Busemann's biplane lifts only on its outer faces, and its channel turns the lift into a quarter-chord load

Busemann’s biplane is two half-diamonds facing each other across a channel, spaced so that the compression from each element’s leading edge lands on the other’s shoulder and is cancelled there by the expansion. At its design Mach number — 1.944 for elements three per cent thick and three tenths of a chord apart — it has no wave drag at all. The essays on it computed that cancellation, the throat that will not let the flow reach it, and a flap that starts the channel below the design point. All of them were for a symmetric pair, which lifts nothing.

An aeroplane’s wing must lift, and the simplest way to make the biplane lift is to tilt it. The tilt does three things at once: the lower element’s inner ramp meets the stream more steeply and the upper element’s less, the outer faces meet the stream at the incidence, and the upper element’s leading edge moves downstream by the gap times the incidence, so the waves no longer land exactly on the shoulders. The question the last essay left was how much lift that buys, how much drag the broken cancellation leaves, and whether the starting problem gets worse. The first answer is the surprising one: the channel buys no lift at all.

The channel, marched with the tilt #

The calculation is the earlier essays’ linear theory, done in the stream’s own frame. Between the elements the disturbance is two families of waves, one running up the Mach lines from the lower element, the other running down from the upper, and at each inner face the flow must follow the surface. The tilt enters only through the surfaces’ slopes, which are the half-diamond’s own slope minus the incidence, and through where the upper element now is: its leading edge sits a distance $G\tan\alpha$ downstream of the lower’s. Each wave that reaches a face is reflected with the face’s own slope added; a wave that reaches the line of a face where the element is absent leaves the channel. Every reflection is kept, marched along the chord on twenty thousand stations.

Outside, the two faces are flat and meet the stream at the incidence: Ackeret’s theory gives a pressure coefficient of $+2\alpha/\beta$ under the lower element and $-2\alpha/\beta$ over the upper, with $\beta = \sqrt{M^2 - 1}$. Those two faces alone give a lift coefficient of $4\alpha/\beta$ per unit of one element’s chord, summed over both elements — exactly what one flat plate of that chord would give, since each outer face is one side of a plate.

The channel lifts nothing at the design point #

The figure’s first surprise is that at the design Mach number the pair’s lift is exactly the outer faces’. Inside the channel the incidence expands the flow over the lower element’s inner face and compresses it under the upper element’s — which would, on its own, lift both elements just as a plate’s two sides do. But each of those disturbances runs across the gap and lands on the opposite face, and at the design Mach number it lands where the opposite face’s own disturbance is the mirror of it. The compression the upper face makes reaches the lower face’s rear half and the expansion the lower face makes reaches the upper face’s rear half, and integrated over the chord the channel’s upward and downward pressures balance to fifteen decimal places.

The pressures show how. On the front half of the chord, the upper face is at higher pressure than the lower by $4\alpha/\beta$, a load pushing the pair up; on the rear half the waves from the leading edges have crossed the gap, and the difference is reversed, $-4\alpha/\beta$, pushing it down. The step happens at the shoulder, exactly where the thickness’s cancellation happened, because the design Mach number is the one at which a wave from a leading edge takes half a chord to cross the gap. It is the same geometric fact applied to a different disturbance: a uniform tilt is cancelled by the same landing that cancels the half-diamond’s compression.

Off the design point the landing misses. At Mach 1.7 the waves cross the gap at a steeper angle and land ahead of the shoulder, the front load lasts less than half the chord and the rear download starts early; the two halves no longer balance and the channel adds 14 per cent to the outer faces’ lift. At 2.2 they land behind it and the channel adds 21. The biplane’s lift curve is therefore flattest at exactly the Mach number at which its drag is least.

The result is half of what two wings would give. Two flat plates far apart lift $8\alpha/\beta$ between them; the biplane at its design point lifts $4\alpha/\beta$ — the same as a single plate of one element’s chord. Mutual interference between the wings of a subsonic biplane costs a few per cent; here it costs half.

Why the cancellation covers the tilt #

The reason is clearest with the thickness taken away. Two flat plates a gap $G$ apart, tilted together by $\alpha$, are a parallel duct inclined to the stream, and inside it the flow must turn through $\alpha$ to follow the walls. Each leading edge sends a wave of strength $\alpha/\beta$ across the duct — an expansion from the lower plate’s upper face, a compression from the upper plate’s lower face — and each wave reflects off the opposite wall with its sign kept, so that behind it the flow is turned again. At the design Mach number a wave takes exactly half a chord to cross. For the first half of the chord each face feels only its own leading-edge wave, and the upper face sits $4\alpha/\beta$ above the lower in pressure; for the second half each face feels the other’s wave as well, which reverses the difference. The duct’s load is a square wave along the chord, up then down, with no mean.

The half-diamonds’ slopes are a second, independent disturbance, and in linear theory the two add. The thickness’s waves cancel each other’s drag at the shoulder; the tilt’s waves cancel each other’s lift at the same place. A biplane designed to cancel its thickness therefore cancels its tilt’s lift without being asked, and the Mach number at which it does both is the single one at which $\beta G$ is half the chord.

The channel carries a couple instead #

The channel’s load does not vanish; only its integral does. Lift on the front half of the chord and download on the rear is a couple, nose-up, of magnitude $\alpha/\beta$ per unit chord — a quarter of the lift times the chord. Added to the outer faces’ lift, which is uniform along the chord and acts at its middle, it moves the whole pair’s centre of pressure forward by a quarter of the chord.

That is a connection worth stopping for. Every thin wing in supersonic flow has its centre of pressure at half the chord, because Ackeret’s pressures are uniform along a flat plate; every thin wing in subsonic flow has it at a quarter. The rearward jump between the two is one of the classic problems of supersonic flight — an aeroplane trimmed subsonically finds its lift moving aft as it goes through the sound barrier, and must carry a download on its tail or pump fuel aft to stay trimmed. Busemann’s biplane, at its design Mach number, has its centre of pressure at the quarter chord, where a subsonic wing has it. With the upper element held in place it is exactly a quarter; with the downstream shift the incidence really makes, 0.232. Away from the design point it drifts back towards the half: 0.30 at Mach 2.2, 0.39 at 1.7.

The biplane does not achieve that by being subsonic; it achieves it by turning the flat plate’s uniform load into a uniform load plus a couple, and the couple is the channel’s waves cancelling in sum but not in place.

What it would mean for trim #

An aeroplane balances its lift against its weight about its centre of gravity, and where the lift acts decides how much download its tail must make to keep it there. A conventional supersonic aeroplane’s wing lift moves aft by about a quarter of the chord as it accelerates through Mach one, and the tail’s download that holds it is pure drag; Concorde moved tonnes of fuel between tanks to shift its centre of gravity instead. A biplane wing at its design point would carry its lift at the quarter chord, where a subsonic wing does, so the trim change between take-off and cruise would be the small one between the subsonic quarter chord and 0.23, not the large one to a half.

The caveat is that the biplane has its quarter-chord centre only at the design Mach number, and drifts aft on either side: at Mach 1.5 it is back at 0.48, nearly where a plate would have it. The transonic acceleration and the climb are therefore no easier than a conventional wing’s; the cruise is. For an aeroplane that spends most of its flight at one Mach number, that is the regime that matters.

What the lift costs #

The broken cancellation shows in the drag. With no lift the pair has none at the design Mach number; with lift it pays, and it pays more steeply than flat plates would.

Two flat plates pay $\beta/8$ times their lift coefficient squared, 0.208 at this Mach number. The biplane pays 0.52 at 1° and 0.45 at 3°, two and a half times as much. A factor of two of that is the halving of the lift already found: the outer faces carry all of the pair’s lift and pay a flat plate’s drag for it, which per unit of lift squared is twice the two plates’. The rest is the channel. The incidence’s waves cancel in the channel’s lift but not in its drag, because the drag weights each face’s pressure by its own slope, and the half-diamond’s slopes differ front and rear; and the upper element’s downstream shift, which moves every landing aft by the gap times the incidence, adds a further 11 per cent at 2°.

The factor is not constant. Drag due to lift falls from 0.52 to 0.45 of the lift squared between 1° and 3°, because the shift grows with the incidence and changes where the reflected waves land relative to each face’s kinks; a linear theory that keeps the geometry exact produces terms in the incidence cubed, and they lower the cost a little. None of this undoes the zero at no lift, which is what the biplane is for.

Still four times a diamond #

A biplane with no drag at zero lift has an infinite lift-to-drag ratio in the limit of no lift, and the ratio falls from there as the drag due to lift grows; its best, 142, comes at a lift coefficient of 0.007, which is too little to fly on. At a useful lift coefficient — 0.084, at 2° — it is 25.5. A diamond aerofoil with the same total thickness, six per cent, pays a fixed thickness drag of 0.0086 before it lifts anything, and its best ratio is 8.3, at a lift coefficient of 0.14. At the biplane’s working lift the diamond gives 7.2. The biplane, at its design point, is three and a half to four times better.

Off the design point the advantage shrinks but does not go: at Mach 1.6 the biplane’s ratio is 17.8, at 2.4 it is 17, still more than twice the diamond’s. The thickness cancellation, which is sharp, accounts for the peak; the channel’s partial cancellation of the incidence, which degrades smoothly, accounts for the rest. Both comparisons count wave drag only: the biplane’s two elements have twice the wetted area of a diamond of one chord, and its friction is correspondingly twice as large, which at the small lift coefficients of supersonic cruise is a large part of what the drag with nothing to rub leaves to be paid for.

The throat barely notices #

The last question was whether incidence makes the channel harder to start. In one dimension the starting limit — Kantrowitz’s, the Mach number at which a channel with a normal shock at its mouth swallows it — depends on the throat over the mouth, and the mouth the stream sees is the gap measured across the stream, $G\cos\alpha$. Tilting by three degrees changes that by a seventh of a per cent: the starting Mach number moves from 2.154 to 2.146. To first order in the incidence the starting limit does not move at all, and the lifting biplane has exactly the starting problem the symmetric one had, with exactly the cure a flap provides.

What the one-dimensional limit cannot say is what the asymmetry does inside the channel. The lower element’s ramp compresses more and the upper’s less, so the flow approaching the throat is not uniform across it, and a channel that starts in one dimension may still unstart locally on its steeper side. That needs the two-dimensional shock structure, which linear theory does not have.

The theory’s reach #

Every number is linear supersonic theory: thin elements, small incidence, no shocks strong enough to matter, a ratio of specific heats of 1.4, and an inviscid flow. The incidence is limited to a few degrees; beyond about five the inner compressions are strong enough that their shocks and the expansions they meet interact nonlinearly, and the cancellation the linear theory finds becomes approximate. Coefficients are per unit of one element’s chord and summed over both elements, so a lift coefficient of 0.084 here is the lift of the whole pair over the dynamic pressure times one chord. The centre of pressure is measured along each element’s own chord from its leading edge. The diamond comparison uses Ackeret’s theory for a symmetric diamond of six per cent at the same lift. Viscosity, which separates the flow behind the shoulders’ expansions and thickens the throat, is absent.

Who flew it #

Busemann proposed the biplane in 1935. Moeckel computed the coefficients of lifting supersonic biplanes in linear theory in 1947, and Licher optimised two-dimensional supersonic wing systems in 1955. Kusunose and his colleagues revived the configuration in the 2000s with computations that included its starting problem and its lift. The quarter-chord centre of pressure, as a consequence of the channel’s couple, follows from the same linear theory; it is read here from the computed pressures.

How it was checked #

At zero incidence the tilted recursion reproduces the symmetric one’s drag at Mach 1.6, at the design point and at 2.4, to $3 \times 10^{-6}$ at the worst, and its lift is zero. With the elements made infinitely thin and five chords apart, so that no wave crosses the gap within a chord, the pair’s lift and drag are two flat plates’ — $8\alpha/\beta$ and $8\alpha^2/\beta$ — to two parts in ten thousand, the residual being the marching grid. The channel’s net lift at the design point is zero to $10^{-16}$ at 3° as at 2°. The model refuses a subsonic stream and an incidence past ten degrees.

Still open: a biplane that cancels its incidence too #

The channel already cancels the incidence’s lift at the design point; it does not cancel its drag, because the half-diamonds’ slopes weight front and rear differently. A pair whose inner faces were shaped for the lifting condition — the lower element’s ramp steeper and the upper’s shallower by the incidence, so that the channel at the design incidence is again a symmetric channel in the stream’s frame — would cancel both. The next calculation would design the inner faces for a stated cruise lift coefficient, ask what the drag due to lift then is, and whether the starting problem, which the asymmetric ramps change, has become easier or harder.