Field

Viscosity

The thin layer next to a surface that ideal flow ignores, and which supplies drag, separation and the wake.

Flow past a cylinder at Re 40. A real fluid past a circular cylinder. At low Reynolds number the flow closes up behind the body much as the ideal theory says; as it rises the flow separates and a region of reversed flow appears behind, which is where drag comes from.

Everything happens in a layer you cannot see

Air has so little viscosity that ignoring it works almost everywhere. Almost everywhere leaves out a film next to the surface, perhaps a millimetre thick, and that film decides drag, stall and whether an aircraft flies at all.

Flow past a cylinder at Re 100. A real fluid past a circular cylinder. At low Reynolds number the flow closes up behind the body much as the ideal theory says; as it rises the flow separates and a region of reversed flow appears behind, which is where drag comes from.

When the flow lets go

Every body asks the air behind it to slow down and climb back up to the pressure it started at. Sometimes the air cannot, and the moment it refuses is separation — the source of most drag, the cause of stall, and the reason a golf ball has dimples.

The two theories, side by side

The exact solution and the real flow, for the same body in the same stream. One is beautiful and predicts nothing has drag; the other is approximate and has a wake in it. Where they agree and where they part is the whole map of the subject.

How thick is thin

The boundary layer has no edge. It approaches the free stream and never arrives, so any thickness quoted for it is a convention — and the three conventions in use measure three different things, one of which is not a height at all.

A wing's two drags, and the lift at which they are equal. Friction drag and induced drag plotted against lift coefficient, with their sum above them. Friction is flat, because a surface costs the same whatever the wing is doing; induced drag rises as the square of the lift. The total is least where the two are equal, and that is also the point of best glide.

The two drags a wing pays

A wing pays for having a surface, and it pays for making lift with a finite span. One of those bills falls as it flies faster and the other rises, so there is a speed at which the total is least — and the condition for it turns out to be that the two are equal.

How much uphill a layer can take

A boundary layer running into rising pressure is climbing a hill on the last of its momentum. There is a definite steepness at which it can no longer do it, and the number is not a rule of thumb — it is where a family of solutions stops existing.

Pressure recovery along the upper surface at 6°. Surface speed and the local Falkner–Skan pressure-gradient parameter, plotted along the upper surface from the nose. The speed peaks near the leading edge and then falls, which is the layer climbing back up to the pressure it started at, and the parameter crosses the separation value where that climb becomes too steep.

Where the straight line stops

Ideal flow will report a lift coefficient at forty degrees of incidence without complaint. What ends the lift curve is the boundary layer refusing to follow the surface, and the estimate of when that happens joins two solvers that have nothing else in common.

The cheapest way to stay up

There is a speed at which an aircraft's drag is least, and a different, slower speed at which its power is least. The ratio between them is the fourth root of a third — a number that does not depend on the aircraft, the altitude, or anything else about the flight.

Every wavelength grows, and the shortest grows fastest. Growth rate against wavenumber for a vortex sheet of zero thickness and for a layer of thickness δ. The sheet's curve rises without limit, which is the model announcing that it has no shortest scale in it; the layer's turns over and dies at kδ = 1, the neutral wavenumber computed from Rayleigh's equation rather than fitted.

The gradient that does both

One line of the boundary-layer equations at the wall says the profile's curvature there equals the pressure gradient. That single sign causes separation and causes instability, and it causes the instability a long way before it causes the separation.

Two drag laws, one derived and one fitted. Flat-plate drag coefficient against Reynolds number, laminar and turbulent, on log axes. The laminar curve is Blasius' similarity solution, solved by shooting; the turbulent one is the 1/7-power correlation, which was fitted to experiment and is drawn dashed-in-kind to keep the difference visible. Their slopes differ — −1/2 against −1/5 — so the gap between them widens rather than staying put.

The cost of going turbulent

A turbulent boundary layer costs several times the friction of a laminar one, and the multiple is not a constant — it rises with Reynolds number, because the two laws have different exponents. That is why laminar flow is worth more on a long fast surface than on a short slow one.

The wall that shakes

Slide a wall back and forth in its own plane and the fluid above it does not follow — a wave travels upwards into the fluid and dies within one wavelength. The depth it reaches is √(2ν/ω), it contains no length from the geometry at all, and the whole thing is one of the very few exact solutions the Navier–Stokes equations have.

$A boundary layer with no x in it. The velocity profile over a porous wall with uniform suction: U(1 − e^{−Vy/ν}), exactly, at every station along the wall. The displacement thickness is ν/V, the momentum thickness is half of it, and the shape factor is two — all of them constants, none of them a function of distance. It is the cleanest demonstration there is that a boundary layer's thickness is a balance rather than an accumulation.$

The layer that stops growing

Blasius' boundary layer thickens as the square root of distance and never stops. Suck fluid through the wall at a uniform rate and it stops immediately — the profile becomes a single exponential with no x anywhere in it, and the friction comes out exactly equal to the momentum of the fluid that was taken away.

20.45 MPa out of a film 25 µm thick. The pressure along a tapered pad, from the closed-form solution of Reynolds' equation, with the same equation's tridiagonal grid solve drawn over it as points. The peak is 20.45 MPa — enough to yield mild steel — and it sits at 69 per cent of the way along rather than in the middle, because the pressure gradient vanishes where the film equals the harmonic mean of its two ends and the harmonic mean is biased towards the thinner one. The pad's own shape is drawn along the top, to a vertical scale of its own. Nothing pumps this oil: the runner drags it into a narrowing gap and the gap does the rest.

Nothing but the shape of the gap

A machine that holds a steel shaft off its bearing with a film of oil twenty-five microns thick has no pump in it, and the pressure it generates would yield mild steel. The mechanism is not the oil and not the speed; it is that the gap narrows.

45° of corner, and an infinite number of eddies in it. The creeping flow in a corner of 45 degrees, drawn from Moffatt's similarity solution. Each eddy turns the opposite way to its neighbours and is 3.17 times smaller and 1.6e+3 times weaker than the one outside it. The contour levels are rescaled inside each eddy, because they differ in strength by three orders of magnitude per step and a single set of levels would show the first and nothing else — which is itself the reason nobody has seen the third. The dividing lines between eddies are drawn where the stream function changes sign, and the dots are the centres, both found from the solved field.

The eddies nobody stirs

A slow flow past the mouth of a sharp corner does not simply fail to get in. The corner fills with an infinite sequence of counter-rotating eddies, each about three times smaller and sixteen hundred times weaker than the last, and the whole structure is decided by one complex number.

$The jet has to be fed from the sides. The velocity field of the plane jet with streamlines integrated through it. The seven central streamlines run down the jet and spread; the ten started at the top and bottom edges bend inwards and join it, which is entrainment and is a consequence of the solution rather than an addition to it. The dashed lines are the half-speed edges, widening as x^{2/3}. The transverse velocity far from the axis is 5.70e-3 m/s at this station, inward on both sides — a jet is a sink as seen from a distance, which is why two parallel jets pull together.$

What a jet keeps, and what it collects

A jet leaving a nozzle into still fluid has no boundary anywhere and one conserved quantity. Its momentum flux is exactly the same at every station downstream; its mass flux is not conserved at all and grows without limit, because a jet is a machine for acquiring fluid it did not start with.

The core spreads and the outside never notices. The swirl velocity at four times a factor of four apart, with the free vortex Γ/2πr drawn behind them. Every curve leaves the free-vortex line at its own core radius and turns over into solid-body rotation inside it; outside the core all four are the same curve, to the precision of the plot. Viscosity has rounded off the singularity and changed nothing else. The peak swirl falls from 52.4 to 6.6 m/s across the four, and the circulation is identical for all of them.

What viscosity cannot take away

Leave a vortex alone in a viscous fluid and every local measure of it falls — the peak spin, the peak velocity, the enstrophy. The circulation round a large loop does not move at all, ever, and the far field is identical to the line vortex it started as.

The current at the surface is 45° from the wind, and nothing sets that angle. The Ekman spiral drawn as a hodograph: each point is the velocity at one depth, and depth runs along the curve. At the surface the flow is at exactly 45 degrees to the wind that drives it — not approximately, exactly, and independently of the wind, the viscosity and the latitude. By one Ekman depth the flow has turned another radian and lost 1/e of its speed; by three it is a hundredth of the surface value and pointing back the way it came. The angle is a property of the equation having two terms in it, and nothing else.

The layer that stops at a depth

Every other boundary layer grows. This one does not — rotation supplies a frequency, the balance against diffusion supplies a length, and the transport that comes out contains the stress on the surface and not the viscosity underneath it.

The wall's condition, on its way to the middle. Five profiles across the half-channel, from just inside the entrance to fully developed, each drawn at the station where it occurs. The march starts from a slab of uniform flow and never assumes a shape: what arrives at the far end is a parabola, with a centre-line speed of 1.4979 times the mean against the exact 3/2 and a momentum flux of 1.1995 against 6/5. Notice what the middle does while the edges are being slowed: it speeds up, because the flow rate is held, and that acceleration is what the entrance's extra pressure drop pays for.

How far before a duct forgets what was fed into it

A pipe is always drawn with its answer already in place. Getting there takes a distance proportional to the Reynolds number, which means a more viscous fluid is done sooner — and the entrance costs a fixed number of dynamic pressures however long the pipe is.

An oscillation with no mean, and the steady flow it drives. The steady second-order velocity through a Stokes layer, in units of U U′/ω. The first-order flow averages to zero at every height; the average of its own nonlinear term does not, and the pale curve is that forcing. Integrating it twice across the layer, with no slip at the wall and no stress at the top, gives a steady velocity that rises through the layer and settles at -0.749998 — Rayleigh's −3/4, which was not put in anywhere. Beyond about five layer thicknesses nothing more happens, which is why the number is a boundary condition for the flow outside.

An oscillation with somewhere to go

Shake a fluid back and forth over a body and it develops a steady circulation that never reverses. The driving flow has no mean at all; the mean of its own nonlinear term does, and integrating that twice across the oscillatory layer gives a slip velocity of exactly three-quarters of U dU/dx over the frequency.

The price of a gradient

Viscosity does not charge for motion. It charges for the rate at which a parcel is being deformed, and a fluid in solid-body rotation at any speed whatever destroys nothing at all. What is charged for is a sum of squares, which is why the bill can be computed twice.

$How far downstream the heat is still being made. The dissipation accumulated from a station ahead of a cylinder to a station behind it, as a fraction of the whole of what is made inside the frame, at five Reynolds numbers. At Reynolds number 1 the fluid has finished paying by about a diameter behind the body. At 100 it has not finished at five, and the curve is still climbing at the edge of the picture — the drag is a force on the body, and the heat it stands for is somewhere else.$

Where the heat of a drag is made

The power it takes to tow a body through a fluid becomes heat, all of it, eventually. None of the interesting words in that sentence are the first four. It is the "eventually" that decides how a wake behaves, how far a disturbance reaches, and why no box drawn round a body contains its own bill.

The cheapest shape the walls allow

The parabola in a pipe is usually presented as what the equations give. It is better understood the other way round — of every profile that could carry that flow between those walls, it is the one that destroys the least energy, and the equations give it for that reason.

A force without the flow that makes it

A sphere carried along by a flow does not travel at the speed of the fluid at its centre. It travels at the average of the flow over its own surface — and getting that result needs no solution of the flow around the sphere at all, only the answer to a completely different problem that everybody already knows.

$A film with nothing useful to show for itself. The pressure under two discs squeezed together, as a fraction of the pressure at the centre. It is a parabola, and it holds an enormous load — but nothing is sliding, so there is no output at all and every joule put in becomes heat in the oil. The two routes to that heat share no arithmetic: one integrates the dissipation function over the film, the other multiplies the force by the approach speed.$

The last of the oil

A wet plate on a table is held down by nothing but the difficulty of getting air in underneath it. The force required to separate two flat surfaces with a film between them goes as the inverse cube of the gap, which means there is no finite energy that removes the last of the film and no finite time in which it drains away.

How hot seven films get. The peak temperature rise inside the film itself, on a logarithmic axis, for seven lubricated contacts with ordinary engineering numbers. A journal bearing runs tens of kelvin above its own housing, which is why lubrication systems are designed around heat rather than around load; a water-lubricated bush runs at a hundredth of a kelvin, because water conducts six hundred times better per unit of viscosity. The air bearing is marked: its gap is smaller than air's mean free path, so its film is not a continuum at all.

The film that heats itself

The oil in a bearing is not at the temperature of the metal around it. It is tens of kelvin hotter, the rise contains no length whatever, and whether the heating runs away or settles down depends on something that is not a property of the oil at all — whether the machine driving it holds the speed or holds the force.

Twice as fast is not twice as thick. The film a plate carries out of water, against the speed it is withdrawn at, both logarithmic. The slope is exactly two-thirds, so doubling the speed thickens the film by 58.7 per cent and never by more. The open marks are outside the range the derivation holds in — above a capillary number of about a hundredth the film is no longer thin against the capillary length, and the measured thickness leaves this line.

What a plate takes with it

Pull a plate out of a bath and it comes out wet. How wet is not set by the plate, the bath or how much liquid there is, but by a competition in a region a fraction of a millimetre long that nobody looking at the plate can see — and the film goes as the two-thirds power of the speed, never as the first.

Seven fluids in one pipe at one pressure gradient. Velocity profiles of power-law fluids in a round pipe, all at the same pressure gradient and the same consistency, scaled to the fastest. A shear-thinning fluid (n below one) is flatter in the middle and steeper at the wall; a shear-thickening one is the reverse. The flattening is often called plug-like, which invites the reading that the fluid is moving more freely — what has actually happened is that all the shear has been pushed into a thin annulus at the wall, which is the expensive place to put it.

A viscosity that depends on the question

Blood, paint, molten polymer and drilling mud have no viscosity. They have a relation between stress and strain rate that is not proportional, so the ratio of the two depends on how hard they are being sheared — and an instrument that reports one number is reporting a property of itself.

$Where Einstein's line stops being the measurement. The viscosity of a suspension of rigid spheres relative to the liquid's, against the volume fraction. Einstein's 1 + 5φ/2 is exact for one sphere and holds while the spheres cannot feel one another, which is up to about five per cent by volume. Batchelor and Green's two-sphere term takes it a little further; beyond about a fifth nothing derived works and the curve drawn is a fit, which diverges at a maximum packing that is itself a measurement.$

A viscosity made of particles

Stir rigid spheres into a liquid and the mixture is thicker, by five halves of the volume fraction. Einstein's coefficient is not an empirical constant — it is a dissipation calculation on one sphere — and doing it as an energy rather than as a stress shows that four-fifths of it comes from somewhere nobody mentions.

$Two-thirds to one, and nothing outside it. The drag on a spherical drop as a fraction of Stokes' drag on a rigid sphere of the same size, against the ratio of the viscosities. It runs from exactly two-thirds for a clean gas bubble to exactly one for a rigid particle and takes no value outside that range, however light or heavy the drop is — the formula has no density in it anywhere. A drop of water in air and a drop of mercury in water sit at opposite ends of it.$

The surface that moves with the flow

A clean gas bubble feels two-thirds of the drag a rigid sphere of the same size would, and the formula has no density in it anywhere. What buys the third is that the bubble's surface is free to move — and real bubbles in ordinary water do not get it, for a reason that is a millionth of a per cent of the water by mass.

A fifth of the drag power is not heat yet. The energy account of towing a flat plate a metre long through air at thirty metres a second. The power it takes is the drag times the speed. The heat made inside the boundary layer is half the free-stream energy times the energy thickness, and it is less — 78.6 per cent of what was paid. The rest has not been destroyed: it is kinetic energy still in the wake, which will become heat somewhere downstream, in fluid that is no longer touching the plate.

The third thickness

A boundary layer has no edge, so every thickness quoted for it is an integral of the profile against some weight. Two of them are famous. The third answers a question the other two cannot — how much of the power spent towing a plate has actually become heat by the time the fluid leaves it — and the answer is 78.6 per cent.

A vortex has no energy. The kinetic energy of a Lamb–Oseen vortex inside a circle, per metre of its length, against the logarithm of that circle's radius. It is a straight line and it does not stop: outside the core the swirl is Γ/2πr, the energy density falls as 1/r², and the area grows as r², so every decade of radius adds the same amount. There is no such thing as the energy of a line vortex without a stated cutoff, and no cutoff is physical.

The energy a vortex cannot have

A line vortex has infinite kinetic energy. Not a large amount — infinite, growing without limit as the logarithm of however far out the counting stops. And it is losing that energy at a rate that is finite, exactly known, and contains no cutoff at all.

$A swimmer that dissipates the same everywhere. Taylor's waving sheet, with the wave drawn along the bottom and the dissipation drawn against height above it. The dissipation function works out at 4μb²c²k⁶y²e^{−2ky} — with no x in it at all, so the sheet is destroying energy at the same rate under every part of the wave and at every instant of the cycle. It peaks one radian of wavelength above the sheet and is gone within about three.$

A swimmer that cannot go backwards

Taylor's waving sheet is the simplest self-propelled object in a viscous fluid, and its arithmetic contains a result that reads like a mistake — the work it does to travel a metre does not depend on how big its waves are. Doubling the amplitude quadruples both the speed and the power, and changes the bill for the journey by nothing at all.

The coefficient Stokes threw away is not a correction. What absorbs sound in dry air, split into the three transport effects that add to make the diffusivity of sound. Shear viscosity is the largest single contribution and it is not most of it; the bulk viscosity — which Stokes set to zero in 1845, saying in the same paper that he had no argument for doing so — is a quarter, and thermal conduction is a fifth. Setting ζ to zero does not make a small error in the absorption; it makes a 24 per cent one.

The viscosity nobody uses

A fluid has two viscosities. One resists a change of shape and appears in every viscous calculation; the other resists a change of volume, was set to zero by Stokes in 1845 in a paper that says he had no argument for doing so, and for carbon dioxide is fifteen hundred times larger than the one everybody quotes.

Stokes' drag stops being the answer almost at once. The damping on a millimetre sphere oscillating in air, divided by Stokes' steady drag on the same sphere, against frequency. The extra term is the sphere's radius over the layer thickness, so it takes over as soon as the layer is thinner than the body — which for a millimetre sphere in air is below one hertz. At a kilohertz the damping is fifteen times the steady value, and the exponent is a half rather than zero.

What a fluid takes out of a swing

The damping a body feels from the air around it is not Stokes' drag, and stops being it far earlier than anybody expects — a millimetre sphere in air is already forty-six per cent above the steady answer at one hertz. Past that the damping rises as the square root of the frequency, and the fluid it is fighting is a shell a fraction of its own size.

The only candidate the far field allows, and the wall it slips past. The general Stokes solution has four constants; the condition at infinity kills two of them and fixes a third, leaving one to satisfy two conditions at the wall. Setting the stream function to zero there uses it up, and the tangential velocity that remains is exactly twice the free stream — for every radius, every speed, and every fluid.

The flow with no solution

Creeping flow past a sphere has a solution and everybody knows it. Creeping flow past a cylinder has none — not a difficult one, not one needing a clever method. The equations, the no-slip condition and the uniform stream at infinity are inconsistent, and the residual is exactly twice the free stream.

The singularity a layer makes for itself

March Prandtl's equations into an adverse pressure gradient and the wall shear reaches zero with an infinite slope at a finite station, and the solution cannot be continued past it. The singularity is real, it is not a numerical difficulty, and it belongs to the boundary condition rather than to the equations.

The core that does not move

Toothpaste in a tube has a region in the middle that is not being sheared at all, and its edge is at a radius you can write down before solving anything. Four things about that core are exact, and one of them is that the fluid does not flow below a threshold — not slowly, at all.

A pure number with no fluid in it. The load a plane slider bearing carries, divided by everything dimensional in it. What is left is a function of the convergence ratio alone — no viscosity, no speed, no size, no material — so the tilt that carries the most load is decided before anything about the machine is known. It is the root of one transcendental equation, at 2.1887048.

The gap that carries the most

A bearing's most consequential dimension is decided by a number with no oil in it, no speed and no size — the root of one transcendental equation, at 2.1887048. And the maximum it sits on is flat enough that the value printed in every handbook is not it.

One shape, at every station and for every jet. The axial velocity of Schlichting's round jet at four distances, each scaled on its own centreline speed and its own half-width. The four curves are one curve: the shape function (1 + eta²/4)⁻² has no parameter in it at all, and every station of every jet — laminar or modelled, weak or strong — collapses onto it exactly.

Exactly similar, and one number short

The round jet has an exact solution of the Navier–Stokes equations, and a turbulent jet is modelled by the same formula with the viscosity replaced by a fitted constant. The shape is identical. Nothing a measurement of the profile can do will tell the two apart.

Four profiles, one drag

The momentum integral is exact and asks nothing about the shape of the velocity profile. Four guesses at that shape span twenty-three per cent in the drag they give — and the two that satisfy the conditions the true profile satisfies are within three, which says the freedom belongs to the family rather than to the constraint.

The stress a pipe knows

A capillary viscometer measures a pressure drop and a flow rate and reports a viscosity. The first half of that inference is a force balance and is exact for any fluid there is; the second needs the slope of a whole flow curve, which is the experiment the instrument was bought to avoid.

A particle released in still fluid, with the history and without. The velocity of a hundred-micron particle released at one metre a second in air, computed with the unsteady Stokes force and with the quasi-steady drag alone. After five particle time constants the one that remembers is still moving nearly four times as fast.

The drag that integrates a whole history

A particle released in still air does not stop exponentially. The unsteady drag on it carries a term that is an integral over everything the particle has already done, weighted by the inverse square root of how long ago — so there is no time constant, and five particle time constants later it is still moving four times faster than the quasi-steady answer allows.

What the fluid at one height is listening to. The weight the fluid two millimetres above a moving wall gives to the wall's velocity a given delay earlier, in water. It peaks at two thirds of a second and has a tail that falls as the delay to the power minus three halves — so the fluid is responding to a broad stretch of the wall's past rather than to a moment of it.

The wall the fluid is listening to

Water two millimetres above a moving wall is responding to what the wall did two thirds of a second ago — most likely. Half of its response is older than four and a half seconds, a tenth is older than two minutes, and the average age of what it is responding to does not exist at all.

The flow stops and the stress does not. Shear stress against time for a fluid sheared at a constant rate for two seconds and then left alone. Nothing is moving after the second mark and the fluid is still stressed: 37 per cent of the peak one relaxation time later, and 0.7 per cent after five.

The fluid that has not finished its last deformation

Shear a polymer solution for two seconds and stop. Nothing is moving and the fluid is still stressed — 37 per cent of the peak one relaxation time later, and still measurably stressed after five. Two histories imposing exactly the same total strain leave it in states differing by a factor of 3.7.

Two external flows that agree where it matters and nowhere else. The velocity just outside the boundary layer, for two pressure distributions, against distance along the surface. They cross at the half-way station with the same value and the same gradient, and they have nothing else in common: one accelerates steadily and the other does most of its accelerating at once.

A layer that is an integral of everything upstream

Two surfaces are given external velocity distributions that agree exactly at one station — the same speed and the same gradient. The boundary layers there differ by 38 per cent in momentum thickness, and the two surfaces separate five per cent of their length apart.

$How long a fluid takes to forget it was not rotating. The fraction of solid-body rotation a container's interior has reached, against time, by the two available routes. The Ekman layers on the end walls pump fluid radially and carry angular momentum inwards in a hundred seconds; diffusion alone would need ten thousand.$

How long a fluid takes to forget it was not rotating

Spin a container of water and the fluid inside reaches solid-body rotation in a hundred seconds rather than the three hours diffusion would need. The shortcut is the thin layers on the end walls, and the advantage they give is exactly the reciprocal of the square root of the Ekman number.

A duct that forgets everything but one number

Whatever is fed into a pipe, what survives a little way down it is one shape. The disturbance's higher modes decay as the square of their mode number, so the sixth is gone in thirteen centimetres where the first survives four and a half metres — and the entrance length is that one mode's decay rate and an arbitrary threshold.

Three profiles that do not depend on the radius. The radial, azimuthal and axial velocities of the flow above a rotating disc, as functions of one similarity variable. The radial one is a jet: fluid thrown outward by the swirl it has picked up, peaking at 0.181 of the local disc speed a fifth of the way through the layer. The azimuthal one falls from the disc's own speed to nothing. And the axial one is the surprise — it does not vanish far from the disc but tends to a constant, so the disc draws fluid down onto itself at 0.8845 times the square root of the viscosity times the rotation rate, at every radius and for ever.

The solution that keeps its nonlinear term

Every exact solution before this one has been exact because the nonlinear term vanished. A rotating disc's does not vanish — at the wall it is the whole of the balance — and the reduction is exact anyway, because the radius divides out of all three momentum equations at once.

One channel, one flux, two flows

Flow between two plane walls meeting at a line has an exact solution. Past a threshold that turns out to be a ratio of gamma functions, it has two — the same wedge carrying the same flux, once outward everywhere and once with the fluid running backwards along both walls, and nothing in the equations chooses.

Six that are symmetries and five that look like them. Each transformation applied to an exact solution, with the Navier-Stokes residual recomputed from the transformed field by finite differences — nothing differentiated by hand. The six symmetries leave the residual at the differencing floor, a few parts in 10^8. The five near-misses leave between 0.048 and 4.3, which is six to nine orders of magnitude larger. The gap is what makes this a test rather than an illustration: a transformation that is nearly a symmetry does not exist here, and every one of the five is something a reader might reasonably believe.

Why the list is this long

Every textbook list of exact solutions of the Navier–Stokes equations is about a dozen long, and the usual explanation is that the equations are hard. It is not the reason. A similarity reduction is a solution invariant under a subgroup of the equations' own symmetries, so the catalogue of possible reductions is the catalogue of subgroups — and that is a finite, countable object.

Where a film stops being a damper and starts being a spring. The stiffness and the damping of a circular gas film against its squeeze number, both in units of ambient pressure times disc area. At the left the stiffness is nothing and the damping is Stefan's incompressible law; at the right the damping has gone and the stiffness has reached π, which is an isothermal gas trapped in the gap with no way out. They cross at σ = 6.23, and neither limit was put in by hand — both fall out of a Bessel function of a complex argument.

A damper that turns into a spring

A film of oil squeezed between two plates resists motion and stores nothing, and that is a property of the oil rather than of the film. Fill the same gap with air and the same equation gives a film that stores and resists nothing — above a squeeze number of six, with nothing changed but the frequency and the gap.

$Below a Stokes number of ten a ball does not come back. The restitution of an impact through a liquid film, as a fraction of the same impact's dry restitution, against Stokes number. Nothing rebounds below the threshold and the recovery above it is a hyperbola: the restitution, as a fraction of its dry value, is one minus the critical Stokes number over the Stokes number, with that critical value 10.18 computed from the film and the dry restitution alone. The measured curve, drawn beside it, is the same expression with ten in place of that number — and ten is what four decades of viscosity and four materials all give.$

A ball that bounces in water and not in oil

A squeeze film cannot be closed with any finite energy, so nothing should ever touch anything. A sphere dropped into a tank nevertheless rebounds, and whether it does is decided by a number near ten that four materials and four decades of viscosity all agree on.

Three strokes, and only the flat one is a theorem. Three cycles drawn in the swimmer's shape space: a square, a circle of the same width, and an out-and-back along the diagonal. The first two enclose area and carry the swimmer forward; the third encloses none and carries it exactly nowhere, which is the scallop theorem with no symmetry argument in it. What the third lacks is not a broken symmetry but an interior.

A stroke is worth the area it encloses

The usual account of swimming without inertia is a symmetry argument about reciprocal strokes, which says what cannot work and nothing about what does. Draw the stroke in the space of the swimmer's own shapes and the displacement is a line integral — so it is an area, it does not depend on how fast the stroke is played, and the scallop theorem is Stokes' theorem.

The only term that turns spin into thrust, and what it is made of. The coupling term of the propulsion matrix against the drag anisotropy, with the geometry held fixed. It is exactly proportional to the difference of the two drag coefficients, so it is zero when they are equal — not small, zero — and no helix of any pitch turned at any rate would move. A real filament sits at 1.66, which is a third of the way from useless to the unreachable limit.

Two drags, or nothing swims

A bacterium turns a corkscrew and goes forward, and the reason is not the corkscrew. It is that a thin filament dragged broadside resists more than the same filament dragged end-on. Make the two resistances equal and the thrust is not small but exactly zero, for every pitch and every rate.

A flow pinned between two guesses

The minimum-dissipation principle says the true flow is the cheapest one the walls allow, so any guessed velocity carries too little. It has a twin that nobody teaches: any guessed stress in balance with the pressure carries too much, and it needs no wall condition at all. Between the two, the flow through a duct with no formula is pinned down to as many figures as anyone wants.

The orbits a rod's axis can be on, seen along the vorticity. The tip of the unit vector along a rod of aspect ratio five, tumbling in a simple shear, seen from along the vorticity axis, for five values of the orbit constant. Every orbit is closed and every one takes the same time. A rod near the centre is spinning about the vorticity axis; a rod on the outer circle tumbles end over end in the plane of shear. The flow never moves a rod from one orbit to another.

A viscosity the flow cannot decide

Spheres stirred into a liquid thicken it by a definite amount. Rods do not. A rod in a shear flow tumbles round a closed orbit, the flow never moves it to another, and the extra viscosity depends on which orbit it is on. The equations of slow flow permit a whole range of values and choose none of them. The smallest amount of noise chooses one, and it does not matter how small.

A film just past its fold lingers where the fold was. The temperature excess at the middle of a sheared film, scaled by the temperature over which the viscosity falls by e, against time in thermal diffusion times, switched on from rest at five stresses either side of the fold. Below the fold the film settles. Just above it the film climbs to the fold's own centre temperature, 1.19, lingers there as if it had found a steady state, and only then runs away — the longer the closer it is to the fold.

A lost steady state still holds the film

Push a self-heating oil film one per cent past the stress at which it can no longer settle and it does not run away at once. It warms to the temperature the vanished steady state would have had, sits there as if nothing were wrong, and only then goes — after a delay that grows as one over the square root of the overload, and during nearly all of which the load could still be taken back.

Where a motor's line crosses the film's own characteristic. The stress a self-heating film carries against the speed it lets one wall slide past the other, both scaled, for every steady state (solid): it rises while the film is cool, peaks at the fold and falls as the film heats and thins. A motor's torque falls along a straight line as its speed rises (dashed), and the film runs where the two cross. A stiff drive crosses once. A soft drive, shallower than the falling limb's steepest slope, 0.0738, can cross three times. A drive held at a fixed stress is a flat line, and at the fold's stress it only touches.

A motor turns the runaway into a jump

A self-heating oil film has a fold at fixed stress and none at fixed speed, and a real motor is neither: its torque falls along a line as its speed rises. Put that line across the film's own torque–speed curve and the runaway disappears for any motor at all. What replaces it depends on the line's slope. A drive softer than one fourteenth of the cold oil's resistance jumps to a hot state and keeps a memory of the load; a stiffer one does neither. And the boundary between the two falls exactly where the motor is turning at a quarter of its no-load speed, which no motor near its rated speed can reach.

A film pulled apart asks for more tension than a liquid has. The pressure below ambient in the film under a sphere moving away from a wall, scaled by the tension the liquid can bear, against the distance from the axis in sphere radii, at the contact gap. The lubrication solution (dashed) asks for four times that tension on the axis. A liquid that cannot give it cavitates: a disc of vapour opens where the demand exceeds the floor, and outside it the pressure is exactly the solution it would have had. Here the disc reaches 0.12 sphere radii.

A torn film still pulls

A sphere bouncing off a wall under liquid has to climb back out through the film it squeezed, and the film pulls it back with a suction no real liquid can supply. Let the liquid cavitate and the obvious guess is that the sphere escapes the torn part of the film for free. It does not. The liquid round the vapour disc goes on pulling, and the disc itself holds the full tension over its area, so a film that tears at five times the contact gap saves a third of what the guess says — and the rebound threshold moves by five per cent where the guess said a third. At an atmosphere, in the liquids the threshold was measured in, it barely moves at all.

Every rotor lands on the same hot state, later and more smoothly. The centre temperature against time, on a logarithmic axis, for a soft drive five per cent past its fold switched on from rest, with rotors of four inertias. Each climbs, lingers and jumps, and each ends at the hot operating point without overshooting it. The time to pass three e-folds is 7, 16, 101, 948 diffusion times for M = 0, 0.1, 1, 10.

A rotor's inertia slows the jump and cannot make it ring

Give the motor driving a self-heating oil film a rotor that has to spin up, and there are two clocks: the film's diffusion time and the rotor's. Two clocks are what an oscillator is usually made of, and this pair cannot make one. Every eigenvalue stays real at every inertia, because the film and the rotor only ever push each other the same way. What inertia does instead is add its own delay to the film's, and in a real machine, where the rotor is hundreds of times slower, the delay past the fold is almost entirely the rotor's.

Below the capillary length, the coat is set by the fibre, not the bath. The coat's thickness in units of ℓ_c Ca^(2/3) against the fibre's radius in capillary lengths, on logarithmic axes: the dynamic meniscus's 1.3376 over the curvature the static meniscus asks of it, 1/b + Z/ℓ_c². Thin fibres take Quéré's 1.3376 b Ca^(2/3), set by their own radius; thick ones take Landau–Levich's 0.9458 ℓ_c Ca^(2/3). The two laws cross at b = 0.71 ℓ_c, where the coat is 0.61 of either; at b = 0.1 ℓ_c it is 0.137 of what the plate's law would give.

A thin fibre coats by its own radius, and beads by it

A plate drawn out of a bath carries a film set by the capillary length, the size at which surface tension and gravity balance. A fibre thinner than that length carries a film set by its own radius instead, often a tenth of what the plate's law promises — and the same radius then decides how quickly that film gathers into beads. The faster the fibre is drawn, the thicker its coat and the shorter the length of it that comes out smooth.

An S-shaped set of states and a slow variable that crosses it. The film's steady mean temperature against the housing's, on the cool and hot branches of operating points, with the line on which the housing is in equilibrium with the film, at a derating of 0.22 per unit. Between housing temperatures 1.24 and 1.59 both branches exist. At the derating the essay uses, 0.22, neither meets the equilibrium line there: on the hot branch the housing is always warming, on the cool branch always cooling, so the film runs round the loop drawn over them, jumping at each fold.

A slow sensor makes the bearing cycle

Derate the motor driving a self-heating oil film on the temperature of its housing, which warms over minutes, and a protection meant to hold the bearing cool instead makes it cycle for ever: jump hot, warm the housing until the hot state is lost, drop cool, let the housing cool until the cool state is lost, jump again. The period belongs to the housing, the amplitude belongs to the film, and the derating strength decides only whether the cycle happens at all.

Five gas films never meet their thermal crossing. Six air films placed by their squeeze number, which decides whether the gas leaves the gap or is trapped in it, and their thermal number, which decides whether the heat of compression leaves. The dashed diagonal is the estimate that the two differ by the Prandtl number. The devices lie between one and eleven decades below it, because heat leaves across the gap and the gas along the disc. Only the levitator at a twenty-micron gap comes within a decade of the thermal crossing at ωh²/κ ≈ 18.

The heat of a squeeze leaves across the gap

Squeeze a film of air and it warms, and whether that heat reaches the walls in time decides whether the gas is compressed isothermally or adiabatically. The natural estimate has the heat leaving the way the gas does, along the disc, and puts the thermal crossing beside the viscous one. It leaves across the gap instead, a distance hundreds of times shorter, so the crossing sits decades away — and when a film does reach it, what it gains is not stiffness but a second band of loss.

On Mars the bulk viscosity is two speeds of sound. The speed of sound in carbon dioxide at 610 pascals and 240 kelvin against frequency, with the bulk viscosity scaled from its value at one atmosphere by pressure alone. Low notes travel at 245 m/s, with the bending vibration keeping up; high ones at 252 m/s, with it frozen. The step is centred near 342 Hz. The Perseverance rover's microphones reported the same thing in 2022: two speeds of sound, near 240 and 250 metres a second, either side of a few hundred hertz.

The largest bulk viscosity is the first to expire

A bulk viscosity is not a separate property of a gas. It is the time a molecule's internal motion takes to catch up with a compression, multiplied by the pressure and by how much heat capacity lags, and read from below that time's frequency. So the coefficient that is largest is the one that stops being a coefficient soonest: carbon dioxide's fifteen-hundred-fold value is ten per cent wrong at 24 kHz, and on Mars it is two speeds of sound in the audible band.

$Smooth below one Mach number, a jump and a tail above it. The velocity change through steady shocks in carbon dioxide, as a fraction of the whole change, against distance in relaxation lengths — the equilibrium sound speed times the relaxation time, 0.737 mm at one atmosphere. Shocks slower than the frozen sound speed, M below 1.041 referred to the equilibrium speed, are smooth throughout: the whole rise is the vibration catching up. Faster ones jump first, at the frozen speed's own shock, and relax afterwards: at M = 1.3 the jump takes 83.9 per cent of the change at once.$

A bulk viscosity holds a shock together until it splits

Carbon dioxide's bulk viscosity, fifteen hundred times its shear viscosity, is a vibrational relaxation seen from below its frequency, and a shock is where that description is tested hardest. Carried through a steady shock, the relaxation reproduces the coefficient exactly for the weakest shocks. At a pressure rise of nine and a half per cent the shock outruns the frozen sound speed and splits into a jump and a tail, and the coefficient then draws a shock that does not exist.

A warm bush stops the bearing hunting

A bearing whose motor is derated on its housing's temperature can hunt, jumping between a cool film and a hot one on the housing's clock. Let the bush warm with the housing, and a second feedback joins the first, of the kind that usually makes things run away. It does the opposite: a warm wall is a stiffer drive, the stiffening closes the S the cycle runs round, and past a bush coupling of about a third the bearing settles, warm and derated, however the protection is set.

Helium wastes least for most stiffnesses; xenon reaches furthest. The film's loss tangent — damping over stiffness — against its stiffness, each gas traced from a tenth of an atmosphere up to the pressure of its greatest stiffness. Raising the pressure buys stiffness and costs loss for every gas. Up to about a hundred newtons per unit relative amplitude helium's film loses least at any stiffness — 0.29 at 50 N against air's 0.4 and xenon's 0.46. Beyond, helium is near its peak of 127 N and the heavy monatomic gases take over: xenon reaches 149 N, and at 120 N loses 0.73 against helium's 0.78.

The gas sets a squeeze film's loss, not its stiffness

A squeeze-film levitator wastes part of every cycle as heat that crosses the gap, and the gas and its pressure decide how much. Fill one levitator with six gases at one atmosphere and its stiffness hardly changes while its loss nearly doubles from helium to xenon. Raise the pressure and every gas stiffens towards a peak near ten atmospheres. Helium wastes least at any stiffness up to a hundred newtons; above that, only the heavy monatomic gases get there.

A thousandth of a monolayer holds a bubble still

A clean bubble rising slowly through water feels two-thirds of a rigid sphere's drag, and real bubbles almost never do, because surfactant swept to the rear holds the surface still over a cap there. Solving the flow with the cap in it shows how little that takes. The drag runs ahead of the area covered — half-way to rigid with a third of the surface held — and the surfactant needed is set by the viscous stress, not by the surface tension. For a bubble a tenth of a millimetre across, a thousandth of a monolayer, spread as a cap, makes it rise within a few per cent of a solid ball.

A weak push stirs the whole fluid; a strong one makes a jet. Streamlines of the exact solution in a plane through the force, for a weak force (jet Reynolds number 1, left) and a strong one (100, right), the force pointing right from the origin. The dashed line is the cone inside which the fluid moves outwards: 89.4° from the axis for the weak force, nearly a hemisphere, and 22.8° for the strong one. Outside it the fluid is drawn back towards the origin from every direction and turned into the jet.

A point force makes a jet only when it is strong

Push on a fluid at one point and there is an exact solution of the full Navier–Stokes equations for what follows, at any strength. Weak, it is Stokes's point force: fluid pushed forward over a whole hemisphere and drawn in behind. Strong, it is Schlichting's slender jet, with fluid drawn in from every direction outside a narrowing cone. One constant joins them, and it says how strong a push must be before the boundary-layer jet everyone uses is right — at the axis by a jet Reynolds number of a hundred, at the edges only by a thousand.

Below 9.8 µm in silicone oil, a gradient of a degree per millimetre outruns gravity. The speed of a clean air bubble in 10 cSt silicone oil against its radius, on logarithmic axes: migrating in a gradient of 1 K/mm, which grows in proportion to the radius, and rising under its weight, which grows with its square. They cross at 9.81 µm and 31 µm/s. The crossing radius, 3|dσ/dT|G/(2ρg), has no viscosity in it: both speeds are set against the same viscous drag.

The warm side pulls a bubble with no force on it

A bubble in a liquid whose temperature varies from place to place moves towards the warm side, with no gravity and nothing pushing it. The surface tension is lower where the liquid is warmer, the surface is pulled towards the cold pole, and the bubble goes the other way. Young, Goldstein and Block's speed comes out of four interface conditions and one more: that the total force on the bubble is zero. That one condition removes the point force a sinking or rising body always carries, so the bubble's disturbance dies a hundred times faster with distance, and the radius at which it balances its own buoyancy has no viscosity in it.

$The same cap costs a rising bubble a quarter of its speed and a migrating one four-fifths. Speed as a fraction of the clean bubble's against the half-angle of a stagnant cap over the rear: for a bubble migrating in a temperature gradient, and for the same bubble rising under its weight. A 90° cap leaves the rising bubble 74 per cent of its speed and the migrating one 21 per cent; a 120° cap, 68 and 4. The rising bubble can only fall to a rigid sphere's two-thirds; the migrating one falls to nothing.$

A cap that slows a rising bubble stops a migrating one

Surfactant swept to the back of a bubble holds the surface still there, and for a bubble rising under its weight the worst this can do is turn it into a solid sphere and cost it a third of its speed. A bubble migrating in a temperature gradient has no such floor. Its own surface is the engine, and a cap removes the engine as well as raising the drag. A cap over the cold hemisphere leaves a rising bubble three-quarters of its speed and a migrating one a fifth; a cap of 120° leaves the migrating bubble four per cent. The surfactant it takes is set by the tension difference that drives the migration, and that is tiny.

A slow bush leaves the bearing hunting, a sixth faster. The film's mean temperature excess over two cycles of a bearing hunting under a derating of 0.2 with a fifth of the housing's excess reaching the bush, for a bush that follows the housing instantly and one ten housing times slow. Both jump between a cool film and a hot one; the instant bush's period is 0.347 housing times, the slow bush's 0.291. The slow bush holds the walls at the cycle's mean, and the film's jumps are the same size.

A slow bush quickens the hunt it cannot start

A bearing's bush warms with its housing and makes the drive stiffer, and past a coupling of about a third it stops a derated bearing hunting. Give the bush a time of its own and the obvious fear is that the lag reopens the hunt. It cannot: whether a bearing hunts is decided by where its equilibrium sits, which no lag can move, and a lag on a positive link cannot make a settled state oscillate. What a slow bush does change is the hunt itself, which runs a sixth faster.

A free jet keeps its momentum; a wall jet keeps the flux of its momentum flux. Each quantity against distance, relative to its value at 2 cm, on logarithmic axes. The free jet's momentum flux is constant and its volume flux grows as x^(1/3). The wall jet's momentum flux falls as x^(−1/4) — the wall's shear takes it — while its volume flux grows as x^(1/4), and the one thing it conserves is Glauert's ∫u(∫u²)dy, the flux, through each height, of the momentum flux carried above it.

A jet along a wall keeps the flux of its momentum flux

A jet in open fluid keeps its momentum and gathers fluid as it goes. Blow the same jet along a wall and the wall takes its momentum away as drag, so it keeps neither. What it keeps is stranger: the flux, through each height, of the momentum flux carried above that height. That one invariant fixes everything else — the jet thickens as the three-quarter power of distance, slows as the inverse square root, and gathers fluid as the fourth root, more slowly than the free jet's cube root.

Hold the nozzle further from the plate and the jet arrives slower but leaves a stronger wall jet. For a laminar jet from a 2 mm slot at 5 cm/s in water, against the nozzle's distance from the plate, each over its value at 10 cm: the arriving centreline speed, which falls as h^−⅓; the inherited invariant F = 3KQ/32, which grows as h^⅓ because the jet has gathered fluid; and the wall jet's peak speed a metre along the plate, which grows as h^⅙. At 2 cm the arriving speed is 1.71 of its value at 10 cm and the far wall jet 0.765; at 2 m, 0.368 and 1.65.

A jet striking a plate leaves a stronger wall jet the further it travelled

A plane jet that strikes a wall splits into two wall jets, and each keeps not the jet's momentum but Glauert's flux of momentum flux. Turned without loss at the stagnation point, each half arrives laid along the wall with its axis on the wall, and its invariant is the momentum it brought times the fluid it had gathered on the way: F = 3KQ/32. Hold the nozzle further away and the jet arrives slower but leaves a stronger wall jet, by the cube root of the distance — and the wall jet takes a distance of about the Reynolds number in jet widths to forget it was ever free.

The paint pushes a scraper off the wall, and roughly as one over the square of the angle it is held at. The force the fluid exerts on a still blade at angle α to a wall sliding into the corner, per unit span over μU ln(R/ε): its component pushing the blade away from the wall and the size of its component dragging it along with the wall, towards the apex. At 10° they are 196 and 22.9; at 45°, 8.56 and 4.88; at 90°, 1.36 and 2.14; at 150°, 0.0757 and 0.924.

A scraper held flat is pushed off the wall it scrapes

A corner stirred from far away fills with an endless sequence of eddies. Drive the corner at the corner instead — let one of its walls slide under a still blade, as paint slides under a scraper — and the eddies vanish: every streamline turns once, and the stress rises into the corner without limit. G. I. Taylor solved it in 1962. The paint pushes the blade away from the wall with a force that grows as one over the square of the blade's angle, so a blade held flat rides up on what it was meant to remove, and the force is finite only because something too small to appear in the problem ends the corner.

A cut deeper than the film's window flips the film twice a cycle; a shallower one leaves it where it is. A protection that trips at a housing excess of 3.0 and resets at 2.6, against the depth of its cut: left, the film's jumps between its branches per protection cycle; right, the cycle's period in housing times. The film has both branches between 0.6496 and 0.7279 of full drive, so a cut past 0.35 leaves it only the cool one. cut 0.3: never resets; cut 0.34: 0 flips, period 1.85; cut 0.36: 2 flips, period 0.767; cut 0.6: 2 flips, period 0.756.

A trip deeper than the film's window flips the bearing twice a cycle

A smooth derating acts on a bearing's housing temperature as it stands. Most real protections trip instead: they cut the motor at one temperature and restore it at a lower one, a hysteresis of their own laid on top of the film's. Whether the oil then jumps between its cool and hot states is decided by one comparison — the cut's depth against the width of the film's window — and past it the protection's own band sets the rhythm. A trip never lowers the film's peak at all; it only protects the housing, and a deep one buys that with a plunge to the cool branch twice in every cycle.

At a levitator's own thermal numbers every gas lifts within a few per cent of the isothermal film; only a thick, slow film lifts more. The mean pressure excess of a trapped film squeezed by a quarter of its gap, over the isothermal film's 0.0328 of ambient, against the thermal number ωh²/κ, for three ratios of specific heats; dots, the six gases of a 20 kHz, 20 µm levitator at one atmosphere: helium 1 at Ω = 0.29; air 1.002 at Ω = 2.4; argon 1.005 at Ω = 2.5; carbon dioxide 1.005 at Ω = 4.7; xenon 1.042 at Ω = 7.9; sulphur hexafluoride 1.003 at Ω = 16. At Ω = 300 a monatomic gas lifts 1.21 times the isothermal film, against the 1.69 a core on the walls' mean temperature would give.

A levitator lifts by its amplitude, not its gas

A squeeze-film levitator holds its load up by the mean of a pressure that swings symmetrically about nothing, and the gas in the gap was expected to decide how much: a gas that cannot shed its heat of compression in time is stiffer, so it should lift more. Solved to every order in the amplitude, with the heat crossing the gap, it barely does. At a levitator's own thermal numbers every common gas lifts within a few per cent of the isothermal film while its losses differ several-fold, and even a gas far too slow to shed its heat lifts only a fifth more, because the walls quietly pump heat out of it and cool its middle.

A round jet held further from the plate arrives slower in proportion and leaves a radial wall jet stronger in proportion. For the round water jet from a 2 mm nozzle, against the nozzle's distance from the plate, each over its value at 10 cm: the arriving centreline speed, the radial invariant F = Kνh/(2π) it hands on, and the radial wall jet's peak speed 5 cm from the stagnation point; dashed, a plane jet's speed and invariant for comparison. At 40 cm the round jet arrives at 0.25 of its speed at 10 cm and hands on 4 times the invariant, and its wall jet runs 2 times as fast; a plane jet's invariant would have grown 1.59 times.

A round jet hands its wall jet a strength proportional to its throw

A round jet striking a plate spreads into a radial wall jet, and like the plane jet before it, the wall jet keeps not the jet's momentum but Glauert's flux of momentum flux — the momentum the jet brought times the fluid it gathered on the way. A round jet gathers fluid at a rate set by the viscosity alone, eight π ν per metre of travel, whatever its strength. So the invariant it hands on grows in exact proportion to the distance it was thrown, where a plane jet's grew as the cube root: hold the nozzle four times further away and the wall jet is four times as strong and runs twice as fast.

Two spheres settle faster together than apart, faster in line than side by side, and still faster at thirty radii than either alone. The settling speed of a pair of equal spheres over a single sphere's, against the distance between their centres in radii: along the line of centres, exactly (the bispherical solution) and by Batchelor's far-field series; across it, by the series; dashed, each sphere's Stokeslet alone, 1 + 3/(2ρ) and 1 + 3/(4ρ). Touching in line the pair falls 1.55 times as fast; at 5 radii, 1.286 in line and 1.154 side by side; at 30, 1.05 and 1.025.

Two spheres settling at a slant keep their slant for ever, and three do not

Two equal spheres settling through a viscous liquid fall faster together than either alone — half as fast again if they touch in line, and still five per cent faster thirty radii apart. Slanted, the pair drifts sideways as it falls, by up to three and a half degrees, and keeps the slant, the separation and the drift for as long as it falls: the two spheres always move with exactly the same velocity. Release a third and the shape is lost at once, the middle sphere drawing ahead and the outer two closing in behind it. What a pair keeps and a trio does not is a statement about symmetry, and reciprocity is half of it.

A sphere two per cent larger overtakes its neighbour round a path that is its own mirror image, and leaves at the offset it came in at. The centre of a sphere two per cent larger than its neighbour, relative to the smaller sphere, as it overtakes it in a viscous liquid, from four horizontal offsets; the circle is the distance at which they would touch. Each path bulges outward as it passes and is symmetric about the smaller sphere's level: from 0.85, 1.5, 2.5, 4 radii the larger sphere leaves at 0.8499, 1.5, 2.5, 4, after closest approaches of 2.27, 2.83, 3.74, 5.15 radii.

A larger sphere overtakes its neighbour and leaves on the line it came in on

Two equal spheres settling together never change their shape. Make one of them two per cent larger and the pair becomes an encounter: the larger sphere sinks past the smaller round a path that bulges out and comes back, slowed sixteenfold at contact but never stopped, and it leaves at exactly the offset it arrived with. The smaller sphere swings seven radii sideways while the larger is above it and gives every radius back while the larger is below. In Stokes flow, meeting a neighbour displaces nothing — unless the spheres touch, and a hundredth of a radius of roughness is enough to make every touching path leave on the same line.

Pressed with one load, a steeper blade leaves a thinner film, and without a land a steep one cannot float at all. The film a loaded scraper leaves on the surface, in micrometres, against the angle its face makes with the surface, at a load of 100 N per metre of span on an ink of 0.1 Pa·s moving at 5 m/s: with a flat land 0.1 mm long under the edge, and with none. With the land the film is 110 µm at 10°, 1.69 at 30° and 0.268 at 60°. Without one it is the whole gap, 145 µm at 8°, and above 44° the gap the load needs is below a ten-thousandth of a nanometre: the edge is in contact.

A pressed blade floats on its land and leaves half the gap behind

A scraper or a doctor blade is not held at a gap; it is pressed on with a load, and it rises until the liquid it lets through carries that load. Without a flat land under its edge the liquid can carry it only through the corner's logarithm, so the gap falls exponentially with the load, and a blade steeper than about forty-four degrees cannot float at a working load at all — its edge is in contact. Give it a land a tenth of a millimetre long and the film becomes half the gap, falls as one over the load, and is a quarter of a micrometre at sixty degrees. The land is also why a blade's film grows as its edge wears.

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