Washout is right at one lift coefficient

The optimum that does not matter found that an untwisted, straight-tapered wing can get within 1.25 per cent of the elliptic wing’s induced drag, at a taper ratio of 0.364, and that the optimum is so flat that anything between 0.25 and 0.51 is within half a per cent of it. It quoted those span efficiencies without saying at what lift coefficient, and it could, because an untwisted wing’s loading has the same shape at every incidence. It ended with a section it did not compute: add twist and the shape starts to change with the incidence, so a twisted wing’s efficiency is a curve rather than a number.

Which part of a wing stalls first is why wings are twisted. A section stalls where its own lift coefficient is highest, a tapered wing’s highest is outboard, and washout — rigging the tips at a lower incidence than the root — moves it inboard, so that the wing stalls where the ailerons are not. This essay computes the curve the first essay promised, and finds that whether washout costs drag depends entirely on which wing it is put on.

Two loadings added #

Prandtl’s monoplane equation is linear in the local incidence. A linearly washed-out wing has an incidence $\alpha$ at the root falling by $\Omega$ to the tip, and so its loading is exactly the sum of two loadings: the untwisted wing’s at incidence $\alpha$, and a pure-twist loading proportional to $\Omega$ — which on its own lifts the root and pushes the tips down, and carries a lift of its own. The first loading’s shape is fixed by the planform; the second’s is fixed too; what changes as the wing flies faster or slower is the ratio of the two.

Write the lift coefficient in place of $\alpha$, and the induced drag becomes a quadratic form,

$$\pi A\,C_{D_i} = C_L^2\,(1 + \kappa_D) - \kappa_{DL}\,C_L\,\Omega + \kappa_{D\Omega}\,\Omega^2,$$

in which $\kappa_D$ is the untwisted planform’s departure from ellipticity and the other two coefficients are fixed by the planform as well. Phillips wrote the induced drag this way in 2004. Everything below follows from it, and the three coefficients come out of two lifting-line solutions.

Three things follow at once. The washout that minimises the induced drag at a given lift coefficient is $\Omega^* = \kappa_{DL}C_L/2\kappa_{D\Omega}$: the best twist is proportional to the lift coefficient it is chosen for. The span efficiency at that point, $1/(1 + \kappa_D - \kappa_{DL}^2/4\kappa_{D\Omega})$, contains no $C_L$, so every design point reaches the same peak. And at any other lift coefficient the excess over the peak is $\kappa_{D\Omega}(\Omega - \Omega^*)^2$, a parabola in the distance from the design point.

A rectangular wing #

A rectangular wing of aspect ratio 8 has a span efficiency of 0.9367 untwisted: its tips carry more load than the ellipse wants, because its chord does not fall as the elliptic loading does. Washout unloads the tips. Two degrees of it make the wing best at a lift coefficient of 0.254, four degrees at 0.509, six at 0.763, and at each of those points the span efficiency is the same, 0.9911 — almost all of the six per cent the planform gave away, recovered.

The curves are also lopsided. Above the design point each falls slowly; below it each falls fast, because at low lift the twist’s own loading — root up, tips down — is a larger share of the total, and that loading on its own is far from elliptic. With six degrees of washout the wing’s span efficiency is 0.68 at a lift coefficient of 0.2. A wing that cruises fast and climbs slowly is asking one twist to serve two points on that curve, and the figure says it cannot.

Why the best twist grows with the lift #

The proportionality has a plain reading. The untwisted loading’s departure from the ellipse is a fixed fraction of its size, so the drag that departure costs grows as $C_L^2$, like all induced drag. The twist’s own loading is fixed in absolute size by the twist and does not grow with the lift at all. At a high lift coefficient the untwisted error is large and a given washout corrects a small fraction of it, so more washout is wanted; at a low lift coefficient the untwisted error is small and the same washout overcorrects it, leaving the twist’s own loading — root up, tips down, nothing like an ellipse — as the dominant shape. The design point is where the correction and the error are matched, and since one scales with $C_L$ and the other does not, the match moves with $C_L$.

That is also why the price of having ends could quote one span efficiency per planform: with no twist there is only one shape, and its error scales with the lift exactly as the ideal drag does. Twist is what breaks the scaling.

A wing that is already nearly elliptic #

On a wing tapered to 0.4, already within 1.3 per cent of elliptic untwisted, the picture is entirely different. Its best washout at a lift coefficient of 0.5 is a tenth of a degree. Four degrees would be best at a lift coefficient of 19.2, a number no wing flies at, so every practical washout sits far to the left of its design point on a parabola that only falls towards the left: at a lift coefficient of 0.4, four degrees costs 7.9 per cent of the span efficiency, and at 1.2 it costs 0.87 per cent. The washout on a well-tapered wing is fitted for the stall and paid for in induced drag at every speed, most at the fastest.

That corrects a sentence the stall essay printed. It said that washout can only lose, because the elliptic loading is the optimum and washout moves away from it, and the figure beside it carried a rectangular curve that rose by five points of span efficiency with four degrees of washout. The argument holds for a wing that starts near the optimum. A rectangular wing starts far from it, on the other side from where washout pushes, and the twist first carries it towards the ellipse and then past.

Which wings want washout, and which want the opposite #

The best twist per unit of lift coefficient, swept across taper ratio, has one sign change. A rectangular wing wants 7.87° of washout for each unit of lift coefficient. The figure passes through zero at a taper of 0.389, and below that the best twist is wash-in — tips at a higher incidence than the root — reaching 8.62° per unit at a taper of 0.1. A sharply pointed wing underloads its tips, and the twist that would make it more elliptic raises them, which is the twist no designer fits, because it is exactly the one that makes the tips stall first.

The zero is not at the untwisted optimum of 0.364. That looks like an inconsistency and is not: the untwisted optimum is the taper at which the span efficiency is greatest over taper, and the zero is the taper at which the efficiency is flat in twist. At 0.364 a small wash-in still helps, by two parts in ten thousand. The two conditions ask different questions of the same quadratic form.

The right-hand panel says how far a straight twist can go. With its best linear twist, no taper reaches an efficiency of one: the best straight twist leaves every planform between 0.9865 and 0.9971. A straight taper and a straight twist are two straight lines, and the elliptic loading is not made of straight lines.

The twist that would be exact #

What would be exact is computable directly. An elliptic loading $\Gamma = 2bU A_1\sin\theta$ is carried when the local geometric incidence is $A_1\big[4b\sin\theta/(a_0 c(\theta)) + 1\big]$, and for a straight taper that is not a straight line along the span. For a rectangular wing the exact twist is all washout, curving down to 5.81° at the tip, against the best straight washout of 3.93° — the straight line is a fit to a curve that steepens towards the tip. For a 0.4 taper it is stranger. The exact twist is wash-in over most of the span, reaching 1.02° at 0.6 of the semi-span, and then a washout that falls steeply to 4.06° at the tip. No straight twist can follow a line that first rises and then falls, so the best one is nearly flat and the wing stays at its untwisted 0.987.

This is also why the exact twist is not built. It is fixed by one lift coefficient, so it is exactly right at that one and wrong at every other — the parabola again — and its steep last tenth of the span makes the tips the least loaded part of the wing at high incidence, which is what washout is for, but its wash-in mid-span moves the stall outboard of the root, which is not.

The parabola #

The excess drag over the elliptic wing makes the trade visible in the units a performance engineer uses. Untwisted, the rectangular wing’s excess grows as $C_L^2$ and reaches 38.7 counts at a lift coefficient of 1.2. With four degrees of washout the excess is a parabola with its least at the design point, 0.509, and at 1.2 it is 16.3 counts — less than half. The two curves cross at a lift coefficient of 0.254, exactly half the design point, and that half is general: the twisted wing’s excess equals the untwisted wing’s where $\kappa_{DL}C_L = \kappa_{D\Omega}\Omega$, which is half of $\Omega$’s own design lift coefficient. Below the crossing the twist costs drag; above it the twist saves.

For an aeroplane that is the arithmetic of choosing washout. Cruise is at the low end of the lift coefficients and climb and landing at the high end; a rectangular-winged light aircraft rigged with four degrees of washout pays a little at high-speed cruise, saves a great deal in the climb, and has the stall where the ailerons are not. Rectangular wings with generous washout are common on training aircraft for all three reasons, and the parabola is the reason the first two are not in conflict.

What the twist buys at the stall #

The drag is half of the bargain, and the other half is the reason anyone twists a wing. At a wing lift coefficient of one, the untwisted 0.4 taper’s sections are most heavily loaded at 0.61 of the semi-span, at 1.06; four degrees of washout move that peak to 0.29 of the semi-span at the same value. The stall then starts nearer the root, ahead of the ailerons, which is the behaviour which part of a wing stalls first was after. The rectangular wing’s sections peak at the root with or without the twist, so its washout is bought for the drag and its stall comes with the planform. On the tapered wing the stall is the whole of what the twist buys, and the drag is the whole of what it costs.

A trainer, in newtons #

Put the rectangular wing on an aeroplane: a span of 11.3 metres and an area of sixteen square metres, aspect ratio 8, carrying a tonne. At a cruise of 55 metres a second its lift coefficient is 0.331; in the climb at 35 metres a second, 0.817. Untwisted, its induced drag is 137.9 newtons in the cruise and 340.5 in the climb. With four degrees of washout, whose design point is 0.509, both lift coefficients are above the crossing at 0.254, and both drags fall: to 132.5 and 324.5 newtons, saving 0.29 kilowatts in the cruise and 0.56 in the climb. The washout fitted to keep the stall at the root costs this aeroplane nothing anywhere in its normal envelope; only a dive at a lift coefficient below a quarter would put it on the losing side of the parabola.

The same four degrees on a 0.4 taper of the same span and area would cost drag at both points, by 7.9 per cent of the span efficiency at a lift coefficient of 0.4. That is why the choice of planform and the choice of twist cannot be made separately. The span is the whole story for the elliptic wing’s drag; for a real wing it is the span and the pair of planform and twist, and a designer who picks a near-elliptic taper for its drag and then adds washout for its stall has paid for the drag twice. The bell-shaped loading is a third choice of the same kind, made for the root bending moment instead, and it too is a twist distribution chosen for one lift coefficient.

How the form was checked #

The form is built from two lifting-line solutions with twenty-four odd harmonics — one degree of incidence and no twist, one degree of washout at zero root incidence — and is then checked against a third, direct solve at 6.3° of incidence and 3.7° of washout, which it agrees with to four parts in $10^{16}$ for both the 0.4 taper and the rectangular wing: the linearity it rests on is exact. With no twist it returns the untwisted 0.364 taper’s 0.98749 at lift coefficients of 0.3 and 1.2 with no difference, as the earlier essay’s single number requires. And the exact elliptic twist, fed back to the monoplane equation as its right-hand side, returns a loading whose higher harmonics are $10^{-17}$ of the first and a lift coefficient of one when one was asked for.

The convention: linear washout, measured at the tip #

Washout is the tip’s geometric incidence below the root’s, varying linearly with the distance from the centre line, positive when the tip is lower; wash-in is negative washout. The section lift-curve slope is $2\pi$ everywhere, the sections are uncambered or share one zero-lift angle, and the span efficiency is the ratio of the elliptic wing’s induced drag to this one’s at the same span and lift. A count is $10^{-4}$ in drag coefficient.

What the picture cannot show #

Lifting-line theory assumes a straight, planar wake and a high aspect ratio, and at aspect ratio 8 it is good to a few per cent in the coefficients and better in the differences between them. It has no viscous drag, and washout changes the profile drag too: a washed-out tip flies at a lower section lift coefficient, which on most sections is a lower profile drag, and on a laminar section can be a higher one — the two drags a wing pays trade against each other here as everywhere. Nothing here says what the stall does; which part of a wing stalls first is about that, and the stall is the reason washout exists. A real wing’s twist is often not linear — it is set by the rib jigs and can be anything — so the straight-twist results are one family, and the exact twist shows what a free one could do.

Who found it, and when #

Glauert’s Fourier form of the lifting line, which makes the twist a right-hand side, is from 1926. The observation that a twisted wing’s span efficiency depends on its lift coefficient is as old as the theory and is in every careful textbook as a caveat. Phillips’s 2004 analysis of twisted wings wrote the induced drag as the quadratic form used here, gave the optimum washout as proportional to the lift coefficient, and showed that a rectangular wing with the right washout can match an elliptic planform’s induced drag closely. The sign change in the best twist with taper, and the wash-in it implies below a taper of about 0.39, follow from the same form.

Still open: one twist for a mission #

The parabola says a twist is right at one lift coefficient, and an aeroplane flies at many. The next calculation chooses the washout that minimises the induced drag averaged over a flight: a weighting of lift coefficients by the time spent at each — a long cruise, a shorter climb, a few minutes near the stall — for a training aircraft and for a glider, and asks how far the mission-optimal washout moves from the cruise-optimal one, whether a rectangular trainer’s usual few degrees are near it, and how much a glider’s thermalling at high lift coefficients justifies washout that its cruise between thermals pays for.