A bending moment never chooses a span

The loading nobody used took the bell-shaped loading, $\Gamma \propto (1-\eta^2)^{3/2}$, and stretched it until its root bending moment was back to the elliptic wing’s. That happened at five quarters of the span, where the induced drag was 64/75 of the elliptic wing’s — fourteen and a half per cent less for the same lift and the same moment at the root. Prandtl’s own 1933 comparison held a different structural quantity and found 22.5 per cent more span for 11.1 per cent less drag. Both are quoted as the optimum, and the earlier essay closed by noting, without solving it, that a real spar is sized by a distribution of moments rather than one.

Solving it properly turns up something the stretched bell hides. Hold the lift and a bending-moment measure, ask for the least induced drag, and let the span be free: there is no answer. The drag keeps falling as the span grows. What makes an answer appear is a second condition, one that neither quoted number mentions — that no part of the wing push down — and the span the answer stops at is fixed by that condition, not by the moment.

The two measures of a spar #

A spar’s weight is set by the bending it carries, and there are two honest stand-ins for it. The root moment is the moment the lift on one side makes about the centre line, $\int_0^s \ell\,y\,dy$. It is right for a spar sized by its worst station alone — a constant section along the span, strong enough at the root. The moment along the span is the bending moment integrated over the semi-span, $\int_0^s M(y)\,dy$. It is right for a spar of uniform depth whose flanges are tapered so that every station is stressed to its limit: the flange area at each station is proportional to the moment there, so the flange volume is the integral.

The second has an identity in it that takes two lines. The moment at $y$ is the lift outboard of $y$ times its arm, $M(y) = \int_y^s \ell(\xi)(\xi - y)\,d\xi$, and integrating that from the root to the tip by parts gives

$$\int_0^s M(y)\,dy \;=\; \tfrac12 \int_0^s \ell(y)\,y^2\,dy,$$

half the lift’s moment of inertia about the centre line — which is exactly the quantity Prandtl held in 1933. His constraint, which reads like a piece of mechanics borrowed from a flywheel, is the flange volume of a fully stressed spar of uniform depth. Computed by quadrature for every harmonic of the loading, the two sides agree to a part in $10^{15}$.

Written in Glauert’s series, $\Gamma = 2bU\sum A_n \sin n\theta$, the lift is $\pi q b^2 A_1$, the drag is $\pi q b^2 \sum n A_n^2$, and both moments are linear in the $A_n$. At a given span the least drag for a given lift and a given moment is a quadratic minimised under two linear constraints, solved exactly. The constraint’s value is the elliptic wing’s, of span $b_0$, carrying the same lift; everything is measured against that wing.

Neither has a best span #

The dashed curves are the free answers, and they only go down. For the moment along the span the calculation can be done in closed form, because the moment of inertia involves only the first and third harmonics — every higher harmonic integrates to zero against $y^2$ — so the best loading uses those two alone and the drag at span ratio $r$ is

$$\frac{D}{D_0} = \frac{1}{r^2} + \frac{3\,(1 - r^2)^2}{r^6},$$

which falls monotonically: 0.889 at $\sqrt{3/2}$, 0.79 at 1.7, 0.23 at four times the span, and to zero in the limit. The root moment’s curve, which uses every odd harmonic, does the same. The structure appears to buy unlimited drag reduction for a fixed spar.

It is not a numerical accident and it is not a paradox; it is the problem being asked carelessly. A long wing whose lift is held and whose moment is held has to carry its lift close to the root, and the cheapest way to do that while still spreading the wake across a wide span is to load the outboard wing negatively. A tip that pushes down reduces the moment about the root, allows more lift inboard, and spreads the trailing vorticity over a longer line — and a longer line is what the span is the whole story says induced drag is paid for.

The tips that push down #

At 1.6 times the elliptic span the free loading for the moment along the span is negative outboard of 1.31 semi-spans, reaching a download of 0.12 of the elliptic wing’s root load near the tip. That is a real aerodynamic loading — tips at negative incidence, pressing down — and the arithmetic is right that it has less induced drag than the bell. It is also something no designer builds, for a reason that has nothing to do with either measure: a tip that pushes down is carrying load the spar must also carry, in the opposite direction, and it is lift paid for twice.

At $\sqrt{3/2}$ the same calculation returns the bell exactly. Its third harmonic is $-1/3$ of its first, to a part in $10^{16}$, and $-1/3$ is the ratio at which $A_1\sin\theta + A_3\sin 3\theta$ stops being positive at the tip: near the tip it behaves as $(A_1 + 3A_3)\,\theta$, and $A_3 = -A_1/3$ makes that coefficient zero.

The stop is where the tip stops lifting #

So the honest problem has three constraints: the lift, the moment, and no section pushing down. Solved at every span with the third included — a convex programme with the load required non-negative at 160 stations and in its tip slope, by an interior-point method — it follows the free curve until the free curve’s tip slope reaches zero, and then stops. Past that span the best non-negative loading on a longer wing is zero over its outboard part, which is a shorter wing with some unused span attached, and it has the drag of the loading at the stop. The solid curves in the first figure are level from there on, and the programme’s own answers, the dots, sit on the level within the truncation of the series.

The tip slope makes the stopping spans exact. For the moment along the span it reaches zero at $r^2 = 3/2$, which follows from the two-harmonic closed form in one line: the moment of inertia fixes $A_1 + A_3 = A_{1,0}/r^4$, the lift fixes $A_1 = A_{1,0}/r^2$, and $A_3 = -A_1/3$ then gives $r^2 = 3/2$. The drag there is $\tfrac23 + \tfrac29 = \tfrac89$. Prandtl’s 22.5 per cent and 11.1 per cent are the span at which his loading’s tips stop lifting — not the place where the trade between drag and structure turns over, since it never does.

For the root moment the slope reaches zero at 1.3320 with sixteen harmonics, 1.3328 with twenty-four and 1.3330 with thirty-two, and the drag there is 0.8437524, 0.8437505 and 0.8437502: converging, slowly because the root moment’s weight has a corner at the centre line, on four thirds of the span for twenty-seven thirty-seconds of the drag. That is 15.6 per cent less induced drag than the elliptic wing, and 1.1 per cent less than the bell stretched to five quarters, which is the one per cent the loading nobody used found between the bell and the true optimum at a fixed span, now taken at the span where the optimum actually stops.

Two loadings, two spars #

The two stopping loadings differ in a way the measures explain. The bell, at $\sqrt{3/2}$, carries 1.09 of the elliptic root load at the centre and falls smoothly to a tangential zero. The root-sized loading, at 4/3, carries 1.12 at the centre and is fuller between a third and two-thirds of the span. The root moment charges each element of lift for its arm and nothing else, so lift at mid-span costs exactly its distance from the centre; the moment along the span charges for the square of the arm, so the same mid-span lift costs relatively more and the bell pushes it further inboard. A spar sized at the root is willing to carry lift farther out, and therefore reaches further for the same moment.

What each spar is carrying #

The moments show the two constraints at work. The root-sized loading meets the ellipse at the root, as it must, and then carries more moment outboard: its spar is no heavier at the root and heavier everywhere else. The bell’s root moment is lower, 0.980 of the ellipse’s, and its moment curve has the same area under it as the ellipse’s — 0.2945 against 0.2945 — which is what holding the moment along the span means. A uniform-depth spar built for the bell has the same flange material as one built for the elliptic wing, spread over 22 per cent more span.

That is also a warning about reading either answer as “the” structural optimum. The root-moment wing is lighter only if its spar is sized by the root alone; if its outboard flanges have to grow with the moment there, as a fully stressed spar’s do, it is the heavier of the two. The two numbers answer two questions about two spars, and a lighter spar turns a box wing into a biplane shows the same thing on a different wing: change what the structure charges for and the aerodynamic optimum moves.

A spar that is only kept from bending backwards #

Forbidding every download is stronger than a structure needs. What a spar cares about is that its moment keep one sign: a small download at the very tip is harmless if the lift inboard of it outweighs it in moment at every station. Solving with that weaker condition — the bending moment non-negative at 120 stations instead of the load — gives a curve that does not stop at 4/3. It keeps falling, but barely: 0.84352 at 1.7 times the span against the stopped 0.84375, three hundredths of a per cent for twenty-seven per cent more span. The no-download condition is not exactly what the structure demands, but the difference is invisible on any wing worth building, and the stopping span it gives is the practical one.

A spar that thins towards the tip #

A real wing is thinner at the tip than at the root, and a spar whose depth falls with the wing’s needs more flange for the same moment where it is shallow: the flange volume becomes $\int M/h\,dy$. That makes outboard moment dearer, and the loading moves inboard to avoid it. At the bell’s span a spar thinning to four-tenths of its root depth gives 0.867 of the elliptic drag against the bell’s 0.889; thinning to a fifth, 0.855. The more the spar thins, the more of the saving the structure can buy — and the further the true optimum is from the bell that a uniform spar produces.

This measure does not stop at a definite span in the same way, and the reason is a limitation worth stating. Tying the spar’s depth to the fraction of the span makes a longer wing’s spar deeper at every station, which costs nothing in this model, because the wing’s own weight, area and profile drag are not in it. Extra span is then structurally free, and the calculation will happily take more of it.

In metres, on a sailplane #

Put numbers on a sailplane of 350 kilograms with an elliptic wing of fifteen metres, thermalling at 25 metres a second at sea level. Its induced drag is $L^2/\pi q b^2$, 43.5 newtons. The same flange material in a uniform spar, spent on the bell, buys a span of 18.4 metres and 38.7 newtons; a spar sized at its root, holding the root moment instead, buys twenty metres and 36.8 newtons. Multiplied by the speed and divided by the weight, the first saving is 3.5 centimetres a second of sink and the second 5.0 — in a thermal rising at a metre or two a second, a few per cent of the climb. The cheapest way to stay up is the reminder that the induced drag is only the larger half of the bill at that speed: the extra three to five metres of wing carry their own skin friction, and the two drags a wing pays trade against each other across exactly this choice.

None of those spans is the end of the trade. Each is where the loading’s tips stop lifting; the extra structure that a still longer wing would need is not in the sum, and neither is its friction. A designer reading 18.4 metres off this calculation is reading the span at which one condition binds, and should know which.

Why the wake does not care how the tip got there #

A wing that leaves the plane records Munk’s theorem: the induced drag depends only on the trace of the wake in the cross-flow plane and the circulation along it, not on where along the flight direction the lift was made. That is why a download at the tip is not a pathology of the arithmetic. Seen from far downstream it is simply a wake whose circulation rises from the tip inwards in a different way, and the theorem prices it as it would any other. What forbids it is structural and practical, which is why no aerodynamic statement of the problem contains it — and why the price of having ends, which holds only the span, never meets the question.

How the spars were checked #

Every integral is taken in the Glauert angle, where the integrands are smooth, so that the identity between the moment along the span and half the moment of inertia is a check to rounding rather than an assumption: $10^{-15}$ for every harmonic. The root measure’s coefficients agree with their exact integrals, $\int_0^{\pi/2}\cos\theta\sin\theta\sin n\theta\,d\theta$, to $10^{-15}$. The bell at five quarters reproduces the earlier essay’s elliptic root moment and its 64/75 exactly. The free moment-of-inertia family matches its closed form to $10^{-15}$, stops at $\sqrt{3/2}$ to $10^{-16}$ with drag 8/9, and the no-download programme at that span returns $A_3/A_1 = -1/3$. The root measure’s stopping span is quoted at three truncations to show its convergence, which is slow and monotone.

The convention: one lift, one elliptic reference #

Every drag here is a fraction of the induced drag of the elliptic wing of span $b_0$ carrying the same lift in the same stream, and every constrained quantity is held at that wing’s value. Span ratios are against $b_0$. The wake is flat and the theory is Prandtl’s lifting line in Glauert’s form, good to a few per cent at the aspect ratios these wings have. “No download” means the section lift is non-negative at every station and the loading’s slope at the tip is non-negative.

What else stops a span #

The wing’s own weight and drag. A longer wing has more skin, more ribs and more wetted area, and its profile drag grows with its area; none of that is here, and in an aeroplane it is what really stops the span. The rest of the structure. Shear webs, torsion boxes and flutter margins size real spars too, and a fully stressed flange is an idealisation of a material used to its limit everywhere. The stall and the yaw. A bell-loaded wing’s outer panels carry little lift and stall late; whether that is a virtue or a waste of wetted area is a question which part of a wing stalls first asks for ordinary planforms. Twist. Every loading here is a loading, and a real wing reaches it with a twist that is right at one lift coefficient only, as one twist is chosen for the speed flown per metre prices across a whole flight. The exact root-moment numbers. The convergence on 4/3 and 27/32 is numerical, to seven figures; the clean values invite a closed form that is not derived here.

Two papers that were one problem #

Prandtl’s 1933 paper held the lift and the moment of inertia of the lift and found the bell; the integration by parts above, which takes a line, shows that his constraint is the flange volume of a uniform fully stressed spar. Robert T. Jones returned to the problem in 1950 at NACA with a bending-moment constraint and reached the same conclusion, and it was ignored again, as the loading nobody used recounts. Later treatments added shear and real spar sizing and found the answer moved, as it does here between the two measures. What none of the quoted numbers says outright is that the span they come with is not a turning point of any trade at all, but the span at which the loading’s tips stop lifting — and that the condition fixing it is a design choice no aerodynamic theorem supplies.

Still open: what does stop the span #

The span here stops at a condition on the loading, and in an aeroplane it is stopped by the wing. The next calculation gives the wing a weight of its own — skin and ribs proportional to its area, a spar proportional to the flange volume — and a profile drag proportional to its wetted area, and minimises the total drag, induced and profile, at a fixed all-up weight, with the loading and the span both free. It asks whether the optimum then sits short of the no-download stop, so that the tips still lift, or beyond it, so that a real minimum-drag wing would carry a download at its tips; and how the answer moves between a sailplane, whose wetted area is small against its span, and a light aeroplane, whose is not.