtl’s bell, for four per cent less drag than the best elliptic wing.

It ended on the assumption that made the arithmetic clean. A spar sized by strength alone thins to nothing at the tips, where the moment does, and no real outer wing is built that way. The outer wing of a sailplane or a transport is sized by stiffness: it must be rigid enough in bending and torsion to keep flutter, the shake that is not resonance, beyond the flight envelope, and to keep the tip’s deflection within what the ailerons and the ground clearance allow. That sets a minimum spar along the span whose weight has nothing to do with the load it carries.

A floor under the spar #

The model keeps everything the earlier essay had — the area held, so the chord and the spar’s depth fall as one over the span; the loading’s first and third harmonics; the spar’s weight lifted with the payload — and adds one thing. At every station the spar is the heavier of two: the strength-sized spar, whose weight per unit span follows the bending moment, and a floor, a fixed weight per unit span set by stiffness. The floor’s size is written $f$: the weight a spar of floor weight would have along the whole semi-span at the reference span, over the payload’s weight, so that it grows in proportion to the span as a weight per metre should.

The bending moment is no longer enough to describe the spar, so it is computed station by station from the loading, $M(\eta) = \int_\eta^1 \ell(\xi)(\xi - \eta)\,d\xi$, and the spar’s weight is $\int \max(\text{strength}, \text{floor})$ along the span. The lift carried is the payload plus that weight, solved as a fixed point, and the drag to be least is the induced drag of that lift on that span and loading, $L^2(1 + 3\varepsilon^2)/x^2$ in the earlier essay’s units. The tips are still forbidden to push down.

What the floor does is easy to see in the figure. Inboard, where the bending moment is large, the strength-sized spar is the heavier and the floor is irrelevant. Outboard, where the moment falls towards zero, the floor takes over. At a floor of 0.4 the crossing is at 47 per cent of the semi-span, so the outer half of the wing carries a spar whose weight does not depend on how much lift that half makes.

The tips get their load back #

That one fact moves the answer. The earlier essay’s best loading shed tip load because a lighter outer load meant a smaller bending moment everywhere inboard of it, and so a lighter spar — every unit of lift moved inboard bought a reduction in flange volume weighted by its distance from the root. With a floor, part of that reduction is no longer available: wherever the floor governs, a smaller moment saves nothing, and the strength-sized part inboard of the crossing still saves, but on a shorter length.

So the best loading retreats towards the ellipse as the floor rises: from −0.136 with no floor to −0.117 at a floor of 0.2, −0.075 at 0.8 and −0.028 at 3.2, by which point the floor governs nearly nine tenths of the semi-span and the loading is all but elliptic. The structural case for unloading the tips is a case about the strength-sized part of the spar, and a stiffness-sized outer wing has removed most of it. The ellipse, which is the answer when the spar costs nothing, is also nearly the answer when the spar costs the same whatever the load, and the bell is the answer only in the narrow case where every part of the spar is paid for by the moment it resists.

The loadings show it at the station a designer would look at. At three quarters of the semi-span the strength-sized optimum carries 48 per cent of the root’s lift per unit span, against the ellipse’s 66; at a floor of 0.8 it carries 56, and at 3.2, 62. The tip load that the first spar essay took away comes back in proportion to how much of the outer wing the floor governs.

Why a bell meets its floor sooner #

How far inboard the floor reaches is set by how fast the bending moment dies away towards the tip, and that is fixed by the shape of the loading there. An elliptic loading falls to zero at the tip as the square root of the distance from it; integrated twice, the moment falls as that distance to the power five halves, and halving the distance cuts the moment by a factor of 5.6 — the computed moments at a tenth and a twentieth of the semi-span from the tip are in exactly that ratio. The bell’s loading arrives at the tip with zero slope, falling as the distance to the power three halves, so its moment falls as the seven-halves power and halving the distance cuts it elevenfold.

So the floor’s reach, for a small floor, grows as the two-fifths power of the floor on an elliptic wing and the two-sevenths power on a bell, and at the same span and the same floor the bell is floor-bound over more of its length. At the best strength-sized span, a floor of 0.1 governs 28 per cent of an elliptic wing’s semi-span and 42 per cent of a bell’s; a floor of 0.2, 37 per cent and 52. The bell’s whole point is a lighter outer spar, and it is the bell’s outer spar that the floor replaces first. That is the mechanism behind the retreat: it is not that the floor penalises the bell, which carries the same lift on the same span, but that the floor removes the saving the bell was chosen for, and removes it from the bell before it removes it from the ellipse.

The quarter does not survive #

The earlier essay’s quarter was a statement about how fast the spar’s weight grows with span — as $b^5$ with the area held, against the lift’s $b^2$ — and the floor’s weight grows differently, in proportion to the span alone. With two parts growing at two rates, the share of the lift spent on the spar at the best span is no longer fixed.

It rises: to 28 per cent at a floor of 0.2, 40 per cent at 0.8 and 68 per cent at 3.2. A wing whose spar is two thirds of its lift is a wing that can barely carry anything, and at a floor a little above six there is no span at which it carries the payload at all. Read the other way, the earlier essay’s quarter was the lowest the spar’s share can be at the best span, reached only by a spar that is sized by its load everywhere. Every requirement that adds weight the load does not set raises it.

What the floor costs in drag #

The floor is not free, and the drag says how much it costs once the wing has made the best of it. At a floor of 0.2 the best wing’s induced drag, at the same payload, is 8 per cent above the strength-sized optimum’s; at 0.8 it is 56 per cent above. Almost none of that is recovered by the loading: the best loading under a floor of 0.8 saves only a per cent and a half over the ellipse under the same floor. The extra drag is the floor’s weight being lifted, and the only lever on it is the floor itself — a lighter way of being stiff.

That is the practical content of the result. Once the outer wing is stiffness-sized, the aerodynamic designer’s choice of loading has little left to give, and the structural designer’s choice of how to provide stiffness — material, section depth, where the torsion box closes — has most of it. A composite outer wing that halves the floor’s weight is worth more to the induced drag than any loading the floor still permits. The loading question that the bell has made famous is, for such a wing, a question about the last per cent.

A weight per metre cannot stop a span #

The surprise is what the floor does not do. A heavier outer wing sounds like a reason to build a shorter one — the earlier essay asked at what floor the outer wing becomes cheaper to delete than to carry — and the best span hardly moves.

Across a range of floors that takes the spar’s share from a quarter to two thirds, the best span changes by less than one and a half per cent. The reason is in the induced drag’s dependence on the span loading, the lift over the span: the drag of a given loading goes as $(L/b)^2$. A spar of fixed weight per metre adds a weight proportional to $b$ to the lift, and so adds a constant to $L/b$ — lengthening the wing adds floor weight exactly as fast as it adds span to spread it over. A constant added to the span loading cannot make a longer wing worse; whatever stopped the span without a floor still stops it, at the same place. The strength-sized spar, whose weight grows as $b^5$ and so adds $b^4$ to the span loading, is what stops the span, and the floor only raises the drag at the span the strength chose.

Written out, with $W_0$ the payload, $w$ the floor’s weight per metre where it governs and $S(b)$ the strength-sized spar, the span loading is

$$\frac{L}{b} = \frac{W_0 + S(b) + w\,\lambda b}{b} = \frac{W_0 + S(b)}{b} + w\lambda,$$

with $\lambda$ the share of the span the floor governs, which changes slowly. The first term is the earlier essay’s, falling as the payload is spread and rising again once the spar’s $b^5$ takes over; the second is a constant. The minimum of the first term is where it was, and the floor shifts the curve up without tilting it. The small movement the figure shows comes from $\lambda$, which grows a little as the span grows and the moment at each station falls relative to the floor — enough to move the span by a per cent, not enough to stop it.

So the outer wing never becomes cheaper to delete than to carry, on induced drag and weight alone. What could make it so is a cost that grows faster than the span: the profile drag of its own area, which the earlier essay charged with the chord held, or a floor that rises towards the tip — and a flutter requirement does rise outboard, because the outer wing’s torsional deflection is what lets it flutter, and the stiffness needed to stop it is set by the tip’s twist rather than the root’s.

What a floor of 0.8 means #

The floor’s number is the weight of a spar of floor weight run along the whole semi-span, at the span the strength-sized wing would choose with no floor, over the payload: a floor of 0.8 means such a spar would weigh about half the payload. That is a heavy floor for a transport, whose outer-wing structure is a few per cent of its weight, and a light one for a sailplane’s wing, whose long, thin outer panels are sized almost entirely by stiffness and whose wings weigh a third of its loaded weight. The two kinds of aircraft therefore sit at different ends of the figures: the transport near the strength-sized optimum, where unloading the tips is still worth a few per cent and is used, and the sailplane near the ellipse, where it is worth a fraction of one and where planforms are drawn to an elliptic loading.

That reading agrees with practice better than the strength-sized model did. Long-span sailplanes are commonly designed close to elliptic loadings and do not wash out their tips heavily; transport wings carry noticeably more washout and inboard-biased loadings, which the bell’s advocates read as partial bells and the strength argument reads as a moment saving that is still available because their outer structure is strength-sized. Both readings are the same calculation at different floors.

What the model holds fixed #

The convention is the earlier essay’s: the wing’s area is held, so its chord and the spar’s depth fall as one over the span; the loading is represented by Glauert’s first and third harmonics, which carry the lift and the flange volume; lengths are in the structural reference length that the material, the thickness ratio and the wing loading set, and drags in units of the best elliptic wing’s without a floor. The floor is uniform along the span. A floor scaled from a flutter speed would be heavier outboard, and that would push the loading towards the ellipse harder than a uniform one. The floor does not depend on the span either, though stiffness requirements for a longer wing are larger; a floor that grew faster than the span would, unlike this one, stop the span. Profile drag is not charged, and the tips are not allowed to push down — the earlier essay showed that rule is needed only to keep a search from wandering past a hump. The bending moment is computed on 1,600 stations, which matches the flange volume’s closed form to a few parts in a hundred thousand; the best loading and span are found by a scan and golden-section searches.

Who argued about the bell #

Prandtl’s loading of 1933 came from a fixed moment about the root, and R. T. Jones’s of 1950 from a fixed integral of moment along the span, which is the flange volume here. Klein and Viswanathan added the spar’s own weight in the 1970s; the stiffness-sized outer wing, and the observation that it takes the tip-unloading argument away, are part of the structural optimisation literature that followed, in which wing designs are sized by strength inboard and by aeroelastic constraints outboard. The model is the earlier essay’s with a floor added; its results are what that addition does, not a survey of any of those designs.

How it was checked #

The bending moment, integrated from the tip on 1,600 stations, gives a flange volume matching $(\pi/32)(1 + \varepsilon)$ to $3 \times 10^{-5}$ for the bell, the ellipse and a tip-loaded wing. With no floor, the scan and searches recover the earlier essay’s exact optimum, $\varepsilon^* = (\sqrt{33} - 9)/24$, to $10^{-8}$, and the spar’s share of a quarter to $10^{-8}$. The best span’s insensitivity to the floor is checked against the span-loading argument: adding a weight proportional to the span leaves the drag of every loading a function of $1/x + \text{const}$, which has no new minimum. The model refuses a span of zero and a negative floor.

Still open: a floor set by the tip’s twist #

The floor here is the same at every station, and a flutter requirement is not: the outer wing twists most, and the stiffness that keeps its twist below what makes it flutter is set near the tip. The next calculation would take the floor from the coupled bending and torsion of a wing of the given span and loading, with the torsional stiffness each station needs to hold the flutter speed above the dive speed, and ask whether a floor that rises outboard stops the span — the uniform floor cannot — and at what flutter margin a sailplane’s best span is set by flutter rather than by strength.