A third surface is worth a square

A canard pays for its stability in induced drag put a wing and one smaller surface on a single calculation and found that the question “canard or tail?” is not about which surface is in front. By Munk’s stagger theorem the least induced drag of two lifting surfaces depends only on how the lift is split between them, and the split is not free: the static margin, the distance the centre of gravity must sit ahead of the neutral point, fixes it. A tail at a margin of a tenth of a chord carries 6.7 per cent of the weight upwards and costs 0.9 per cent more induced drag than an elliptic wing; a canard at the same margin must carry 19 per cent and costs 12.8 per cent more.

That essay ended on the aircraft that has both. Canard, wing and tail give two shares of the lift to choose, and the moment balance about the centre of gravity is still one equation. So one share remains free after the aircraft is trimmed, and it can be spent on drag. This essay spends it, and finds that the whole comparison reduces to a single square — which says precisely when the third surface earns its place and when it is carried for other reasons.

Three wakes, one quadratic #

The drag calculation is the earlier essay’s with a third wake added. Far behind the aircraft, in the plane where only the trailing vortices remain, there are three straight lines: the wing’s, of span 1; the canard’s, 0.35 of that span and a twenty-fifth of the span above; and the tail’s, also 0.35 of the span and a twenty-fifth below. Munk’s stagger theorem is why their positions along the stream — canard 3.5 chords ahead, tail 3.5 chords behind — do not appear: the induced drag of a lifting system at fixed circulations is decided entirely in that far plane. The wing has an aspect ratio of 8.

On that plane the least induced drag with the canard carrying a share $f_c$ of the lift, the tail $f_t$, and the wing the rest, every loading otherwise free to take its best shape, is a quadratic form minimised under three linear constraints. Its value is therefore exactly a quadratic in the two shares:

$$D(f_c, f_t) = a + b f_c + c f_t + d f_c^2 + e f_t^2 + h f_c f_t .$$

Six solves fix the six coefficients, and fresh solves at four other splits then reproduce the fitted form to two parts in $10^{14}$. For this geometry the bowl is nearly round — $d$ and $e$ are both 4.45 and the cross term $h$ only 0.016 — because the canard and tail are the same span at the same distance from the wing’s wake, and they see the wing identically and each other hardly at all.

The bowl, and the lines trim draws across it #

The first figure is the bowl seen from above. Its foot, the split of least drag with nothing asked of it, is at 1.87 per cent of the lift on each small surface, where the whole system pays 0.9969 of an elliptic wing’s induced drag. Rings of equal extra drag run round it: half a per cent, two per cent, five and ten. The two small surfaces at their foot shares extend the lifting system a little in height, which is the same reason a biplane beats a monoplane of the same span — a slightly non-planar wake carries its lift more cheaply.

Why the foot is not at zero #

It is worth asking why the best split puts any lift on the small surfaces at all, since each has a third of the wing’s span and a surface of short span is an expensive place to carry lift. Carrying all of the lift itself, with the wing’s net load zero, the canard would cost the system 5.3 times an elliptic wing’s induced drag, and the tail the same. The reason is that neither is alone. Each sits a twenty-fifth of a span from the wing’s wake, and a small load on it changes the shape of the whole system’s far wake as seen in the Trefftz plane: its trailing vortices lie inboard of the wing’s tips and slightly off the wing’s plane, and a lifting system whose wake has some height carries its lift at lower cost than a flat one of the same span. A box wing with a light spar is the extreme of the same effect, where the height is a fifth of the span and the saving a third.

Here the height is small, so the saving is small — three-tenths of a per cent at the foot — and the price of pushing the small surfaces’ shares beyond it rises fast, because their short spans make the bowl steep. That steepness is the curvature $d$ of 4.45 that appears in everything below. It means a share of 10 per cent on either surface already costs 2.8 per cent of an elliptic wing’s induced drag more than none at all, and a share of 20 per cent costs 14.5. The foot is shallow and the walls are steep, and the whole question of three surfaces is how far from the foot trim forces the aircraft to sit.

Trim draws a line across the bowl. Taking moments about the centre of gravity, with the canard at $x_c = -3.5$ chords from the wing’s aerodynamic centre and the tail at $x_t = +3.5$,

$$f_c x_c + f_t x_t = x_{cg} + \frac{C_{m0}}{C_L},$$

in which $x_{cg}$ is the centre of gravity’s position and $C_{m0}$ the wing section’s own pitching moment about its aerodynamic centre — the one point on the chord about which the moment does not change with incidence, and which a camber line fixes along with the lift — divided by the lift coefficient because a fixed moment needs a smaller balancing force when the lift is larger. With the canard and tail at equal and opposite arms the line runs at forty-five degrees: every unit of lift moved onto the canard must be matched by a unit more on the tail.

The tail aircraft is the point where the line crosses the axis of zero canard load. The three-surface aircraft with the same centre of gravity sits somewhere on the same line and can slide along it, and the best place to stop is the point of the line nearest the foot. Three lines are drawn: an uncambered wing at a lift coefficient of 0.5, a cambered cruise section with $C_{m0} = -0.15$, and a wing whose moment is $-0.3$, which is what a section with a flap deployed carries: a flap moves the lift curve sideways and the load aft, and its moment grows with the deflection.

One assumption makes this comparison fair, and it has to be named. Adding a canard moves the neutral point forward, since a surface ahead of the centre of gravity is destabilising, and at the same margin the centre of gravity would have to move forward with it. That would change the right-hand side and compare two different aircraft. Here the neutral point is held — the tail enlarged, or the canard’s contribution otherwise compensated — so that the canard enters purely as a second trimming surface, which is the case its advocates make for it.

The saving is a square #

The second figure computes the saving — the tail aircraft’s drag less the least drag on its trim line — for twenty-four aircraft: six wing moments from zero to $-0.3$ at four margins from a twentieth to three-tenths of a chord. Plotted against the tail aircraft’s own trimmed share $f_{t0}$, all twenty-four fall on one parabola.

The parabola has a closed form, and it comes from the bowl’s geometry alone. For a round bowl of curvature $d$ with its foot at $(f^*, f^*)$ and a trim line of slope one through $(0, f_{t0})$, the point of the line nearest the foot is a perpendicular distance $|f_{t0} - 2f^*|/\sqrt 2$ from the axis crossing, and the saving is

$$\Delta D = \frac{d}{2}\,(f_{t0} - 2f^*)^2 .$$

It matches the lattice to 0.2 per cent of the largest saving, the difference being the bowl’s slight asymmetry. Two things follow from its shape.

It is zero where the tail aircraft already carries twice the foot share, 3.7 per cent. There the trim line passes through the axis at its own nearest point to the foot, and the canard has nothing to do. And it grows as the square of the distance from there, so a tail aircraft trimming near that value gets almost nothing from a canard, while one trimming far from it gets a great deal. The canard’s worth is not a property of the canard at all. It is a property of how badly the tail aircraft trims.

Where a tail aircraft actually trims #

So the question becomes where real tail aircraft trim, and the third figure answers it for four wings. With no pitching moment of its own, the wing trims with the tail lifting upwards, 8 per cent at a margin of a twentieth of a chord and 1 per cent at three-tenths; the line crosses twice the foot share near a margin of 0.2, and there the canard is worth nothing. With a mildly cambered cruise section, $C_{m0} = -0.08$, the tail carries between 3.5 per cent up and 3.6 per cent down across the same range. With $C_{m0} = -0.15$ it carries a download at almost every margin.

The flapped wing is the one that looks as though it should need the canard most, and does not. Its moment of $-0.3$ is divided by a landing lift coefficient of 1.6, so the moment per unit lift is only $-0.19$ — smaller than the cambered cruise wing’s $-0.3$ per unit lift at $C_L = 0.5$. A flapped wing at landing trims closer to the foot than an aft-loaded cruise section does at cruise.

Putting this beside the saving: at a margin of a tenth of a chord the uncambered wing saves 0.19 per cent of its induced drag by using a canard; the cruise camber of $-0.08$ saves 0.06 per cent; $-0.15$ saves under one per cent; and only a wing moment of $-0.3$ at cruise lift saves several per cent — 4.5 at this margin.

What a canard does across the loading range #

The fourth figure turns the same calculation into the question an operator asks, because the centre of gravity is not fixed: it moves as fuel burns and as passengers and cargo are loaded, and an aircraft is certified for a range of it. The tail aircraft’s drag against the margin is a parabola, lowest where its trim line happens to pass near the foot. The three-surface aircraft’s is much flatter, because as the centre of gravity moves the trim line slides across the bowl and the best point on it stays near the foot.

For the mildly cambered wing the two curves never differ by more than 2.3 per cent anywhere between a margin of a fiftieth and four-tenths of a chord. For the heavily cambered one they differ by 3.2 to 11.6 per cent. This is the strongest honest argument for a third surface: not a saving at the design point, where a tail aircraft can be balanced close to the foot, but a flat drag across a wide range of centre-of-gravity positions, where the tail aircraft’s parabola climbs at the ends.

The ceiling, and the skin under it #

A canard is not free. It has its own wetted area, and its skin friction is paid at every lift coefficient. The fifth figure measures the saving in the currency the skin is paid in — drag coefficient on the wing’s area — and sets it beside the skin friction of a canard of a tenth of the wing’s area at a profile-drag coefficient of 0.008: eight drag counts.

The saving in coefficient terms has a ceiling. At low lift the tail aircraft’s trimmed share is dominated by the moment term, $f_{t0} \approx C_{m0}/(C_L\, l_t)$, and the elliptic induced drag the percentage is taken of is $C_L^2/\pi A$. The two powers of $C_L$ cancel, and the saving tends to

$$\Delta C_{D} \to \frac{d}{2}\left(\frac{C_{m0}}{l_t}\right)^2 \frac{1}{\pi A},$$

a constant set by the wing’s pitching moment and the tail arm alone. It equals the canard’s skin friction where $|C_{m0}| = l_t\sqrt{2\pi A\, S_c c_{d0}/d}$, which for this geometry is 0.33. A wing moment of $-0.1$ saves between a fifth and half a count; $-0.2$ between a tenth of a count and two and a half, falling as the lift rises; $-0.4$, nearly eleven counts at low lift, the only case drawn that clears the canard’s eight.

That is the arithmetic behind the answer the earlier essay suspected. At cruise, a tail aircraft’s small trimmed load is already close enough to the bottom of the bowl that a third surface can only add wetted area. It becomes worth its skin only for wings whose pitching moment is as large as a heavily flapped or strongly aft-loaded section’s, and then its value is mostly in flattening the drag across the centre-of-gravity range rather than lowering it at one point.

The case the advocates make, and what it rests on #

The usual argument for three surfaces is that the tail need no longer push down, so the wing no longer carries the tail’s download as extra lift. That argument counts lift and not drag. What matters is the least induced drag of the whole system at the split trim requires, and a small download on the tail, or a small upload, costs almost nothing when it is near the bowl’s foot, because the bowl is flat at the bottom. The download’s cost is second order in its size — which is precisely what the square says.

Nor is induced drag the whole bill at cruise. An aircraft flying at its best lift-to-drag ratio pays equal shares of induced and profile drag, and faster than that the profile share dominates, so a saving of a per cent of the induced drag is half a per cent of the total at best and less at the speeds airliners actually cruise. The canard’s skin friction, by contrast, is all profile drag and grows in share as the aircraft goes faster. The comparison in the ceiling figure is made in drag coefficient precisely so that the two bills are counted in one currency; at the speed for least power, where induced drag is three-quarters of the total, the third surface looks its best, and even there the numbers above hold.

What the argument gets right is that large downloads do cost, and grow as the square. An aircraft whose wing has a large nose-down moment, or whose loading moves its centre of gravity through a wide range, spends much of its time far from the foot if it has only a tail, and there a canard earns real induced drag. The Piaggio P.180 Avanti’s foreplane carries flaps of its own that deploy with the wing’s, to take the pitching moment of the wing’s large flaps — which is the case the figures single out; the reduced tail load and lighter structure also cited for it are not induced-drag arguments at all.

What was checked #

The ledger holds four checks. With the canard removed, the three-surface calculation reproduces the earlier essay’s tail aircraft, share and drag, within $2 \times 10^{-5}$ at three margins; with the tail removed, its canard aircraft. The fitted bowl reproduces fresh solves to $2 \times 10^{-14}$. The trimmed optimum satisfies its moment equation exactly, and moving along the trim line by five hundredths either way raises the drag. And the saving matches the closed form to 0.2 per cent of the largest saving computed. The break-even moment, 0.333, comes from the closed form with the bowl’s own curvature.

What the Trefftz plane and the trim leave out #

The neutral point, held by assumption. Holding it needs either a larger tail or a canard whose destabilising contribution is cancelled some other way, and each costs area and weight this calculation does not price. The canard’s wake also lands near a tail of the same span, and a real design separates them in height.

Profile drag beyond the canard’s skin. A surface carrying more lift at a higher lift coefficient pays more profile drag, and the three-surface optimum puts lift on a small, highly loaded canard. The saving here is an upper bound for that reason as well.

Stall. A canard must stall first for the aircraft to recover, and a canard trimming at its induced-drag optimum is not necessarily one that stalls first. The two requirements do not share a free variable.

Structure and interference. Weight, the fuselage, and the interference between a canard’s wake and the wing root are not in a Trefftz plane of three straight lines.

The convention: drag against the elliptic wing #

Induced drag is quoted as a multiple of an elliptic wing’s of the same span and total lift; drag counts are ten-thousandths of the wing’s drag coefficient. Shares are fractions of the total lift. Positions are in wing chords from the wing’s aerodynamic centre, positive aft; the static margin is the neutral point’s distance behind the centre of gravity; $C_{m0}$ is the wing section’s pitching moment about its aerodynamic centre, nose-up positive.

Who worked it out #

Munk’s stagger theorem of 1921 is what removes the surfaces’ streamwise positions. The minimum-induced-drag trim of multi-surface aircraft was treated by Kroo and McGeer in the 1980s, who showed that the least trimmed induced drag of two- and three-surface aircraft differs little at practical margins; Selberg and Rokhsaz compared three-surface and conventional configurations in the same decade and reached a similar conclusion, with the advantage appearing in wide centre-of-gravity ranges.

Still open: where the canard’s wake meets the tail #

The comparison held the neutral point by assumption, and the assumption hides the one place the three surfaces interact strongly. A canard’s trailing vortices pass the wing and arrive at the tail, and a tail of the same span, near the same height, sits in their downwash: in this geometry the canard raises the tail’s downwash gradient from 0.41 to 0.54, taking nearly a quarter of the tail’s effectiveness. The next calculation places the tail at a height the designer chooses, computes the canard’s downwash at the tail from the same horseshoe fields, and asks how much tail area it takes to hold the neutral point at each height — which is the wetted-area bill the square above has to be set against, and which decides whether the best three-surface aircraft has a T-tail.