Where the reaction to a wing's lift is

There is a question about aeroplanes that will not go away, and it is usually put as: does the ground carry the weight of an aircraft flying over it? One side says obviously yes — the momentum has to go somewhere and the earth is where it goes. The other says obviously no — the pressure disturbance under a cruising airliner is far too small to measure, and the momentum flux through any surface between the aircraft and the ground is a fraction of the weight.

Both sides compute. Both computations are correct. They are computing different limits of the same conditionally convergent integral, and the argument is about the shape of a contour rather than about a fluid.

The theorem is not in doubt #

Take any closed contour in the fluid that does not touch the body. The force on whatever is inside it is

$$\mathbf{F} = -\oint \left[p\,\mathbf{n} + \rho\,\mathbf{u}(\mathbf{u}\cdot\mathbf{n})\right] ds,$$

which is Newton’s second law written for a control volume and nothing more. This site established the method on an aerofoil and has used it since to get the force on things whose surfaces it never touches — a jet, a disc, a bend in a pipe.

Applied to a two-dimensional lifting body it gives $L = -\rho U \Gamma$, the Kutta–Joukowski theorem, and the computation here returns that on circles of radius two, ten, a hundred and a thousand chords, and on rectangles four hundred times longer than they are tall. Eight figures, every time. There is nothing wrong with the theorem and nothing surprising about the total.

The split is another matter entirely #

The two terms in that integral have separate physical names. $p\,\mathbf{n}$ is the pressure on the face; $\rho\mathbf{u}(\mathbf{u}\cdot\mathbf{n})$ is the momentum carried through it. A reader who asks what holds the aeroplane up is asking which of them is doing it.

The answer is that it depends on the box.

The reason is arithmetic rather than physics. Each term separately falls off like $1/r$ — the disturbance from a vortex is $\Gamma/2\pi r$, and the pressure perturbation is $\rho U$ times that — while the contour it is integrated over grows like $r$. So each integral separately is $O(1)\times$ the number of radians of contour that face a given way, and which part of the contour faces which way is the shape of the box.

Take the box tall and thin and almost all of the contour is vertical: the flux through the sides carries the force. Take it wide and flat and almost all of it is horizontal: the pressure on the top and bottom carries the force. Take it square, or circular, and the two share it exactly.

That last one is worth pausing on. On every circle, at every radius, the pressure carries precisely half. It is a clean result and it is not a general one: it is a property of circles.

Which is the aeroplane, and which is the wind tunnel #

The two famous positions in the argument are now identifiable as two contours.

The wide flat box is an aeroplane over the ground. The contour is the ground beneath it and a matching surface above, closed a long way off to the sides. On that box the force appears almost entirely as pressure on the horizontal faces, and the pressure on the ground under an airliner does indeed integrate to its weight — spread so thinly that no instrument sees it, but there.

The tall thin box is a wind tunnel with the model in it. The contour is the walls, close in on either side, extending far upstream and downstream. On that box the force appears almost entirely as momentum flux through the inlet and outlet, which is exactly what a tunnel’s wake survey measures.

Neither box is wrong. They are answers to different questions, and the disagreement between them is not resolvable by better measurement because the individual terms do not converge as the contour is enlarged. Only the sum does.

What can be felt from far away, and it is very little #

The reason the total survives while the parts do not is a statement about how disturbances decay, and it decides more than this argument.

So a distant contour can feel exactly two things about whatever is inside it: the circulation and the net mass flux. A doublet — which is what thickness is, in the far field — exerts no force at all, and the computation confirms it to $10^{-9}$ at every radius tried.

Everything else about a body is invisible out there. Its thickness, its camber, its incidence, the shape of its nose: all of it is a rearrangement of higher multipoles, all of which decay faster than the perimeter grows. Two bodies with the same circulation and the same displacement are indistinguishable from a long way off however different they look, which is why a body can be made out of nothing and why a wing’s far-field can be replaced by a single vortex without losing anything a distant observer could check.

That indifference is worth stating as a claim rather than an aside. Two bodies with the same circulation exert the same force on the fluid outside any contour that encloses them both, whatever they are made of and however they acquired it. It is why a lifting line can stand in for a wing, why an aerofoil’s far field is a single vortex plus a doublet, and why the whole of the wake-plane analysis can be done without any reference to the surfaces that shed it.

The converse is the useful half. Since a distant contour feels only Γ, no measurement made far enough away can distinguish a good wing from a bad one at the same lift. Everything a designer cares about — the peak suction, the recovery, whether the layer stays attached — lives in the terms that decay faster than the contour grows, and is invisible out there by construction.

The one force that is not circulation #

There is a surprise in the decay figure, and it is the source. A source disturbs the flow like $1/r$, which is the same rate as a vortex, so it must be felt at infinity — and it is:

$$F_x = -\rho U Q.$$

Mass introduced into a stream at rest has to be accelerated up to the free stream by the fluid around it, and the reaction on whatever is doing the introducing points upstream. A source in a stream is a thrust of $\rho U Q$, computed here to six figures on contours of every size, and it is the only force in this whole calculation that has nothing to do with circulation.

It is also the momentum theorem’s account of every jet engine ever built. An engine takes air in at flight speed, adds fuel and throws it out faster; strip away everything but the mass addition and what is left is a source in a stream, pushing forward. The actuator disc is the same statement dressed as a machine.

The question that started it, answered with numbers #

The essay has settled why the argument is hard and left the argument itself alone, which is unsatisfying when the arithmetic is available. It is available, and both sides turn out to be saying something true and quantitative.

Take an airliner of 250 tonnes — a weight of about $2.4\times10^6$ newtons — cruising at eleven kilometres. The pressure disturbance it lays on the ground has to integrate to that weight, and the lateral scale over which a lifting system’s disturbance spreads at a distance $h$ is of order $h$ itself, so the footprint is some hundreds of square kilometres. Divide and the peak perturbation is of order

$$\frac{W}{\pi h^2} \approx 6\times10^{-3}\ \mathrm{Pa},$$

which is six hundredths of a millionth of atmospheric pressure, and the same change a barometer sees on being raised half a millimetre. Ordinary wind produces pressure fluctuations of order a pascal at the ground, so the aircraft’s signature is a hundred times below the noise before any instrument is considered.

Both sides of the argument are exactly right. The ground does carry the weight; the integral is the weight, to the last digit, on the wide flat contour. And nothing on the ground could ever detect it, because the same weight spread over the same footprint is a peak that falls as $1/h^2$ while the integral does not move — which is this essay’s whole theme, arriving as a fact about barometers.

Two honest qualifications belong with those numbers, and both make the aircraft’s claim on the ground weaker rather than stronger.

The disturbance takes time to arrive. The pressure field of a lifting body is not established instantaneously at a distance; the information travels at the speed of sound, so the ground eleven kilometres below learns about the aircraft some thirty seconds after it is overhead, by which time the aircraft is eight kilometres further on. For steady level flight over many minutes that lag changes nothing about the integral, and for a manoeuvre it means the footprint is smeared along the track rather than sitting under the aeroplane.

And the momentum in the wake is genuinely going somewhere else. A wing’s trailing vortices carry downward momentum, they descend at a few metres a second, and they decay long before reaching the ground from cruise altitude — their momentum ending up spread through the atmosphere as heat and large-scale motion rather than delivered to the earth. So the correct statement is not that the weight is transmitted to the ground as a moving pressure spot; it is that the atmosphere as a whole is in equilibrium with the aeroplane inside it, the ground closes the system underneath, and the book balances over the whole of it rather than at any one place.

Which is a fair summary of what a conditionally convergent integral does to an argument. The total is a fact and the location is a choice, and a question phrased as where is the weight carried has smuggled a contour into itself before anybody has answered anything.

What the picture cannot show #

Three limits, and the first is the largest.

This is two-dimensional and steady, and it is the wrong dimension for the question that started it. A real wing is finite, its wake is a pair of trailing vortices rather than a bound one, and the momentum in that wake is genuinely carried downwards and away. The conditional convergence survives the change — the three-dimensional integrals have the same character — but the physical picture does not translate line by line, and this essay’s figures should not be read as a settlement of the aeroplane-and-ground question. What they settle is why the question is hard: two correct computations of the same quantity can disagree about its parts.

There is no wake here at all. A real body leaves a momentum deficit that does not decay with distance, and that deficit is the only way drag can be seen from far away — which is why surveying a wake to get a drag works in a wind tunnel and why this site refuses to do it on its own coarse grid, where the answer varies by more than its own magnitude between stations.

And the pressure gauge is arbitrary. The pressure used above is the perturbation from the free stream, which is the only choice that makes the integrals converge. Adding a constant to the pressure adds a constant times $\oint \mathbf{n}\,ds$ to the force, and that integral is exactly zero round any closed contour — which is why nothing sucks, and which is the same statement as the pressure’s absolute level being unmeasurable.

The generalisation worth carrying #

The pattern here is not about lift. It is that a quantity can be perfectly well defined while its decomposition is not, and that this happens whenever a conserved quantity is collected over a region that is being taken to infinity.

The same structure appears in the energy of a vortex pair, whose kinetic energy integral diverges logarithmically until the two circulations are made to cancel; in the added mass of a body, where the integral converges only because the disturbance is a doublet; and in the lattice sums that a wind tunnel’s images produce, where a truncation that keeps whole groups of images converges and one that does not, does not.

The practical rule that falls out of it is short. When an argument turns on how a total divides, check whether the parts converge before checking who is right.

What a control volume is actually for #

It is worth separating two uses of the same tool, because this essay’s difficulty belongs to only one of them.

A control volume drawn close to a body, with its faces in a region where everything is known or measured, is an entirely reliable instrument: what crosses each face is finite, the integrals converge briskly, and the force that comes out is the force. That is how this collection got the thrust of a jet, the loss across a sudden expansion and the force on a bend, and none of those results has any of the trouble described above.

The trouble begins when a face is taken to infinity in order to avoid having to know what is on it. That is a legitimate manoeuvre and it is what makes the Kutta–Joukowski theorem so useful — the force is obtained without any reference to the surface, its shape, its incidence or its boundary layer. What is bought with distance is ignorance of the body; what is paid for it is that the integrand is now decaying at exactly the rate that keeps the answer finite and its parts indeterminate.

So the practical rule is not avoid large contours. It is that a large contour buys a total and not a breakdown, and any statement about which part of the total is doing the work has to name the contour it was made on. A wind-tunnel engineer surveying a wake and a meteorologist integrating a pressure field over the ground are both right, they are using different contours, and the disagreement between them is not about aerodynamics at all.

Who worked it out, and when #

Kutta and Joukowski had the theorem by 1906 and each proved it on a contour of his own choosing; neither had reason to notice that the split depends on the contour, because neither was asking what holds the aeroplane up. The question in that form is more recent and belongs mostly to teaching rather than to research — it appears whenever somebody has to explain flight to somebody who is not going to accept circulation as an answer.

Prandtl and Tietjens set out the conditional convergence properly in their 1934 textbook, in a passage about how far downstream a control surface must be taken before the momentum integral means anything, and it has been rediscovered by every generation since. The surprising connection is with Lanchester, who in 1907 described the lift as being carried by “the whole of the air” and was criticised for vagueness. He was not being vague. He was declining to specify a contour.

Where the ladder goes next #

Two rungs, in different directions.

The first is the three-dimensional version, where the wake is a trace in a plane rather than a point in one, and where the same integrals decide what a wake costs. The conditional convergence there is the reason the Trefftz plane is taken at infinity rather than at any finite station.

The second is what happens when the far-field condition is applied at a finite distance because there is a wall there. That is a wind tunnel, its correction factor is exactly an eighth for a closed circular section, and the sign of it reverses when the wall is replaced by an open jet — which is the clearest demonstration in the subject that the condition at the edge is doing the work.