The tail a canard needs costs more than the canard saves

A third surface is worth a square asked whether an aircraft with a canard, a wing and a tail can beat an ordinary tail aircraft in induced drag. It can: with both small surfaces available, one share of the lift stays free after trim, and the split can slide along the trim line to the point nearest the least-drag foot. The saving came out as a square — half the drag bowl’s curvature times the square of how far the tail aircraft trims from twice the foot share — and it was small: a fifth of a per cent of an elliptic wing’s induced drag for an uncambered wing, 4.5 per cent for a wing with a flapped section’s pitching moment.

That comparison rested on an assumption, stated plainly there: the neutral point was held. A canard ahead of the centre of gravity is destabilising, so adding one moves the neutral point forward, and at the same static margin the centre of gravity would have to move forward too. That would change the trim and compare two different aircraft. Holding the neutral point made the canard a pure trimming surface — “the tail enlarged, or the canard’s contribution otherwise compensated”. This essay computes what the enlargement costs, and asks whether a tail at a different height would cost less.

What holding the neutral point asks of the tail #

The aircraft is the previous essay’s, which a canard pays for its stability in induced drag first built with two surfaces. A wing of aspect ratio 8; a canard and a tail each a third of its span, 3.5 wing chords ahead of and behind its aerodynamic centre; the canard a tenth of the wing’s area, in a plane a twenty-fifth of a span above the wing’s. The tail aircraft has a tail of a fifth of the wing’s area a twenty-fifth of a span below the wing, and its neutral point sits a third of a chord behind the wing’s aerodynamic centre, the one point on the chord about which the wing’s moment does not change as it pitches. That is the neutral point every aircraft here is held to.

The neutral point is the average of the surfaces’ positions, each weighted by how much lift it adds when the aircraft pitches up — its lift slope times its area, reduced by the downwash gradient it sits in. The canard adds 0.54 of that weight at 3.5 chords ahead. To keep the average where it was, the tail must add enough weight at 3.5 chords behind to cancel the canard’s pull and to cover the canard’s share of the denominator, which dilutes the tail’s existing weight. The arithmetic gives the tail’s required weight directly, and the tail’s area follows from its lift slope and from how much of it the downwash takes away.

The downwash is where the tail’s height enters, and it has two sources. The surface in the wake computed the wing’s, since the air a lifting wing must push down is still moving down when it reaches the tail: the tail sits in the downwash of the wing’s trailing vortices, and its effective lift slope is reduced by the factor one minus the downwash gradient. Now the canard’s trailing vortices pass over the wing and arrive at the tail as well. Both are computed from the same horseshoe fields as before, averaged across the tail’s own span. Enlarging the tail keeps its aspect ratio, 4.9, so a bigger tail is wider as well as deeper. Holding its span fixed instead would let its lift slope saturate as its chord grew, and at the reference height no area at all would then restore the neutral point — an artefact of the choice rather than a property of tails.

Two and a half times the tail #

The first figure is the answer. At the reference height the tail aircraft needs its tail of 0.2 of the wing’s area, by construction. The same aircraft with the canard needs 0.496 — two and a half times the tail. Removing the canard’s wake at the tail — a fiction that separates the two effects — the requirement is 0.419. So of the extra 0.30 of the wing’s area, 0.22 pays for the canard’s own lift ahead of the centre of gravity, and 0.08 for its wake arriving at the tail.

The height moves both curves. Raising the tail takes it out of the wing’s downwash, so every aircraft needs less tail: the tail aircraft 0.154 of the wing’s area a quarter of a span up, and the three-surface one 0.354. Lowering it below the reference height does the same, more slowly. The worst place for the tail is between the two wakes’ planes, where it sits in the strongest downwash of both, and the heights within a sixtieth of a span of either plane are not drawn: there the tail’s wake coincides with another in the Trefftz plane and the least-drag calculation has no solution.

Where each wake is strongest #

The second figure shows the two sources separately, for a tail of the reference span. Each peaks at its own wake’s height. The wing’s gradient is 0.40 at the reference height and about 0.43 in the wing’s own plane. The canard’s is 0.15 at the reference height, rises to about 0.31 in the canard’s own plane, and falls off on either side. The previous essay’s last section quoted the pair together — 0.41 without the canard and 0.54 with it — and the figure places that on its curve.

A quarter of a span above the wing both have fallen, to 0.235 from the wing and 0.065 from the canard. A trailing vortex’s downwash falls off with the square of the distance from it across the tail’s span, and the two wakes lie close together, a twenty-fifth of a span apart, so a tail well above both escapes both. Below the wing the wing’s wake is between the tail and the canard’s, and the canard’s contribution falls more slowly.

Most of the bill is the canard’s lift #

The third figure builds the tail up in steps. First the tail aircraft; then the canard’s lift ahead of the centre of gravity, with its downwash on the wing, which slightly reduces the wing’s lift slope and so helps; then the canard’s wake at the tail. At the reference height the canard’s wake is 26 per cent of the extra area; a quarter of a span up, 13 per cent.

That split matters for what the tail’s height can do. The larger part of the bill is not an interaction at all. It is the lever arm: a surface of a tenth of the wing’s area and a lift slope of 5.4 per radian, 3.5 chords ahead of the centre of gravity, destabilises by its own weight, and nothing the tail’s height does changes that part. A T-tail escapes the canard’s wake and most of the wing’s, which is what makes it smaller; but the canard’s own lift ahead has to be paid for at any height.

The bill against the saving #

The fourth figure prices both in drag counts — ten-thousandths of the drag coefficient — at a lift coefficient of 0.5 and a static margin of a tenth of a chord. The bill is the skin friction of the extra tail area and of the canard itself, at a profile coefficient of 0.008 on every planform. The saving is the previous essay’s induced-drag square, recomputed with the tail’s wake at its actual height and its actual span.

At the reference height the bill is 31.7 counts and the savings are 1.5 counts for an uncambered wing, 0.8 for a cambered cruise section with a moment of −0.15, and 5.1 for a wing whose moment is −0.3, as a section with a flap deployed carries. Moving the tail away from the two wakes lowers the bill and raises every saving, and at 0.3 span above the wing the bill is 22.9 counts and the savings 3.5, 5.5 and 20.4. The saving grows with height for a reason the previous essay gave: a lifting system whose wake has height carries its lift at lower cost, and a tail well above the wing gives the whole system’s wake more of it — the effect that makes a box wing with a light spar worth a third.

A T-tail comes closest #

The fifth figure subtracts. The three-surface aircraft is compared with a tail aircraft whose tail is at the same height, so that the height’s own benefit, which both enjoy, is not credited to the canard. The advantage is negative everywhere drawn. The best height is the highest, 0.3 span above the wing — a T-tail on a fin about two and a half wing chords tall — and there the three surfaces still cost 19 counts more than the tail aircraft for an uncambered wing, 17 for a cambered cruise wing, and 2.5 for the flapped one.

So the previous essay’s still-open question has an answer on both counts. The wetted-area bill the square has to be set against is large — ten to thirty times the saving for a cruising wing — and the best three-surface aircraft does have a T-tail, because the T-tail escapes the canard’s wake and most of the wing’s. But even with it the third surface does not pay for a cruising wing, and comes close only for a wing whose pitching moment is a flapped section’s.

The same bill under water #

A keelboat meets the same account with its two underwater foils. The rudder pays for the keel’s wake found the rudder flying in the flow the keel has already turned, so that the best division of side force between them was not the one that minimised either foil’s drag alone — the same interaction that here makes the tail pay for the canard’s wake. The difference is which way the account runs. On the boat, the second surface sits in the first one’s wake and the designer’s question is how to share a fixed side force; here the second surface is added ahead of a tail already sized, and the question is what adding it costs the surface behind. Both come down to the downwash one lifting surface leaves for another, and both are settled in the Trefftz plane, where the wakes’ positions decide the induced drag and the surfaces’ order along the stream does not.

What that says about three-surface aircraft #

The best-known three-surface aircraft is consistent with that. The Piaggio P.180 Avanti has a small foreplane, a wing, and a T-tail, and its foreplane carries flaps that deploy with the wing’s: it earns its keep in the configuration whose pitching moment is large, where the flapped wing’s nose-down moment has to be trimmed by something, and it sits in front of a tail placed high enough to escape its wake. What the calculation says is that it cannot also be a cruise-drag device of this size, and that the lore attributing the aircraft’s efficiency to the foreplane’s lift is attributing it to the wrong surface.

The calculation also says what would change the answer. The bill is mostly the canard’s lever arm, which scales with its lift slope times its area times its distance ahead of the centre of gravity. A canard half the size costs roughly half the extra tail; a canard nearer the wing costs less but saves less, since where a surface sits along the stream does not change the pair’s total and the saving is set in the Trefftz plane. The saving is not proportional to size in the same way — it depends on how far the tail aircraft trims from the foot, which does not depend on the canard at all — so a smaller canard closes the gap faster than it loses the saving, up to the point where it is too small to carry the share the square asks of it.

What was checked #

The canard’s downwash at the tail at the reference height, with the reference tail, is exactly what the previous essay’s calculation used: the difference is at the last bit. Far behind the wing, its downwash per unit lift coefficient across the tail is the elliptic wing’s far-wake value, twice its own induced downwash, 2/πA, within 1.5 per cent — the unevenness of a loading stepped in forty pieces. The tail area found at each height puts the neutral point back to its target to rounding, and at the reference height without a canard it is 0.2 exactly.

One trap is worth recording. The wakes are rows of discrete trailing vortices, and exactly in a wake’s own plane each one has a singular downwash, so a tail placed exactly in a wake’s plane sampled a scatter of near-singular values — 0.35 or 1.51 where the neighbouring heights give 0.43 and 0.33. A continuous sheet has no such singularity. The calculation gives the sheet a thickness of a two-hundredth of a span, where the curve is already flat, and leaves out of the drag figures the heights at which the tail’s wake coincides with another.

What the calculation leaves out #

The fin. A T-tail sits on a fin tall enough to put it there, and the fin has area, weight and its own skin friction, none of which is counted. A fin 2.4 chords tall is a substantial structure, and its bill is set against the T-tail’s advantage, which here is 7 counts for the tail aircraft and 18 for the three-surface one.

A wake that sinks and rolls up. Both wakes are flat and straight, at the height they were shed. In the air a wing’s wake sinks behind it and its edges roll up into two vortices, and by the tail — 7 chords behind the canard, 3.5 behind the wing — both have moved; the tail’s height against the real wakes is somewhat different from its height against these.

Fixed arms and a fixed canard. The canard’s size and position are the previous essay’s. A design study would move the wing relative to the centre of gravity, size the canard to its job, and let the tail’s arm change. The calculation says which way each of those should go, not where they end.

The canard’s own trim drag at other lifts. Everything is at a lift coefficient of 0.5. A canard’s share of the lift changes with the lift coefficient through the moment term $C_{m0}/C_L$, as the previous essay showed across the loading range.

Who worked it out #

The downwash at a tail and the neutral point it implies are the classical longitudinal stability of the 1930s and 40s, and the horseshoe estimate of the downwash gradient is in every stability text since. The Trefftz-plane view of a multi-surface aircraft’s induced drag is Munk’s stagger theorem of 1921 extended to non-planar systems; its application to canard and three-surface layouts was argued in the 1980s, when several three-surface designs flew, by Kroo and McGeer among others, whose conclusion — that a canard’s induced-drag advantage is small and easily consumed by its other costs — the calculation here reaches by a different route.

Still open: a canard sized to its job #

Every number here is for a canard of a tenth of the wing’s area, which is large for a surface whose purpose is to trim. The next calculation lets the canard’s area and its arm vary with the tail’s height, asks for the combination that makes the three-surface aircraft’s total drag least at the flapped wing’s moment, and compares it with the tail aircraft at its own best tail height — so that the question the lore poses, whether a small foreplane can pay for itself in the configuration where trim is hardest, is answered for the canard that would actually be built rather than the one the previous calculation borrowed.