Stable air takes back the momentum a wing pushes down

Behind a finite wing the wake face keeps its share drew a box round a wing and asked which of its faces carries the weight. Round a two-dimensional section, the face across the wake empties with distance: far behind, the whole lift crosses the faces above and below as pressure. Behind a finite wing it does not empty, because the trailing vortices run downstream for ever, and the downward momentum the wing gave the air between them crosses every face drawn across the wake. A box as tall as it is wide sends exactly half the weight through its back face, however far back that face is put.

That result needed the air to be the same density at every height. The real atmosphere is not. Under almost every weather condition but the strongest daytime convection, air further up is lighter than it would be if lifted adiabatically from below — the atmosphere is stably stratified — and a parcel pushed down is lighter than the air it arrives in. The trailing vortices push a whole column of air down. The essay ended on what the atmosphere does about that, and the answer turns the face behind the wing back into the section’s.

The column a vortex pair carries #

The trailing vortices of an elliptically loaded wing roll up a distance $b_0 = \pi b/4$ apart, the spacing where the wake ends up found from the sheet’s invariants, and each induces on the other a downward velocity $w = \Gamma/(2\pi b_0)$: the pair sinks. It does not sink alone. In the frame moving with the pair, the streamlines close round both vortices in an oval, and the fluid inside it travels with the pair, while the fluid outside parts to let the oval through. This is Lamb’s oval, and it is computed here rather than quoted, by following the pair’s streamfunction in its own frame to the closed streamline that passes through the two stagnation points on its axis.

The oval is 2.087 times the spacing wide and $\sqrt3$ times it tall, and its area is $A = 2.850\,b_0^2$, a number that moves by two parts in ten million when the computation is refined. For a wide-body airliner with a sixty-metre span, $b_0$ is 47 metres and the oval is a stream tube about a hundred metres across and eighty high, carried downwards at a couple of metres a second.

It matters that the oval, and not the two vortices, is what the atmosphere acts on. Vortices move each other treated a pair as two points translating together, and for the pair’s kinematics that is enough. For its momentum it is not: the impulse $\rho\Gamma b_0$ is the momentum of all the fluid the pair sets moving, and the part of that fluid that travels with it — the oval — is the part that changes its height, and so the part whose density stops matching its surroundings. Fluid outside the oval is pushed aside and returns to its own level once the pair has passed; fluid inside is carried bodily down and stays displaced. The stratified atmosphere sees the wake as a falling body of a definite size, and the size is a property of the pair’s own streamlines.

Lighter than where it arrives #

In air whose buoyancy frequency is $N$ — the frequency at which a displaced parcel would oscillate — a parcel carried down a distance $z$ without mixing is lighter than its new surroundings by $\rho N^2 z/g$. The oval is such a parcel. Saffman’s model of 1972 takes it to keep its original density, so after a descent $z$ it feels an upward force $\rho N^2 A z$ per unit length of wake, and that force is the rate at which the pair loses its impulse $\rho\Gamma b_0$:

$$\rho b_0\frac{\mathrm{d}\Gamma}{\mathrm{d}t} = -\rho N^2 A z,\qquad \frac{\mathrm{d}z}{\mathrm{d}t} = \frac{\Gamma}{2\pi b_0},\qquad \ddot z + \omega^2 z = 0,\quad \omega = N\sqrt{\frac{A}{2\pi b_0^2}}.$$

With the oval’s computed area, $\omega = 0.6735\,N$. The pair’s descent is a quarter of an oscillation: it sinks a distance $w_0/\omega$ and stops after $\pi/(2\omega)$, and at that moment its circulation, and so its impulse, is gone. The equations marched numerically land on the closed form to machine precision, with the circulation at the stop $10^{-14}$ of its start.

A few hundred metres down, tens of kilometres back #

For a wide-body airliner at cruise — three meganewtons, sixty metres of span, 250 metres a second in air of density 0.4 — the pair starts down at 2.15 metres a second. At a buoyancy frequency of 0.01 per second, typical of the troposphere, it sinks 319 metres and stops 233 seconds later, 58 kilometres behind the aeroplane. In the stratosphere, where $N$ is nearer 0.02 and a cruising airliner usually is, it sinks 160 metres and stops 29 kilometres back, well inside the distance a following aeroplane on the same track covers in a few minutes; in weakly stable air, $N = 0.005$, it sinks 638 metres and stops 117 kilometres back. In neutral air the line would go on climbing off the figure.

Those depths are the right order. Pilot guidance has long said that a large aeroplane’s vortices sink a few hundred feet a minute and level off some hundreds of feet below its flight path; the model says the levelling off is the atmosphere’s stability returning the momentum, and how far down it happens is the descent speed divided by a frequency the weather sets.

The face behind the wing empties #

The far face’s share of the weight is the pair’s impulse times the share the box’s shape gives it, $(2/\pi)\arctan(H/W)$, so in stratified air it falls with the impulse: $(2/\pi)\arctan(H/W)\cos(\omega X/U)$ at a distance $X$ behind. A square box behind an airliner at $N = 0.01$ sends 0.48 of the weight through its back face ten kilometres behind the wing, against the neutral atmosphere’s 0.50, and nothing at all from 58 kilometres on. The dial shows the emptying move closer as the air grows more stable.

Beyond that distance the box round the wing is the box round a section again. Every newton of the weight crosses the faces above and below, as the pressure the stratified air holds in place, and the momentum argument’s picture — air thrown downwards, carrying the weight away — is true for the first few minutes behind the aeroplane and false after them. Air must be pushed down was right that the total is fixed and wrong about which face keeps it; the stratified atmosphere moves the answer from one face to another along the wake.

How much rests on the oval’s size #

The one number the model takes from the pair’s geometry is the oval’s area, and the frequency goes as its square root. A pair that carried only half its oval — because the outer parcels had already been shed, or because a real core’s streamlines close differently from a point vortex’s — would stop after 1.41 times as long and sink 1.41 times as deep; one that carried twice as much would stop sooner and shallower by the same factor. That is the honest size of the model’s uncertainty, and it is smaller than the spread of real atmospheres: the buoyancy frequency varies by a factor of four between a weakly stable morning and the stratosphere, and the depth varies with it in inverse proportion. The order of magnitude — a few hundred metres for an airliner, a few minutes, tens of kilometres — survives any reasonable oval; the precise depth for a given flight does not, and the model should be read for the first and not the second.

The same four minutes for every aeroplane #

The surprising part is that the time does not depend on the aeroplane. The stop comes after $\pi/(2\omega) = 2.33/N$ seconds for a light single and an airliner alike: 233 seconds at $N = 0.01$. The reason is the one that makes a pendulum’s period independent of its swing. The buoyancy restoring force grows in proportion to the depth, the pair’s inertia — its impulse — is in proportion to its circulation, and the ratio of the two is fixed by the oval’s geometry, which is the same for every pair in units of its own spacing. The aeroplane sets how fast the pair starts down and how far it goes; the air sets how long the momentum is kept.

So the distance behind the aeroplane at which the wake has given its momentum back is the flight speed times 2.33/N, about fourteen kilometres behind a light single at 60 metres a second and 58 behind an airliner at 250, and the depth is the descent speed times 1.485/N. A forecaster who knows the night’s stability knows how long every wake above the airport will keep falling before knowing which aeroplanes will land; the traffic decides only how far each one falls in that time.

Neutral air is the boundary between two behaviours #

The neutral atmosphere of the earlier box is the case $N = 0$, where $\omega$ vanishes, the stop recedes to infinity and the pair keeps its impulse for ever. It is a boundary rather than a typical state. In a convective layer — air warmed from the ground on a sunny afternoon, $N^2 < 0$ — a parcel carried down is heavier than its new surroundings, and the same equation gives a pair that accelerates downwards exponentially at a rate $0.6735|N|$ instead of stopping. Low down on a hot day the wake of a landing aeroplane is pushed towards the ground faster than in still air, by the same mechanism that stops it at altitude on a stable night. The face behind the wing then carries more than its neutral share for a while, and the momentum argument’s picture is, for once, an under-statement.

Where the energy goes #

The pair’s crossflow carries a kinetic energy per unit length that the drag a wake keeps however it rolls up identified with the induced drag: the work the wing does on the air per metre flown. In neutral air that energy stays in the pair until viscosity and the bending instability dissipate it. In stratified air it is converted as the pair descends into the potential energy of the displaced oval, a heavier atmosphere lifted and a lighter column sunk; at the stop the pair’s kinetic energy of translation is gone and the potential energy is at its largest. What then radiates it away is the same internal-wave field that carries the momentum: the induced drag’s energy ends as waves in the stable air, not as heat in the wake.

Weight over span squared decides the depth #

The descent speed is $\Gamma/(2\pi b_0)$, and $\Gamma$ is the weight over $\rho U b_0$, so the depth goes as the weight over the span squared, over the air density and the speed. At $N = 0.01$ a light single’s wake sinks 58 metres, a regional jet’s 216, a wide-body’s 319 and the largest airliner’s 299: the largest aeroplane’s longer span spreads its weight over a wider pair and a slower descent. The ordering is the one the wake-turbulence categories already follow by weight, with the span as the correction that makes a very large aeroplane less extreme than its weight alone suggests.

What it means for the bending of the wake #

The trailing pair has a second clock. A wake ends by bending: each vortex bends in the other’s strain, the bend grows by a factor of e every thirty-three seconds behind an airliner, and within a couple of minutes the pair has pinched itself into rings. That is shorter than the stratified stop at $N = 0.01$, about four minutes, and comparable to it at $N = 0.02$, two minutes. In the troposphere’s usual stability, then, the wake usually breaks into rings before buoyancy has returned all its momentum, and the rings carry what is left; in the stratosphere’s stronger stability the two clocks run together, and a pair can be stopped before it links. Observed wakes are said to descend less and break up sooner in strongly stratified air, which is the second case; the model here gives the first half of the reason, and the bending gives the rest.

Where the weight finally arrives #

The momentum the pair loses does not vanish. Buoyancy is the difference between the oval’s weight and the pressure gradient of the air round it, and as the oval is pushed back up, the surrounding air is pushed down by the same force: the impulse passes from the pair to a pressure field in the stratified air, which a stable atmosphere carries away as internal gravity waves. Those waves spread the force over a wide region and, over the depth of the atmosphere, deliver it to the ground, which must in the end carry an aeroplane’s weight whatever happens in between. The stratification does not change that destination. It changes the route: from a narrow column of moving air behind the aeroplane, which a wing over the ground would feel as an image’s pressure, to a broad pressure field that reaches the ground kilometres from the track and minutes after the aeroplane has passed.

The convention: the box, the clock and the air #

Distances are along the flight path behind the aeroplane, times since it passed. The buoyancy frequency is uniform with height and the pair’s spacing is held at $b_0 = \pi b/4$ throughout. Shares are fractions of the weight for a box of stated height and width, as in the essay on the neutral case. Aeroplane figures are stated, not measured: weights, spans, cruise speeds and air densities typical of each class.

What Saffman’s model leaves out #

What happens after the stop. The model’s oscillation would carry the pair back up, which real wakes do not do; they lose their oval’s fluid by detrainment as they sink, so the impulse is returned faster and the rebound never comes. Greene’s 1986 model adds detrainment and gives descents somewhat smaller than these. Viscosity and turbulence, which decay the circulation on their own clock. Baroclinic vorticity at the oval’s edge, which is how the buoyancy force acts in the fluid and which eventually forms a layer of opposite-signed vorticity round the pair; the model replaces it with a force on the oval as a whole. A real atmosphere, whose stability varies with height and whose wind shears the wake. And the Richardson number of the wake’s own edge, which decides whether the stratification also suppresses the mixing that would otherwise erode the oval.

And the ice that makes the wake visible. A contrail is ice formed in the engines’ exhaust and drawn into the vortices. As the pair descends it is compressed and warmed, and the ice inside it can sublimate while the fluid left above the pair, which never descended, keeps its ice. That is why many contrails show a fading lower part and a persisting curtain above it. The model follows the pair’s motion and its impulse, not the ice, so the stopping depth it gives is where the vortices end, not where the visible trail does.

Who modelled it #

Scorer and Davenport observed contrails’ vortex pairs stopping in stable air in 1970. Saffman’s model of the buoyant oval is from 1972, and Crow’s study of how the pair’s descent and its bending interact in stratification from 1974; Greene’s detraining model of 1986 is the one wake-vortex forecasts long used. The calculation here takes the simplest of them, computes the oval rather than borrowing its area, and reads the result in the box the earlier essay drew: the face across the wake keeps its share of the weight only for as long as the air lets the pair keep its momentum.

Still open: a pair that loses fluid as it sinks #

The model keeps every parcel of the oval. A real oval sheds its outer fluid into a trailing wake as it descends — the lighter parcels it carries are left behind in a vertical sheet — and its area shrinks, so the buoyancy that brakes it is smaller than Saffman’s but the impulse it has to return falls too. The next calculation lets the oval detrain at a rate set by the buoyancy jump across its edge, as Greene’s model does, and asks whether the stop moves earlier or later and the depth smaller or larger, and whether the detrained sheet, which carries some of the impulse out of the pair, keeps the face behind the wing from emptying entirely.