Behind a finite wing the wake face keeps its share

Air must be pushed down, and the usual sum is wrong drew a rectangle round an aerofoil section and split the lift between the rectangle’s four faces. The popular momentum argument says a wing holds itself up by throwing air downwards, and it computes the momentum leaving through the face behind the wing. That face turned out to carry about three per cent of the lift, with the wrong sign. The faces above and below carried most of it, as pressure and as momentum crossing them, because a wing’s influence spreads up and down much further than it trails behind. The conclusion was that the argument is right in its total and wrong in its picture: the wing does deflect air downwards, but the face the story keeps is the smallest of the four.

That box was round a section — a wing of infinite span. A real wing has tips, and its tips shed trailing vortices that run downstream as far as anyone cares to look. The drag a wake keeps however it rolls up put those vortices in a plane behind the wing and found the induced drag there as the kinetic energy of their crossflow. The same plane can be asked the momentum question. A face behind a finite wing is crossed by a flow that is still moving downwards between the vortices and upwards outside them. How much of the weight does it carry?

The answer is not “a few per cent”, and it is not “all of it”. It is whatever the shape of the face says.

The box round a finite wing #

The wing here is a horseshoe vortex, the simplest finite wing lifting-line theory has. A bound vortex of circulation $\Gamma$ lies across the span, and two trailing vortices run from its ends straight downstream. The legs are $b_0 = \pi b/4$ apart, which is where an elliptically loaded wing’s trailing sheet rolls up to, as where the wake ends up found from the sheet’s invariants. The lift is $\rho U \Gamma b_0$, the relation a wake that says what made it read back out of the vortices.

The box runs from far upstream to a face a distance $X$ behind the wing, with a width $W$ and a height $H$. The flow is linearised about the free stream, as lifting-line theory is everywhere, so the pressure on a face is $p = -\rho U u$ and the vertical momentum crossing a face normal to the stream is $\rho U \int w\, \mathrm dA$. The lift is then the sum of two things: the downward momentum crossing the face behind, and the pressure difference between the faces above and below. The upstream face, far ahead, carries nothing. The side faces carry no vertical force at this order.

Two facts make the calculation short. The trailing legs lie along the stream, so they induce no streamwise velocity and no pressure; the faces above and below feel only the bound vortex. And every integral reduces to one dimension in closed form. A straight segment’s induced velocity integrates across the width exactly. A semi-infinite leg’s velocity at a distance $X$ downstream of its start is its two-dimensional value times $\tfrac12\bigl(1 + X/\sqrt{X^2+r^2}\bigr)$, and that factor integrates across the height to an arctangent. A two-dimensional vortex’s downwash integrates over a rectangle to logarithms and arctangents.

Far behind, the share is the face’s own shape #

The first figure puts the face far behind the wing and changes only the box’s shape, keeping its area at thirty-six square spans. A box sixteen times as tall as it is wide sends 0.96 of the weight through the face behind it. A square box sends 0.501. A box sixteen times as wide as it is tall sends 0.04. The rest, in each case, is pressure on the faces above and below. The points lie on

$$\frac{2}{\pi}\arctan\frac{H}{W},$$

the share a large face takes of a two-dimensional vortex pair’s momentum. The span has dropped out: once the box is large against it, the trailing pair looks like a dipole, and a dipole’s downwash integrates over a rectangle to a number that depends on the rectangle’s aspect alone.

The reason is visible in the downwash itself. Between the two vortices the air moves down; outside them, above and below and to the sides, it moves up, more weakly and over a larger area. A tall, narrow face catches the downward column and very little of the upward return flow beside it. A wide, flat face catches the column and all of the return flow on either side, and the two nearly cancel. A square face catches exactly half.

So “how much downward momentum does the wing leave behind it per second” has no answer until the face is named. The arctangent itself is not new. Where the lift reaction is drew rectangles round a section’s bound vortex and found the split between pressure and momentum running from three per cent to ninety-seven with the rectangle’s aspect, and the momentum with no value traced it to the momentum of an unbounded flow being conditionally convergent. Those rectangles surround the wing. The face here does not: it lies across the wake, as far behind the wing as one likes, and the share it carries never changes. For a section the same face carries nothing, which is the next figure’s subject. Behind a finite wing the dependence on the face is the whole of the story, because the far face never empties.

A section’s face empties; a finite wing’s does not #

The second figure moves the face backwards. For a square box four spans across, the face carries 0.74 of the lift a tenth of a span behind the wing. Two spans behind it carries 0.585, and fifty spans behind 0.502, settling on the half the first figure gave. The tall box settles on 0.845 and the wide one on 0.157.

The dashed curve is the same square box round a two-dimensional section of the same lift, a bound vortex with no ends. Its face carries 0.97 of the lift at the same tenth of a span’s distance behind, where the bound vortex’s downwash fills it. By two spans’ distance it carries a half, and by fifty 0.025. A section’s downwash behind it is the bound vortex’s alone, and that falls away with distance. Far behind a section, the weight is carried entirely by pressure above and below — which is what the earlier essay’s small, negative downstream term was showing in a box only a few chords long.

A finite wing never lets the face empty. Its trailing vortices keep their circulation and their spacing as far back as the wake survives, and every face across them carries the same share of momentum. Two spans behind the wing, the square box’s face has lost the bound vortex’s contribution and kept the trailing pair’s. This is the honest form of the momentum argument: a finite wing does leave a column of air moving downwards behind it, and it goes on carrying a fixed share of the weight past every plane behind the wing, for as long as the vortices last.

The settling is quick on an aeroplane’s scale. The square box’s face is within five per cent of its far value by five spans behind and within two per cent by ten. Behind an airliner of sixty metres’ span that is six hundred metres — under three seconds of flight — after which every face of that shape carries the same share for as long as the pair keeps its spacing. The tall box settles more slowly, because its face reaches further from the wake’s centre and the bound vortex’s influence there decays over distances set by the face’s height rather than by the span.

This has a practical side. Weighing what is missing measured a section’s drag from a rake drawn across its wake, and noted that a rake of finite extent misses what lies beyond its ends. A rake asked for the lift behind a finite wing would face the same problem, much worse: the downward momentum it reads depends on the rake’s own height and width through the arctangent above, so a survey plane that stops short of the upward return flow reads more than the weight and one that takes in a wide band of it reads less. Lift is measured on a balance for good reason.

The face just behind the wing #

The third figure holds the box square and changes its size. Far behind the wing, a box only 0.9 spans across sends 0.54 of the weight through the face behind it. One of twenty spans sends 0.5001. The excess for the small box is the span showing: the face is close enough to the two vortices to see them as two, not as a dipole.

With the face one span behind the wing the picture changes. The share rises with the box’s size, from 0.57 for the smallest box to 0.73 for one twenty spans across. A big box with its face only one span back has the face nearly level with the wing compared with its own height, and close to the wing the bound vortex adds its own downwash to the trailing pair’s. What sets the share is not the face’s distance in spans but its distance compared with its height. A face that is near on that measure carries the bound vortex’s contribution as well as the wake’s.

That is a third way the momentum picture can mislead. “The air leaving the back of the box” depends on where the back is, in units of how tall the box is. None of these are wrong numbers. They are the same lift, divided differently by different boxes.

What every box agrees on #

Every one of these boxes recovers the whole lift: the momentum through the face behind plus the pressure difference above and below adds up to $\rho U\Gamma b_0$, for tall boxes and wide ones, near faces and far ones. The total is the wing’s. The split is the box’s.

There is one more thing every box agrees on, and it comes from the same plane: the induced drag. The energy of the crossflow far behind the wing is fixed by the trailing vortices alone, and the earlier essay showed that rolling the sheet up does not change it. Momentum through a face depends on the face. Energy per unit length of wake does not. So the drag a wing pays for its lift is a cleaner statement about the wing than “how much air it throws down”, which needs a box to be stated at all.

The stream tube the momentum story imagines #

There is a version of the momentum argument that gets both numbers right, and it is the one textbooks of propeller theory would recognise. Imagine the wing takes a column of air of cross-section $A$ and turns it downward to a final velocity $w_\infty$. The downward momentum it gives per second is $\rho U A w_\infty$, which must be the lift. The energy it gives per second is $\tfrac12 \rho U A w_\infty^2$, which must be the induced drag times the speed. The two together fix the column’s area: $A = L^2/(2\rho U^2 D_i)$.

The fourth figure computes that area for three span loadings, using the lift and the induced drag of lifting-line theory. For an elliptic loading it is exactly the circle whose diameter is the span, $\pi b^2/4$. For a parabolic loading it is 0.889 of that circle, and for a triangular one 0.721. Those are the loadings’ span efficiencies — the ratios of their induced drag to the elliptic loading’s at the same lift and span, inverted. So the momentum picture of a wing as something that turns a tube of air downwards is correct in its numbers: the tube is the circle on the span, shrunk by the loading’s efficiency, and its final downwash is twice the downwash at the wing.

It is still a picture. No face behind the wing sees a tube of that area moving down with nothing beside it; every face sees the descending column and the rising air around it, in proportions set by the face. The tube is a summary of two integrals that every box agrees on, the impulse and the energy, written as though they belonged to one column of air. That is why it gives the right drag. The span is the whole story found induced drag set almost entirely by the span, and the tube says why in one line: the air a wing can affect is the circle its span sweeps out.

What this does to the argument about lift #

The momentum explanation of lift has two versions, and this calculation sorts them.

The weak version says a wing that is held up must give the air a downward momentum at the rate of its weight. That is true, and it is Newton’s third law applied to a control volume; it does not care which faces the momentum crosses. What holds a wing up states the same fact through the circulation, and the two are one theorem seen from two sides.

The strong version says the momentum leaves behind the wing, as a column of downward-moving air, and that this column is where the lift is. For a two-dimensional section that is false: the face behind carries almost nothing far back. For a finite wing it is partly true and box-dependent: a tall face behind the wing carries nearly all of it, a square one half, a wide one almost none. Nothing physical distinguishes those faces. The weight of an aeroplane is not “in” the air behind it. It is shared between the air behind and the pressure field above and below, in proportions the observer chooses by choosing the box, and it ends up, as a wing over the sea found, pressing on the ground or the water beneath, spread over a width about the wing’s height.

Twenty boxes that all close on the lift #

The fifth figure is the ledger. The shares were computed for twenty boxes — tall, square, wide, small and very flat, with the face a quarter span, one span, four spans and infinitely far behind — and in every one the momentum through the back and the pressure on the top and bottom add up to the lift to two parts in a thousand million million. The two terms come from different integrals: the pressure from the bound vortex’s streamwise velocity over the horizontal faces, the momentum from both vortices’ downwash over the vertical face. Their sum being exact is the box’s momentum balance holding. The closed-form integral of the pair’s downwash over a rectangle matches a direct two-dimensional quadrature to four parts in a million, the quadrature’s own error near the singular vortex cores. Boxes four hundred spans across match $(2/\pi)\arctan(H/W)$ to two parts in ten million. The model refuses a box of negative width, a face at the bound vortex itself and a pair of zero spacing.

What a straight, linear wake leaves out #

A wake carried straight back. The trailing legs run straight downstream from the wing’s tips and do not sink, roll up or bend. Far behind, a real pair descends, and a wake ends by bending found it pinched into a chain of rings a couple of minutes behind an airliner. The faces here are inside the stretch where the pair is still a pair.

A rolled-up pair from the start. The legs are placed at the rolled-up spacing $\pi b/4$, where the sheet would be by a few spans behind. A face just behind the wing would see the flat sheet instead; the far shares are the same, because the sheet and the pair have the same impulse.

Linear theory. Pressure is linear in the disturbance and the momentum flux through the top and bottom faces is dropped as second order. For an aeroplane’s lift coefficients those are the right approximations everywhere except very close to the wing.

From Lanchester’s momentum to Trefftz’s plane #

Lanchester and Prandtl put the momentum of the trailing vortices at the centre of finite-wing theory between 1907 and 1918. Munk’s analysis of 1919 and Trefftz’s of 1921 made the far-wake plane the place to compute induced drag. The observation that the momentum of an unbounded flow is conditionally convergent, and that the well-defined quantity is Kelvin’s impulse, is in the classical treatments of a vortex pair, Lamb’s and Batchelor’s, and has been rediscovered in arguments about lift ever since. The stream-tube reading of induced drag is the actuator-disc theory of propellers, turned on its side.

Still open: the box round a wing in a stratified atmosphere #

Everything here has the air above and below the wing the same density. In a stably stratified atmosphere the column moving downwards behind a wing carries lighter air into denser surroundings, and buoyancy pushes it back up. The impulse $\rho U\Gamma b_0$ per unit time is then not conserved along the wake: the atmosphere takes it back, and the downward momentum the wing gave the air is returned to the wake as the pair slows. The next calculation follows the pair’s impulse down through a stratified layer, and asks how far behind an aeroplane the air it pushed down has been pushed back, and where, in that atmosphere, the weight finally arrives.