No elevon reaches past a quarter chord

More lift than weight trimmed a conventional aeroplane exactly. With the centre of gravity ahead of the neutral point, as stability requires, the tail pushes down, the wing carries 106 per cent of the weight, and the induced drag pays for the difference. The tail’s virtue in that calculation was its arm: a small load three or four chords behind the centre of gravity makes a large moment, so the download it needs is small.

A wing with no tail has no such arm. It must make its own nose-up moment, against the nose-down moment its stable centre-of-gravity position creates, out of the section it already has. It can do that in two ways. It can deflect a surface at its trailing edge upwards — an elevon, the combined elevator and aileron of every flying wing — or it can be built with a camber line whose own pitching moment is nose-up, a reflexed line that curls upward at the back. This essay prices both in the theory that prices them exactly, and the answer turns on one surprising number: the arm through which an elevon acts is never more than a quarter of the chord.

Two functionals and a hinge #

Thin-aerofoil theory writes the camber line’s slope as a cosine series in the angle $\theta$, where $x/c = \tfrac12(1-\cos\theta)$: $dz/dx = B_0 + \sum B_n\cos n\theta$. Then

$$C_l = 2\pi\alpha + \pi(B_1 - 2B_0), \qquad C_{m,c/4} = \frac{\pi}{4}(B_2 - B_1).$$

Three numbers out of a camber line made the point that lift and moment read only the first three coefficients, and that everything else about the shape is invisible to them. Here the point is that they read different combinations, so a change of shape can buy moment and lift in any proportion. A pure second harmonic, $dz/dx = B_2\cos 2\theta$, changes the moment by $\tfrac{\pi}{4}B_2$ and the lift by nothing.

An elevon is a much more constrained change: a flat piece of the chord, from a hinge at $x/c = 1-E$ to the trailing edge, rotated through an angle $\delta$. Glauert worked out its series in closed form. With the hinge at $\theta_f$, $\cos\theta_f = 2E - 1$, a downward deflection gives

$$\Delta C_l = 2(\pi - \theta_f + \sin\theta_f)\,\delta, \qquad \Delta C_{m,c/4} = -\tfrac12\sin\theta_f(1-\cos\theta_f)\,\delta.$$

The ratio of the two is the price of trimming with it: the lift given away per unit of moment bought. Its reciprocal has a physical meaning. If a load $\Delta C_l$ makes a moment $\Delta C_m$ about the quarter chord, it acts at a distance $-\Delta C_m/\Delta C_l$ behind it. That distance is the elevon’s arm, and it is the number to compare with a tail’s.

The arm #

The figure is the arm against the elevon’s share of the chord, and it is short everywhere. A fifth-chord elevon’s load acts 0.185 chords behind the quarter chord; a tenth-chord elevon’s, 0.217; half-chord, 0.097. A flap of the whole chord is simply a change of incidence and acts at the quarter chord itself, with no arm at all.

The surprising end is the other one. As the elevon shrinks to a tab at the very trailing edge, the arm does not grow towards three-quarters of a chord, which is where the tab is. It grows towards a quarter, exactly. Near $E = 0$ the hinge angle is close to $\pi$; writing $\varepsilon = \pi - \theta_f$, the lift is $2(\varepsilon + \sin\varepsilon) \approx 4\varepsilon$ per radian and the moment $\tfrac12\sin\varepsilon(1 + \cos\varepsilon) \approx \varepsilon$, and their ratio tends to $\tfrac14$. No elevon of any size acts through more than a quarter chord.

The reason is that a deflected flap does not load only itself. It changes the circulation of the whole section — the Kutta condition now applies at a trailing edge that has moved — and the circulation it adds is distributed along the whole chord with most of it near the leading edge, as incidence’s is. What a flap does found the same thing from the lift side: a flap is mostly a change of the section’s effective incidence. Seen from the moment side, that is the statement that its load sits far forward of the flap.

What the elevon’s load looks like #

The picture makes the arithmetic visible. Ask both camber changes for the same nose-up moment, 0.02 in coefficient. The fifth-chord elevon must rise 1.79°. Its added loading is negative along the whole chord — it unloads the section — most strongly near the leading edge, where the loading of any circulation change is concentrated, with a logarithmic spike at the hinge. Integrated, it takes away a lift coefficient of 0.108. The moment it makes is the small difference between the leading edge’s unloading and the trailing edge’s, which is why it needs so much lift to make a little moment.

The second-harmonic reflex loads the front half of the chord and unloads the back half by the same amount. Its integral is zero, its moment is the 0.02 asked for, and it costs no lift at all. It is a pure couple. That is not an approximation: in thin-aerofoil theory the lift reads $B_1$ and $B_0$ and the second harmonic has neither.

So the two ways a tailless wing can make its moment are opposite ends of one scale. A hinge buys moment by spending lift, at a rate of at least four to one. A shaped camber change can buy moment and spend nothing. The difference is not in the aerodynamics of the section but in what shapes a mechanism can make.

What trim costs in lift-curve slope #

The arm becomes a cost the moment the wing has to be stable. With the centre of gravity a static margin $h$ ahead of the quarter chord — the section’s aerodynamic centre — a lift $C_l$ makes a nose-down moment $hC_l$ about it, and the wing is trimmed only if the section’s quarter-chord moment equals $hC_l$. If the control that supplies that moment gives away $q$ units of lift per unit of moment, then trimming at lift $C_l$ costs $qhC_l$ of lift, and the trimmed section’s lift-curve slope is

$$\frac{dC_l}{d\alpha}\bigg|_{\text{trim}} = \frac{2\pi}{1 + qh}.$$

For a fifth-chord elevon $q = 5.4$. At a static margin of five per cent of the chord the trimmed wing keeps 0.787 of its slope; at ten per cent, 0.649. The tail, three chords back, has $q = 1/3$ and keeps 0.984 and 0.968. A twentieth-chord elevon, close to the quarter-chord limit, does a little better than the fifth-chord one, and a two-fifths-chord elevon noticeably worse. A camber that could change its second harmonic in flight would keep all of the slope at any margin.

This is the quantitative content of the flying wing’s best-known handicap. Every unit of stability, measured as static margin, is paid for at the elevon’s poor exchange rate, so flying wings are flown at small margins — a few per cent of the mean chord — and still carry less lift at a given incidence than the same wing with a tail would. A canard pays for its stability in induced drag found the other configurations’ bills; the tailless wing’s is in incidence.

The two shapes #

The two camber changes that make the same moment look similar from a distance and differ in exactly the way the loadings did. The elevon’s trailing edge rises 0.63 per cent of the chord, from a hinge at eighty per cent. The reflex rises over the first seventh of the chord, dips through the middle and rises over the last seventh, ending 0.85 per cent of the chord below its leading edge. The reflex’s rise at the back is what a designer recognises as reflex; its rise at the front is the part that pays for the absence of a lift loss.

Real flying-wing sections are built with a reflex near the trailing edge and an ordinary camber ahead of it, which is a mixture of the two: some lift-neutral couple, some elevon-like trailing-edge deflection. Where in that mixture a given section sits is exactly its $q$, and the theory says the forward part of the reflex is what lowers it.

A worked tailless glider #

The numbers become concrete on a small flying wing of the kind built as a model or a light glider: a mean chord of 0.3 metres, a fifth-chord elevon, a static margin of five per cent of the chord — fifteen millimetres — and a landing lift coefficient of 0.8. Trimming that lift needs a nose-up moment coefficient of 0.04 about the quarter chord, and the elevon makes 0.64 per radian, so it must rise 3.6°. In doing so it takes away a lift coefficient of 0.216, more than a quarter of what is being asked for, and the wing must fly at 9.3° instead of 7.3° to make it up. If the section stalls at twelve degrees, the margin to the stall has shrunk from nearly five degrees to under three.

The same wing with a tail one metre behind its centre of gravity — three and a third chords — needs a tail download of 0.012 in the wing’s lift coefficient for the same moment, and flies at 7.4°. The difference of almost two degrees of incidence is the price of the elevon’s arm being 0.185 chords, or 56 millimetres, against the tail’s thousand. That is why the classic tailless designs use small margins and generous sweep, and why their pilots learn that the landing is where the configuration shows.

Where the quarter chord comes from #

It is worth seeing that the quarter is not a coincidence of the flap formula. The quarter-chord point is where the lift of any incidence change acts, which is where the lift acts on every thin section: the loading of an incidence change has the shape $(1+\cos\theta)/\sin\theta$, heavy at the leading edge, and its centroid is at a quarter chord. Glauert’s loading for a flap splits into two parts. One has exactly the shape of an incidence change — moving the trailing edge moves the stagnation line the Kutta condition fixes — and acts at the quarter chord. The other is centred on the hinge but spreads along the chord as a slowly decaying logarithm. For a tab of a ten-thousandth of the chord, integrated numerically, the two carry equal lift to four figures, and the second’s centre is not at the tab but at 0.750 of the chord — half a chord behind the quarter-chord point. Equal lifts at arms of nothing and of half a chord put the total a quarter of a chord back. Every larger elevon spreads its local load further forward, which shortens the arm, until the whole-chord flap is all incidence change and no arm at all.

The same structure is behind the control that works backwards: an aileron’s local load twists the wing, and the twist is an incidence change of the whole section that can outweigh the aileron’s own contribution.

A fixed reflex trims one speed #

A real reflex is built into the wing and cannot change. It supplies one moment, so it trims exactly one lift coefficient, and the elevon trims the rest. The figure follows that through. A tail trims a lift coefficient of 1.0 at 9.27°, barely above the untrimmed 9.12°. A fifth-chord elevon alone needs 11.6°. A reflex built to trim at 0.5 needs no elevon there and flies at the changing reflex’s 4.56° — the free trim at one speed — but at a lift coefficient of 1.0 it needs 10.3°, because the elevon is trimming the difference between 1.0 and 0.5 at its usual rate. At low lift the built-in reflex supplies more nose-up moment than trim needs, the elevon goes down to cancel it, and that adds lift: 0.1 in lift coefficient at −0.07°, against the changing reflex’s 0.91°.

A built-in reflex, in other words, is a choice of where on the speed range the elevon is centred, and it buys nothing at the other speeds. A designer who sizes it for cruise gets cruise for free and the landing at full price — and landing is where lift at a given incidence matters most, because the incidence is limited by the stall and by the ground.

The elevon’s zero moves, its slope does not #

The deflection schedule says the same thing in the quantity a pilot or a flight-control system sees. With no reflex the elevon rises in proportion to the lift, 4.48° per unit of lift coefficient, reaching 4.48° up at a lift coefficient of one. A reflex built for 0.5 shifts the whole line: the elevon is neutral at 0.5, 2.24° up at 1.0 and 2.24° down at zero lift. The slope is unchanged, because it is set by the static margin and the elevon’s moment per degree, and neither has anything to do with the reflex. The reflex chooses where zero is.

That has a consequence for the stability argument. The elevon’s deflection per unit of lift coefficient is the quantity a stick-fixed analysis calls the elevator’s trim gradient, and it is fixed by $h$ alone for a given elevon. A tailless wing cannot use its reflex to become more stable or less; it can only use it to fly at a neutral elevon at one speed.

Five checks on the exact solution #

Glauert’s closed forms are checked against an independent quadrature of the flap’s own slope at four chord fractions, split at the hinge so the kink is a node, and agree to $3\cdot10^{-13}$. The second-harmonic reflex, integrated the same way, has a lift of $10^{-18}$ and a moment of exactly the 0.02 asked for. The fifth-chord elevon’s closed-form loading, integrated along the chord through its logarithmic singularity at the hinge, gives back its lift and moment to four parts in a million. A flap of the whole chord gives a lift of $2\pi$ per radian and no moment, as an incidence change must. And an elevon of a millionth of the chord has an arm of 0.25000 chords. The tests also refuse a chord fraction of zero, of one, and a negative one.

What thin-aerofoil theory leaves out #

Viscosity. An elevon’s effectiveness falls well below Glauert’s value once its deflection is more than a few degrees, because the boundary layer separates at the hinge’s corner. The arm is less affected, since it is a ratio, but the moment per degree, and therefore the deflection needed, are both optimistic.

Three dimensions and sweep. A real flying wing is swept, and its elevons sit near the tips, far behind the centre of gravity along the root chord. Sweep gives the wingtip elevons an arm the two-dimensional section does not have — which is why swept flying wings exist and unswept ones are rare — and the section result here is the part of the trim that sweep has to make up for.

Thickness. Thin-aerofoil theory has no thickness, and real elevons sit on a thick section whose trailing-edge angle changes their effectiveness.

Stall. The trimmed slope is a slope. Whether the lost lift is lost at the stall depends on the section’s maximum lift at the trimmed incidence and with the elevon raised, which inviscid theory does not compute; the usual measured result is that trimmed maximum lift of a tailless wing is markedly lower than its untrimmed value, the same direction as the slope.

The convention: moments about the quarter chord, margins as fractions of it #

Moments are about the quarter-chord point, the section’s aerodynamic centre in thin-aerofoil theory, and positive nose-up. A static margin $h$ is the centre of gravity’s distance ahead of that point, in chords. Elevon deflections are positive downward, so trim deflections are negative. A tail is represented only by its arm: a load at $l_t$ chords behind the centre of gravity makes a moment $l_t$ times its lift, and its downwash, which more lift than weight and the surface in the wake both account for, is ignored here because it changes the tail’s lift and not its arm.

Glauert, and the flying-wing designers #

Glauert published the flap solution in 1927 and it has been the standard estimate of a plain flap’s effectiveness since. The lift-neutral property of the second harmonic follows directly from Munk’s and Glauert’s Fourier form of thin-aerofoil theory, and the reflexed sections of tailless aircraft — the Horten brothers’, Northrop’s, and the hang-glider and model designers’ after them — were designed to a zero or slightly positive pitching moment by exactly that freedom. The observation that a trailing-edge control’s centre of pressure stays near the front of the chord is old; the quarter-chord limit is the exact form of it.

Still open: the arm that sweep provides #

The section result says how little an elevon can do on its own. A swept flying wing works anyway, because sweep puts the elevons’ spanwise stations behind the centre of gravity by a distance the section does not have. The next calculation carries the same pricing into a swept wing’s lifting line: an elevon spanning the outer third of a wing swept by twenty, thirty and forty degrees, its lift and moment from the span loading it changes, and its effective arm about the wing’s own aerodynamic centre. It asks at what sweep the outboard elevon’s arm reaches the tail’s few chords, and how much of that the tip’s own reduced lift-curve slope takes back — which would say how much of a flying wing’s sweep is there for compressibility and how much is there to make its elevons into a tail.