A strut inside a box wing is invisible to the wake, and halves the spar

A box wing’s fins earn their keep in the spar treated the two wings and the two tip fins of Prandtl’s best wing system as one frame. Joined at the tips, the spars hand part of the root’s bending moment to a couple across the gap — tension in one wing, compression in the other — and a couple on a lever arm the size of the gap costs far less material than bending. With fins as stiff as the spars the couple took three tenths of the moment, and it could never take more than a third, because a joint at the tip can only guide a tip. The essay ended by asking about the obvious next member: a strut joining the wings part of the way out, at what station it should go, and whether what it saves in the spar outweighs what it costs in the air.

Both halves of the answer are cleaner than expected. In the air it costs nothing but its friction — not approximately, but exactly, in the theory of induced drag — because the air inside Prandtl’s box moves straight down. In the spar it saves half the material, but not in the way a strut is supposed to: it carries almost no load along its length, and works entirely through the stiffness of its joints.

The air inside the box moves straight down #

The induced drag of any wing system is decided in the Trefftz plane, the cross-section far behind it, where the wake is a sheet of trailing vorticity whose trace is the wing system’s outline seen from behind. Munk’s theorem says the drag at a given lift is least when the wake moves as if it were rigid: the velocity normal to the trace at every point equals a uniform downwash $w_0$ times the vertical component of the trace’s normal. On a horizontal wing that is $w_0$; on a vertical fin, whose normal is horizontal, it is zero.

Prandtl’s box is a closed trace, a rectangle, and its inside is a region of the Trefftz plane bounded entirely by that trace. The flow inside is a potential flow — the wake’s vorticity is all on the sheet — and its velocity normal to the boundary is known all round: $w_0$ times the normal’s vertical component. A uniform downwash, potential $-w_0z$, satisfies that condition on every side of the rectangle. Neumann’s problem for the inside of a closed curve has one solution up to a constant, so the uniform downwash is the flow. Inside the best box, the air moves straight down at $w_0$, with no sideways component anywhere.

That is the whole aerodynamic argument for a strut. A vertical strut anywhere inside the box is a surface whose normal is horizontal, and the optimum asks of it a normalwash of $w_0$ times zero. The flow already there supplies exactly that, with no circulation on the strut. So the box with a strut has the same optimum loading as the box without, the strut carries no load at the optimum, and the induced drag is unchanged. A wake that is rigid inside its own outline cannot be altered by anything placed in that outline.

The figure checks the argument against the lattice solution the earlier essays used, summing the wake’s shed vortices at points across the gap. At the centreline the sidewash is zero by symmetry. At a fifth of the semi-span it is under five hundredths of a per cent of the downwash, at two fifths under a quarter of a per cent, and it grows towards the fins to a per cent at three fifths. That growth is the lattice’s: the optimum loading has singular behaviour at the box’s corners, which a finite lattice resolves only approximately, and its errors decay away from the corners into the interior. The exact flow has none.

The same argument says something about the box that the earlier essays did not draw. The wake’s whole disturbance is outside the box: the inside is carried bodily downwards, so the box’s drag is spent entirely on the air around it, and the inside’s uniform motion is the reason a closed wake beats an open one of the same span. Anything placed inside — a strut, a second fin, a fuselage — is in a region of the wake that already moves as the optimum wants.

The strut the frame wants #

The structure is the earlier essay’s frame: two spars of the same depth, a hundredth of the span, clamped at the centreline, joined at the tips by fins as stiff in bending as the spars, loaded by the best box’s own lift, which is equal on the two wings. To it is added one vertical strut each side, joined rigidly to both spars at a station along the span. The spars’ material is measured by their flange volume: at each station each spar’s two flanges carry its bending moment over its depth, plus or minus half its axial force, and the flange area is whichever of those is larger, integrated along the span for both spars. That measure counts the bending a strut moves outboard of itself, which the root moment alone, the earlier essay’s measure, would not see.

The best station, for a strut as stiff as a spar, is four tenths of the semi-span, and there the spars’ flange volume is exactly half of what it was without the strut. A strut near the root saves little, because it holds the spars where they bend least, and one near the tip saves little, because the tip fins already do its job there: at a tenth of the semi-span it saves 16 per cent, at nine tenths 7. A stiffer strut does a little better and wants to be a little further in — ten times stiffer, it saves 53 per cent at 38 hundredths of the semi-span — and a strut a tenth as stiff saves much less.

The moments show why four tenths is the place. Without the strut the spar is a cantilever with a guided tip: its moment is largest at the root and reverses near the tip, where the fin holds it straight. The strut adds a second guided point. Inboard of it the spar bends between two held ends, outboard of it it is a shorter guided cantilever, and the station that balances those two spans — so that neither the root nor the outer bay carries the largest moment alone — is where the total material is least. Put the strut at two tenths and the root moment falls further, to half its unstrutted value, but the long outer bay bends hard and the total saving is only 35 per cent.

The gap hardly matters, for a reason #

The fins’ share of the moment depended on the gap, because the couple they make has the gap for its lever arm. The strut’s saving hardly does. At a gap of a tenth of the span the best strut, still at four tenths of the semi-span, leaves 0.487 of the spars’ flange volume; at a fifth, 0.500; at three tenths, 0.519. The unstrutted boxes differ by more: their flange volumes are 0.925, 0.942 and 0.947 of the monoplane’s.

The reason is in what the strut does. It does not make a couple across the gap — it carries no axial force, so the spars’ axial forces are what the fins alone made — but turns each spar’s single long bay into two short ones by holding its slope at the strut’s station. That is a statement about each spar’s length, not about the distance between them, and a spar’s bending material falls with the square of its unsupported length whatever its neighbour is doing. The gap enters only through the stiffness of the frame’s panels, a second-order effect, and the best station is set by balancing the two bays’ moments, which is a statement about the loading along the span — the same at every gap because Prandtl’s optimum loading changes little with it.

That separation is the useful design fact. The gap is chosen for the induced drag, which falls as it grows, and for the fins’ couple; the strut’s station is chosen for the spar and stays near four tenths. Neither choice constrains the other, so a box can be laid out for its wake first and strutted afterwards without either giving ground.

A strut that is not pushed or pulled #

A strut between two wings is usually drawn as a truss member: it carries the lift of one wing to the other, in tension or compression, and it is pinned at its ends because a truss member needs no more. In Prandtl’s best box that picture fails at once. The two wings carry equal lift, so under load they deflect by the same amount at every station, and a vertical member between them is neither stretched nor squeezed. The strut’s axial force is a hundred-thousandth of the lift, rounding error on a computed frame.

What the strut does is resist the spars’ rotation. Both spars turn at its station by the same angle as they bend; a strut rigidly joined to both forces their slopes to be connected through its own bending, and that turns the two spars and the strut into a frame — the closed panels of a Vierendeel girder, the bridge form with no diagonals that holds its shape by the stiffness of its corners. The figure shows the dependence plainly: a strut with three thousandths of a spar’s bending stiffness, close to a pin-jointed rod, saves one per cent; as stiff as a spar, half; and stiffer than that hardly more, because the spars’ own stiffness then limits the frame.

So a strut for a box wing is not a rod. It must be a member deep enough to bend as stiffly as a spar, joined to both spars with joints that carry moment. That is heavier than a rod and it is also larger in section, which is what the air charges for. A diagonal strut would be a truss member and would carry axial force, but it would cross the box at a slant, and a slanted surface inside the box has a normal with a vertical component, so the optimum would ask it to carry load — a diagonal is not invisible to the wake.

What the strut costs, and when it pays #

In the Trefftz plane the strut is free. On the aircraft it is a surface the air rubs along, and the second of a wing’s two drags charges for its wetted area whether or not it carries any load. What it buys is spar weight, and spar weight is worth induced drag: at a fixed span and speed the induced drag goes as the square of the weight, so removing a share of the weight removes twice that share of the induced drag.

The balance is a ratio of small numbers, and it comes out near a third. A box spar that is six per cent of the aircraft’s weight, halved by the strut, removes three per cent of the weight and six per cent of the induced drag; at cruise, with induced drag at four tenths of the total, that is 2.4 per cent of the whole drag. Two struts across a gap of a fifth of the span, each with the wing’s chord, add a fifth to the wings’ wetted area and so a tenth to the aircraft’s, which at six tenths friction is six per cent of the drag. They break even at four tenths of the wing’s chord. A strut that is to bend as stiffly as a spar needs depth, so its chord cannot be arbitrarily small; but a strut a third of the wing’s chord and a spar’s depth is a buildable member, and on a box wing whose spar is a large share of its weight — a long-span, high-aspect ratio design, which is where box wings are proposed — it pays.

The convention, and what it leaves out #

The box is Prandtl’s best at a gap of a fifth of the span, with the earlier essays’ lattice of 320 segments for its loading and drag; every flange volume is relative to the box without a strut or to the elliptic monoplane of the same span, lift and spar depth, whose flange volume the unstrutted box needs 0.942 of — slightly less than the monoplane by this measure, where the root-moment measure of the earlier essay gave 0.78, because the fins’ reversal near the tip puts material back that the root moment does not count. Stations are fractions of the semi-span from the centreline. The frame is linear and elastic, the spars a hundredth of the span deep, the struts and fins built of the same material as the spars with the bending stiffnesses stated.

What the model leaves out is mostly in the joints. A rigid joint between a strut and a spar is a heavy fitting, and its weight is not counted; neither is the strut’s own material, which is set by its bending and so by the moments it carries at its ends. The strut is assumed to carry no load in the air, which holds at the optimum; off it — at another lift coefficient, in a roll, in a gust — the strut does carry side force, and its section must be shaped for that. The interference drag where the strut meets each wing, often larger than the friction of a short member, is not charged, and it is the likeliest thing to move the break-even.

Who joined two wings #

Wire-braced biplanes of the first two decades of flight carried their lift through trusses of struts and wires, and their struts were compression members between unequally loaded wings. Prandtl showed in 1924 that a closed rectangular wake has the least induced drag of any system of its span and height. Joined-wing and box-wing designs of the 1980s onward, Wolkovitch’s and Frediani’s among them, studied the structure of closed systems with members between the wings, and found that how those members are joined decides what they do. The invisibility of an interior strut follows from Munk’s condition and the uniqueness of the interior Neumann problem; it is not stated in those terms in the sources this essay draws on, and is computed and checked here.

How it was checked #

The frame’s statics — bending plus the couple across the gap equal to the lift’s moment about the root — close to $10^{-11}$, with and without the strut, and the frame refuses a strut outside the span, on a root or a tip, or with no stiffness. The interior sidewash, from the optimum’s own shed vortices, is under a quarter of a per cent of the downwash at the two strut stations nearest the centre and falls towards the centreline as the corner errors decay, the exact solution’s being zero. The monoplane’s flange volume is integrated from its elliptic loading by quadrature, and the box without a strut reproduces the earlier essay’s frame to the digits it printed.

Still open: a box whose wings lift unequally #

Everything here rests on Prandtl’s box lifting equally on its two wings, which is what makes a vertical strut idle along its length and the interior flow uniform. A box wing with a staggered or unequal pair — the upper wing smaller, or set behind the lower for balance — has an optimum that lifts unequally, and then its two spars deflect differently and a strut between them is pushed or pulled. The next calculation would solve the optimum for a box whose upper wing is shorter or carries a stated share of the lift, ask whether the interior flow stays uniform — it does if the trace is still closed and the optimum still Munk’s — and find what a strut does when its two ends no longer move together: whether it becomes a truss member, cheaper than a Vierendeel post, and moves to a different station.