More lift than weight

The first diagram in every book about flight has four arrows: lift up, weight down, thrust forward, drag back. Two of them are equal and opposite in level flight, and the equation $L = W$ is the foundation of every performance calculation there is.

It is also a collapse — the whole configuration onto one number — and like every collapse in this collection it discards something. Here the discarded thing is that there are two lifting surfaces, and the equation that decides how they share the load is the one about moments rather than the one about forces.

Two equations, solved #

The aeroplane has a wing at $x_w$ and a tail at $x_t$, a centre of gravity at $x_g$ between them, and a nose-down zero-lift pitching moment $M_0$ from the wing’s camber. Level flight requires

$$L_w + L_t = W,\qquad L_w(x_g - x_w) + L_t(x_g - x_t) + M_0 = 0.$$

Two linear equations in two unknowns, so the split is determined and there is nothing to approximate. The residual of the solve is $1.6\times10^{-17}$, which makes the numbers below arithmetic rather than a model.

At a quarter-chord static margin the wing carries 11,672 N against a weight of 11,000 and the tail carries −672 N. The wing is lifting 106.1 per cent of the weight and the tail is pushing down on the other 6.1.

Move the centre of gravity aft and the excess falls, reaching zero where the tail carries nothing and going negative — the tail lifting — behind that. The excess is not small and it is not a correction: it is a direct consequence of the moment equation, and every aeroplane with a conventional tail has it.

Why the tail carries down #

The reason is the camber, and it is worth spelling out because it is the same argument the aerodynamic centre rests on.

A cambered wing has a nose-down pitching moment about its aerodynamic centre, and that moment does not depend on incidence. Something must balance it. If the centre of gravity is at the wing’s aerodynamic centre, the wing’s own lift has no moment arm and the tail must supply the whole balancing moment — a download, because the moment is nose-down.

Moving the centre of gravity forward makes it worse: now the wing’s lift also produces a nose-down moment about the centre of gravity, and the tail must push down harder still.

And moving it aft eventually reverses the sign. But the centre of gravity cannot be moved aft indefinitely, because of the second thing the tail is for.

The neutral point, computed twice #

Stability is the constraint on how far aft the centre of gravity may go, and the boundary is the neutral point.

Forward of the neutral point the slope $\mathrm dC_M/\mathrm d\alpha$ is negative: a gust that raises the incidence produces a nose-down moment that removes it, and the aeroplane returns to trim. Behind it the slope is positive and it does not. At the neutral point the moment does not depend on incidence at all.

The neutral point is computed here two ways — from the closed form with the tail volume in it, and by differentiating the whole-aeroplane moment numerically — and they agree to seven decimals. The reason for doing both is the term everybody leaves out: the downwash derivative $\mathrm d\varepsilon/\mathrm d\alpha$, which is 0.382 for this wing, and which reduces the tail’s contribution by nearly forty per cent. The tail sits in the wing’s wake and a calculation that forgets it puts the neutral point about thirteen per cent of chord too far aft.

The drag of carrying it twice #

Now the part the force equation cannot see at all.

The induced drag of two surfaces carrying $L_w$ and $L_t$ is not the induced drag of one surface carrying their sum. Each pays

$$D_i = \frac{L^2}{\pi q b^2}$$

for its own load on its own span, so a download on a short tail is expensive twice over: the load is squared, and the span is small.

With no interference between the wakes, relaxing the static margin from a fifth of a chord to a twentieth saves 9.32 per cent of the induced drag. That is a large number — comparable with what a winglet buys, and available for nothing but moving the ballast aft.

The least-drag aeroplane is unstable #

Following that curve to its minimum gives an answer that is worth sitting with.

The least-drag centre of gravity is 0.40 chords behind the neutral point, where the tail carries 1,391 N upwards rather than downwards. At that point the pair behaves like a biplane: the two surfaces share the total lift in proportion to their spans, in exactly the sense Munk’s theorem means, and the total induced drag is minimised by loading both.

And an aeroplane there is statically unstable in pitch. It will not hold an attitude by itself, it diverges from any disturbance, and it is flyable only with a control system fast enough to catch it.

That is a real trade rather than an artefact, and it is the reason relaxed static stability exists: a fly-by-wire aeroplane can be balanced aft of the neutral point because the computer does what the tailplane was doing, and the drag saved is the number above. It is also why a delta-canard fighter has its centre of gravity behind the neutral point by a substantial margin and cannot be flown with the computers off.

What the model does not supply #

The conclusion above depends on a coefficient the model does not contain, and the honest thing is to compute where it stops holding rather than to pick a default.

The interference term is $2\sigma L_wL_t/\pi q b_wb_t$, and its sign follows the product of the loads. At σ = 0 the least-drag aeroplane is unstable. Raise σ and a download becomes worth something — the cross term is negative when the loads have opposite signs — and above σ = 0.35 the minimum jumps forward of the neutral point and the trade runs the other way.

A real tail sits in the wing’s downwash by construction, so σ is not zero. It is also not a number this two-term model can supply, and a proper answer needs the two wakes’ actual positions in the Trefftz plane. So the model’s conclusion depends on a coefficient the model does not have, and saying so is the result.

What is not sensitive to σ is everything before this section: the load split, the excess over the weight, and the neutral point are all consequences of the two-equation trim and do not involve the drag model at all.

What the tail is actually for #

It is worth separating the three jobs the tail is doing, because the essay has now touched all of them and they are usually run together.

Trim. Balancing the wing’s zero-lift pitching moment and the moment of its lift about the centre of gravity. That is a force the tail must produce, it is usually a download, and it is what the first figure computes.

Stability. Making $\mathrm dC_M/\mathrm d\alpha$ negative, so that a disturbance is opposed. That is a slope rather than a force, and a tail can be providing plenty of stability while carrying almost no load — which is exactly the condition at the neutral point minus a small margin.

And control. Producing a moment on demand to change the trim, which is what an elevator is for and which sets the size of the tail more often than either of the other two. The critical case is usually rotation at take-off with the most forward centre of gravity, where the tail must have enough authority to lift the nose at a speed well below flying speed.

Those three requirements pull in different directions. Trim wants the download small, so the centre of gravity aft. Stability wants it forward. Control wants a large tail, which makes the download expensive. The tail volume that comes out is a compromise among three constraints, and a canard or a tailless configuration is a different compromise rather than an escape from one.

The trim drag of real aeroplanes #

Some numbers from outside the model are worth putting beside it.

Trim drag on a conventional transport is typically one to three per cent of the total drag in cruise, which is small and is worth chasing: a per cent of cruise drag on a long-haul aeroplane is a great deal of fuel over a fleet’s life. It is managed by moving fuel aft in flight — the Concorde and the A340 both had trim tanks in the tail for exactly this — which shifts the centre of gravity towards the neutral point as the flight progresses and the static margin can be relaxed.

The number is larger for a configuration with a small tail volume or a large zero-lift moment, which is why a highly cambered wing costs twice: once in its own profile drag and again in the download needed to balance it.

And it is negative for a canard, where the forward surface must lift up to balance a nose-down moment. That is often quoted as a canard’s advantage and is only half the story: the canard must stall before the wing for the configuration to be safe, which forces it to run at a high lift coefficient, and the induced drag of a small highly loaded surface is exactly what the $L^2/b^2$ above is about.

Where else two surfaces share a load #

The trim problem’s structure — a total that is fixed and a split that is decided by something else — shows up several times in this collection, and the comparison sharpens what is peculiar about the tail.

A biplane shares its lift between two wings, and Munk’s stagger theorem says the total induced drag does not depend on the fore-and-aft spacing at all. Two wings and it does not matter where computes that, and the optimum split there loads both surfaces upwards — which is exactly the state the least-drag aeroplane above wants and cannot have.

A wing with a winglet shares its load between a horizontal and a vertical surface, and the vertical one earns its keep by extending the wake trace out of the plane. Same arithmetic, and no moment constraint forcing a download.

And formation flight shares a lift between two aeroplanes. The lift beside a wing shows a pair tip to tip costing exactly half what the two cost apart, which is the $L^2/b^2$ scaling with the span doubled.

In every one of those the split is chosen to minimise drag. The aeroplane is the only case where a second equation forces the split, and the second equation is about moments. That is why trim drag is a cost rather than an optimisation: nothing is being traded away, it is being spent on a constraint.

Four points, not one #

The neutral point computed above is one of several, and the differences between them are what an aircraft is actually certificated against.

It is the stick-fixed neutral point: the elevator is held at a fixed deflection while the incidence changes. Let the control go and the elevator floats to whatever angle makes its own hinge moment vanish — and under a gust it floats in the direction that reduces the tail’s change of lift. The tail is therefore less effective with the stick free than with it held, and the neutral point moves forward. That gap is why an aerodynamic balance or a horn on a control surface is not merely a lightening of the stick forces: it changes where the aircraft’s stability boundary is.

And in a manoeuvre the aircraft has a pitch rate, so the tail — being behind the centre of gravity — is swept through the air at an angle the wing is not. That extra incidence is stabilising and exists only while the aircraft is turning, so the boundary in a pull-up sits aft of the level-flight one. The distance to it is the manoeuvre margin, and it is what sets the stick force a pilot feels per unit of load factor.

Which is the quantity the regulations actually bound. Not the static margin, but the force per g — a number that is a property of the wing, the tail, the hinge moments and the gearing at once.

What is left out #

Only induced drag. Which is the term the trade above is about, and is a minority of the total: on a transport in cruise the induced drag is around a third of the whole and the trim component of it is a few per cent of that. Profile drag, compressibility drag and the tail’s own parasite drag are absent, and for a real aeroplane the profile drag of the tail is comparable with its induced drag.

Elliptic loading on each surface separately, which no real tail has.

A rigid aeroplane, so no aeroelastic effects — and a swept wing’s twist under load moves its own aerodynamic centre, which moves the neutral point.

Steady, level, one flight condition. The static margin that matters is the worst case over the whole loading envelope and the whole flight, and the aft limit is usually set by the most aft loading at the most aft fuel state.

And no elevator. The trim above is achieved by an unspecified moment; a real aeroplane deflects a surface, which changes the tail’s own lift curve and adds its own drag. That deflection is also what makes the tail’s stalling incidence a design case: a tail at a large download with a large elevator deflection is close to stalling on its lower surface, and a tailplane stall in that condition is unrecoverable because the nose drops and the download that was holding it up disappears.

The number a structure is sized by #

There is a second consumer of the wing’s lift, and it wants a different answer from the one the performance engineer wants.

A performance calculation asks for the drag of an aeroplane weighing $W$, and $L = W$ is a perfectly good statement of the equilibrium it is in. A structural calculation asks what the wing carries, and the answer is 106.1 per cent of the weight in the case computed here — because the tail is pulling down and the wing is making up the difference. At a limit load factor of 2.5 that six per cent is six per cent of a much larger number, and it arrives in the root bending moment where a per cent is a real mass of aluminium.

The two figures are not a disagreement about the aeroplane. They are two different questions, and the collapse that makes them look like one question is the same collapse this essay is about: writing $L = W$ for the whole aircraft and then using $L$ as though it were the wing’s.

The direction of the error is the unhelpful one. A wing sized on $L = W$ is sized light, and the configuration in which the discrepancy is largest — a small tail volume, an aft-loaded aerofoil, a forward centre of gravity — is exactly the one a designer is pushed towards by cruise efficiency. The static margin is where the two calculations meet: it sets the download, and the download sets both the trim drag and the excess the spar has to carry.

Where the arithmetic came from #

The two-equation trim is as old as the tailplane, and its modern form dates from the aerodynamic-centre concept in the 1920s. What is more recent is the idea that the static margin is a variable rather than a fixed requirement.

Relaxed static stability arrived with digital flight control in the 1970s, and the aeroplanes that exploit it — the F-16, the Airbus family, every modern fighter — are the ones for which the drag computed above was worth the certification argument. The number that made the case is exactly the one in the drag figure: a few per cent of induced drag, which on a fighter is the difference between a turn sustained and a turn lost.

What this leaves #

$L = W$ is true of the aeroplane and false of the wing, and the difference is a download on a short span that is paid for twice. The residual the collapse discards is the second surface, and its price is a few per cent of the induced drag and the whole of the stability.

The next essay takes the collapse of a wing onto a line, and asks where a line stops being a reasonable model of a wing: where the line stops being a line.