A lighter spar turns a box wing into a biplane

A wake that closes on itself solved for the best loading on a box wing — two wings joined at their tips by vertical fins — and found Prandtl’s result of 1924: at a gap of a fifth of the span the system has two-thirds of the induced drag of an elliptic monoplane of the same span and lift, and the drag keeps falling as the gap grows, with no floor. It ended by naming the obvious next question. Every drag in it was the least drag at a stated lift and nothing else, and no real wing is free to put its lift anywhere.

The loading nobody used asked that question of a flat wing. Lift carried further out bends the root harder, and the root’s moment is what sizes the spar, so a designer minimising weight as well as drag should constrain the moment. With the moment held and the span left free, the answer is Prandtl’s bell-shaped loading of 1933: a longer wing, lighter at the tips, with less drag than the ellipse at the same structural cost.

This essay asks the same question of the box, and answers it twice. The first answer is the one the box’s own numbers give, and it is flattering: a box wing pays about half what a monoplane pays for each unit of moment it gives up. The second comes from asking what the fins have to do with that, by taking them off — and it is that they have very little to do with it, because a lighter spar is paid for with exactly the saving the fins provide.

The problem, stated once for three wings #

Far behind any system of lifting surfaces, the induced drag is the kinetic energy of the crossflow in a plane across the wake, and that energy is a quadratic form in the circulations shed along the wake’s trace. That is Munk’s result, and a wing that leaves the plane used it as a tool: the least drag at a given lift is the minimum of a quadratic form under a linear constraint. Adding a second linear constraint — on the moment of the lift about the centreline — adds one row and one column to the same system. Nothing else changes, and nothing about how the wake later rolls up enters.

Three traces are solved on the same lattice. For the monoplane the trace is a straight line. For the biplane it is two parallel lines a gap apart, and the biplane matters here only as the box with its fins removed. For the box it is a rectangle, which brings one complication with it: a circulation that runs round the whole box sheds nothing and carries no force, so the system is singular until that gauge is fixed, which is done here at a mean circulation of zero. The constant mode carries no moment either, so the bending constraint leaves the gauge exactly as free as before.

The moment constrained is the moment of the lift about the centreline, summed over both wings. That is the aerodynamic quantity the monoplane problem uses, which makes the three comparable. How that moment is shared between the upper and lower spars, and how much the fins relieve it by carrying load from one to the other, depends on the structure joining them, and is a separate calculation this one does not make.

What every wing pays #

The picture has three facts in it, and they are worth taking one at a time.

The monoplane’s parabola has its minimum at the ellipse, by definition, and rises symmetrically either side: loading the lift further in or further out both cost drag. The box minima sit to the right: a box wing’s unconstrained optimum bends its root harder than the ellipse does — 6 per cent at a gap of a tenth, 8 per cent at a fifth, 10 per cent at two-fifths. And the box parabolas are flatter. That third fact is the one that looks like the result, because it decides what a box wing costs when the moment is held down.

That every curve is a parabola is no accident. The drag is a quadratic form, the constraints are linear, and the least value of a quadratic under a moving linear constraint is itself quadratic in how far the constraint has moved. It is the same sum of squares that makes the elliptic optimum so flat near its bottom, seen from further out.

Eight, for the monoplane #

Plotted against the square of the moment reduction, every penalty is a straight line through the origin, so the added drag is a single coefficient times the reduction squared. For the monoplane the coefficient is eight, to within a hundredth of a per cent at four reductions from 5 to 30 per cent:

$$\frac{D}{D_\text{ell}} = 1 + 8\,(1 - f)^2,$$

where $f$ is the moment as a fraction of the elliptic one. A monoplane made to bend its root 10 per cent less than the ellipse pays 8 per cent more induced drag at the same span; 20 per cent less, 32 per cent more.

The number has a quick approximate derivation that shows where it comes from. Write the circulation as the ellipse plus a third harmonic $\sin 3\theta$ in the usual spanwise angle of the lifting line. The lift sees only the first harmonic. The moment sees both — $\tfrac13$ of the first’s amplitude and $\tfrac15$ of the third’s — so holding the moment to $f$ fixes the third harmonic at $\tfrac53(f - 1)$ of the first. The drag is the sum of each harmonic’s amplitude squared times its order, so the added drag is $3\cdot\tfrac{25}{9}(1 - f)^2 = 8.33\,(1-f)^2$. The full optimum can use every odd harmonic rather than only the third, and does slightly better: the lattice, which is free to use all of them, returns 8.00.

The box’s coefficients are a little under half of that at practical gaps: 4.9 at a tenth of the span, 3.9 at a fifth, 3.2 at two-fifths. As the gap shrinks towards nothing the box becomes two coincident wings, which is a monoplane again, and its coefficient climbs back towards eight: 6.0 at a gap of a twentieth. Read at face value, the figure says that moving a box wing’s lift inboard is cheaper by a factor of two, and it is tempting to credit the closed loop — to say that circulation a monoplane would shed as it peaks its load at the root can instead be carried round the corners through the fins. The way to test that is to take the fins off.

Four, for any two wings #

Most of the halving belongs to having two wings, and there is a one-line reason. Put two wings so far apart that neither feels the other’s downwash. Each carries half the lift, and because induced drag goes as the square of the lift, each has a quarter of the monoplane’s elliptic drag. Each carries half the root moment as well, so a reduction to $f$ of the total is the same fractional reduction on each, and each pays eight times its own drag times $(1-f)^2$. Two of them pay $2 \times \tfrac14 \times 8 = 4$. Two wings far apart pay exactly half a monoplane’s penalty, and the lattice gives 4.00 at a gap of two spans.

Brought closer, the two wings do feel each other. Each sits in the other’s downwash, which raises the whole system’s drag above the far-apart half and raises the penalty too, until at zero gap they are one wing and the coefficient is eight again. The biplane’s curve runs between those limits: 6.3 at a gap of a twentieth, 5.5 at a tenth, 4.7 at a fifth, 4.2 at two-fifths, 4.0 at four-fifths. It does not matter, for this, where the two wings sit fore and aft, for the reason the stagger theorem gives: only the wake’s cross-section enters.

So at a gap of a fifth the monoplane’s eight becomes 4.7 by adding a second wing, and 3.9 by adding the fins. The second wing does four-fifths of the work. The fins do the rest, and they do more of it as the gap opens — 4.2 against 3.2 at two-fifths, 4.0 against 2.9 at four-fifths — which is where the box’s coefficient passes below the four that no pair of separate wings can reach. That part is real, and it needs explaining. The explanation turns out to undo it.

Where the fins’ saving lives #

The unconstrained box carries a nearly flat circulation that stays high right out to the tips, where it turns the corner into the fins. That is what the fins are for. A biplane’s wings must each take their circulation to zero at the tip and shed it there as a concentrated trailing vortex; a box’s wings can end with circulation still on them, because the fin carries it round to the other wing. The fins lift nothing — lift is circulation times a horizontal length, and a fin has none — but they move where the wake’s vorticity sits, and that is the whole of their saving. At a gap of a fifth, the free box has 0.067 less drag than the free biplane.

That saving is made at the tips, and load at the tips is moment at the root. So it is exactly the first thing a moment constraint takes away. Held to the elliptic monoplane’s moment, the box’s fins carry a side force of 0.037 in units of the lift, against 0.049 free, and its lead over the biplane has fallen to 0.045. At 90 per cent of the moment the side force is 0.021 and the lead 0.015. At 80 per cent the side force is 0.006, the lead is 0.001, and the loading in the figure has pulled so far inboard that there is almost nothing at the tip for a fin to carry round.

This is why the box’s parabola is shallower, and why the shallowness flatters it. A penalty coefficient is measured from each wing’s own unconstrained optimum. The box starts further to the right, at 1.08 of the monoplane’s moment against the biplane’s 1.03, and further down, with the fins’ saving in hand. Pulling its moment in, it spends that saving first — and a saving being spent looks, on a plot of added drag, like a penalty that is unusually small. By the time the moment reaches 80 per cent both wings arrive at the same drag, and the box has travelled further to get there. A longer run for the same rise is a smaller coefficient. It is not a cheaper redistribution.

So the fair comparison is not between coefficients measured from different starting points. It is between the three wings at the same moment, and on that comparison the answer is plain: at the monoplane’s own moment the box still leads the biplane by two-thirds of its free advantage; at a spar a fifth lighter it leads by a thousandth.

The downwash that marks the optimum #

There is a check on whether the lattice has found a true optimum, and it comes from Munk. At the least-drag loading under a set of linear constraints, the downwash normal to the wake is a combination of the constraints’ own weightings: a constant for the lift, whose weighting is the same everywhere, and a term proportional to the distance from the root for the moment, whose weighting is that distance. So the optimum downwash on the wing is a straight line, $w = a + b\,|y|$.

The computed downwash lies on a straight line to within a small fraction of its range, with the only departures at the outermost points where the wing’s lattice meets the fin’s. That is not something the solver was told. It was told to minimise a quadratic form, and a straight-line downwash is what minimising that form under those constraints produces. The line’s slope is the price of the moment constraint made visible: at the free optimum it is flat, and the harder the moment is held down, the more steeply the downwash has to rise towards the tips to keep lift away from them.

The spar both of them can carry #

The design question a box wing’s advocate would actually ask is how much structure its drag advantage can be exchanged for. Holding the drag at the elliptic monoplane’s and minimising the moment instead gives the answer directly. At a gap of a fifth of the span, a box wing can bend its root 21 per cent less than an elliptic monoplane and have the same induced drag. At a gap of a tenth it can bend it 15 per cent less; at two-fifths, 28 per cent; at three-fifths, 31.

And a biplane of the same gap can do exactly the same: 0.7922 against the box’s 0.7921 at a fifth of the span, and within a point of it at every gap computed, from 0.890 against 0.886 at a twentieth to 0.675 against 0.668 at four-fifths. The break-even moment is a statement about how much the second wing saves, because by the time the moment has come down that far the fins have handed their saving back.

That is a sharper result than the one the coefficients suggested, and it cuts in both directions. For the box, it means that an aircraft built to the drag of a monoplane with a lighter spar would get nothing from its fins that a strut-braced biplane of the same span and gap would not also get. For the biplane, it means an old configuration was carrying an exchange rate between drag and structure that its drag at the unconstrained optimum — 0.738 of the monoplane, well behind the box’s 0.669 — never showed. What the fins buy is a lower drag at a heavier spar. Whether they buy anything at a lighter one depends on the structure, which is where this calculation stops.

The checks #

The monoplane’s coefficient of eight is the check that the constrained solver is right, because the flat wake has no corners and no gauge, and its lattice reproduces the continuous result to within a hundredth of a per cent. The biplane at a gap of two spans returning four is the second check, and it is a check on the same solver applied to a trace in two pieces.

The box has one known defect, and it is carried forward rather than hidden. A circulation round the box should cost nothing, and on this lattice it moves the drag by 0.40 per cent — the reciprocity defect at the box’s corners that the closed-wake essay measured and bounded, and that refining the lattice does not remove. Every box number here is quoted at the mean-zero gauge. The box’s lead over the biplane at 80 per cent of the moment, a thousandth of the monoplane’s drag, is smaller than that defect, and so that claim is made in the form the defect cannot overturn: the lead has fallen from 0.067 to within the lattice’s own uncertainty. The break-even moments are safer. The gauge carries no moment, so it can move them only through the drag, and near the break-even the box’s drag changes by about 1.8 for each unit of moment — so a 0.4 per cent drag defect moves the break-even by about 0.002, a fifth of a point, against an agreement of a ten-thousandth.

What the lattice cannot show #

The structure. The moment constrained is the lift’s moment about the centreline. A box wing’s upper and lower spars are joined at the tips, so the structure is statically indeterminate: how the moment divides between the two spars, and how much the fins take by carrying load between them, depends on their stiffnesses. This is where a box could still earn its fins — structurally rather than aerodynamically — and deciding whether it does needs a structural model.

Wing weight beyond the root. The root moment is a proxy for spar weight. The real weight depends on the moment distribution along the whole span and on the depth of the spar, and two thin wings have less depth each than one thick one.

Profile drag. A box wing has two wings and two fins, so more wetted area than the biplane and much more than the monoplane. Only induced drag is computed, and the other half of the bill favours whichever wing has less surface.

The span. The span is held fixed, which is the fair comparison for a vehicle with a span limit — a stand at an airport — and not the comparison Prandtl’s bell loading makes. The span is the whole story for induced drag, and letting it grow at fixed moment is a different calculation.

Viscosity at the corners. The wing–fin junctions are where a real box wing’s interference drag lives, and a lattice of line vortices has nothing to say about them.

Who worked it out #

Ludwig Prandtl gave the best wing system in 1924, as the box, and gave the bending-moment problem for a flat wing in 1933, with the bell-shaped loading as its answer. Robert Jones solved a version of the monoplane problem with a structural constraint in 1950 and reached a similar conclusion by a different route. Box and joined-wing configurations have been studied as aircraft many times since, most visibly in Luciano Demasi, Aldo Frediani and colleagues’ work on the PrandtlPlane from the 1990s onwards. What the lattice adds here is the decomposition: that the halved exchange rate belongs to the second wing, and that the fins’ own saving is the first thing a lighter spar spends.

Still open: the fins as structure #

The aerodynamics has now said what the fins do under a moment constraint, and it is very little. What it has not said is what they do to the moment itself, because the moment constrained here is the lift’s, not the spar’s.

The next calculation joins the two spars at their tips with fins of a stated stiffness, loads them with the constrained optimum’s lift, and solves the statically indeterminate beam for the moment each spar actually carries at the root. If the tip joint lets the upper wing’s spar relieve the lower’s, the structural moment of a box could sit well below its aerodynamic moment while a biplane’s cannot — and the optimisation would then have to be rerun with that structural moment as the constraint. Whether the fins earn their wetted area in the spar rather than in the wake is a matter of computing it.