What is refuted here, and by what

Air parting at the leading edge has to meet again at the trailing edge, so the longer upper path forces the flow over the top to go faster.

What decides it: Surface length and surface speed measured on the solved field over the same section, compared as ratios that depend on no arbitrary release point. The build asserts the two ratios differ by more than two per cent, and refuses to draw the figure if they ever agree.

Tested in The story about air meeting up again, at the figure it turns on.

A wind turbine is inefficient because it only converts about half of the wind's energy. A better design, or a better generator, would get closer to all of it.

What decides it: No device of any kind can take more than 16/27 of the kinetic energy flux through the area it occupies, and the argument contains no device. Air that has given up all its energy is air at rest, and air at rest does not leave; the mass flow through the disc is proportional to how little the air has been slowed, and the energy taken from each kilogram to how much. The product has a maximum, located here by golden-section search at an induction factor of 0.33333333 with a power coefficient of 0.5925925926, and the closed form is exactly 16/27. A modern rotor reaching 0.45 to 0.50 is therefore at 76 to 84 per cent of what physics allows, not at half.

Tested in The most a disc can take, at the figure it turns on.

Above the critical Reynolds number the laminar solution for pipe flow breaks down and no longer satisfies the equations.

What decides it: The parabolic profile is substituted back and integrated: its flow rate agrees with the closed form to better than one part in a million at every Reynolds number, including 100,000. It satisfies the Navier–Stokes equations there exactly, as it does everywhere. What changes at 2300 is which solution the flow occupies, not which solutions exist.

Tested in The solutions stop being chosen, at the figure it turns on.

Sound travels more slowly in thin air, which is why an aircraft at altitude is closer to the speed of sound than one at sea level.

What decides it: The speed of sound is √(γRT), in which no pressure and no density appear separately. At the tropopause it is 295.1 m/s against 340.3 m/s at sea level, and every bit of that difference is the 71.5 K temperature drop. Air at the same temperature carries sound at the same speed at any pressure whatever, because squeezing it raises the density in exactly the proportion that keeps the ratio fixed.

Tested in What a signal travels at, at the figure it turns on.

A wing lifts because of its shape: the curved upper surface is what makes the difference.

What decides it: Lift computed twice by unrelated routes on a Joukowski section — once as ρUΓ, once by integrating the surface pressure without ever mentioning circulation. What sets the number is the circulation the sharp trailing edge selects, and a shape with no circulation makes no lift at all.

Tested in What actually holds a wing up, at the figure it turns on.

An aircraft going supersonic makes a bang at the moment it breaks the sound barrier, and if you missed it you missed it.

What decides it: The cone is attached to the aircraft and travels with it, so the pressure jump sweeps the ground continuously along the whole supersonic track. At Mach 2 the cone half-angle is 30.000000°, measured off the drawn wavefronts rather than quoted, so an aircraft at 11 km drags a boom carpet whose leading edge lies 19 km behind it and never stops arriving.

Tested in When the warning cannot arrive, at the figure it turns on.

The boundary layer is a definite thickness, and the number can be quoted.

What decides it: The Blasius profile solved by shooting and its three thicknesses integrated from it. The layer approaches the free stream and never arrives, so every thickness quoted is a convention — and the momentum thickness is not a height at all, which is why θ = 2f''(0) is an identity the code does nothing to arrange.

Tested in How thick is thin, at the figure it turns on.

Lift needs an aerofoil: a curved section with a sharp trailing edge.

What decides it: A spinning circular cylinder, which has none of those things, solved in closed form. Its surface-pressure integral gives a lift of 2.994 against the theorem's 3.000, and a drag of 10⁻¹⁶.

Tested in Lift with no wing at all, at the figure it turns on.

In a steady flow nothing is accelerating — the picture is the same at every instant, so the fluid in it is not being thrown about.

What decides it: The convective term computed by central differences on the exact field and integrated along a path. Along the streamline into the nose the integral is −0.394 31 and the change in ½q² between its ends is −0.394 31; nothing in the code makes them agree.

Tested in Steady does not mean nothing is happening, at the figure it turns on.

A wind-tunnel photograph shows the flow field.

What decides it: One solved field rendered as four instruments would record it — smoke, tufts, surface oil, pressure taps — from a single field object, so the four cannot disagree about the flow, only about which function of it each measures. None of the four is the velocity field.

Tested in What a photograph of a flow shows, at the figure it turns on.

The head loss where a duct widens is friction — the fluid rubbing along the extra wall area and through the turbulent eddy in the corner.

What decides it: The loss is computed here without a viscosity, a Reynolds number, a wall roughness or a wall length appearing anywhere. Momentum across the step gives one pressure rise, the energy balance gives another, and the difference between them is the loss — half the square of the velocity change, exactly, and matching the closed form to 1.1e-16 over four hundred area ratios. The dissipation is real and is of course ultimately viscous, but its SIZE is fixed by the geometry alone: a step of the same area ratio in honey and in helium loses the same fraction of the dynamic head. Friction on the extra wall is a separate and much smaller term that this calculation does not contain.

Tested in A loss with no viscosity in it, at the figure it turns on.

Fluid going round in circles is rotating, and fluid running in straight parallel lines is not.

What decides it: A pair of assertions rather than one, on two flows with identical streamline pictures. Solid-body rotation is refused at a worst vorticity of 1.400, which is 2Ω exactly; a free vortex over the same grid passes at 1.6 × 10⁻⁷, which is round-off. A straight shear is refused as well.

Tested in Spin is not the same as going round, at the figure it turns on.

If you pull a harder vacuum downstream, more gas will come through the hole. There is always some gain left.

What decides it: Below a pressure ratio of 0.528 the throat is sonic and the mass flux is stuck at 0.6847 p₀/√(RT₀) — computed two ways, from the closed form and by assembling ρ*a* out of the sonic state ratios, agreeing to 1e-12. Every choked case in a swept solve returns the identical throat condition, and a sweep in which the flow creeps up as the back pressure falls is refused.

Tested in The throat that stops listening, at the figure it turns on.

The closure problem is a technical difficulty that a sufficiently clever derivation will one day remove.

What decides it: The counting forbids it. In three dimensions the symmetric moment of order n has C(n+2, n) independent components, and its own transport equations introduce the moment of order n+1, which has more. Six needs ten, ten needs fifteen, fifteen needs twenty-one, and the gap widens at every rung — this is arithmetic on binomial coefficients and contains no fluid mechanics at all. What can be derived is an exact hierarchy; what cannot be derived from inside it is a place to stop.

Tested in The ladder that never closes, at the figure it turns on.

The energy a hydraulic jump destroys is lost to friction with the river bed, so a smooth concrete channel loses less than a rough one.

What decides it: The bed does not appear in the calculation. Equating the specific force q^2/gh + h^2/2 on the two sides gives the depth ratio; putting those depths into the specific energy gives a loss of (h2-h1)^3/4h1h2, which agrees with the difference of the two energies to nine figures here. No bed friction, no roughness and no channel length is used anywhere. At the case drawn, an arriving Froude number of 5.05, the jump destroys 49.5 per cent of the energy that arrived, and it would destroy the same fraction over glass. The dissipation happens inside the roller, between the fluid and itself.

Tested in The shock in a river, at the figure it turns on.

A wing works by deflecting the air that strikes its underside, so the lift is the rate at which that air's momentum is turned — which makes the lift proportional to the square of the angle.

What decides it: This is Newton's own theory and it is wrong by a factor of thirty-six at the angles aircraft fly at. The impact picture gives C_L = 2 sin²α cos α, which at 5° is 0.0151 against thin-aerofoil theory's 2πα = 0.5483 — and the two curves have different SHAPES, one quadratic and one linear, so halving the incidence doubles the discrepancy. The error is not the momentum argument, which is correct and is used elsewhere on this site; it is the assumption that only the air actually striking the surface is involved. A real wing turns a mass of air many times greater than the flux through its own frontal area, because the pressure field reaches far above and below it. Newton's formula becomes exact only where the shock lies on the body and the deflected air really is confined to a thin layer, which is hypersonic flight.

Tested in The theory that forbade flight, at the figure it turns on.

Lift acts at the centre of pressure, and that is a property of the section.

What decides it: A sweep rather than a demonstration: forty-one candidate reference points along the chord, the pitching moment about each computed at four incidences, and the slope of moment against incidence taken. It crosses zero once, at 25.6% of chord — while the centre of pressure moves with incidence and runs off to infinity at zero lift.

Tested in Where the lift acts, at the figure it turns on.

A flow with circulation round it must be spinning, so an irrotational flow cannot have any circulation.

What decides it: A free vortex, with the line integral of velocity round a square enclosing its centre computed at 2.000000 — the vortex's own strength — while the vorticity at every point the field is defined on comes out at 10⁻⁸ or smaller. The theorem is not violated: the loop encloses a point that is not in the domain.

Tested in Circulation is vorticity, added up, at the figure it turns on.

The low pressure over a wing sucks it upwards — the air above pulls on the surface.

What decides it: A fluid at rest or in motion exerts only a normal compressive stress: it pushes. What is called suction is a smaller push against a larger one from underneath, and the arithmetic says so twice. First, the numbers: on a solved Joukowski aerofoil at 6° and 60 m/s the lowest absolute pressure anywhere on the surface is 96.5 kPa against an ambient of 101.3 — a dip of 4.8 per cent, with the whole atmosphere still pressing on the wing. Second, the identity: ∮n dS round any closed body is zero, computed here at 3×10⁻¹⁵ against a perimeter of 8.3, so adding a constant to the pressure field changes no force. The lift computed from gauge pressure and from absolute pressure agree to twelve significant figures. A quantity whose sign depends on where zero was put cannot be a mechanism.

Tested in Nothing sucks, at the figure it turns on.

A shock is a place where the equations of fluid mechanics break down.

What decides it: The jump conditions follow from applying the same conservation laws to a control volume containing the discontinuity. Substituting the resulting ratios back into mass, momentum and energy gives residuals of 0, 0 and 1.4e-16 at Mach 2 — so the discontinuity is a solution of the conservation laws rather than a failure of them.

Tested in The jump the equations allow, at the figure it turns on.

Below the speed of sound there is no supersonic flow to worry about, and shocks are a supersonic-aircraft problem.

What decides it: A ten per cent thick section at two degrees has a critical Mach number of 0.631 — computed as the crossing of its own solved suction peak, corrected for compressibility, with the pressure coefficient at which the local flow reaches Mach one. Above that free-stream Mach number there is supersonic flow on a subsonic aeroplane, and it has to be closed by a shock.

Tested in The pocket on top of the wing, at the figure it turns on.

Dimples make a golf ball go further by reducing the friction between the ball and the air.

What decides it: They increase it. A tripped boundary layer has several times the skin friction of a laminar one at the same Reynolds number, which is the flat-plate result this site derives elsewhere. What dimples do is trip the layer early so that it separates late, which narrows the wake and cuts the PRESSURE drag by far more than the friction rose. On a sphere the total drag coefficient falls from about 0.5 to about 0.15 through that transition. Integrating two trajectories at 70 m/s with the same launch and the same spin, the tripped ball goes 411 m against 192 m — and the smaller of the two contributions to drag is the one that went up.

Tested in The drag that falls as it speeds up, at the figure it turns on.

Expansion shocks do not exist because they would violate conservation of energy.

What decides it: A jump from Mach 0.8 satisfies mass, momentum and energy with residuals of 0, 0 and 0 to machine precision, computed by substitution rather than assumed. What it does is lower entropy by 0.01638 R and raise total pressure by 1.65 per cent. The second law is the only thing standing in its way, and the build refuses such a jump on that ground alone.

Tested in The only law that forbids it, at the figure it turns on.

Streamlines that curve mean the fluid is rotating, and streamlines that run straight mean it is not.

What decides it: Both halves are false and the same figure refutes both. A free vortex has circular streamlines and zero vorticity everywhere outside its core: a parcel carried round it does not turn at all, it is sheared, and a paddle wheel dropped into it holds its bearing. A parallel shear flow has perfectly straight streamlines and vorticity equal to the shear rate: the parcel turns at half that rate the whole way along. The velocity gradient measured at a point of this site's own cylinder solution comes out with a vorticity of order ten to the minus nine everywhere outside the body, while the stretching rate reaches a third of U over a — so the curvature of the picture and the rotation of the fluid are independent quantities, and only one of them is visible.

Tested in What a parcel does in the first instant, at the figure it turns on.

A wing flies by the Coandă effect: the air follows the curved upper surface because of the effect, and following the curve is what produces the lift.

What decides it: The pressure gradient across a curved streamline is minus rho q squared kappa — Euler's equation resolved normal to the flow — and this site computes both sides by routes sharing no arithmetic, agreeing to a part in a hundred thousand on the cylinder and on both surfaces of an aerofoil. It holds ABOVE the wing and BELOW it, on the concave side and the convex side, in every flow in this collection. An identity that holds everywhere cannot be the reason for anything in particular: naming it where a jet meets a wall labels the observation rather than explaining it, and the lift itself is fixed by the circulation, which the Kutta condition sets at the trailing edge and which this identity says nothing about.

Tested in The effect that explains nothing, at the figure it turns on.

A cricket ball swings because of the Magnus effect — the bowler puts spin on it and the spin drags air round with it, producing a sideways force.

What decides it: There is no spin in the model that produces the swing, and a conventional swing bowler's ball spins about an axis along its own flight path, which produces no Magnus force at all. The mechanism is asymmetric separation: the angled seam trips the boundary layer on one side, that side holds on to about 120 degrees while the other lets go at about 80, and the pressure distributions no longer cancel. Integrating that distribution with a measured wake pressure gives a side-force coefficient of 0.29485 towards the later-separating side — computed twice, by quadrature and in closed form, agreeing to a part in ten million — and a ball with equal separation angles gets exactly zero to fourteen decimal places.

Tested in A ball that swings without spinning, at the figure it turns on.

Water goes down the drain anticlockwise in the northern hemisphere and clockwise in the southern, because of the Coriolis force.

What decides it: The planet's rotation does supply a circulation to a bathtub — f times the area, computed here two ways that agree to a part in a million — and concentrating it at a one-centimetre drain gives a swirl of 1.3 millimetres per second. A residual motion of one centimetre per second at the rim, which is what filling a bath leaves behind and what viscosity takes about four hours to remove, gives half a metre per second at the same radius: 388 times larger. The Rossby number of a bath is three thousand, which says the same thing in one number. The effect is real and it is not zero; it is simply not what decides the outcome, and the experiments that DO see it took a symmetric tank, a full day of settling and a plug pulled from a distance.

Tested in The bath does not know the hemisphere, at the figure it turns on.

A turbulent boundary layer separates sooner than a laminar one, because turbulence is disorderly and disorder makes a flow leave a surface.

What decides it: It separates considerably later, and the reason is momentum rather than order. A turbulent profile carries far more momentum close to the wall than a laminar one at the same thickness, because the mixing brings fast fluid down to it. The Falkner-Skan family locates laminar separation at beta = -0.198838, by bisecting on the gradient until the wall shear reaches zero; a turbulent layer survives adverse gradients several times steeper. The whole practice of tripping a layer deliberately exists because of this.

Tested in The gradient that does both, at the figure it turns on.

Turbulence is a hard computational problem that faster machines will eventually solve.

What decides it: The count rises as Re^(9/4) in memory and Re^3 in work, so an airliner wing at Re = 5·10⁷ needs about 2·10¹⁷ grid points against the roughly 10¹² of the largest simulations ever run — five orders of magnitude in memory and about eleven in work. A machine a thousand times faster than today's buys a factor of ten in Reynolds number. Moore's law at its historical rate would need something over a century to close the gap for one wing at one instant, and the design process needs thousands of cases.

Tested in The grid nobody can build, at the figure it turns on.

A wing in ground effect rides on a cushion of air compressed between it and the ground.

What decides it: The model here has no compressibility in it at all — the density is constant everywhere — and it still produces 74% more lift at a fifth of a chord above the ground. What changes is the circulation the tangency condition forces the section to carry, from 0.5476 in free air to 0.9533 near the ground, and the mechanism survives the removal of the thing the popular explanation names.

Tested in The wall that pushes back, at the figure it turns on.

A shock destroys energy, which is why supersonic flight is inefficient.

What decides it: The stagnation temperature is identical on both sides of a shock, checked independently on each side and agreeing to 1e-12: no energy has gone anywhere. What falls is the stagnation pressure, by 27.9 per cent at Mach 2 — a loss of availability rather than of energy, which is a different quantity with a different accounting.

Tested in What a shock costs, at the figure it turns on.

A fluid layer heated from below starts to convect as soon as the top is colder than the bottom.

What decides it: It does not move at all until the Rayleigh number exceeds a definite value, because a displaced parcel loses its buoyancy by conduction and its momentum by viscosity while it travels. The threshold is an eigenvalue: minimising Ra(a) = (pi squared plus a squared) cubed over a squared by golden-section search puts it at a = 2.221441 and Ra = 657.5114, which are pi over root two and 27 pi to the fourth over four to fourteen digits. Below it the layer conducts and is perfectly still.

Tested in A threshold with a closed form, at the figure it turns on.

A sailing boat is pushed along by the wind, so it cannot travel faster than the wind that is pushing it.

What decides it: A boat is not pushed; it is a pair of aerofoils working against each other in two different fluids, and its speed is set by a velocity triangle rather than by the wind's speed. Solving the force balance gives an apparent wind angle equal to the sum of the rig's drag angle and the foil's, and a boat speed of sin(beta - lambda)/sin(lambda) times the true wind. For drag angles of 14 and 6 degrees that exceeds one wherever the course is more than 40 degrees off the true wind, and peaks at 2.9238 times the wind speed on a broad reach — found by search and matching one over sin lambda exactly. Ice yachts and foiling catamarans do this routinely; the model says any boat with good enough foils will.

Tested in Faster than the wind that drives it, at the figure it turns on.

Mixing is a matter of stirring hard enough — a vigorous laminar flow mixes fluid, and turbulence is only a faster version of the same thing.

What decides it: A steady two-dimensional flow cannot mix at any strength whatever, because its pathlines are its streamlines and a particle released on one is confined to that curve for ever. Measured here: a material line in a steady vortex grows in length linearly, at 1.20 times its original length per period, constant to a tenth of a per cent over the second half of the run. The same line in a flow that merely switches between two such vortices grows exponentially at 1.07 per period, and a pair of neighbouring particles separates at 1.23 per period — a factor of a thousand in under six. What separates the two cases is not vigour but time dependence.

Tested in No randomness, and it mixes anyway, at the figure it turns on.

Lowering a flap makes a wing more responsive: it produces more lift per degree of incidence.

What decides it: Lift curves solved at flap deflections of 0, 10 and 20 degrees. The zero-lift angle moves from 0.00° to −6.12° to −12.60°, and the slope goes 0.10944, 0.10890 and 0.10765 per degree — a spread of 1.6%, in the direction of slightly *less* response rather than more. What the flap buys is a shift, not a tilt.

Tested in What a flap does, and what it does not, at the figure it turns on.

A jet breaks up into a spray when it is fast enough — the faster the jet, the finer the droplets, and the number that decides it is the jet's Reynolds number.

What decides it: The number that decides which of the four break-up regimes a nozzle is in is the GAS Weber number, which contains the density of the surrounding air rather than of the liquid. The same 1 mm water jet at the same 40 m/s has a gas Weber number of 26 in air and 0.03 in a vacuum chamber at a thousandth of an atmosphere — the first atomises and the second is still in Rayleigh break-up, making drops twice the jet's diameter. The Reynolds number is the same in both cases. Speed matters because it appears in the Weber number, and it is the air that makes the difference between a stream and a mist.

Tested in Where a jet stops being a jet, at the figure it turns on.

A wing's curved upper surface forms a constriction with the air above it, like half a venturi. Continuity forces the air through the narrower channel to speed up, and Bernoulli's theorem then says its pressure must fall — which is where lift comes from.

What decides it: Three computations on exact solutions. A symmetric section at zero incidence has a genuine constriction — its streamtube narrows by a fifth at mid-chord — and makes exactly no lift, because the identical thing is happening underneath and the argument cannot tell the two sides apart. A flat plate of zero thickness has no upper surface, no curvature and no channel at all, and lifts 2π sin α. And the "constriction" over a real section is still measurable several chords above it, in air that is nowhere near anything solid, because the tube's upper boundary is a streamline rather than a wall — so its position is part of the solution that already contains the lift.

Tested in Not half a venturi, at the figure it turns on.

A sketch of a flow pattern is a matter of judgement — as long as the streamlines look plausible and go the right way round the body, the drawing is as good as any other.

What decides it: The pattern is constrained by an integer. Walking once round a closed curve and counting how many complete turns the velocity vector makes gives a whole number, and that number must equal the sum of the indices of the critical points inside, counting a centre as +1 and a saddle as −1. Checked here on four solved fields by two routes that share no arithmetic: a single cell of a cellular flow gives 1 and 1, four cells give 3 and 3, a vortex pair gives 1 and 1, a separation bubble gives 0 and 0. A sketch whose two numbers disagree contains a stagnation point nobody drew, or one that cannot exist. And in an incompressible plane flow only two kinds of critical point are possible at all — a sketch with a spiral in it is wrong before any physics is consulted.

Tested in The count a pattern cannot break, at the figure it turns on.

To get to a mark upwind as fast as possible, point the boat as close to the wind as it will go — every degree closer to the mark is a degree less distance to sail.

What decides it: Pointing closer is slower over the ground, and the trade has a maximum well off the closest course. The quantity to maximise is the component of boat speed towards the mark, V_b cos beta, and differentiating sin(beta - lambda) cos(beta) gives cos(2 beta - lambda) = 0, so the best beat is at 45 degrees plus half the apparent wind angle. For a lambda of 20 degrees that is 55 degrees off the true wind, found here by golden-section search at 55.000000 and matching the closed form to six figures — against a no-go zone that ends at 20 degrees. A boat sailing at 30 degrees is pointing far better and arriving later.

Tested in The fastest way is not the straight one, at the figure it turns on.

A tracer that diffuses more readily will spread out along a pipe faster — high diffusivity means fast spreading, in a flow as much as in still fluid.

What decides it: In a pipe the opposite is true above a Péclet number of about seven, and the calculation says so with a slope. The effective diffusivity along the pipe is D(1 + Pe²/48) with Pe = Ua/D, so at large Péclet number it is U²a²/48D — inversely proportional to the molecular diffusivity. Computed here across four decades of D at a fixed flow, the effective value falls with a slope of exactly −1. The mechanism is why: dispersion happens because molecules sample streamlines of different speeds, and a molecule that crosses the pipe quickly spends less time being carried at the wrong speed.

Tested in Two slow things make a fast one, at the figure it turns on.

A vortex is carried along by its own swirling flow, so a strong one moves faster than a weak one.

What decides it: A point vortex has no velocity of its own at all, whatever its strength. Its field is antisymmetric about its own centre — the flow on one side is exactly the reverse of the flow on the other — so there is no velocity to evaluate there and nothing to carry it. The stepper here leaves the self-term out by construction rather than by neglect, and a lone vortex integrated for five hundred steps moves a distance of exactly zero. What does move a vortex is other vortices, or a wall, which is another vortex wearing a disguise. A pair of opposite sign travels at Γ/2πd and the stepper reproduces that closed form to fourteen decimal places.

Tested in Vortices move each other, at the figure it turns on.

An aeroplane stalls when it goes too slowly. The stalling speed is a property of the aircraft, it is printed in the handbook, and staying above it is what keeps the wing flying.

What decides it: A wing stalls at an angle, and the angle does not move: 16.5 degrees here, unchanged by weight, by altitude, by load factor and by limit load, which the solver requires to machine precision across every variation. The *speed* at which level flight puts the wing at that angle moves with all four. Twenty-one per cent more weight raises it by exactly ten per cent; thin air at sixteen thousand feet raises the true speed by nineteen; a sixty-degree banked turn raises it by forty-one, because the accelerated stall goes as the square root of the load factor. And at the never-exceed speed this aircraft could pull 9.6 g before the wing let go — so it can stall at any speed it can reach, and staying above a number in a handbook does not prevent it.

Tested in An angle, not a speed, at the figure it turns on.

A vortex ring travels forward because the air inside it is being pushed through, like a jet.

What decides it: The ring's translation is induced by the ring itself, and the mechanism is curvature rather than any flow through it. The same Biot-Savart integrator that gives the ring a speed of 0.53 Gamma over R at a cutoff of a fiftieth of the radius returns exactly zero on a straight filament of the same circulation — magnitude under 10⁻¹⁴, for every length and every cutoff, because dl and r are parallel at every element and the cross product vanishes identically. A ring with nothing passing through it, in a fluid at rest, still moves. What travels with it is a closed body of fluid carried along by the induced field, which the meridional streamlines show as a region that returns on itself rather than as a stream passing through.

Tested in A ring moves because it is bent, at the figure it turns on.

Flow in an artery can be estimated from the pressure with Poiseuille's law — pressure difference times πR⁴ over 8μL — since the vessel is a tube and the flow is laminar.

What decides it: Both hypotheses hold and the answer is still wrong by a factor of twenty-four. Poiseuille's law assumes the profile has had time to develop, and whether it has is decided by the Womersley number α = R√(ω/ν), which is about 18 for the human aorta. Solving the unsteady problem exactly gives a flow amplitude of 4.3 per cent of the quasi-steady value at α = 13 and less at 18, a phase lag of 83 degrees rather than zero, and a velocity profile whose maximum is at three-quarters of the radius rather than on the axis. Every one of those is a difference in kind, and the number that produces them contains the frequency, which Poiseuille's law has no place for.

Tested in Too fast for a profile, at the figure it turns on.

Turning a supersonic flow always produces a shock.

What decides it: Turning it away from itself produces a fan of Mach waves and no discontinuity at all. Across a 10° expansion at Mach 2 the flow reaches Mach 2.385 with p/ργ unchanged to 1e-12, so the total pressure is preserved exactly — against a 10° compression at the same Mach number, which keeps 98.46 per cent of it.

Tested in Turning the other way is free, at the figure it turns on.

A bearing works because oil is slippery — the film's job is to keep the metal surfaces apart and reduce friction, and any film thick enough to separate them will do.

What decides it: Two pads with the same oil, the same speed and the same clearance carry entirely different loads: a converging film supports 641 kilonewtons per metre of width and a parallel one supports zero. Not a small load — zero, because the term that drives the pressure in Reynolds' equation is the gradient of the film thickness, and a parallel film has none. The solved pressure field is what separates the surfaces, and it exists only where the gap narrows. A film that merely fills the space is not a bearing.

Tested in Nothing but the shape of the gap, at the figure it turns on.

Aerodynamics proves that a bumblebee cannot fly. The calculation has been done and the wings are too small to hold the animal up, which shows that the theory has limits.

What decides it: The calculation used the animal's forward speed as the velocity over its wings, which is the arithmetic of a fixed wing. Done that way the required lift coefficient ranges from 0.88 for a hawkmoth to 26 for a fruit fly — a spread of a factor of twenty-nine across four animals, because the number is governed by a speed that has nothing to do with how any of them makes its lift. Done with the velocity the wings actually see, the requirement spreads by a factor of 1.9 and every animal lands between what a steady aerofoil gives at those Reynolds numbers and what a revolving wing with an attached leading-edge vortex gives. And in a hover the fixed-wing calculation has no answer at all: with no forward speed there is no dynamic pressure and no coefficient however large will do, which the solver refuses rather than returning a large number.

Tested in The bee that cannot fly, at the figure it turns on.

A sphere and a cylinder behave much the same in an ideal flow — the geometry is the same in cross-section, so the surface pressures and the forces are the same to within a small factor.

What decides it: Solving both exactly gives a fastest surface speed of 1.5U on the sphere against 2U on the cylinder, a minimum pressure coefficient of −1.25 against −3, a disturbance at five radii of 0.4 per cent against 4, and an added mass of half the displaced fluid against all of it. Every one of those is in the same direction and none of them is a small factor. The deepest difference is not a number at all: a loop drawn round a cylinder cannot be shrunk to a point without crossing it, so its circulation is free; a loop round a sphere slides off over the pole, so the circulation must be zero and no ideal flow can give a body of revolution any lift whatever.

Tested in Three dimensions are kinder, at the figure it turns on.

Cavitation is the water boiling because the propeller heats it — friction on a fast-moving blade raises the local temperature past the boiling point.

What decides it: Nothing is heated. Liquid boils when its pressure falls to its vapour pressure, and that can be reached by lowering the pressure at constant temperature just as well as by raising the temperature at constant pressure. A body in a stream lowers the pressure at its own surface by exactly the amount Bernoulli's theorem requires — a coefficient of minus three at the shoulder of a cylinder, computed here as minus 2.999200 by walking the site's own solved field — so the condition is sigma equals minus C_p,min and it contains no temperature at all. The vapour pressure of water at 20 degrees is 2.3 kilopascals, and a body needs a dynamic pressure of about 33 kilopascals to reach it at that pressure coefficient, which is 8 metres per second.

Tested in When a body tears the water, at the figure it turns on.

A flow whose every linear mode decays is stable, so a disturbance in it can only get smaller.

What decides it: Modal decay bounds the behaviour as time goes to infinity and says nothing about what happens first. The model operator here has eigenvalues −1/Re and −2/Re, both negative at every Reynolds number, and the worst-case disturbance in it grows by a factor of Re²/16 in energy before decaying — a hundredfold at Re = 40, sixty thousandfold at Re = 1000, with the exponent fitted at 1.99 over four decades rather than quoted. The mechanism is that the eigenvectors are nearly parallel, so a small disturbance can be the difference of two large ones that decay at slightly different rates. A NORMAL operator with the same two eigenvalues, run through the same routine, never exceeds a gain of one at any time — which is the control that makes the argument mean something.

Tested in Every mode decays and it grows anyway, at the figure it turns on.

Separation destroys lift. A wing stalls when the flow leaves the upper surface, so a design that separates deliberately is a design that has given up on lift.

What decides it: On a slender delta with sharp leading edges the separation is where most of the lift comes from. The flow cannot round an edge that sharp, so it leaves it along the whole span, rolls into a pair of vortices sitting over the upper surface, and their suction adds a term quadratic in incidence to the linear one an attached flow gives. Computed here: the two are equal where tan α = AR/2 — 26.6 degrees at an aspect ratio of one — and the total peaks at 1.68 at 49 degrees, where a conventional wing of aspect ratio six has stalled at 15 degrees with a maximum of 1.19. What separation destroys is the lift of a wing that was relying on the flow staying attached.

Tested in Lift out of a failure, at the figure it turns on.

The pitting on a cavitating propeller is corrosion — the vapour and the dissolved gases attack the bronze chemically, and a better alloy solves it.

What decides it: The damage is mechanical, and its scale can be computed. An empty cavity collapsing under one bar reaches a wall speed that would exceed the speed of sound in water at R/R0 = 0.0312, and the pressure in the liquid a bare 1.59 radii outside the closing bubble is already 5182 times ambient at that radius — over five thousand atmospheres, computed here by scanning Rayleigh's pressure field rather than quoted. That is far above the yield stress of any propeller alloy. Better alloys help, in the sense that harder ones last longer, and no alloy is immune; the remedy is to stop the cavitation rather than to armour against it.

Tested in The bubble that hammers, at the figure it turns on.

The fluid in a sharp corner is stagnant — the main flow cannot get into it, so nothing happens there and a corner is a dead volume.

What decides it: The creeping-flow equations have exactly one family of solutions near a corner, and below an opening of 146.31 degrees every one of them oscillates: the stream function changes sign at radii in geometric progression, which is an infinite sequence of counter-rotating eddies reaching the vertex. Nothing is stagnant. What is true is that the motion becomes unmeasurable very fast — each eddy is some hundreds to tens of thousands of times weaker than the one outside it — so "stagnant" is a statement about instruments rather than about the fluid, and the difference matters for anything carried by the flow rather than seen in it.

Tested in The eddies nobody stirs, at the figure it turns on.

The mean flow is what is left when the fluctuations are averaged away, so it is the steady flow the body experiences.

What decides it: Averaging an oscillating potential flow past a cylinder gives a mean velocity that is exactly zero everywhere and a mean pressure that is not: the coefficient reaches −2.00 at the shoulders and averages −1.00 over the surface. A body in that flow is loaded steadily by a flow whose average delivers no momentum at all, and the averaged momentum equation is satisfied only with a term the mean velocity cannot supply — computed here as the divergence of ⟨u u⟩, which is a Reynolds stress in a flow with no turbulence, no randomness and no viscosity.

Tested in The mean is not the flow, at the figure it turns on.

Circulation is conserved in fluid flow — it is one of the conservation laws, like mass and energy, so the circulation in a flow cannot change.

What decides it: Kelvin's theorem conserves the circulation round a MATERIAL loop in a fluid that is inviscid, barotropic and driven by conservative body forces, and each of those three can fail. A material circle in a diffusing vortex does not move at all and loses 89 per cent of its circulation in a minute, because vorticity crosses it. A rectangle in a fluid at rest whose density falls sideways acquires circulation at 2.57 m²/s per second, out of nothing. A ring contracting on a rotating planet gains 0.19 m/s of swirl. Every vortex in nature exists because one of the three hypotheses failed somewhere, and naming the theorem without its hypotheses gets the physics backwards.

Tested in What survives being wound up, at the figure it turns on.

Turbulence transfers energy from large eddies to small ones and eventually to heat. That is what a cascade is, and it is what turbulence does.

What decides it: It is what THREE-DIMENSIONAL turbulence does, and the reason is a term that vanishes identically in a plane. Differencing (ω·∇)u on a plane flow gives exactly zero — not a small number, no number at all, because the vorticity points out of the plane and nothing varies along it — while the same expression on Burgers' vortex gives 0.0748, which is as large as the vorticity there. With that term gone the enstrophy is conserved as well as the energy, and two conservation laws applied to three modes give an exact answer: 80 per cent of any energy transferred goes to the LARGER scale. Two-dimensional turbulence organises itself into bigger and bigger vortices, which is what the atmosphere at planetary scale and a soap film both do.

Tested in The cascade that runs backwards, at the figure it turns on.

A duct chokes because the passage is too small — choking is a geometric condition, and a constant-area pipe cannot choke.

What decides it: A constant-area pipe with friction chokes, and so does one with heat added, and neither has a throat anywhere in it. What chokes a flow is the second law rather than the geometry: both processes drive the entropy up, both have their entropy maximum at exactly Mach 1, and a flow can only move towards a maximum it cannot pass. Sweeping the entropy over Mach number on each line and locating the peak numerically puts it at 1.000 on both to within the resolution of the sweep, while each line's own invariant — the stagnation temperature for Fanno, the momentum flux p + rho u² for Rayleigh — is recomputed from the state and holds to a part in 10⁹. A subsonic flow speeds up along either line, a supersonic one slows down, and both stop at the same place.

Tested in Two ways to choke, at the figure it turns on.

A jet slows down because the surrounding fluid pushes back on it — the air ahead has to be shoved out of the way, and that drag is what brings the jet to a stop.

What decides it: The momentum flux of a free jet is constant along it, computed by quadrature on the solved profile at five stations spanning a factor of sixty in distance and flat to nine figures. Nothing is pushing back: the surroundings are at rest and at uniform pressure, so no force acts on the jet at all. The centreline slows because the same momentum is being shared with more and more fluid, and the mass flux — which is what grows — rises as the cube root of distance. A jet does not lose momentum; it dilutes it.

Tested in What a jet keeps, and what it collects, at the figure it turns on.

A dimensionless group of order one marks the boundary between the two regimes it separates, so a flow with the group well below one is safely in the first regime and one well above it is safely in the second.

What decides it: It marks where the two terms the group compares are equal, which is a different statement and is usually not the interesting one. Computed for fourteen groups on this site: the continuum equations with a no-slip wall are one per cent wrong at a Knudsen number of 1/594; a quasi-steady lift is one per cent wrong at a reduced frequency of 0.0061; Stokes' law at Reynolds number one, where the two terms are exactly equal, is sixteen per cent low. Two of the fourteen have thresholds nowhere near one in the other direction: a layer heated from below convects at Ra = 1707.762 and nothing happens at Ra = 1.

Tested in What "of order one" is worth, at the figure it turns on.

A wing makes no lift when it is level, and lift begins when the pilot raises the nose. Zero incidence is the same thing as zero lift.

What decides it: Zero lift happens at the zero-lift angle, which is a property of the camber line and is computed here from the line's own slope with no flow in the calculation at all. A four per cent circular arc lifts nothing at −4.58 degrees and is already at a lift coefficient of 0.50 when it is level; a reflexed line lifts nothing at a positive incidence and has to be pointed upwards before it stops. The two angles are the same only for a symmetric section, where the integral is zero by symmetry. Checked against the closed form α₀ = −2m for a parabolic arc, which the quadrature reproduces to ten decimal places, and against the exact conformal map, which agrees to within a fiftieth of a degree at every thickness from eighteen per cent down to one.

Tested in Where lift starts, at the figure it turns on.

A numerical method is a way of getting an approximate answer when the exact one is out of reach. What it computes is the same thing the exact theory computes, only less accurately.

What decides it: Set up the panel system without the Kutta condition and it has N equations for N + 1 unknowns. Every solution of that underdetermined system is a legitimate ideal flow: the surface is a wall to two parts in a thousand at every collocation point, the equations of motion hold, and the lift coefficient runs from −0.60 to +0.65 across the family at a single incidence, exactly linearly in the circulation with slope 2Γ/Uc. The method has not approximated a unique answer badly; it has found that there is not a unique answer, which is a fact about the physics that the exact map hides by having the Kutta condition built into the choice of vortex strength before anything is solved.

Tested in Nothing but the edge, at the figure it turns on.

The discharge coefficient of an orifice is an empirical fudge factor — a number measured for each shape and looked up in a table, because the real flow is too complicated to compute.

What decides it: Two orifice geometries have contraction coefficients known exactly and neither was measured. Borda's re-entrant mouthpiece gives one half from momentum on the vessel, with no flow field computed anywhere: the pressure over the whole wall is the undisturbed hydrostatic value, so the force is known, and the only contraction that balances it is a half. A sharp slot in a plane wall gives π divided by π plus two, which is 0.61102, from Kirchhoff's free-streamline solution — a conformal map in which the shape of the jet's surface comes out of the arithmetic rather than going into it. The momentum theorem, applied afterwards to that solution as a test it was never told about, closes to a part in a million.

Tested in The hole that halves the flow, at the figure it turns on.

A stratified shear flow becomes turbulent when the Richardson number falls below a quarter — Ri = 1/4 is the transition, in the way that Re = 2300 is for a pipe.

What decides it: The theorem runs the other way and only the other way. Miles and Howard proved in 1961 that Ri > 1/4 EVERYWHERE is sufficient for stability; it says nothing whatever about what happens below a quarter, and flows with minimum Richardson numbers well under it are routinely observed to be stable. The quarter is a guarantee of stability rather than a threshold of instability, and the difference matters in exactly the cases where the number is used — inferring turbulence in the ocean or the night-time atmosphere from a measured profile. Computed here: the parcel-exchange energy balance crosses at 0.2500 for every stratification and every displacement, and what it establishes is that below that value the exchange is merely energetically POSSIBLE.

Tested in The number that stops the mixing, at the figure it turns on.

Turbulence is random motion, so it has to be described statistically for the same reason Brownian motion does — there is a stochastic element in the physics.

What decides it: There is no stochastic term in the Navier–Stokes equations. Given the initial and boundary data the solution is determined, and the same data run twice gives the same flow — which is exactly what a deterministic simulation demonstrates every time it is repeated. What produces the statistics is sensitivity: the Lorenz system computed here separates two trajectories differing by 10⁻⁸ into completely different states within a few tens of time units, with a Lyapunov exponent that is measured rather than assumed. The randomness is in the data and in the amplification, not in the equations.

Tested in The randomness that is not in the equations, at the figure it turns on.

A supersonic nozzle is designed by its area ratio: pick the exit Mach number, look up A/A*, and make a duct that expands by that much.

What decides it: The area ratio is necessary and nowhere near sufficient, and the figures here show three ducts with identical area ratios of which only one works. A quasi-one-dimensional theory has no shape in it at all — it relates one number to another — while the real flow is two-dimensional and full of waves that reflect off the walls unless the walls are curved to absorb them. The designed contour delivers a stream uniform at the design Mach number and parallel to 10⁻¹⁵ of a degree; a cone of the same area ratio delivers one still diverging at its wall angle, which costs a real rocket nozzle about one per cent of its thrust; and a duct that reaches the area too soon delivers the right average Mach number with oblique shocks in it.

Tested in The wall that cancels its own waves, at the figure it turns on.

Viscosity destroys circulation — a vortex in a real fluid decays, so the circulation round it falls with time, and an old vortex is a weaker vortex.

What decides it: The exact solution has the vorticity obeying the heat equation, and diffusion moves a conserved quantity about rather than destroying it. Integrating the vorticity over the plane by quadrature at four times spanning four decades returns the same circulation to a part in 10⁹. What falls is the peak vorticity, as 1/t, and the peak swirl, as 1/√t — both local measures. At a hundred millimetres from the axis the swirl velocity is 1.592 m/s at every time while the core is smaller than that, which is Γ/2πr exactly. Viscosity rounded off the singularity at the centre and changed nothing outside it.

Tested in What viscosity cannot take away, at the figure it turns on.

The continuum equations hold while the Knudsen number is small, so a flow with Kn below the usual continuum boundary of 0.01 can be computed with the ordinary Navier–Stokes equations and a no-slip wall.

What decides it: Poiseuille flow with Maxwell's slip condition is exact, and it gives a flow rate 1 + 6Kn times the no-slip one. So the no-slip answer is one per cent wrong at Kn = 1/594, five per cent at 1/114 and six per cent at the ladder's own continuum boundary of 0.01. The equations are fine; it is the wall condition that goes first, and it goes two orders of magnitude before anything a reader would call rarefied.

Tested in Where a fluid stops being one, at the figure it turns on.

The slot in a slotted flap works by ducting high-energy air from underneath the wing up through the gap, where it is blown into the boundary layer on the flap and re-energises it so that the flow stays attached.

What decides it: The whole lift increment is computed here in a flow that has no boundary layer anywhere in it — no viscosity, no wake, no mixing region and no energy to transfer. A two-body panel solve at four degrees with the flap at thirty gives a system lift coefficient of 2.76 against 0.70 for the main element alone, and the main element is carrying 3.98 times the circulation it carries by itself. That increment cannot be a boundary-layer effect, because there is no boundary layer for it to be an effect on. What the slot does do is let two elements each satisfy their own Kutta condition while sitting in each other's velocity fields, and the resulting interference is an inviscid one.

Tested in A slot is not a nozzle, at the figure it turns on.

A wing whose motion is slow compared with the time air takes to pass it — a reduced frequency well below one — can be treated as quasi-steady, carrying at each instant the lift its instantaneous angle of attack would give in steady flow.

What decides it: Theodorsen's function is the exact transfer function for that assumption in linear theory, and it is C = 1 only in the limit. Its magnitude is one per cent below one at k = 0.0061 and fifteen per cent below at k = 0.1, which nobody would hesitate to call quasi-steady; its phase reaches one degree of lag at k = 0.003. The reduced frequency at which the two terms the group compares are equal is 1.086, so the accuracy threshold is 178 times lower than the balance the number is named for.

Tested in Slow enough to be steady, at the figure it turns on.

A jet hitting a plate at an angle pushes it along as well as into it — the water is sliding across the surface, so it must drag the plate in the direction it is going.

What decides it: An inviscid fluid has no mechanism to exert a force along a surface: pressure acts normal to it by definition, and there is nothing else. The momentum sum resolved along the plate is therefore zero, and it is zero to the last bit of the arithmetic at every angle from grazing to square on, because that statement is what the mass split was DERIVED from rather than a consequence checked afterwards. The dragging force a real plate feels is skin friction in the two sheets, which is a boundary-layer effect of order one over the square root of the Reynolds number and is not in this model at all.

Tested in What a jet cannot push sideways, at the figure it turns on.

A flow either has stagnation points or it does not, and a picture of the streamlines shows what the fluid is doing.

What decides it: The same ideal flow past a cylinder has exactly two stagnation points measured in the tunnel frame and none anywhere measured in the frame of the undisturbed air — the surface speed there is the free-stream speed at every angle, to two parts in a thousand million. The two fields differ by one constant vector, every force and every pressure is identical, and neither picture is the flow. What the fluid is doing that both observers agree about is the strain rate, the pathlines and the pressure; what belongs to the camera is the streamline pattern.

Tested in The picture belongs to whoever is watching, at the figure it turns on.

Better measurements and bigger computers will eventually make turbulent flows predictable.

What decides it: The horizon is logarithmic in the initial error, exactly: T = ln(tolerance/error)/lambda. So a tenfold better measurement buys ln(10)/lambda more time — the same increment every time, with no accumulation. For the flow computed here that is 24.5 time units per decade, so making the forecast twice as long needs the initial state nine hundred times better and making it ten times as long needs a factor with twenty-six zeroes. There is no amount of measurement that changes the exchange rate, because the exchange rate is the flow's own stretching.

Tested in An hour for every tenfold, at the figure it turns on.

A water turbine cannot be a hundred per cent efficient even in principle, because the water has to keep moving to get out of the machine and it carries kinetic energy away with it.

What decides it: The water has to leave the BUCKET, which is not the same as having to move in the ground frame. At a bucket speed of half the jet speed, a bucket that reverses the flow completely sends the water back at V/2 relative to itself while itself travelling forward at V/2 — so the absolute exit velocity is exactly zero, computed here as 6e-17 of the jet speed, and the shaft has taken every joule the jet carried. It falls out of the bucket dead. The reasons a real Pelton wheel reaches 90 per cent rather than 100 are the bucket's 165-degree turn, friction on the bucket surface, and windage — each of which is a defect of the machine rather than a limit on the idea.

Tested in Half the jet speed takes everything, at the figure it turns on.

Stokes' law holds while the Reynolds number is small, and since the Reynolds number compares inertia with viscosity, "small" means comfortably below one.

What decides it: Oseen's correction gives C_D = (24/Re)(1 + 3Re/16), so the relative error in Stokes' law is 3Re/16 over one plus that. Inverting exactly: one per cent at Re = 16/297 = 0.0539, five per cent at 0.281, and at Re = 1 — where the two terms the Reynolds number compares are exactly equal — Stokes' law is 15.8 per cent low. The threshold is a factor of nineteen below the value the number is named for.

Tested in How small is small enough, at the figure it turns on.

A swept wing works because the air meets it at an angle, so the wing effectively sees a slower airflow and can fly faster before shock waves form.

What decides it: The wing does not see a slower airflow. For an infinite yawed wing the equations of motion split exactly: the flow in the plane normal to the leading edge is the same two-dimensional problem it would be at no sweep at all, driven by the normal component of the free stream, and the spanwise velocity is a passive scalar carried along by it. Checked here in the form that has no wriggle room: on a flat plate the chordwise and spanwise boundary-layer profiles are the same function of the similarity variable to two parts in a hundred million, computed by code that shares nothing — one a shooting solution of a nonlinear third-order equation, the other a single pass through a linear second-order one. A swept flat plate therefore has no crossflow at any sweep angle whatever. The section feels exactly the wind it would feel unswept, at U cos Λ, and carries a spanwise flow it never learns about.

Tested in The wind a swept wing feels, at the figure it turns on.

A measurement of the flow at a distance from a body, made accurately enough, determines the body — since the flow is the solution of a well-posed problem with the body as its boundary.

What decides it: It determines the body's first few multipoles and nothing else. A circular cylinder and a Rankine oval 1.17 radii long are built here with the same doublet strength; the largest difference between their velocity fields falls as the −4.04 power of the radius while the disturbance either makes falls as r⁻², so at six radii the two flows differ by one per cent of the disturbance and at forty-eight by 0.016 per cent. No measurement outside a few radii distinguishes them, and no measurement at infinity distinguishes any two bodies with the same three leading coefficients.

Tested in What the far field remembers, at the figure it turns on.

The vorticity distribution determines the flow, so computing the vorticity everywhere is computing the flow.

What decides it: A Rankine vortex and the same vortex with a uniform stream added have vorticity fields equal at every point to eleven decimal places, and velocity fields differing by six-tenths of the core's peak speed. Every local measurement of vorticity, divergence and rate of strain is identical in the two. The vorticity fixes the flow only up to a field carrying neither vorticity nor divergence, and which of those the fluid is doing is settled at the boundary — which is why a vortex method needs boundary conditions and not merely a vorticity field.

Tested in Every flow is two flows, at the figure it turns on.

A model that reproduces the measured lift is a model of lift.

What decides it: Newtonian impact theory with its constant tuned at five degrees reproduces a NACA 2412's lift coefficient exactly at five degrees — to the last bit of double precision, on a mechanism this collection has already shown to be false. It is out by 72 per cent at two degrees and 181 at twelve, its lift-curve slope at the tuning point is 2.8 times too steep, and it gives exactly no lift at zero incidence where the section gives 0.23.

Tested in A right total from a wrong picture, at the figure it turns on.

Refining a grid always improves a calculation, so a first cell placed closer to the wall than the guidance suggests is at worst wasteful.

What decides it: A wall function assumes its point lies in the logarithmic region, so moving the point below that region invalidates the assumption rather than resolving anything. Measured against the closure that supplies the function, a first cell at y⁺ = 30 gives the friction to two per cent, at y⁺ = 10 to eighteen per cent, and at y⁺ = 1 to sixty-four — the error growing monotonically as the grid is refined. The remedy is not a finer grid but a different wall treatment, and choosing between them requires knowing a y⁺ that is not known until after the calculation has run.

Tested in What a code says to a wall, at the figure it turns on.

A flow that oscillates symmetrically about zero can transport nothing on average, since every forward half-cycle is cancelled by a backward one.

What decides it: The cancellation holds for the velocity and fails for its own nonlinear term. Inside an oscillatory boundary layer the streamwise fluctuation and the transverse velocity it drives are correlated, so ⟨u∂u/∂x + v∂u/∂y⟩ is not zero anywhere in the layer; integrating that forcing twice across the layer, with no slip at the wall and no stress at its edge, leaves a steady velocity of −0.74999953 U dU/dx over ω. Round a cylinder that slip changes sign four times and drives four permanent recirculating cells.

Tested in An oscillation with somewhere to go, at the figure it turns on.

An orifice plate wastes pressure because the flow through the hole is turbulent and the eddies downstream dissipate energy — it is a viscous loss, and a smoother plate or a cleaner fluid would waste less.

What decides it: The permanent loss is computed here with no viscosity anywhere in the calculation and comes out at 73.5 per cent of the reading at a diameter ratio of one half. It is the Borda–Carnot head of the jet expanding suddenly back into the pipe — the exact gap between what the momentum balance allows and what the energy balance would permit — and it is the same expression, to the last decimal place, that a sudden enlargement in a plain pipe produces. The eddies dissipate the energy but they do not decide how much: that is fixed before any eddy exists, by the fact that a jet expanding into a larger area cannot recover its own dynamic pressure. Polishing the plate changes nothing.

Tested in The price of knowing the flow rate, at the figure it turns on.

A drop is held in shape by surface tension until its weight overcomes it, so drops smaller than the capillary length — Bond number below one — are spheres and larger ones are not.

What decides it: Integrating Young–Laplace along the meridian gives an aspect ratio that departs from one by a per cent at Bo = 0.0079, five per cent at 0.054 and ten at 0.134. At Bo = 1, where weight and tension are exactly comparable, the drop is already half as tall as it is wide. For water the one-per-cent threshold is a drop 0.49 mm across, whose Bond number is a hundred and twenty-six times below the one the folklore quotes.

Tested in The size a drop is allowed, at the figure it turns on.

A jet is quiet compared with a loudspeaker of the same size because the acoustic sources in it are weak.

What decides it: They are not weak; they cancel. Summing four point sources of alternating sign as exact spherical waves — with no expansion in the compactness anywhere — gives a field that is 80 per cent of a single source's inside the cluster and 9.3·10⁻⁵ of it at one radian of wavelength, for kd = 0.01. The near field is enormous and almost entirely reactive; what escapes is the residue of a cancellation, and each order of cancellation costs exactly one power of kd — measured here as 1.000 and 1.999 for a dipole and a quadrupole.

Tested in The sound is what does not cancel, at the figure it turns on.

An elliptical planform is the best wing shape. It has the lowest induced drag of any wing, so a designer who could afford to build one would.

What decides it: It has the lowest induced drag, and that is checked here as a comparison rather than quoted: three planforms are brought to the same lift and none beats the elliptic one. But elliptic loading gives every section on the wing the same lift coefficient, so every section reaches its maximum at the same instant — the wing stalls from root to tip simultaneously, with no warning and with the ailerons going at the same moment as everything else. That is the worst possible stall behaviour, and it is a direct consequence of the property that makes the planform optimal. The four per cent of induced drag a 0.4-taper wing gives away buys a stall that starts inboard and can be felt.

Tested in Which part of a wing stalls first, at the figure it turns on.

Whether a quantity inside a region is increasing or decreasing is a fact about the flow.

What decides it: Three control volumes of identical size, drawn at the same point of the same flow at the same instant and moved at three different velocities, give rates of −0.356, −0.540 and +0.157 for the same dye. Each reconciles with the flux of the relative velocity across its own boundary to two parts in a hundred million, and all three reconcile with one material rate of zero. The sign of the answer is a property of the box.

Tested in A rate of change that will not hold still, at the figure it turns on.

Wind-tunnel wall corrections are empirical adjustments — rules of thumb that tunnel operators apply from experience, and a source of uncertainty that better instrumentation would remove.

What decides it: For a body on the centreline of a two-dimensional tunnel the correction is exact. Two parallel walls are an infinite row of image doublets at spacing 2h, and the speed-up they produce at the model is computed here twice: by summing the row term by term, where it is a zeta function of two in disguise and converges like one over N, and by summing it in closed form, where an infinite line of doublets has a cotangent for a potential and no summation appears at all. Both give (pi squared over 12) times the square of the size ratio, agreeing to five figures. What is genuinely uncertain in a real tunnel is the WAKE blockage, which needs the drag that is being measured, and the model's own shape distortion — which is fourth order and which no standard correction touches at all.

Tested in The instrument in the answer, at the figure it turns on.

Elliptical span loading is the minimum-induced-drag distribution, so it is the loading a wing should aim at.

What decides it: Elliptical loading minimises induced drag at a given lift on a given span, which is the constraint a wing does not have. A wing is limited by the bending moment its structure can carry, and the loading that minimises drag at given lift and given root bending moment is a different curve. Computed here by minimising the drag quadratic form under two constraints rather than one: the bell loading has a span efficiency of exactly three quarters and a root bending moment of exactly four fifths of the elliptic wing's, so at five quarters of the span it carries the same lift for the same root moment at 64/75 of the induced drag — a saving of 14.7 per cent over the loading it is supposedly worse than.

Tested in The loading nobody used, at the figure it turns on.

The two-thirds law's constant, about 4.02, is an experimental number of the same kind as the von Kármán constant — a measurement that theory has not been able to produce.

What decides it: It is a gamma function. With a one-dimensional spectrum C ε^(2/3) k^(−5/3) held over an unbounded inertial range, the second-order structure function is 2C(εr)^(2/3) times the integral of x^(−5/3)(1 − cos x), and that integral is −Γ(−2/3)cos(π/3) exactly. Computed here by quadrature it is 2.009205 against the closed form's 2.009204, so the constant is −Γ(−2/3) = 4.018408. What is measured rather than derived is C itself, which is a different number and is genuinely empirical.

Tested in The one exact result, at the figure it turns on.

Viscosity destroys energy wherever a fluid is moving fast relative to its surroundings, so the fastest part of a flow is where the losses are.

What decides it: The dissipation function contains the rate of strain and not the velocity. The fluid on the centreline of a pipe is the fastest thing in it and is dissipating exactly nothing, because it is not being deformed; every joule the pipe loses is made at the wall, where the fluid is barely moving. Half of a Poiseuille flow's dissipation is made in the outer 29 per cent of the gap.

Tested in The price of a gradient, at the figure it turns on.

The rate of strain says how fast a fluid is being pulled apart, so two flows with the same strain rate stretch material lines equally fast.

What decides it: Pure strain at α and simple shear at γ = 2α have identical rate-of-strain magnitudes. At αt = 5 a material line has grown by 148 in the first and 10.1 in the second, and the ratio grows without bound afterwards. The strain rate says how fast a line aligned with the extensional axis is stretching; whether a line can stay aligned is decided by the rotation, and simple shear rotates the material out of alignment exactly as fast as it stretches it.

Tested in Longer, with nothing pulling it, at the figure it turns on.

A flame is a slow combustion wave and a detonation is a fast one, so the same theory gives both speeds.

What decides it: The Chapman–Jouguet detonation speed follows from the conservation laws and a tangency condition, needs nothing about the chemistry beyond how much heat is released, and is checked here against a closed form to fifteen digits. The deflagration branch has no such selector: every point on it satisfies the same three laws, from a wave at almost the speed of sound down to one at almost nothing. A real laminar flame runs at half a metre a second — 218 times slower than the slowest speed conservation rules out — and what puts it there is diffusion.

Tested in The other branch of the same curve, at the figure it turns on.

A falling raindrop is teardrop-shaped, drawn out to a point at the back by the air rushing past it.

What decides it: Matching the Legendre component of a sphere's own potential-flow pressure distribution against the curvature change it produces gives a distortion of −3We/32 in the second Legendre mode: the poles come in and the equator goes out. A falling drop is therefore OBLATE — a bun, flattened along the direction of travel and fore-and-aft symmetric — and at the Weber numbers real raindrops reach it is measurably so. A teardrop has a fore-and-aft asymmetry that nothing in the pressure field supplies.

Tested in The drop that is not a tear, at the figure it turns on.

A winglet works by blocking the high-pressure air underneath the wing from escaping round the tip, which stops the tip vortex forming and so removes the drag it causes.

What decides it: The whole calculation here is Munk's theorem, which says the induced drag depends only on the circulation in one cross-section of the wake far downstream. That cross-section contains no tip, no leaking air and no escape route — it is a curve in a plane. A winglet of fifteen per cent of the semi-span reduces the induced drag to 0.793 of the flat wing's at the same span and the same lift, and the calculation that produces that number cannot represent blocking, because there is nothing in it to block. What has changed is that the wake reaches further from the flight path in a direction other than sideways, and the winglet is a lifting surface carrying its own circulation, which the figure shows going right round the corner.

Tested in A wing that leaves the plane, at the figure it turns on.

The pressure surge when a valve shuts is the water's momentum being converted to pressure, so it depends on how long the pipe is: a longer pipe holds more water, more momentum, and therefore a bigger bang.

What decides it: Joukowsky's pressure is rho times the wave speed times the velocity change, and no length appears in it. The control volume that gives it rides the wavefront, so it contains only the fluid the wave has reached so far, and it closes whatever the pipe behind it is doing — the method of characteristics run here on a 600 m pipe returns a head rise agreeing with rho a delta V to fifteen figures without being told the formula. Length decides two other things: how LONG the pressure lasts, which is 2L over a, and whether closing the valve more slowly helps at all, which it does not until the closure outlasts that return time.

Tested in Stopping water costs more than moving it, at the figure it turns on.

An inviscid theory cannot produce drag, so the drag of a bluff body is a purely viscous effect and needs the Navier–Stokes equations to compute at all.

What decides it: Kirchhoff's free-streamline solution is a solution of Laplace's equation with no viscosity anywhere in it, and the pressure integral over the front of a normal flat plate gives C_D = 2π/(4+π) = 0.8798 — computed here two ways, by quadrature and in closed form, agreeing to machine precision. Feeding in the measured base pressure instead of the theory's assumed one gives 1.94 against a measurement of about 1.9. What viscosity supplies is not the drag; it is the pressure in the wake.

Tested in Drag in the theory that forbids it, at the figure it turns on.

A vortex is a region of high vorticity, so contouring the vorticity shows where the vortices are.

What decides it: A parallel shear layer has vorticity peaking at its centre and Q negative at every point of it — the strain is exactly as large as the rotation at every point of a parallel shear. A vorticity threshold calls the whole layer a vortex; the criterion built on whether the local flow pattern closes on itself calls none of it one. For a Lamb–Oseen vortex the two disagree at a definite radius, and the enclosed circulation runs from ten per cent to ninety-nine as the vorticity threshold is moved through values nobody can choose between.

Tested in Where a vortex stops, at the figure it turns on.

Which reflection a shock makes is decided by the flow conditions, so the same wedge angle at the same Mach number always gives the same picture.

What decides it: At Mach 4 the three-shock solution first exists at 20.85° and the two-shock solution stops existing at 25.61°. Between them both satisfy the conservation laws, both are stable, and the equations admit two answers. Raising the wedge angle through the domain keeps a regular reflection to 25.61°; lowering it keeps a Mach reflection down to 20.85°. The conditions do not decide; the history does, and the domain is 9.5° wide at Mach 7.

Tested in When a shock cannot bounce, at the figure it turns on.

A wing has to be curved on top: the shape is what makes the lift, and a flat or symmetric section could not fly.

What decides it: A flat plate has upper and lower surfaces of identically equal length — the excess is zero as an identity — and a lift-curve slope of exactly 2 pi per radian. A symmetric section twelve per cent thick has two surfaces that are the same curve reflected and reaches a lift coefficient of 0.4 at 3.65 degrees. A four per cent cambered section flown inverted, with its longer surface underneath, makes the same 0.4 at 7.80 degrees. And a section with a wavy upper skin has twice the path-length excess of a cambered one and makes half the lift.

Tested in The wing that is flat, and flies, at the figure it turns on.

As the viscosity of a fluid is reduced, the rate at which it dissipates energy falls in proportion, so a fluid of very small viscosity dissipates very little.

What decides it: The dissipation is fixed by the large scales at about u³/L and contains no viscosity at all. What the viscosity decides is the scale at which the dissipating happens: the Kolmogorov length falls as Re^(−3/4) and the velocity gradient there rises as Re^(1/2), so ν(u_η/η)² returns the same ε at every viscosity — computed here across six decades and constant to the last bit. The consequence is the one d'Alembert's paradox has met from the other side: a drag coefficient that flattens rather than falling as the Reynolds number rises.

Tested in The limit that is not the value, at the figure it turns on.

A hydraulic ram gets something for nothing — it needs no fuel and no electricity, so the energy to lift the water must be coming from the pipe's own pressure, and a well-designed one could in principle deliver most of its supply to any height.

What decides it: Every drop that arrives at a ram has fallen through h and every drop it delivers is lifted through H, so the delivered fraction can never exceed h/H. That is conservation of energy with every loss set to zero, it holds whatever the valve timing, beat rate, air-vessel size or drive pipe length, and the solver refuses an operating point above it rather than returning an efficiency above one. A ram working across a head ratio of ten cannot send more than a tenth of its supply, and a real one at 65 per cent efficiency sends 6.5 per cent. Nothing is free: the other 93.5 per cent falls through h and leaves at the waste valve, having done the lifting.

Tested in A pump with no engine, at the figure it turns on.

A biplane's two wings interfere with each other and each one spoils the other's flow, so a biplane pays an aerodynamic penalty which it accepts in exchange for a lighter, braced structure.

What decides it: The interference is real and it is a saving. Two wings of equal span carrying half the lift each, with a gap of a fifth of the span, cost 0.742 of what a single wing of that span costs carrying the whole lift — because their wake is not in one plane, and by Munk's theorem the shape of the wake in its own cross-section is the only thing the induced drag knows about the arrangement. The interference factor read out of that ratio is 0.484, against 0.485 in Prandtl's published chart, computed here by a method that was never fitted to it. The structural case for a biplane is the famous one; the aerodynamic case is that a non-planar system beats a planar one of the same span.

Tested in Two wings and it does not matter where, at the figure it turns on.

An incompressible flow is one of a fluid whose density does not change, so a flow with a varying density is compressible.

What decides it: Ideal flow past a cylinder carrying a density that is constant along each streamline has a divergence of 2.6 × 10⁻¹⁰ — which is the differencing — and a material rate of change of density of 3.2 × 10⁻¹⁰, while the density itself varies by a factor of three across the field. Continuity says ∇·u = −(1/ρ)Dρ/Dt, so incompressible means each parcel keeps its own density, and says nothing whatever about whether two parcels have the same one.

Tested in Incompressible is not a property of the fluid, at the figure it turns on.

An inviscid flow that starts irrotational stays irrotational, so a shock cannot make vorticity.

What decides it: Crocco's theorem is T∇s = ∇h₀ − u × ω, and stagnation enthalpy is conserved across any shock. A curved shock gives every streamline a different entropy rise, so ∇s is non-zero with ∇h₀ zero and the identity forces u × ω to be non-zero. Computed for a parabolic bow shock at Mach 6, the vorticity just behind it peaks at 1,894 per second and scales as the inverse of the nose radius with a fitted exponent of −1.00000. A straight shock from a wedge gives every streamline the same entropy and makes exactly none.

Tested in The spin a shock leaves behind, at the figure it turns on.

Birds fly in a V so that each one can shelter in the slipstream of the bird in front, the way a cyclist drafts. The bird behind is in calmer air and has less drag to overcome.

What decides it: A wake behind a lifting wing is not calmer air and there is no useful drafting in it — the region directly behind a wing is a downwash, which makes a following wing's job harder rather than easier. What is beneficial is beside the tip, where the wake induces an upwash, and the saving is an induced-drag saving rather than a shelter from a wake. The arithmetic here says by how much: two wings with their tips touching pay exactly half of what the two pay flying alone, computed to six digits, because the pair is then one wing of twice the span carrying twice the lift, and induced drag goes as the square of span. Pull them apart and the saving decays; a bird in trail rather than beside would gain nothing at all.

Tested in The lift beside a wing, at the figure it turns on.

A particle whose density matches the fluid's is a perfect tracer, and a bubble — being much lighter than the liquid around it — follows the flow even better.

What decides it: The relaxation time that includes the added mass and the pressure gradient of the undisturbed flow is (2ρ_p + ρ_f)d²/36μ, which for a neutrally buoyant particle is exactly three halves of the usual formula's and for a bubble is d²/36ν where the usual formula gives zero. The radial drift in a vortex goes as (1 − β)τ with β = 3ρ_f/(2ρ_p + ρ_f), which is zero only at β = 1 — so a neutrally buoyant particle is the perfect tracer and a bubble is the least perfect one there is, drifting inward at the same rate a heavy droplet drifts out.

Tested in The tracer that is not one, at the figure it turns on.

A flow photograph shows the flow; the exposure time is a property of the camera rather than of the measurement, and a longer one simply gives a cleaner picture of the same thing.

What decides it: Averaging a Kármán street over exactly one shedding period returns the infinite-time mean to 2.8·10⁻¹⁴. Averaging over 1.42 periods leaves a residue of 5.1·10⁻², and the residue between the zeros falls only as one over the exposure — a factor of 3.8 across four more periods. The resulting long-exposure profile is smooth and symmetric and no instantaneous profile resembles it.

Tested in The shutter is part of the answer, at the figure it turns on.

The critical Rayleigh number is a property of a fluid layer heated from below, so a single number describes the onset of convection.

What decides it: Three numbers describe it and the difference between them is a factor of 2.6, produced by nothing but the boundary conditions. Two free surfaces give 657.408 against the exact 27π⁴/4 = 657.511; one rigid and one free gives 1100.697; two rigid walls give 1707.757. The fluid, the depth and the temperature difference are identical in all three, and only what the surfaces are permitted to do has changed.

Tested in The threshold the walls decide, at the figure it turns on.

A pulsating pipe flow keeps its parabolic profile while the Womersley number is below one, since below that the viscous term dominates the unsteady one.

What decides it: Sweeping Womersley's exact solution in frequency, the flow amplitude falls one per cent below the quasi-steady Poiseuille value at α = 0.911 — which is very nearly the value the number is named for, and makes this the best-behaved threshold on the site. The phase does not follow it: the lag reaches one degree at α = 0.324 and one per cent of a right angle at 0.307, three times lower. At α = 1 the flow is 1.4 per cent short and 9.5 degrees late, and a calculation that checks only the amplitude passes while being a tenth of a radian out.

Tested in Where the parabola goes, at the figure it turns on.

The vortices behind an aircraft come off the wingtips, so they are a wingtip's span apart — and a device on the tip is what decides how strong they are.

What decides it: They are inboard of the tips, always, and the amount is computable before anything rolls up. Vorticity is shed all along the span in proportion to the rate at which the circulation changes, and the rolled-up core sits at the centroid of what each half shed. For elliptic loading that centroid is at πb/8 from the centreline, so the pair ends up πb/4 = 0.7854 of the span apart — checked here against the closed form and computed by quadrature for loadings that have no closed form. A bell loading, which sheds further inboard, rolls up to 0.586 of the span; a nearly rectangular loading, which sheds almost everything at the tips, to 0.978. The spacing is a property of the loading and not of the tip.

Tested in Where the wake ends up, at the figure it turns on.

Inviscid flow is potential flow: with no viscosity there is nothing to generate vorticity, so an inviscid flow has none.

What decides it: Kelvin's theorem says vorticity is not generated in an inviscid barotropic flow with conservative body forces; it says nothing about vorticity that was already there. A uniform shear approaching a cylinder carries vorticity −K, and the exact steady solution ψ = U(r − a²/r)sinθ + Kr²/4 − (K/4)(r² − a⁴/r²)cos2θ is inviscid, rotational, and produces a lift of 2πρUKa² on a body with no circulation round it. The drag is still zero.

Tested in Inviscid does not mean irrotational, at the figure it turns on.

A disturbance in a flow is carried along by the flow, so it travels at the speed of the fluid.

What decides it: For a conserved quantity whose flux depends on its own density, a disturbance travels at dq/dc — the slope of the flux curve — while the material travels at q/c, the slope of the chord from the origin. For a wide river under Manning's formula the ratio is exactly 5/3 at every depth, so a crest arrives 3.58 units of distance ahead of the water that raised it. For traffic the wave speed changes sign at half the jam density while no vehicle ever goes backwards.

Tested in A wave nothing in it travels with, at the figure it turns on.

A turbulence or sea-state specification is complete when the spectrum is given: two records with the same spectrum are statistically the same, so they load a structure the same way.

What decides it: Two 600-second records are synthesised from one amplitude spectrum with different phases. Their variances agree to 3·10⁻¹⁵ and their autocorrelations to 2·10⁻³. Their crest factors are 2.83 and 3.11, their peak quadratic drag loads differ by 20 per cent, and the range of their running integrals by 29. A third record with the same spectrum and every phase set to zero is a single impulse with 31 times the peak load.

Tested in The same statistics, and a different load, at the figure it turns on.

The velocity in Darcy's law is how fast the fluid moves through the pores, so dividing the distance by it gives the time a tracer takes to cross the bed.

What decides it: It is the flow rate divided by the WHOLE cross-section, solid included, and no fluid particle anywhere in the bed has that speed. The fluid occupies only the fraction epsilon of that area, so its average speed is larger by exactly 1/epsilon — two and a half times for a typical packing — and a residence time computed from the superficial velocity is wrong by that factor before any other consideration. The two speeds differ by the porosity and by nothing else, which makes it the easiest correction in fluid mechanics to apply and the one most often dropped.

Tested in A velocity nobody has, at the figure it turns on.

A stratified shear flow is stable when its Richardson number exceeds a quarter and unstable when it does not.

What decides it: Two things are wrong with it. The theorem runs one way only — Ri > 1/4 everywhere is sufficient for stability, and Ri < 1/4 somewhere does not imply instability. And "its Richardson number" is not one quantity: on a tanh shear layer with a matching density profile, the minimum gradient value, three bulk conventions and a depth average span a factor of 17.3, and the last of them grows without limit as the measurement domain is widened. The quarter belongs to the first, which is local, and to none of the others.

Tested in Five numbers, one name, at the figure it turns on.

Flutter happens when the airflow excites the wing at its natural frequency. It is a resonance, like a wine glass and a singer, and the cure is to keep the excitation away from the structure's own frequencies.

What decides it: A steady airstream oscillates at no frequency at all, so there is nothing to be resonant with. What the computation shows instead is that the two structural frequencies are driven together by the aerodynamic coupling — a factor of 2.30 apart at rest and 1.88 apart at the flutter speed — and that the flutter frequency, 95.6 radians per second, is neither of the two the structure started with. The energy route confirms it: the work the air does over a cycle is exactly proportional to the sine of the phase between pitch and plunge, is exactly zero in phase and in antiphase however large the motion, and changes sign at the same speed the eigenvalues cross to three decimal places. Keeping the two natural frequencies apart is in fact a real cure, and it works by making them harder to coalesce rather than by avoiding an excitation that does not exist.

Tested in The shake that is not resonance, at the figure it turns on.

A thinner lubricating film runs cooler, because there is less oil being sheared and the metal is closer on both sides.

What decides it: The peak temperature rise in a sheared film is μU²/8k, which contains the viscosity, the sliding speed and the conductivity and contains no gap at all. Halving the film doubles the shear rate, quadruples the dissipation per unit volume, and halves both the volume being heated and the distance the heat has to travel — and the four factors cancel exactly. A one micron film and a one millimetre film at the same sliding speed reach the same temperature.

Tested in The film that heats itself, at the figure it turns on.

The Coriolis effect cannot steer a draining bath, so the experiments that showed a hemisphere dependence in a tank must have been measuring something else.

What decides it: Both hold, and the number that separates them is how many turns the Coriolis contribution completes before the vessel empties. A bathtub's Coriolis velocity at the drain gives a rotation period of 570 seconds and the tub empties in 56, so it gets through 0.098 of a revolution. A 1.8 metre tank draining through a 4.8 mm hole has a period of 3.5 seconds and takes 108 minutes to empty: 1,873 revolutions. The remaining requirement is that the swirl left by filling has decayed below the Coriolis velocity, which takes 3.9 hours.

Tested in The bath that was only ever a wait, at the figure it turns on.

A quantity that is independent of the Reynolds number in the high-Reynolds-number limit can be treated as independent of it at any Reynolds number large enough to be called high.

What decides it: Two cases from this collection's own solvers. The Blasius profile's shape contains no Reynolds number at all, so its limit is reached at every Reynolds number — that is complete similarity. The overlap layer's best power-law exponent falls from 0.131 at Re_τ = 2,000 to 0.102 at 10⁶, and the sequence fits a/ln Re_τ: driving it below a hundredth needs a Reynolds number with 114 in its logarithm on the mildest fit and 3,270 on the fit the data actually prefer. The limit exists and nothing ever gets there.

Tested in A limit nothing reaches, at the figure it turns on.

The downwash at the tailplane is twice the downwash at the wing, so the tail's effective incidence falls at twice the induced rate. That is the standard factor and it is what tail sizing uses.

What decides it: Two is the value at infinity, and the field approaches it from above rather than from below. Computed here by Biot–Savart over eighty horseshoes: the induced angle at the lifting line is 1/πAR to within a few per cent, far downstream it is exactly twice that, and at two and a half chords behind — which on a wing of aspect ratio eight is a third of a span, and is where a tailplane actually is — it is 2.46 times the value at the wing. The standard formula therefore understates the downwash at a real tail by about a quarter. Raise the tail a tenth of a span out of the wake and the same station reads 1.95, which is below the far-field figure — so the height matters as much as the distance, and neither appears in the formula.

Tested in The surface in the wake, at the figure it turns on.

Darcy's law fails at high flow rates because the flow in the pores becomes turbulent.

What decides it: The departure begins at a pore Reynolds number of about ten and the two terms of Ergun's equation are equal at 85.7, both of which are far below any transition to turbulence in a passage — a pipe holds its laminar solution to Re 2300. What ends the linear law is INERTIA rather than turbulence: the fluid is accelerated into each pore and decelerated out of it, and that costs a pressure proportional to the square of the velocity with no viscosity in it at all. The second term of Ergun's correlation contains no viscosity, which is the evidence. Real pore-scale turbulence arrives much later and adds nothing qualitatively new to the pressure drop.

Tested in Where Darcy stops, at the figure it turns on.

A plate pulled out of a liquid twice as fast comes out twice as wet, because it spends half as long draining.

What decides it: The film thickness goes as the two-thirds power of the withdrawal speed, so doubling the speed thickens it by 58.7 per cent and not by 100. There is no regime in which it is proportional: the exponent comes out of matching a flat film to a static meniscus and is the same two-thirds for a plate, for a fibre and for a bubble in a tube, which have nothing else in common.

Tested in What a plate takes with it, at the figure it turns on.

The cellular flow ψ = A sin kx sin ky is the standard model of a recirculating region, and being an exact steady solution of the inviscid equations it is the flow a fluid in such a region approaches as the viscosity becomes small.

What decides it: Integrating the steady vorticity equation over the region inside a closed streamline gives F′(ψ)∮|∇ψ|dl = 0 for any non-zero viscosity, however small. The contour integral is measured here at 5.48 on a streamline of length 6.89 and is positive on every closed streamline, so F′ = 0 and the vorticity must be uniform. The cellular flow has F′ = 2k², so it is an exact solution of the inviscid equations and the limit of no viscous flow whatever.

Tested in The vorticity nothing decides, at the figure it turns on.

A rough wall changes the turbulence, so the logarithmic law's slope — the von Karman constant — is different on a rough surface.

What decides it: It cannot be. The argument that produces the logarithm forbids any length from entering the overlap region, so it forbids the roughness height as surely as it forbids the viscous length, and the slope is whatever it was. What roughness can change is the intercept, and that is the whole of its effect: the profiles at roughness Reynolds numbers of 0, 20, 100 and 400 differ by 11.1 wall units of velocity and by nothing else, and their velocity defects are identical.

Tested in A second length at the wall, at the figure it turns on.

Helicity measures how twisted a flow is, so a flow with more swirl in it has more helicity.

What decides it: It measures how linked it is, which is a different quantity and an integer one. Two rings brought arbitrarily close together have zero helicity if they are not linked, and two linked rings pulled arbitrarily far apart have twice the product of their circulations. The Gauss integral returns 0, 0, 1, −1, 1 and 0 on six configurations to 1.6 parts in ten thousand, and the pair that looks most linked in a drawing — a unit ring through a ring of radius 3 — has a linking number of exactly zero.

Tested in The knot a flow cannot untie, at the figure it turns on.

The pressure drop across a bed keeps rising as the flow through it is increased — faster and faster, since the resistance is quadratic at high flow.

What decides it: It stops, and where it stops is exact. A control volume round the bed says the pressure drop cannot exceed the buoyant weight of the solid it contains per unit area, because there is nothing else for the force to be carried by once the walls have stopped taking any. Past that point the bed expands rather than resisting harder and the drop is CONSTANT — flat on a gauge, over a factor of six in velocity in the computation here. The rising branch is a correlation; the flat one is a weight and an area and nothing else.

Tested in The bed that weighs itself, at the figure it turns on.

Decaying turbulence forgets its initial conditions, so the decay exponent is universal.

What decides it: It forgets almost all of them and never forgets one. The exponent is 2p/(p+2), where the conserved large-scale quantity is the energy times the integral scale to the power p, and p is decided by the shape of the spectrum at the smallest wavenumbers — a k² spectrum gives p = 3 and an exponent of 6/5, a k⁴ spectrum gives p = 5 and 10/7. Two boxes with the same energy, the same integral scale and the same spectrum everywhere except the first decade below the box size are 3.63 times apart after three decades of time and further apart for ever after.

Tested in What decay never forgets, at the figure it turns on.

A suspension is thicker than its liquid because the particles get in each other's way, so the correction should be small until they are close enough to touch.

What decides it: Einstein's coefficient is computed for a single sphere in an unbounded fluid, with no neighbours anywhere, and it is already 2.5 times the volume fraction. What thickens the mixture is what one particle does to the flow around it — a disturbance decaying as 1/r² — and the departure from the dilute law begins at about five per cent by volume, where the mean spacing is still nearly three diameters.

Tested in A viscosity made of particles, at the figure it turns on.

Slender-body theory gives the wave drag of a body from its area distribution, so it can be applied to any slender shape.

What decides it: The integration by parts producing the drag integral discards a boundary term that vanishes only for a body that closes. Applied to a Sears–Haack forebody cut off square at the same length and volume, the routine returns 4.28 × 10⁻³ — which is *less* than the Sears–Haack body's 8.15 × 10⁻³, and a reader shown that number would conclude that the way to reduce wave drag is to cut the tail off. The theory has to refuse it, and this site's does.

Tested in The least drag a volume can have, at the figure it turns on.

A flow picture that looks convincing has been integrated well enough.

What decides it: Two integrations of the same incompressible flow, at the same step size and to the same time, produce patches of dye that are indistinguishable as pictures. One of them has an exactly unimodular tangent map at every step size, because the implicit midpoint rule's Cayley transform is exactly unimodular for a traceless matrix; the other's determinant is wrong by six parts in a hundred thousand and gets worse the coarser the step. Nothing about the drawings says which is which, and the only thing that does is a conserved quantity carried alongside.

Tested in The area that must not move, at the figure it turns on.

An aeroplane flying through gusts experiences, on average, the lift its mean angle of attack would give it, since the gusts are as often upward as downward.

What decides it: Only where the lift curve is straight. Averaging a stated lift curve by quadrature over a normal gust distribution of three degrees about a mean of ten gives a mean coefficient of 0.795 against the 0.837 the mean angle promises — a five per cent deficit, with nothing stalled and no gust taking the wing past the stall angle. At five degrees of gust the deficit is fourteen per cent. The drag runs the other way for the same reason, exceeding the drag at the mean by half the polar's curvature times the variance.

Tested in The lift at the mean angle, at the figure it turns on.

A bubble rises faster than a solid sphere of the same size because it is lighter.

What decides it: The Hadamard–Rybczynski drag law contains no density at all. A bubble's advantage over a rigid sphere is a factor of exactly 2/3 in the drag, and it comes from the surface being free to move rather than from what is inside it. A drop of mercury in water — denser than everything around it — gets almost none of the same advantage, because its interior is so viscous that its surface barely moves.

Tested in The surface that moves with the flow, at the figure it turns on.

A water-surface profile is the solution of an ordinary differential equation, so it can be integrated from either end and the answer is the same.

What decides it: The same equation, the same starting depth of 1.6 times the normal depth and the same integrator, differing only in the sign of the step. Integrated upstream the profile relaxes onto normal depth and stays there for sixty kilometres; integrated downstream it departs exponentially and reaches fifty metres of depth in a channel two metres deep. On a steep bed both statements reverse.

Tested in The section that decides the river, at the figure it turns on.

The power spent overcoming a body's skin friction is the rate at which it heats the fluid, so the drag of a flat plate and the dissipation in its boundary layer are the same number.

What decides it: The dissipation in the layer up to a station is half the free-stream energy density times the energy thickness; the drag up to the same station is the momentum thickness times ρU². Their ratio is δ₃/2θ, which for a Blasius layer is 0.786 — so 21.4 per cent of the towing power is still in the fluid as kinetic energy when it leaves the plate, and will not become heat until much further downstream.

Tested in The third thickness, at the figure it turns on.

Murray's law is a statement about vessel radii, so the angles at a branching junction are set by something else — packing, growth, or convenience.

What decides it: The same cost that gives the cube law gives the angles, because at the optimum each segment's cost per unit length is proportional to the square of its radius and the junction is then a three-force balance. Moving the branch point in the plane until the total cost is least gives angles agreeing with the closed-form triangle of forces to five parts in a billion, at four different degrees of asymmetry.

Tested in The angle a junction chooses, at the figure it turns on.

The equations of ideal flow are linear, so a flow can be built by adding simpler ones and the pressures and forces of the parts add too.

What decides it: C_p(A+B) − C_p(A) − C_p(B) = −1 − 2u_A·u_B/U², an identity checked here to the last digit at ten points and reaching 1.92 — larger than the whole range of a suction peak. And a cylinder in a stream has no lift, a point vortex alone has no lift, and their sum has ρUΓ: two flows with no force each adding to the only force in aviation.

Tested in The one thing that does not add up, at the figure it turns on.

Nusselt, Reynolds and Prandtl numbers are the three dimensionless groups of forced convection, and a heat-transfer problem is specified by choosing them.

What decides it: Two of the three can be chosen and the third cannot. Setting Re and Pr and solving the Blasius energy equation determines Nu to as many digits as anybody wants — 0.292680360 times the root of the Reynolds number at Pr = 0.7, by quadrature and by an independent shooting solve that agree to 10⁻¹³. There is no experiment in which the Nusselt number is held fixed and something else observed, because it is not an input to anything: it is h, made dimensionless, and h is what the apparatus reports.

Tested in The number that is an answer, at the figure it turns on.

Two rotors of the same diameter, the same solidity and the same tip-speed ratio have the same performance, because the actuator disc that represents them is the same disc.

What decides it: Computed at a fixed solidity of 0.05 and a tip-speed ratio of 7, a two-bladed rotor returns C_T = 0.629 and a twenty-bladed one 0.693 — a spread of 1.101 in thrust and 1.166 in power. The whole of the difference is Prandtl's tip loss, which depends on the gap between blades rather than on the total blade area, so it is a function of the blade count at fixed solidity. The actuator disc cannot express it because the blade count is not one of its variables: it has no blades and therefore no gaps between them.

Tested in A disc that knows no blades, at the figure it turns on.

C-mu = 0.09 is a constant of turbulence, measured once and applicable generally.

What decides it: In a shear layer in local equilibrium, production equals dissipation, which makes the constant the square of the ratio of dissipation to strain times energy — so 0.09 is the statement that the strain measured on the turbulence's own time scale is ten thirds. That is one flow. Eleven per cent above it the same closure gives a first normal stress below zero — a variance below zero — and at the strains a stagnation region reaches, ten to fifty, it reports a production up to 648 times the dissipation.

Tested in The constant that makes a variance negative, at the figure it turns on.

How much a blob of fluid has been stretched is decided by how hard it was strained and for how long, so integrating the strain rate along its path gives the answer.

What decides it: Two histories are built here from the same two pieces — a simple shear and a pure straining flow, each of strain rate root two — applied in the two possible orders. The integral of the strain rate along the path is identical in the two cases, to zero difference rather than to a tolerance. The stretch of the most-stretched material direction comes out 6.087 one way and 2.818 the other. An integral of the rate cannot distinguish two histories that differ only in their order, and the answer does.

Tested in Two strainings, and the order they came in, at the figure it turns on.

A fluid's resistance to deformation is described by one number, its viscosity, and everything else about its response to a flow follows from that.

What decides it: There are two independent ways to deform an element — change its shape and change its volume — and they have separate coefficients. The second, the bulk viscosity, multiplies the square of the divergence, so it does nothing at all in an incompressible flow and is a quarter of the absorption of sound in dry air. For carbon dioxide at acoustic frequencies it is about fifteen hundred times the shear viscosity.

Tested in The viscosity nobody uses, at the figure it turns on.

The roll-up of a vortex sheet can be computed by putting enough point vortices on it and integrating; the answer converges as the number of points is increased.

What decides it: The discrete system's growth rate is exactly pi·m(1 − m/N), whose largest value is at m = N/2 — the grid scale. Round-off at 10⁻¹⁶ therefore grows at pi·N/4 and reaches the size of the physics before the physics has finished. Refining from 64 points to 256 multiplies the computed peak curvature by four hundred, and it is the grid mode being measured rather than the sheet. Removing every Fourier coefficient below 10⁻¹³ removes the seed, and the same refinement then moves the answer by four per cent.

Tested in A sheet that cannot stay a sheet, at the figure it turns on.

A shock's thickness and its jump conditions are two aspects of the same physics, so a model that gets the structure right is needed to get the jump right.

What decides it: Seven dissipation models — Prandtl numbers from 0.25 to 2, viscosities from constant to linear in temperature — are integrated through the same Mach 2 shock. Their thicknesses span a factor of 2.3 and their interiors differ in the total enthalpy by up to 7.8 per cent, and the states they connect agree to seven parts in ten billion. The jump is three conservation statements and contains no transport property at all.

Tested in The jump does not ask what made it, at the figure it turns on.

A dye field shows where the fluid is going, so the pattern in it can be read forward from the velocity that is there now.

What decides it: A conserved scalar's value at a point is its initial value at the origin of that point's own back-trajectory, and the map from now to then is not a function of the present velocity field. Reconstructing a stripe pattern six time units on by carrying it along the flow map is exact — it moves by 3·10⁻⁷ when the integration is refined. Reconstructing it by displacing the pattern through the instantaneous velocity times the elapsed time is wrong by 1.89 of a range of 2 at its worst and by 0.58 on average, which is most of the way to no information at all.

Tested in A scalar is a record of where its fluid was, at the figure it turns on.

A body oscillating slowly in a fluid feels the steady drag appropriate to its instantaneous velocity, so its damping is Stokes' 6πμa and does not depend on frequency.

What decides it: The unsteady Stokes solution gives a damping of 6πμa(1 + a/δ) with δ = √(2ν/ω), so the correction is the body's size against the layer thickness. For a millimetre sphere in air the layer is 2.2 mm at 1 Hz — the "steady" answer is already 46 per cent low there — and 22 µm at 10 kHz, where the damping is forty-seven times Stokes'.

Tested in What a fluid takes out of a swing, at the figure it turns on.

The Grashof number is the Reynolds number of a natural-convection flow, so the transition to turbulence happens at the same value.

What decides it: Gr is the *square* of a Reynolds number built from ν/x, which is not a velocity anything in the apparatus has. The flow's own Reynolds number is a result of the solution: Squire's closed form gives Re_x = 80√Gr/√(240(20/21 + Pr)), which at Gr = 10⁸ in air is 39,900 on the scale velocity and 5,916 on the profile's own peak. So a plate that is transitioning at Gr ≈ 10⁹ is doing so at a Reynolds number of about twenty thousand — a number in the same range as a forced boundary layer's, and three orders below the Grashof number quoted for it.

Tested in A speed nobody imposed, at the figure it turns on.

A compressor blade row can be designed from isolated-aerofoil data, because each blade is an aerofoil at a known incidence.

What decides it: At a solidity of one, a row of flat plates at 30° stagger meeting a flow at 45° turns it by 14.16°. The isolated-aerofoil calculation — 2π on the inlet incidence, converted to a turning through Δtanβ = σC_L/2 — predicts 34.9°, which is two and a half times too much. The lift slope per unit of mean-flow incidence has fallen to 67 per cent of 2π by that solidity and to 37 per cent by σ = 2. The blade's incidence is not measured from the inlet: it is measured from the vector mean of inlet and outlet, and half the row's own turning is already present at the blade.

Tested in A row is not a set of aerofoils, at the figure it turns on.

A stratified shear layer becomes unstable when its Richardson number falls below a quarter.

What decides it: Miles' theorem says the converse: above a quarter, everywhere, no instability is possible. Below a quarter it says nothing at all, and nothing is not a prediction. Plane Couette flow between walls with a uniform stratification has a Richardson number of 0.05 at every height — a fifth of the threshold — and a hundred and twenty-five searches over wavenumber and starting guess find no unstable mode, because its velocity profile has no inflection point and Rayleigh's criterion already forbids instability with no stratification at all.

Tested in Sufficient, and not necessary, at the figure it turns on.

A liquid is incompressible for practical purposes, so a flow of water at a few metres a second can never be a compressible flow.

What decides it: Wood's formula gives the sound speed of a mixture from two mixing rules that go opposite ways — density by volume, compressibility by volume — so a mixture takes the heavy phase's inertia and the light phase's springiness. One per cent of air by volume takes water from 1,481 metres a second to 119, and a ten-metre-a-second flow that was at Mach 0.007 is at Mach 0.084. At twenty per cent it is at Mach 0.34.

Tested in Slower than either of them, at the figure it turns on.

A mean flow can only come from a mean force, so an oscillation with no mean cannot transport anything.

What decides it: The transport produced by an oscillation is the average of the parcel's displacement dotted into the velocity gradient it meets, and both factors average to exactly zero on their own. In a linear deep-water wave the two terms of that average are each 0.500000000000 of the answer and their sum matches the closed form to twelve figures. The control is the decisive part: an oscillation of the same amplitude and frequency with no spatial gradient in it drifts at 2·10⁻¹⁵ of the orbital speed, which is zero, and one carrying only the horizontal half of the gradient carries 0.502 of the drift.

Tested in A drift made of two things that average to zero, at the figure it turns on.

A packed bed is characterised for process purposes by its mean residence time — bed volume over volumetric flow — and two beds with the same mean behave the same way.

What decides it: Six beds are given the identical mean residence time and Peclet numbers from 4 to 400. The first hundredth of an inlet step reaches the outlet at 0.20 of the mean in the loosest and 0.85 in the tightest; the window of inlet history the outlet mixes runs from 1.53 mean times down to 0.18; and a first-order reaction at the same mean time converts 0.769 in one and 0.863 in the other against plug flow's 0.865.

Tested in The outlet is the inlet, a while ago, at the figure it turns on.

A slenderness expansion has an error of order ε², because the transverse derivatives are smaller than the longitudinal ones by ε and they enter squared.

What decides it: True of one of the three cases here and false of the other two. A rectangular duct's departure from the parallel-plate answer is first order in the aspect ratio, with coefficient 0.630249, because what the reduced problem left out is a pair of side walls rather than a gradient. A prolate spheroid's added-mass error is ε²ln(1/ε), whose local exponent drifts from 1.88 to 1.73 across the range a computation can reach and reaches two only in a limit that would need ε = 10⁻⁴³. Only the long water wave has the exponent the folklore claims.

Tested in One group, three exponents, at the figure it turns on.

The multipole expansion of a body's far field converges everywhere outside the body, so a few terms always describe the flow on its surface.

What decides it: The expansion converges outside the circle containing the body's singularities, which is not the same as outside the body. A Rankine oval of fineness 6.7 has its waist at a half-thickness of 0.158 against a source at 1, so the surface passes inside that circle and the series diverges there — the error grows from 3.95 at one term to 6·10⁶⁴ at eighty-one. A fat oval of fineness 1.17 converges to machine precision in seven.

Tested in The part of the flow inside the body, at the figure it turns on.

A transport barrier is a feature of the flow, so it can be located from a snapshot of the velocity field the way a vortex core or a stagnation point can.

What decides it: The barrier is a crest in a field of finite-time stretching exponents, and that field is a property of a window of history. Here the crest at an eight-unit window stands 2.4 times its own field's median and correlates with the instantaneous rate of strain of the same flow at the same moment at 0.28 — so a snapshot is not a blurred version of it. Changing the window from two units to twelve moves the crest by 1.04 in a domain two wide, and computing the same field backwards in time gives a field correlating with the forward one at −0.04, which is a different set of curves.

Tested in A boundary that only exists over a window, at the figure it turns on.

The jump conditions give the state behind a shock, so the temperature immediately behind one is the temperature they predict.

What decides it: They give the state after the gas has equilibrated. A shock is a few mean free paths thick, and vibrational relaxation takes thousands of collisions, so the gas leaves the shock with its vibrational modes still in their pre-shock population and with all of the shock's energy in translation. At Mach 6 in air that frozen temperature is 2,382 K against an equilibrium 2,059 — 13.6 per cent higher — and the gas takes 0.4 mm to relax, which is thousands of times the shock's own thickness. At Mach 10 the gap is 22 per cent.

Tested in A gas that has not finished being shocked, at the figure it turns on.

A self-similar solution's exponents follow from dimensional analysis, so any problem with a similarity solution has exponents that can be obtained by counting.

What decides it: Sedov's blast wave is of that kind: energy, density and time give R ∝ (Et²/ρ)^⅕ and the two-fifths is the same for every gas, because γ is dimensionless and no exponent formed by counting dimensions can contain it. A converging shock is not: energy is not conserved in the focusing region, the group that would have fixed the exponent does not exist, and the answer comes from a solvability condition instead. Computed here, the spherical exponent runs from 0.790 at γ = 1.1 to 0.667 at γ = 2 — a span of 0.123 — and lands on Guderley's 0.717174 at γ = 1.4 to four figures.

Tested in An exponent dimensions cannot give, at the figure it turns on.

A boundary-layer computation that fails at separation has run into a numerical difficulty, and a finer grid or a better scheme will get through it.

What decides it: The failure is a property of the equations with a prescribed pressure, and refinement confirms it rather than removing it. Over a factor of eight in the step the separation station moves by 0.108 per cent — so it is the equations' station and not the grid's — while the measured exponent of the wall shear falls steadily from 0.7375 towards Goldstein's one half. What removes the singularity is not a better scheme but a different boundary condition: prescribe the displacement thickness, or let the pressure interact, and the same equations march straight through.

Tested in The singularity a layer makes for itself, at the figure it turns on.

A simulation that resolves the smallest scales of the velocity field resolves the flow, so its scalar transport is resolved too.

What decides it: At Schmidt numbers above one the scalar's smallest structure is smaller than the velocity's, by exactly the square root of the Schmidt number. For dye in water that is a factor of forty-five in length and ninety thousand in three-dimensional grid points. On a model spectrum at that Schmidt number, ninety-five per cent of the scalar variance sits above the Kolmogorov scale and seventy-seven per cent of its destruction happens below it.

Tested in The scalar has its own cascade, at the figure it turns on.

A measured mean velocity profile characterises the flow it came from, so two flows with the same profile are the same flow for any practical purpose.

What decides it: Two flows are constructed here with mean profiles agreeing to 3.6·10⁻¹⁷ across the channel. One is a plain shear; the other carries a zero-mean disturbance. The second transports 0.069 of a dynamic head of momentum across the shear, dissipates 1.40 times as fast, and does both as an exact square of a disturbance amplitude the mean profile does not contain. Sweeping the phase between the disturbance's two components takes the transport from its full value to zero while the mean profile and the fluctuation amplitude are held fixed.

Tested in What a mean profile cannot tell anybody, at the figure it turns on.

Once the waves in a shock tube have passed, the gas between them is in one uniform state.

What decides it: It is at one pressure and one velocity and in two states. The contact surface separates gas that has been through the shock from gas that has been expanded, and matching the pressure and the velocity is all that a contact surface does. At a diaphragm pressure ratio of ten the two sides differ by a factor of 1.99 in temperature and the same in density, and by 694 J/kg/K in entropy — which is the record of the shock and is the reason the surface cannot heal.

Tested in A surface that remembers the diaphragm, at the figure it turns on.

A ball's swerve grows late in its flight because the spin has decayed and the ball has "run out of pace", so the two fall away together.

What decides it: Integrating the spin along the trajectory shows it decaying as an exponential in distance flown with a constant of 1,202 metres for a golf ball, so a 206-metre drive keeps 84 per cent of its launch spin. Over the same flight the speed falls from 70 to 30 metres a second, so the spin ratio — which is what the lift coefficient is a function of — rises from 0.096 to 0.189. The swerve grows because the speed fell, not because the spin did.

Tested in The ball that never forgets its spin, at the figure it turns on.

A fluid cannot pull, so nothing can raise water more than about ten metres by lowering the pressure above it — the atmosphere's push is all there is.

What decides it: A pump from a free surface does stop at 10.1 metres, because the water under it reaches its vapour pressure with somewhere to boil. A column with no such place does not stop. Gravity at 9.79 kPa a metre and Poiseuille friction at 10.02 kPa a metre, in 40 µm conduits carrying sap at a quarter of a millimetre a second, put the top of a hundred-metre column at −1.98 MPa absolute, two megapascals below zero. The floor under it is the pressure a pit-membrane pore lets air through, −2.81 MPa for a 50 nm pore, and that column would reach it at 142 metres.

Tested in Where a liquid does pull, at the figure it turns on.

A numerical solution of the lifting-line equation is more trustworthy than the closed-form formula, because it makes fewer approximations.

What decides it: It makes exactly the same approximations, and it reports no sign of them. Solving the monoplane equation on twelve odd harmonics at aspect ratios from 0.5 to 20 reproduces 2π/(1 + 2/AR) to 2.4 × 10⁻¹⁶ at every one of them — including aspect ratio 0.5, where the correct slope is 1.14 and the formula gives 1.26, forty-one per cent high at AR = 1. Nothing in the solve grows, nothing fails to converge, and no residual increases. A model outside its range does not report that it is outside its range.

Tested in Where the line stops being a line, at the figure it turns on.

The flow past a body is determined by the equations and the boundary conditions, so the circulation round it is determined too.

What decides it: Five circulations from −8 to +7 are put on the same cylinder in the same stream. Every one satisfies Laplace's equation, the tangency condition on the surface and the condition at infinity; every one has a pressure drag of zero to 6·10⁻¹⁶; and each has a different lift, equal to rho·U·Gamma to eight figures. The equations do not choose, because the region outside a body is not simply connected and the potential is not single-valued.

Tested in The constant a hole leaves behind, at the figure it turns on.

Wall models differ mainly in how they treat the buffer layer, so the uncertainty in a predicted skin friction comes from the buffer layer.

What decides it: Three standard wall models agree to 1.5 per cent at y+ = 10 and give friction factors 6.6 per cent apart at a friction Reynolds number of two thousand. A control with no buffer layer at all is 19.6 per cent out in the profile and 2.6 in the friction. Measured in the logarithmic region the three models carry additive constants of 5.28, 5.63 and 5.00, and shifting them to a common value collapses the friction spread by a factor of ten at that Reynolds number and a hundred and thirty at fifty thousand.

Tested in Three buffer layers, one friction, at the figure it turns on.

Everything about a steady flow can be read off its fields, because nothing in a steady flow changes.

What decides it: The age of the fluid at a point is a well-defined quantity with its own transport equation — its material derivative is exactly one — and in a steady flow it is a steady field which no measurement of velocity, pressure or density determines. In the contraction computed here the local rate of change of velocity is exactly zero at every point and every instant, and a parcel leaves at 4.000000000 times the speed it entered at, having spent ln(4)/3 of the reference time inside. Neither the four nor the transit time is recoverable from a reading taken at a point.

Tested in How long the fluid has been in there, at the figure it turns on.

The Doppler shift is a change in frequency caused by the motion of a source, and it is a separate effect from the propagation delay.

What decides it: It is the propagation delay, differentiated. The arrival time at a fixed observer is the emission time plus the distance from the source's position then divided by the sound speed, and the ratio of arrival intervals to emission intervals is the reciprocal of that mapping's slope. Computing the slope numerically and comparing it with the closed-form Doppler factor gives agreement to 4·10⁻⁹. Above Mach one the mapping is no longer monotone: it folds, two emission times arrive together, and the fold is the sonic boom.

Tested in The sound now is the source then, at the figure it turns on.

A patch of turbulence that has survived a long way down a pipe has established itself, so it is more likely to survive than a newly formed one.

What decides it: A puff's decay is a Poisson process: its survival probability is an exponential in time, so the hazard rate is constant and the probability of lasting one further mean lifetime is the same for a new puff and for one that is three lifetimes old — identical to the last bit of double precision. What is not memoryless is the population. Puffs also split, the splitting time falls with Reynolds number where the decay time rises, and where the two curves cross at 2040 the flow acquires a memory it did not have below.

Tested in A puff that does not know how old it is, at the figure it turns on.

Friction in a duct slows the flow down, so a long enough duct brings a supersonic flow back to rest.

What decides it: Friction drives the Mach number towards one from either side: a subsonic flow speeds up and a supersonic one slows down, because what friction does is raise the entropy and the entropy is maximum at the sonic point on the Fanno line. A duct entered at Mach 0.3 reaches Mach 0.94 after 5.3 friction lengths and produces 451 J/kg/K on the way, and adding more duct does not push it past one — it reduces the mass flow instead. The supersonic branch runs the other way and stops at the same place.

Tested in A duct that cannot be run backwards, at the figure it turns on.

Friction in a siphon's hose spends pressure on the way up to the crown, so a longer, rougher or narrower hose brings the crown nearer to breaking.

What decides it: Friction can only raise the crown pressure. Substituting the outlet's energy balance into the crown's gives the crown pressure exactly as atmospheric, less ρg times the crown's height above the outlet, plus ρg(1 − φ)Δ, with φ the share of the hose's whole resistance lying before the crown, which cannot exceed one. Eighteen metres of 12.5 mm hose six metres above its outlet hold the crown at 51.5 kPa against the frictionless 23.0; the margin is 29.0 kPa in a 6 mm hose and 24.6 in a 50 mm one while the flow changes 263-fold, and the draining tank returns every pascal of it by the end.

Tested in The margin friction lends a siphon, at the figure it turns on.

Wagner's function gives the lift after a sudden change, so it is the right thing to convolve a gust with.

What decides it: They are two different problems with two different indicial functions. Wagner's Φ(0⁺) is a half — half the steady lift arrives instantly — and Küssner's Ψ(0⁺) is zero, because a wing entering a sharp-edged gust has met the gust only at its leading edge. Solved here with one code and two right-hand sides, at one semichord the two are 0.464 and 0.057, a factor of 8.2 apart. Using Wagner's function for a one-semichord gust overestimates the peak load by 68.9 per cent, and by 27.9 per cent at two.

Tested in Two answers to one question, at the figure it turns on.

A wind measurement made over a surface is a measurement of that surface, provided the instrument is well above the roughness elements.

What decides it: A change of surface starts an internal boundary layer which grows as the fetch to the four-fifths power, and only its lowest tenth is in genuine equilibrium with the new surface. Above that, the flow is still the old surface's. A measurement at ten metres over crops needs 958 metres of uniform fetch before it is measuring the crops — 96 times its own height — and the ratio runs from 36 to 214 across the heights and roughnesses anybody works at, so the hundred-to-one rule is the middle of a factor of six.

Tested in How far downwind a surface is remembered, at the figure it turns on.

A detonation is a shock with the heat release attached to it, so the two happen in the same place.

What decides it: They are separated by an induction zone in which the gas has been shocked and has not yet begun to react. Its length is the post-shock flow speed times an Arrhenius induction time, and the Arrhenius dependence makes it exponentially sensitive to the shock's own strength: at Mach 5 in the model here it is 1.93 mm, a one per cent stronger shock shortens it by 15 per cent, and a ten per cent stronger shock by a factor of 3.4. A front whose local strength varies at all therefore has an induction length that varies enormously, which is why real detonations are cellular.

Tested in A gas that has not decided to react yet, at the figure it turns on.

Making a wing stiff enough in torsion to avoid divergence makes it safe from static aeroelastic problems.

What decides it: Reversal comes first, and it comes first at every elastic-axis position behind the aerodynamic centre: for the wing computed here the reversal dynamic pressure is 0.4998 of the divergence one, and the *ratio* does not depend on the stiffness at all because both are proportional to it. Worse, a wing with its elastic axis at the quarter chord has an infinite divergence speed and a perfectly finite reversal pressure — twice the ordinary wing's, and still there — because what drives reversal is the aileron's own nose-down moment rather than the moment of the incidence.

Tested in The control that works backwards, at the figure it turns on.

The method of images can be applied to any corner by reflecting the vortex in both walls repeatedly and keeping enough of the resulting images.

What decides it: Reflection in two walls generates a dihedral group, and the group is finite only when the angle is pi over a whole number. At 1.1 radians the orbit never closes: the image count goes from nine to thirty-three as the construction is iterated, and the worst leak through the walls goes from 0.086 to 0.152 — it does not improve. The same construction at a quarter of pi closes at eight images with a leak of 10⁻¹⁶.

Tested in The corners that can be done with mirrors, at the figure it turns on.

A quantity that follows from an exact conservation law is a quantity the conservation law determines, so measuring it settles the flow it came from.

What decides it: An exact constraint is one linear functional imposed on a space of profiles, and it determines a second functional only when the second lies in the span of the first. Four wake profiles are built here with identical deficit integrals — identical drag, to three parts in a hundred million million — whose peak deficits differ by a factor of four and a half, whose half-widths differ by five, and whose kinetic energy differs by a factor of three. The drag is exact and the wake is not measured at all.

Tested in Exact in the total, free in the profile, at the figure it turns on.

The dissipation of turbulent energy is set by the large scales, so it follows a change in the forcing immediately.

What decides it: It is set by the large scales through a cascade, and the cascade takes a turnover to deliver. Stepping the strain rate by a factor of 2.5 multiplies the production by 6.28 immediately — production is an eddy viscosity times the strain squared — and moves the dissipation by 0.93 per cent and the turbulent energy by 0.68. The ratio of the two departs from its equilibrium value by 520 per cent, falls to a tenth of that departure in a third of a turnover and to a hundredth in three quarters, and the standard scaling with the integral scale held where it was moves by a factor of 530 in the meantime.

Tested in A dissipation that lags its production, at the figure it turns on.

A shock destroys stagnation conditions, so both the total pressure and the total temperature fall across one.

What decides it: The total temperature is unchanged across an adiabatic shock, exactly. Computed from the static states on both sides at seven Mach numbers from 1.2 to 8, the ratio is 1.000000000000000 in every case. The total pressure over the same range falls from 0.9928 to 0.0085, and its logarithm is the entropy rise divided by minus the gas constant — the same number written twice. One quantity records energy and the other records irreversibility, and a shock adds no energy and a great deal of irreversibility.

Tested in Two totals, one of which a shock cannot touch, at the figure it turns on.

The cushion under a wing settling onto a runway is a squeeze film of air, the same viscous film that carries a bearing, and it grows stronger as the gap closes for the same reason.

What decides it: The viscous squeeze force on a closing plate is μVc³/h³ and the inertial one is ρV²c³/24h², and their ratio is the gap Reynolds number over twenty-four. They are equal at h = 24ν/V, which for a wing closing at a metre a second in air is 0.36 mm. At every gap a wing flies at, the cushion is the air's inertia and not its viscosity. The two films differ in kind as well as size: a loaded plate under water lands through the inertial film in 0.108 s and, through the viscous one, is still 1.6 μm clear after ten thousand seconds.

Tested in A cushion that changes its physics, at the figure it turns on.

Kutta–Joukowski is a result about aerofoils, so a body that is not aerofoil-shaped needs something else.

What decides it: Two circles of different radii, two ellipses of different fineness and two Joukowski sections with opposite emphasis on camber and thickness are all set to a circulation of exactly two and the pressure integrated round each. The six lifts agree to three parts in a hundred billion. The theorem is about a resultant force on a closed contour in a steady irrotational stream, and the only thing about the body that enters it is the circulation the body happens to carry.

Tested in One formula, and it does not ask what the shape is, at the figure it turns on.

A collapse of experimental data onto one curve, plotted against a dimensionless group, is evidence that the group governs the process.

What decides it: Two hundred and forty synthetic points are generated with a drag coefficient of exactly 0.44 and twelve per cent scatter, so there is no Reynolds-number dependence in them whatever. Plotted as F/(mu U d) against the Reynolds number — both perfectly legitimate pi groups forming a complete set — they fall on a straight line of slope 0.9985 with a coefficient of determination of 0.9985. The ordinate contains the abscissa, and the collapse is arithmetic.

Tested in The groups are not the only groups, at the figure it turns on.

An eddy viscosity is an approximation whose error is a matter of getting the constant right.

What decides it: Its error is structural rather than numerical: it asserts that the stress responds to the strain instantaneously, and the stress takes about a turnover to respond. Computed as a frequency response, the eddy-viscosity hypothesis is the value at zero frequency — a magnitude of one with no phase — and the real response falls to 0.71 at one turnover, 0.20 at five and 0.01 at a hundred, with a phase lag reaching 45 degrees at one turnover. No choice of constant recovers any of that, because a constant has no frequency in it.

Tested in A closure with no memory at all, at the figure it turns on.

A surface close to an accelerating body always makes it harder to accelerate, because the fluid between them is trapped — the cushion is something nearness does.

What decides it: It is something a solid wall does. A plate a tenth of a chord from a wall borrows 1.966 times its free added mass; the same plate a tenth of a chord from a boundary held at constant pressure borrows 0.677 times, and as the gap closes the first grows without limit while the second tends to exactly one half. A circle centred in a tunnel 1.25 radii wide borrows 3.25 times its free mass with solid walls and 0.59 with an open jet. The difference is the sign of the image each boundary requires.

Tested in The borrowed mass the boundary decides, at the figure it turns on.

A surge tank cushions water hammer, so a valve behind one can be shut as fast as its actuator allows.

What decides it: For any closure faster than the penstock's own round trip the tank changes nothing: shut in 0.2 or 0.4 seconds, the valve takes 381 metres of head with the tank and 381 without, because the wave has not yet reached the tank when the peak is made. What the tank does is move the closure time at which the rise starts to fall — from the whole line's 4.4 seconds to the penstock's 0.4 — so a four-second closure behind it makes 26 metres, which the bare line needs between twenty and forty seconds of closure to match.

Tested in A tank that turns a hammer into a swing, at the figure it turns on.

A small particle in a fluid has a drag proportional to its velocity, so it slows down exponentially with a time constant set by its mass and the viscosity.

What decides it: The unsteady Stokes force has three parts and only one of them is proportional to the present velocity. The history term is an integral over the particle's whole acceleration record with a kernel falling as the inverse square root of the elapsed time, and here it grows to 39 per cent of the total force. Its effect is to oppose the deceleration, so the particle travels 12 per cent further rather than stopping sooner, and its velocity decays as a power law rather than an exponential — at five particle time constants it is 3.8 times what the quasi-steady answer gives.

Tested in The drag that integrates a whole history, at the figure it turns on.

Dissipation happens at the smallest scales, so a measurement of it at one point is independent of a measurement made a few hundred smallest scales away.

What decides it: A multiplicative cascade gives the logarithm of the coarse-grained dissipation a variance proportional to the logarithm of the ratio of scales, and gives two points a correlation equal to a ratio of logarithms. In a flow with four decades of inertial range the correlation is still 0.25 at a thousand smallest scales, and it falls to a half only at the geometric mean of the smallest and the largest — a hundred smallest scales, two decades up. A flow with more decades correlates over more of them rather than fewer.

Tested in A dissipation correlated across every scale, at the figure it turns on.

Morison's equation is an empirical sum of two forces, so the split between them is a matter of curve-fitting rather than of physics.

What decides it: The two terms are different in kind. The inertia term is the added mass times the acceleration — a function of the present state, whose impulse over a manoeuvre depends only on the endpoints. The drag term is a quadratic in the velocity with a coefficient that depends on the wake, and the wake was made by the previous half cycle. Their amplitude ratio is exactly Cd·KC/(π²Cm), which the computation reproduces to nine figures, and it crosses one at a Keulegan-Carpenter number of 16.45 — so one number decides which kind of memory a body is living in.

Tested in Two forces, and only one of them remembers, at the figure it turns on.

A surface that cannot hold a shear stress makes no vorticity, so the flow round a clean bubble is the ideal flow, and like every ideal flow it has no drag.

What decides it: The ideal flow round a sphere leaves a tangential stress of −3μU sin θ/a on the surface, computed here from the potential's own second derivatives and exact to 2×10⁻¹⁵; a clean surface cannot hold it, so the real flow carries vorticity 2κuₛ = 3U sin θ/a there. And the ideal flow, though irrotational, is strained everywhere: its viscous dissipation, integrated over the whole exterior, is 12πμaU², so the bubble's drag is 12πμaU and its drag coefficient 48/Re — not zero, and not a rigid sphere's either.

Tested in The vorticity a clean surface cannot refuse, at the figure it turns on.

Friction in a hydro tunnel is only a loss, so a smoother or wider tunnel is better in every respect, the stability of the plant included.

What decides it: Under a governor holding the turbine's power, friction is the only damping the tank's swing has, and the smallest stable tank is set by it. For the same two-kilometre tunnel carrying two metres a second at a hundred metres of head, Thoma's criterion asks for a tank 2.02 metres across with ten metres of friction, 2.78 with five, 4.33 with two and 6.09 with one. Cutting the loss from five metres to one multiplies the smallest stable tank's area by 4.8.

Tested in The better tunnel needs the bigger tank, at the figure it turns on.

Added mass is the mass of the fluid a body drags along with it, which is why it scales with the body's size.

What decides it: A flat plate has no volume at all and a broadside added mass of pi rho a squared per unit span, which does not move by a part in ten trillion as the thickness is taken from a tenth of the chord to a millionth of it. The ratio of added mass to displaced mass therefore diverges as the reciprocal of the thickness, reaching a million at a thickness ratio of a millionth. Added mass is twice the kinetic energy of the disturbance over the square of the speed, and the disturbance of a translating body is a dipole whose strength is set by the body's outline rather than by what is inside it.

Tested in The mass a body has to borrow, at the figure it turns on.

A wake becomes self-similar a few tens of diameters downstream, so a measurement made there is independent of what made it.

What decides it: The approach to self-similarity is algebraic rather than exponential. Two initial profiles with identical momentum deficits — a slab and a pair of separated lobes — are marched down a far-wake equation here, and their difference falls as the 0.96 power of the distance. They agree to twenty per cent only after 217 initial half-widths, to ten after 457 and to five after 933. The momentum deficit, which is the drag, is conserved to four parts in 10¹¹ throughout — so the wake keeps one number exactly and loses the rest slowly.

Tested in A wake that keeps the drag and forgets the body, at the figure it turns on.

A flow at given conditions is whatever the equations and the boundary conditions give, so two identical experiments at identical conditions produce identical flows.

What decides it: Above the critical driving the neutral curve admits a band of wavelengths rather than one, and a narrower band inside it is stable against slow adjustment — narrower by exactly one over the square root of three. A container of finite length quantises the admissible wavelengths, so at twice critical an apparatus twenty gaps long has eight distinct steady stable flows available at exactly the same conditions. Which one it settles in is decided by the acceleration history, and by nothing in the final conditions at all.

Tested in The state a machine was started into, at the figure it turns on.

A gas-textured surface slips according to how much of it is gas: make the solid fraction small enough and the liquid slides almost freely.

What decides it: For stripes, the along-stripe slip length is (L/π) ln sec(πφ/2), which grows only as the logarithm of the solid fraction. A surface nine-tenths gas slips 0.591 of a period, ninety-nine hundredths gas 1.322, and 99.99 per cent gas 2.788: each tenfold cut in the solid adds the same 0.733 of a period. The slip is set by the period, and doubling the period doubles it at any gas fraction.

Tested in Twice as slippery along as across, at the figure it turns on.

A nozzle's exit Mach number and pressure fix its thrust, and the cone angle of its divergent section is a matter of length and weight rather than of performance.

What decides it: A conical section sends its gas out as a source flow, and only the axial part of that momentum pushes. The share that does is the exit disc's area over the spherical cap's, (1 + cos α)/2: a 15° cone keeps 0.9830 of its momentum and a 30° cone 0.9330, by quadrature as well as in closed form. Integrating momentum flux and pressure over the flat exit plane — where nothing is uniform — gives the same thrust as the cap with that factor, to 2.8×10⁻¹⁴. At an area ratio of 25 in vacuum the 30° cone gives 6.37 per cent less thrust than an ideal nozzle with the same exit.

Tested in The jet a cone sprays sideways, at the figure it turns on.

A boat overpowered in a breeze must bear away: depowering worsens its drag angle, and the best course to windward is 45 degrees plus half that angle, so the optimum moves off the wind as the wind rises.

What decides it: Solved on every course, with the sail flattened just enough to hold the heeling moment at the righting moment, the best beat found by search moves closer to the wind — from 53.98 degrees in light air to 49.60 in 6.5 m/s — while 45 degrees plus half the λ the boat has on that course moves away, to 55.51. The rule is exact only for a λ that does not depend on the course; a flattened rig's λ rises as the boat bears away, from 20.29 degrees at 35 to 27.04 at 80 in 8 m/s. Only once the sail is nearly flat does the optimum turn and follow, reaching 63.55 degrees in 20 m/s against the rule's 68.66.

Tested in A breeze the boat cannot use, at the figure it turns on.

Measuring an aerofoil's lift and moment accurately enough determines its camber line, since thin-aerofoil theory relates the two.

What decides it: The theory's lift depends on the mean slope and the first cosine coefficient of the camber line's slope; the moment about the quarter chord on the first and the second. Two camber lines differing only in the fourth coefficient — and differing by eight tenths of a per cent of the chord, which is forty per cent of the section's own camber — have lifts agreeing to four parts in 10¹⁶ at every incidence and moments agreeing to the same.

Tested in Three numbers out of a camber line, at the figure it turns on.

A fluid's stress is set by how fast it is being deformed now, so a fluid that is not moving is not stressed.

What decides it: A viscoelastic stress is a convolution of the strain-rate history with a relaxation kernel, so it survives the flow that produced it. Here a fluid sheared for two seconds and then left alone retains 37 per cent of its peak stress one relaxation time later, 5.0 per cent after three, and 0.67 per cent after five — with the strain rate exactly zero throughout. Two histories imposing the same total strain of two, one in a fifth of a second and one over four, reach peak stresses of 18.1 and 4.9, and long after both have stopped the slow one has left twelve times more stress behind.

Tested in The fluid that has not finished its last deformation, at the figure it turns on.

A sailing boat's hydrodynamic drag angle is a property of its hull and keel, so the angle measured on the beat describes the boat on every course.

What decides it: The keel is a wing whose lift is the rig's side force, and its lift coefficient is set by the course. In 3 m/s the hull and keel's drag angle is 9.86 degrees at 35 degrees off the wind, 13.44 at 45, 31.54 across the wind and 69.72 at 130. A polar drawn with the drag angle the boat has on its best beat agrees with the solved boat on that course only: across the wind it predicts 2.079 of the wind's speed where the boat makes 1.054, and at 130 degrees 2.235 where it makes 0.755.

Tested in A keel flies wherever the course puts it, at the figure it turns on.

A spinning cylinder's circulation is built up gradually by the fluid's viscosity, so in a less viscous fluid it acquires less of it.

What decides it: The circulation round the cylinder's surface is 2πa²Ω the instant it spins, whatever the viscosity, because no slip makes the fluid touching it turn with it. The spin-up also lays down an equal and opposite ring of vorticity just outside, so the circulation round a larger circle starts at zero and rises only as that ring diffuses past it — half-way at two radii by νt/a² = 0.665, at five by 7.75 and at twenty by 143. Viscosity sets those times; the value arrived at is 2πa²Ω in every fluid.

Tested in A wall puts in exactly its own speed, at the figure it turns on.

A choked nozzle is sonic at its throat, so the throat's area and the stagnation conditions fix the mass flow it passes.

What decides it: With wall friction the flow can pass smoothly through Mach one only where (1/A)dA/dx equals (γ/2)·4f/D, which is in the divergent section. At a friction length 4fL/Dₜ of 0.2 the sonic point sits 2.7 per cent of the way down the divergent section, the throat is at Mach 0.965, and the choked mass flow is 0.9805 of the frictionless value; at a friction length of 1 the throat is at Mach 0.852 and the mass flow 0.9086. Along each solution the mass flow is constant to 7.5×10⁻⁹, so the reduction is the solution's, not the arithmetic's.

Tested in Friction moves the sonic point past the throat, at the figure it turns on.

Lorenz's three-mode truncation is a rough sketch of convection everywhere above onset, reasonable in shape and wrong in detail, and it gets steadily worse as the heating increases.

What decides it: It is not rough at onset: steady rolls computed with 2, 20 and 42 temperature modes all give Nu − 1 = 2(r − 1) to leading order, with slopes of 1.9980, 1.9983 and 1.9983 at r − 1 = 0.001, and the rolls carry only 0.17 per cent more excess heat than Lorenz at r − 1 = 0.01. Nor does it degrade gradually: its heat flux has a ceiling of three that the rolls pass by r = 5, where they carry 3.323, and at r = 30 they carry 5.970 against Lorenz's 2.933. The truncation is exact to first order and then wrong by a growing factor.

Tested in The truncation that cannot carry three times the heat, at the figure it turns on.

A panel method's source and vortex strengths are physical quantities of the flow, so a distribution that reproduces the surface pressure is the distribution the flow has.

What decides it: Two rings of point sources at different depths inside a 4:5 ellipse both reproduce the exact surface speed to a part in ten thousand, and their strengths differ by a factor of a hundred. Across the whole family of admissible placements the largest strength runs from 0.27 to 77,000 for one unit free stream, and the least-squares matrix's condition number spans thirteen decades. Nothing anywhere in the fluid distinguishes the fields they produce.

Tested in The inside a flow does not decide, at the figure it turns on.

The lag in unsteady aerodynamics is a refinement, so a quasi-steady flutter calculation gives roughly the right answer with roughly the right mechanism.

What decides it: The quasi-steady model of this collection's own typical section, with no structural damping, has a positive real part at every airspeed down to one metre a second — it predicts a wing that flutters at walking pace, which is an artefact rather than an answer. With Theodorsen's deficiency in place the same section is stable to 129.5 m/s with no structural damping at all, and to 131.0 with two per cent. The quasi-steady figure with that damping is 80.8, which is 62 per cent low, and the deficiency at the flutter point has magnitude 0.726 and a phase of −15.5 degrees — not a correction to one.

Tested in The lag that makes flutter possible, at the figure it turns on.

A boundary layer responds to the pressure gradient it is in, so two layers meeting the same gradient at the same external speed are in the same condition.

What decides it: A layer's thickness is an integral of everything upstream of it. Thwaites' method gives the momentum thickness as the fifth power of the external velocity integrated from the leading edge and divided by the sixth power of the local one, so two distributions matched at a station and differing before it give different layers there. Two are computed here whose external velocities agree at the station to 0.0000 per cent: their momentum thicknesses differ by a factor of 1.38, their shape parameters differ accordingly, and they separate at 0.528 and 0.584 of the surface.

Tested in A layer that is an integral of everything upstream, at the figure it turns on.

The no-slip condition is a property of fluids, so a gas is at rest at a solid wall.

What decides it: It is a limit rather than a property. A molecule striking a wall last collided about a mean free path away and arrives carrying the gas velocity from there, so the mean velocity at the wall is proportional to the mean free path times the local velocity gradient. That slip is exactly proportional to the Knudsen number — the computation reproduces the proportionality to a part in 10⁹ — and it is 2.0 per cent of the centreline velocity at a Knudsen number of a hundredth, carrying six per cent more flow through the channel, and 17 per cent at a tenth, carrying sixty per cent more.

Tested in Slip is a memory of one mean free path, at the figure it turns on.

A convecting layer settles into the pattern that carries heat most efficiently, which is why a given fluid heated to a given degree always shows the same pattern.

What decides it: In the amplitude equations with g = 0.1 and λ = 2, rolls and hexagons are both stable between ε = 0.0100 and 0.0400. At ε = 0.025 rolls carry an excess heat flux of 0.0250 and hexagons 0.0199 in the same units, yet a layer started nearly hexagonal settles to hexagons of amplitude 0.0814 and stays there; a layer heated slowly upwards keeps hexagons until ε = 0.0448, and one cooled slowly downwards keeps rolls until 0.0085. The pattern is set by history inside the window, and the one kept is not the one that carries more heat.

Tested in Hexagons remember how the heat was turned up, at the figure it turns on.

The largest pressure a sudden valve closure can produce is the Joukowsky rise, ρaΔV above the steady pressure, so a line designed for that rise is designed for the worst case.

What decides it: Where the returning wave takes the line to vapour pressure, the pulse after the cavity closes is 2ε⌈1/ε⌉ − 1 Joukowsky rises above the steady head, with ε the margin to vapour pressure in rises. In 600 metres of pipe with a wave speed of 1200 m/s and 25 metres of steady head, shutting off 0.36 m/s gives a Joukowsky head of 69.1 metres and a pulse of 121.3 metres; the excess can reach twice the margin, 70.2 metres, and a grid solver told nothing of the closed form agrees to a millimetre at ten velocities.

Tested in Twice the margin, on top of the hammer, at the figure it turns on.

A cone in a supersonic stream behaves like a wedge of the same angle, with a slightly weaker shock from three-dimensional relief, so the wedge's oblique-shock tables give its pressure and its detachment angle to a reasonable approximation.

What decides it: Solved from the Taylor–Maccoll equation at Mach 2, a 10° cone's shock sits at 31.21° against the wedge's 39.31°, turns the flow crossing it through only 1.48° and leaves 85 per cent of the turn to an isentropic compression behind it; the cone's surface pressure coefficient is 0.1045 against the wedge's 0.252, 41 per cent of it; and the cone stays attached to a half-angle of 40.69° where the wedge detaches at 22.97°. At 20° the cone keeps 99.0 per cent of the total pressure and the wedge 89.3.

Tested in A cone finishes its turn after the shock, at the figure it turns on.

A conserved quantity that drifts in a computation is telling you how well the physics is being conserved.

What decides it: The circulation round a material loop in a point-vortex flow has an exact value that owes nothing to any quadrature: a vortex cannot cross a material loop, so the circulation is the sum of the strengths inside, and that sum is computed here by winding number rather than by integration. Over fourteen time units, in which the loop's perimeter grows six-fold, that sum changes by exactly zero while the line integral of the velocity round the same polygon is wrong by 39 per cent. The drift is a property of the measurement and refines away with the loop's resolution, at a different rate from the trajectory error and independently of it.

Tested in The drift was the instrument, at the figure it turns on.

A wing's lift is a function of its incidence, so a measured lift curve describes what the wing will do at any incidence it reaches.

What decides it: Near the stall the lift depends on a separation point which takes time to move, so the wing's lift is a function of the incidence and of the incidence's recent history. Pitched sinusoidally about twelve degrees at a reduced frequency of 0.1, the model here returns 0.22 more lift on the upstroke than on the downstroke at exactly the same incidence, reaches a peak lift 1.14 times the largest the static curve offers, and encloses a loop whose area is the work done on the wing over a cycle. Nothing in the static curve contains any of it.

Tested in Two lifts at one incidence, at the figure it turns on.

The fluid in a spinning container is set into rotation by viscosity, so the time it takes is the time viscosity needs to diffuse across the container.

What decides it: Diffusion is available and is not what happens. The thin Ekman layers on the end walls pump fluid radially outwards, and the interior is drawn down into them and returned — so angular momentum is carried by a secondary circulation rather than by diffusion. The time is H/√(νΩ) rather than H²/ν, and their ratio is the reciprocal of the square root of the Ekman number exactly: a hundred for a 0.1 m tank at one radian a second, 1,118 for a centrifuge rotor, and 342 for the ocean.

Tested in How long a fluid takes to forget it was not rotating, at the figure it turns on.

A multistage compressor built from stages that each work well at their design flow will work well at any shaft speed, provided the throttle is set to give each of them a sensible flow.

What decides it: One mass flow passes every stage, so no throttle setting can give each its own. In an eight-stage machine designed for a pressure ratio of 6.94, each stage tolerating flow coefficients from 0.82 to 1.30 of its design value, the range of mass flow that keeps all eight inside their limits is 0.772 to 0.867 of the design flow at 90 per cent speed and closes completely at 80.06 per cent; at 75 per cent the first stage stops stalling only above 0.654 of the design flow and the last starts choking at 0.607.

Tested in Matched at one speed and at no other, at the figure it turns on.

An air-data system's true airspeed is only as good as its temperature probe, because true airspeed is the Mach number times a speed of sound that depends on nothing but the temperature.

What decides it: The Mach number is taken from the two pressures, and its sensitivity to them grows as 1/M². At 11 km, with pressure sensors good to 25 Pa and a probe to 0.5 K, the pressure sensors put 1.20 per cent into the true airspeed at Mach 0.3 against the temperature channel's 0.11, and still 0.13 against 0.11 at Mach 0.85; the two contributions are equal only at Mach 0.93. At sea level they are equal at Mach 0.52.

Tested in Three readings, and the one each answer leans on, at the figure it turns on.

A steady swell in the open ocean carries surface water, and anything floating in it, steadily downwave at the Stokes drift speed.

What decides it: On a rotating planet the Coriolis force acts on the drift and drives an Eulerian current that answers it. With no friction, the float's mean velocity under an 8 s, 1 m swell at 45° keeps its size, 0.049 m/s, and turns at the inertial frequency, so the float goes round a circle of radius 479 m every 16.9 hours and makes no headway. With any eddy viscosity the steady Eulerian transport is exactly minus the Stokes transport, so the depth-integrated mass transport the drift promised is zero.

Tested in The drift a rotating planet takes back, at the figure it turns on.

Viscosity only ever damps disturbances, so a flow that is stable without viscosity is stable with it.

What decides it: Plane Poiseuille flow has no inflection point and no growing wave without viscosity. With it, a two-dimensional wave of wavenumber 1.0205 grows above a Reynolds number of 5772.22, and at Re = 10⁴ the wave of wavenumber one grows with cᵢ = 0.00374. The energy it grows on is drawn by a Reynolds stress concentrated at the critical layer, which viscosity creates: along the fastest-growing wave, production per unit wave energy falls from 1.05 × 10⁻² at Re 2 × 10⁴ to 4.6 × 10⁻³ at 10⁶ as the viscosity is reduced, and the unstable band narrows towards zero wavenumber.

Tested in The profile Rayleigh cleared and viscosity did not, at the figure it turns on.

A wake is turbulence left over from an aeroplane, so what it can be measured for is how dangerous it is, not what made it.

What decides it: The rolled-up pair carries two numbers that between them determine the aircraft's lift: the circulation of a core and the separation of the pair. For elliptic loading the separation is exactly π/4 of the span, so the span comes out of the second alone, and the weight is rho V Gamma b'. Inverting the relations returns the aircraft to 10⁻¹⁶, and the descent rate is a third measurement that checks the other two. What limits the reading is not the principle but the decay: at a 1.6-minute half life the recovered weight is 42 per cent of the truth after two minutes and 18 after four.

Tested in A wake that says what made it, at the figure it turns on.

Momentum theory gives a rotor's induced velocity at any rate of climb or descent; it is only less accurate in some conditions than in others.

What decides it: In axial flight the momentum balance has two branches, v(V + v) = 1 for climb and hover and v(V + v) = −1 for the windmill-brake state, in units of the hover induced velocity. The first is a streamtube only for V ≥ 0 and the second has a real root only for V ≤ −2, so every descent slower than twice the hover induced velocity has no momentum answer at all. For a rotor loaded at 350 N/m² at sea level that is every descent rate from zero to 23.9 metres a second.

Tested in Between hover and twice the hover inflow, at the figure it turns on.

At a Prandtl number of one the static temperature across a laminar boundary layer follows the Crocco–Busemann relation — total enthalpy linear in velocity — for a wall at any temperature, and the Reynolds analogy St = Cf/2 holds with it.

What decides it: Both are exact only at zero pressure gradient. At a plane stagnation point (β = 1) in a Mach 5 stream at 220 K, over a wall at half the edge total enthalpy, the linear relation puts the temperature peak at 759 K against the solved 684 K and is 158 K out at η = 0.81, and it cannot produce the solved profile's 206 K band below the stream's temperature. The analogy factor 2St/Cf is 0.463 there and 2.96 at β = −0.18.

Tested in The gradient the heat never hears, at the figure it turns on.

In a batch settling test the sediment builds up from the floor behind one sharp front that rises until it meets the falling interface, and settling is then complete.

What decides it: With a Richardson–Zaki hindered-settling flux and a start at φ₀ = 0.1, a single front straight to packing would rise at 0.084 of the particle speed and finish settling at t = 1.86 column-heights per particle speed, and it conserves solids exactly. The admissible solution has a shock rising at 0.148 only as far as φ = 0.317, with a graded fan beneath it; the interface meets it at t = 1.66 and then slows, with 23 per cent of the bed's settlement still to come at t = 10 and 5 per cent at t = 1000.

Tested in The column the chord rule cannot settle, at the figure it turns on.

A venturi's flow rate always rises when its downstream pressure is lowered, because the pressure difference driving the flow has grown.

What decides it: From 5 bar upstream, a venturi with a 5 mm throat in a 25 mm pipe passes more water as the downstream pressure falls — until 4.25 bar, where the throat reaches the vapour pressure of 2.34 kPa and the flow reaches 0.608 L/s. From there to a vacuum downstream it passes 0.608 L/s and not a drop more. The critical downstream pressure is the vapour pressure plus 85 per cent of the upstream margin over it, and 85 per cent is the diffuser's recovery.

Tested in The venturi that stops listening downstream, at the figure it turns on.

The final period of decay has a universal exponent of 5/2, because the equations are linear there and a linear equation has no invariant to choose between.

What decides it: The linear equation solves exactly, mode by mode, and integrating it gives K ∝ t^(−(m+1)/2) where k^m is the spectrum at wavenumbers smaller than any eddy. That is the same m the turbulent stage read. Running the exact solution at m = 2 gives 1.4999870 and at m = 4 gives 2.4999783, against 3/2 and 5/2 — two answers a decade of measurement would separate without effort. The 5/2 everybody quotes is the k⁴ case, and it is quoted because it is the case a laboratory can reach.

Tested in Universal, and one of five, at the figure it turns on.

A rotor's vibration comes from the difference between the advancing and retreating sides, so a rotor in hover — which is axisymmetric — has nothing to vibrate about.

What decides it: A hovering rotor is axisymmetric in its mean flow and is not axisymmetric in its wake: each blade lays down a tip vortex, and the next blade meets it one passage later. At five per cent of the radius the encounter produces 5.5 degrees of induced incidence, reversing sign as the blade passes, within a width of a few per cent of the radius. The peak goes as the reciprocal of the miss distance to the power 1.0000, so halving the clearance doubles the load, and the forcing appears at the blade-passing frequency and its harmonics — twenty per second for four blades at 300 rpm.

Tested in A blade that flies through what it shed, at the figure it turns on.

A variable-speed drive cuts a pump's power as the cube of the flow, so slowing a pump to half its flow saves seven eighths of its power.

What decides it: The cube law is the affinity laws applied to a system whose head falls as the square of the flow, which is a system with no static lift. For a pump drawn for 0.1 m³/s against 40 metres, slowed to half its flow, the shaft power is 0.125 of design with no static lift, 0.250 with 30 per cent of the design head as lift, 0.398 with 60 per cent and 0.564 with 90 per cent, because the head cannot fall with the flow and the pump has been moved away from its best efficiency.

Tested in The specific speed a pump spends its life at, at the figure it turns on.

A supersonic aeroplane's boom reaches the ground wherever the aeroplane is flying faster than sound.

What decides it: The air near the ground is warmer, so sound is faster there and the boom's rays bend back upward. From 11 km on a standard day nothing lands below Mach 1.153; at Mach 1.1 the ray straight under the track turns at 4.0 km and climbs back to flight height 65 km farther on. A 40 m/s headwind at flight height over calm ground raises the cut-off to Mach 1.289.

Tested in The boom that turns back before the ground, at the figure it turns on.

A pump has to push fluid along the channel; a wall that only moves in and out cannot produce a steady net flow.

What decides it: A channel whose half-width carries a sinusoidal wave of amplitude 0.7 of its mean, travelling at speed c, delivers a time-mean flow of 0.590 of c times the mean half-width against no pressure, although no point of the wall moves along the channel. In the frame of the wave the walls are streamlines, and continuity makes the laboratory flow rate past any station q + ch at every instant; its average, q/ca + 1, is the pump's output.

Tested in A wave on the wall is a pump, at the figure it turns on.

The no-slip condition holds wherever a liquid touches a solid, so the liquid at the edge of a spreading drop is held still against the surface like all the rest of the liquid in contact with it.

What decides it: At a moving contact line it cannot hold. Stokes flow in the liquid wedge gives a wall stress that falls as one over the distance from the line — 5,740 Pa a nanometre from a 30° line of water moving at 1 mm/s, 5.7 Pa a micron away — and its integral, the force needed to move the line, diverges logarithmically. With a slip length of one nanometre the force out to a millimetre is 6.93 × 10⁻⁵ N/m; with ten nanometres it is 5.61 × 10⁻⁵. With no slip at all there is no finite answer, and the drop could not spread.

Tested in The drop a no-slip wall would never let spread, at the figure it turns on.

A wing's lift response to a gust is its response to being pitched — Theodorsen's function — so however short the gust, half the quasi-steady circulatory lift survives.

What decides it: A wing flown through a sinusoidal gust has Sears' function for its response, not Theodorsen's. The two agree below a reduced frequency of about a tenth; above it Theodorsen's levels off at one half and Sears' falls as 1/√(2πk), to 0.71 of Theodorsen's at k = 1, 0.25 at k = 10 and 0.11 at k = 50. For an airliner wing of 5 m chord at 230 m/s the root-mean-square gust lift in turbulence of 762 m scale is hardly affected, but the load spectrum at an 8 Hz torsion frequency is 3.88 times below quasi-steady and 1.42 times below what Theodorsen's function gives.

Tested in The gusts that cancel along the chord, at the figure it turns on.

The decay exponent of grid turbulence measures which quantity the large scales conserve.

What decides it: It measures that and the closure together, and cannot separate them. Holding the dissipation coefficient fixed gives 6/5 and 10/7 for the two candidate invariants; letting it fall as one over the Taylor-scale Reynolds number, which is what the near field of a grid does, gives 3/2 and 5/2 for the same two. Batchelor's invariant with a constant coefficient is 1.4286 and Saffman's without one is 1.5000 — five per cent apart, both inside the published band of 1.15 to 1.45, and produced by different physics at both ends.

Tested in The constant that travels, at the figure it turns on.

The Navier–Stokes equations can be solved exactly only where the nonlinear term disappears.

What decides it: The rotating disc's convective terms do not disappear anywhere. At the wall F''(0) = −G(0)² = −1 exactly, so the entire curvature of the radial profile at the surface is the centrifugal term; through the outer layer the balance is carried by the axial convection H F', which is nonlinear because H is the integral of F. The convective and viscous groups reach the same magnitude, and mid-layer the three convective terms cancel to a residue 96 times smaller than any of them. The reduction is exact because every term scales as Ω²r and the radius divides out, not because anything vanishes.

Tested in The solution that keeps its nonlinear term, at the figure it turns on.

A body that has been accelerated violently has left more disturbance in the fluid than one brought up to the same speed gently.

What decides it: In an ideal fluid the force on a body is its added mass times its acceleration, so the impulse left behind is the added mass times the velocity reached — a function of the endpoint and of nothing else. Six acceleration laws are computed here, including one that overshoots and comes back and one that goes backwards before it goes forwards. Their peak forces span a factor of 39.3 and their impulses agree to 2.8·10⁻¹³ of themselves. The energy left in the fluid is the same story: half the added mass times the square of the final speed, in every case, to nine figures.

Tested in Everything about the start, except one vector, at the figure it turns on.

A blade in a turbomachine is designed for the flow it meets, so its unsteady loading is set by its own geometry and its own operating point.

What decides it: Its unsteady loading is set by the row in front of it. A blade in the second row passes through the wakes of every blade in the first row once per revolution, so its incidence carries a comb of harmonics at the upstream blade count and its multiples, and nothing at all at its own row's count. With thirty upstream blades the first harmonic is at order thirty, the sixth is 144 times smaller, and how far the comb extends is decided by how much the upstream wakes have spread — the fourth harmonic falls from 0.53 of the first to 0.0028 of it as the wake width goes from five per cent of a pitch to thirty.

Tested in A row that meets the row before it, at the figure it turns on.

Kolmogorov's four-fifths law and Yaglom's four-thirds law are two separate exact results with two different constants in them.

What decides it: They are the same constant. Both follow from an isotropic radial flux in separation space whose divergence is a constant sink, which integrates to 4Q r/d and gives 4/3 in three dimensions for the scalar and for the velocity's mixed third moment alike. The four-fifths is that four-thirds multiplied by 3/5, and the 3/5 is the exact isotropic relation between the mixed moment and its longitudinal component evaluated at an exponent of one — 3/(n+4), computed by differencing at 0.599999999999982. It is a fact about tensors, not about fluids.

Tested in The fraction that is really four thirds, at the figure it turns on.

An exact solution of the Navier–Stokes equations settles what the flow does.

What decides it: Not when there is more than one. In a wedge of 0.2 radians a flux of 24.71 is carried by a profile with a centreline value of 105.1 running outward everywhere, and by a second profile with a centreline value of 979.9 whose fluid runs backwards along both walls to a depth of −333.9. Both satisfy the full equations exactly, both satisfy every boundary condition, and their fluxes agree to nine parts in a hundred thousand. The reduction is exact and the answer is not unique.

Tested in One channel, one flux, two flows, at the figure it turns on.

A supersonic aeroplane lays a carpet of boom behind it, and each place under the track hears it once.

What decides it: Only in level flight at constant speed, where every ray carries the same invariant and the arrival map is the aeroplane's own motion. Accelerating through the transonic from 15 km at one metre per second squared, the map falls before it rises: the first ray to reach the ground lands 56.0 km past the point of Mach one, and rays launched a few seconds later land 48.8 km — nearer. Every kilometre of the 7.2 km between them is reached by two separate rays, and on the fold itself neighbouring rays converge onto one line where geometrical acoustics predicts an infinite pressure.

Tested in The carpet an accelerating aeroplane folds, at the figure it turns on.

A cavitating venturi isolates the pipe upstream of it from pressure disturbances downstream, because once the throat is choked nothing downstream can reach the flow.

What decides it: The isolation lasts exactly as long as the throat's cavity. A valve shut downstream sends a 14.8-bar surge to the venturi; the cavity absorbs it and the upstream pipe stays at 5 bar while the choked flow and the downstream column's return fill it. A 5 mL cavity closes 8 ms after the surge arrives and 13.8 bar — 93 per cent of the surge — passes into the upstream pipe. Only a cavity large enough to survive the downstream wave's round trip weakens the pulse, and a 256 mL one still passes 7.1 bar.

Tested in A choked throat buys time, not silence, at the figure it turns on.

There is a best area ratio for a jet pump, and a well-designed one is built at it.

What decides it: There are at least three questions an area ratio answers and they do not share an answer. With representative losses the shut-off head peaks at R = 0.800, which is exactly 1/(1 + Kt + Kd) and independent of the nozzle loss; the best efficiency, 33.4 per cent, is at R = 0.275; and the flow ratio at free delivery has no maximum anywhere, rising from 0.20 at the head optimum to 13.5 at a nozzle a fiftieth of its throat. A jet pump is built for one of the three, and is poor at the others.

Tested in The nozzle that is best at one thing, at the figure it turns on.

A flow is determined by the shape of the body, the speed of the stream and the equations, so two experiments in the same tunnel with the same model at the same speed measure the same flow.

What decides it: The exterior of a body is a region with a hole in it, so Laplace's equation and the wall condition have a one-parameter family of solutions rather than one. Six members are computed here: each lets exactly nothing through the surface — the largest normal velocity anywhere is 0 to double precision — while their lifts run from 0 to 37.7 and their peak suctions from −3 to −63. Nothing in the present statement of the problem selects a member. Kelvin's theorem does, from the initial condition, and two start-ups ending in identical present conditions are computed with lifts of 0 and 18.8.

Tested in Nothing in the present picks the flow, at the figure it turns on.

Momentum theory gives the induced velocity from the thrust, so a rotor's inflow can be computed from its instantaneous loading at every moment of a manoeuvre.

What decides it: Momentum theory is a statement about a steady state, and the state takes time to establish. The air the disc has to accelerate has an apparent mass, and putting that mass into the momentum balance gives a first-order lag: a six-metre rotor stepped from 20 to 26 kN reaches 63 per cent of its new induced velocity after 33 milliseconds, against a prediction of 31 from the apparent mass alone. That is 5.07 per cent of the time the wake takes to convect one radius, and the fraction is the same at every rotor size in the family — so the lag is a property of the model rather than of the machine.

Tested in The inflow that takes time to arrive, at the figure it turns on.

The second-order structure function's measured exponent of about 0.70, rather than 2/3, is evidence of intermittency.

What decides it: A model spectrum with no intermittency in it — two smooth factors, its two shape constants solved from the energy and the dissipation rather than fitted to any exponent — gives a local slope of 0.7093 at a Taylor Reynolds number of 200, 0.6957 at 1,000 and 0.6737 at 10,000, all of them at the flattest point of the curve. The departure is a finite-Reynolds-number effect that falls roughly as the reciprocal square root of the Reynolds number, and it is the same size as the departure attributed to anomalous scaling.

Tested in A relation with no turbulence in it, at the figure it turns on.

There are so few exact solutions of the Navier–Stokes equations because nobody has found the others yet.

What decides it: A similarity reduction is a solution invariant under a subgroup of the equations' symmetry group, and that group has eleven generators — verified here by transforming an exact solution and re-differencing the equations from the result, which leaves 3·10⁻⁸ for each symmetry against 0.048 to 4.3 for five transformations that look like symmetries and are not. Of the subsets of those eleven, 66 close under the bracket at dimension three, which is what a reduction to an ordinary differential equation needs. The catalogue is finite, it is short, and it was short before anybody looked.

Tested in Why the list is this long, at the figure it turns on.

A jet pump that is short of flow can be given more by lowering the pressure it discharges into, or by raising the pressure of its motive supply.

What decides it: Neither works once the throat entry has reached vapour pressure. With an 8-bar supply and a suction at 60 kPa absolute, a jet pump with a nozzle 0.28 of its throat entrains more as its discharge pressure falls, until 294 kPa; below that its flow ratio is 0.675 whatever the discharge, down to the suction pressure. And the cavitation parameter is (Pₛ − pᵥ)/(Pₘ − Pₛ), so raising the motive pressure shrinks it: at a 2.3-metre suction lift a 6-bar supply is just able to reach the best-efficiency point, and a 10-bar supply cannot reach it from any lift at all.

Tested in The wall the suction puts in the curve, at the figure it turns on.

Because the Coriolis force cancels the Stokes transport of a steady swell, waves carry no net mass anywhere on a rotating planet except inside the surf zone.

What decides it: The cancellation needs water deep enough that no stress reaches the floor. Integrating the steady equations over depth gives the Coriolis force on the net transport equal to the bottom stress, so a floor that holds a stress keeps transport. With an eddy viscosity of 0.01 m²/s at 55° north the Ekman depth is 12.9 m, and an 8-second, one-metre swell keeps 90 per cent of its Stokes transport over 10 m of water, 45 per cent over 20 and 8 per cent over 40 — a depth-mean drift of 4.3 km a day in the first case.

Tested in The floor that gives the drift back, at the figure it turns on.

The largest pressure anywhere on a hypersonic vehicle is the stagnation pressure behind the bow shock, which is what modified Newtonian theory's maximum pressure coefficient represents.

What decides it: That is true only of streams that reach the surface through the bow shock alone. Where an incident shock from another part of the vehicle crosses the bow shock, the stream between them reaches the surface through two oblique shocks and a normal one and loses far less total pressure. At Mach 8 the worst such jet stagnates at 8.6 times the bow-shock stagnation pressure, a pressure coefficient of 15.9 against the modified Newtonian ceiling of 1.83; at Mach 12, 12.4 times.

Tested in The spot a local theory cannot see, at the figure it turns on.

Blood pressure falls continuously from the heart to the periphery, since the flow loses energy to friction along the way, so the pulse pressure measured at an arm or a leg is a lower bound on the pulse pressure in the aorta.

What decides it: The mean pressure falls; the pulse does not have to. A wave reflected from a load that raises the pressure — a narrowing or a stiffer continuation — returns in step with the outgoing wave near the reflecting end. For a 50 cm tube with a reflection coefficient of 0.6 the pressure pulse at the far end is 1.88 times the pulse at the entrance while the flow pulse there is 0.26 of it; with a coefficient of −0.5 the pressure pulse falls to 0.45 and the flow pulse rises to 1.29. The peripheral pressure pulse overstates the central one whenever the reflection is positive.

Tested in The pulse that grows as it leaves the heart, at the figure it turns on.

The sonic boom at the edge of the carpet is the same bang as under the track, only quieter.

What decides it: The edge ray at Mach 1.8 from 15 km has travelled 2.68 times as far as the ray under the track, almost all of the extra distance through the lowest and densest air, and it arrives nearly level rather than at 50 degrees. Ray theory keeps 0.65 of the under-track overpressure on it and then puts silence one step further out — a cliff no real carpet has. A shock's rise time lengthens with the distance it has aged over, so the edge signature is a slower event as well as a somewhat smaller one, and the sharp edge is where diffraction, not the rays, decides what arrives.

Tested in The edge is a rumble, not a quieter bang, at the figure it turns on.

The entrained stream in a critical-mode ejector is choked in the ordinary sense: it reaches Mach one at the narrowest part of the annulus the primary jet leaves it.

What decides it: Holding the entrained stream at Mach one where its pressure matches the jet's gives a flow that is not the largest the machine can pass. Drawing the entrained gas a little slower raises the common pressure, and the supersonic jet beside it, expanding less, occupies less of the tube; the annulus grows faster than the flux through it falls, and the entrained flow keeps rising until the entrained gas is at Mach 0.886. That maximum satisfies the compound-choking condition for both streams together to 3·10⁻⁸ and entrains 1.6 per cent more than the sonic point; in a narrower tube, 9.6 per cent more.

Tested in A choke that belongs to two streams, at the figure it turns on.

A peristaltic pump, like the waving sheet it mirrors, has an efficiency that does not depend on the amplitude of its wave.

What decides it: The sheet's cost per metre is amplitude-free because its power and its speed both go as the amplitude squared. The pump's useful power and its wall power do not scale together. Its best efficiency is 9φ²/8 for a shallow wave — 4.5 per cent when the wave closes a fifth of the channel — rises to exactly 2 − √3 at half closure and 82 per cent at nine-tenths, and approaches one as the wave closes the tube, with a shortfall of 1.88 times the remaining gap.

Tested in The pump that is better the more it squeezes, at the figure it turns on.

Photographing a flow from more directions sharpens the reconstruction gradually, so a few views give a blurred but essentially correct picture of the field.

What decides it: Below a threshold number of views the missing information is not blur, it is a null space: patterns of density that every view records as exactly zero. Four views of a 16 × 16 field determine 79 of its 256 independent patterns and leave 177 invisible, so that two fields differing by any of them give identical pictures. The best reconstruction from those four perfect views is 43 per cent wrong; eight views leave it 16 per cent wrong; at 14 views the null space vanishes and the error with perfect pictures is zero.

Tested in Four cameras and a field they cannot see, at the figure it turns on.

Turbulent heat transfer follows the flow, so a correlation fitted to air and water — Nu proportional to Re⁰·⁸ Pr⁰·⁴ — will serve for any fluid once the Prandtl number is put in.

What decides it: The correlation assumes the eddies carry the heat. In a liquid metal they do not: at Reτ = 2000 the turbulent heat diffusivity never exceeds molecular conduction anywhere in the pipe for Pr = 0.005, and only by a factor of two for Pr = 0.01. The computed Nusselt number for a liquid metal at a Reynolds number of 131,000 is 16.2, against 45.3 from the air-and-water correlation — and it follows Lyon's liquid-metal correlation instead, which depends on the Péclet number, not on the Reynolds number.

Tested in The heat the eddies do not carry, at the figure it turns on.

The signature at the edge of a sonic-boom carpet has aged over a path nearly three times as long as the one under the track, so it is a much longer and weaker N-wave.

What decides it: The age is not the path. Integrated along the refracted ray with the tube's amplitude and the air's density, the edge ray at Mach 1.8 from 15 km has 1.47 times the under-track age over 2.53 times the path — less than even a uniform medium would give — because 58 per cent of its age is gathered above the tropopause in the first 17 per cent of its path. Its N-wave is 21 per cent longer, not 59, and ageing takes its overpressure from 0.68 of the under-track value, where spreading left it, to 0.56.

Tested in A boom is aged in the thin air it starts in, at the figure it turns on.

Because the four-fifths law has no adjustable constant, the largest value of −Dₗₗₗ/(4r/5) measured in decaying grid turbulence is the dissipation rate.

What decides it: Only at a Reynolds number no grid reaches. The Kármán–Howarth balance of a decaying flow keeps a viscous term that takes the law back at small separations and a decay term that takes it back at large ones, and between them −Dₗₗₗ/((4/5)εr) peaks at 0.63 at a Taylor-scale Reynolds number of 100 and 0.75 at 200. Read as a dissipation, that is an underestimate by 37 and 25 per cent, and the shortfall closes only as about the two-thirds power of the Reynolds number.

Tested in The decay inside the four-fifths law, at the figure it turns on.

A box wing pays less drag than a monoplane for a lighter spar because its fins carry round the closed loop the circulation a monoplane would have to shed.

What decides it: Remove the fins. A biplane of the same span and a fifth-span gap pays 4.7 times the squared moment reduction against the box's 3.9 and the monoplane's 8, so most of the halving is the second wing. And as the moment is held down the fins unload: their side force falls from 0.049 to 0.006 at 80 per cent of the monoplane's moment, the box's lead over the biplane falls from 0.067 to 0.001, and the two break even with the monoplane at the same moment, 0.792.

Tested in A lighter spar turns a box wing into a biplane, at the figure it turns on.

Two coaxial vortex rings travelling the same way leapfrog: each in turn is drawn through the other, indefinitely in an ideal fluid.

What decides it: Only above a ratio of radii. Started in one plane with cores six hundredths of the larger radius, a pair whose smaller ring is 0.30 of the larger's radius separates for good — the smaller ring widens to 0.52 and the larger shrinks to 0.90, and the gap keeps opening — while a pair at 0.50 exchanges places every 8.3 time units. The dividing ratio, 0.340, follows from energy and impulse alone and matches the marched threshold to the width of the search.

Tested in Two rings leapfrog only if they start alike, at the figure it turns on.

Kirchhoff's ellipse is an exact solution of Euler's equations, so an elliptical vortex of any elongation will keep its shape and turn steadily.

What decides it: Only up to an aspect ratio of three. Contour dynamics started from an ellipse of aspect ratio 4 with a three-lobed bump of a ten-thousandth grows the bump at 0.1045ω, within 1.6 per cent of Love's 0.1061ω — more than fifty-fold in one turn of the ellipse — and a bump of three thousandths has pulled one end into a filament within a turn. The same bump on an ellipse of aspect ratio 2.5 never exceeds 1.8 times its starting size.

Tested in Past three, an ellipse is a shear layer, at the figure it turns on.

The streamlines in a picture of a flow are the paths the particles take.

What decides it: One unsteady field, integrated two ways with the same integrator: frozen at an instant for the streamlines, advanced in time for the pathlines. Only the treatment of time differs, and the two families come out visibly different — they coincide only when the flow is steady.

Tested in Streamlines are not the paths particles take, at the figure it turns on.

Bernoulli's equation explains the pressure difference across a wing: the air goes faster over the top, so the pressure there is lower.

What decides it: The theorem's hypotheses, taken one at a time to the places it is routinely invoked — between two streamlines, through a fan, inside the boundary layer, across a shock. Four of the five rows break one.

Tested in Where Bernoulli's equation applies, at the figure it turns on.

Bernoulli's equation still works above Mach 0.3 as long as you use the local density.

What decides it: The quantity Bernoulli conserves is p + ½ρU², and it is conserved because ρ was pulled out of an integral. With ρ varying, the integral gives the enthalpy instead, and p + ½ρU² is simply not constant: at Mach 0.85 the exact stagnation pressure exceeds the incompressible estimate by 19.4%, and the discrepancy is the term that was dropped rather than an error that can be absorbed into a local value.

Tested in Energy instead of pressure, at the figure it turns on.

A scale model in a wind tunnel reproduces the flow over the full-size aircraft.

What decides it: The Reynolds and Mach loci computed from their definitions at sea-level air. Full scale and quarter scale are parallel on the logarithmic axis, displaced by exactly log 4, so they never intersect: the gap is 0.60 in log₁₀ Re and 2.11 in Ma, and closing one opens the other.

Tested in The model that cannot be matched, at the figure it turns on.

The Reynolds stress is a stress, in the same sense that the viscous stress is.

What decides it: It has the dimensions of a stress and it enters the averaged equation in the same place as one, and there the resemblance stops. A viscous stress is a molecular momentum transfer whose relation to the local rate of strain is a property of the fluid, measured once and valid everywhere. The Reynolds stress is the mean momentum flux carried by the resolved motion itself, has no constitutive relation of any kind, and takes different values in different flows at the same rate of strain. Treating the analogy as though it were a derivation is what produces an eddy viscosity, and an eddy viscosity is a fitted quantity that varies by orders of magnitude within one flow.

Tested in What averaging costs, at the figure it turns on.

The dynamic pressure is ½ρU², and a pitot-static system measures it directly.

What decides it: ½ρU² is the leading term of a series whose corrections are all positive. The exact ratio (p₀ − p)/½ρU² is 1.0227 at Mach 0.3, 1.1939 at Mach 0.85 and 1.6573 at Mach 2, verified monotone across that whole range, so a speed inferred from ½ρU² alone reads high — and the series 1 + M²/4 + M⁴/40 tracks the exact value to within 0.2 per cent up to Mach 0.9.

Tested in What the airspeed indicator believes, at the figure it turns on.

Streamlines drawn close together mean fast flow, so a picture of streamlines can be read as a picture of speed.

What decides it: The exact statement is that the flow rate between two streamlines is fixed, so speed times gap is constant along a streamtube. Measured on the solved field at four stations, the gap times the mid-tube speed comes out at 0.161, 0.170, 0.166 and 0.160 against an exact flow rate of 0.160 — the rule is a limit for a thin tube, not an identity, and in three dimensions or a compressible flow it is not even that.

Tested in The number on a streamline is a flow rate, at the figure it turns on.

The logarithmic law of the wall is derived from the Navier-Stokes equations.

What decides it: It is derived from Prandtl's mixing-length closure, which is an assumption added to those equations because they do not determine the mean profile by themselves. The derivation is exact given the assumption — integrating the closure numerically here returns 1/kappa at 0.4076 against the 0.41 that went in — and the assumption is a guess about eddy size with no derivation behind it. A separate dimensional argument reaches the same logarithm without the mixing length, and it too requires an assumption: that the profile in the overlap region depends on neither viscosity nor the outer scale.

Tested in A guess with a constant in it, at the figure it turns on.

The most efficient speed to fly at is the one where drag is least.

What decides it: Drag is a force and staying airborne costs power, which is force times speed. Setting the derivative of power to zero gives a speed of 3^(−1/4) = 0.7598 times the minimum-drag speed; at that speed the drag is 15.5% higher and the power is 12.3% lower. Which speed is efficient depends on whether the aim is distance or time.

Tested in The cheapest way to stay up, at the figure it turns on.

The pressure coefficient is the pressure, in convenient units.

What decides it: The same cylinder at three tunnel speeds. The gauge pressure at the shoulder is −1.50, −24.00 and −384.00 as the speed goes 1, 4, 16; the coefficient is −2.999999 in all three cases. One of those quantities is a fact about the shape and the other is a fact about the tunnel, and they cannot both be called the pressure.

Tested in The number that does not depend on the tunnel, at the figure it turns on.

An elliptical wing is the efficient shape, which is why the Spitfire had one.

What decides it: The lifting-line solve gives a span efficiency of 1.0000 for elliptic loading, 0.9869 for a tapered wing and 0.9368 for a rectangular one — so the planform is worth at most 6% of the induced drag. Over the same comparison, changing the aspect ratio from 4 to 28 at a fixed lift coefficient changes the induced drag by a factor of seven.

Tested in The span is the whole story, at the figure it turns on.

Lift is the Coandă effect: the air follows the curved upper surface, and that is what makes the low pressure there.

What decides it: The mechanism named is real and is the normal momentum balance across curved streamlines. What separates the two flows is where the integral of that balance comes from. For a wall jet on a convex surface everything outside the jet is at rest and contributes nothing, so all of the deficit is made within the jet's own thickness — two per cent of the radius. For a cylinder in a stream, half of the deficit is made beyond a third of a radius from the surface and a tenth beyond 1.67 radii.

Tested in The effect that is real, and where it stops, at the figure it turns on.

Bernoulli's equation says that where the flow is fast the pressure is low, so a faster region of a flow always has a lower pressure than a slower one.

What decides it: In a parallel shear flow — an exact steady solution of the Euler equations — the static pressure is uniform, exactly, everywhere. The fast side and the slow side are at the same pressure, and the total pressure varies across the layer by 3.2 dynamic heads. The statement that fast means low pressure holds along a streamline and holds across streamlines only if the flow is also irrotational, which is a second and stronger hypothesis.

Tested in Four Bernoullis and one name, at the figure it turns on.

Turbulent mixing can be treated as a diffusion with an enhanced coefficient, in the same way that molecular mixing is a diffusion with a small one.

What decides it: Only after a parcel has forgotten its own velocity. Taylor's integral gives the mean square displacement as 2u²∫(t−τ)R(τ)dτ, which is u²t² at short times — a straight line, not a random walk — and 2u²Tt only once t is several correlation times. The running diffusivity computed here reaches half its limiting value after 1.7 correlation times and is still six per cent short after twelve, so a diffusive description applied to a plume within a few eddy turnovers of its source overstates the spreading by a factor that depends on how long ago the source was.

Tested in How far a parcel gets, at the figure it turns on.

The turbulence intensity measured across the edge of a jet peaks where the turbulence is strongest, so the profile of the measurement is a profile of the turbulence.

What decides it: A conventional average across an intermittent edge mixes turbulent fluid with irrotational fluid moving at a different mean speed, and the variance acquires a term γ(1−γ)(ΔU)² that is the variance of the switching rather than of the turbulence. Computed here from a two-state model whose conditional intensity is flat by construction, the measured intensity peaks twenty per cent above it at γ = 0.64 — a peak with no counterpart anywhere in the turbulent fluid.

Tested in Turbulent some of the time, at the figure it turns on.

A viscoelastic material has a relaxation time, and the Deborah number formed from it says whether the material behaves as a solid or as a liquid.

What decides it: The statement is exactly true of a Maxwell fluid, which has one relaxation time by construction, and the crossover it predicts sits at λω = 1 with no coefficient in it. A real polymer has a spectrum, and summing the same two formulas over a Rouse chain gives moduli that never separate: the loss tangent settles at 1.04 instead of falling as 1/λω, so the ratio of the two moduli at λ₁ω = 100 is 1.0 rather than 100. The crossover stops being a point and becomes a graze 2.98 decades wide, thirty-one times wider than the single time's 0.095, and where it sits is decided by the number of modes rather than by any property called "the" relaxation time.

Tested in A solid, if it is not given time, at the figure it turns on.

The hydraulic diameter lets one friction correlation cover any duct shape.

What decides it: In turbulent flow it is good to about ten per cent, and it is the reason the idea is worth having. In laminar flow it is not a correlation with scatter, it is a different number for every shape, known exactly: 64 for a circle, 96 for parallel plates, 160/3 for an equilateral triangle, 56.908 for a square. Using the circular value on parallel plates under-predicts the pressure drop by a third, and the error is not scatter — it is a factor of 1.8 across the shapes, computed to four figures.

Tested in A number that is only the shape of the hole, at the figure it turns on.

Fitting Morison's equation to a measured force record determines the drag and inertia coefficients of the body.

What decides it: Fitting the peak does not. At KC = 10 there is a one-parameter family of (C_D, C_M) pairs whose maximum force is identical to fifteen digits, and the energy they dissipate spans a factor of exactly 4 across it. Worse, over the lower half of that family the inertia coefficient does not move at all: when the peak is the inertia peak it occurs where the velocity is zero, so it carries no information whatever about the drag coefficient. The phase of the record separates them; its maximum cannot.

Tested in Long enough to make a wake, at the figure it turns on.

A thicker aerofoil has a higher lift-curve slope, because potential-flow theory says the slope rises with thickness.

What decides it: Potential flow does say that, and the Joukowski map gives it exactly: 2π at zero thickness and 7.098 per radian at 16.9 per cent, which is 2π(1 + 0.766 t/c). Measurements say the opposite — the slope of a real section falls with thickness, by roughly seven per cent at fifteen per cent thickness — because the boundary layer thickens towards the trailing edge and decambers the section. Thin-aerofoil theory, which has no thickness term at all and returns 2π for every section, is closer to the measurement than the exact inviscid answer is.

Tested in The half that carries nothing, at the figure it turns on.

The Kutta condition is a law of fluid mechanics, so a real aerofoil's circulation is determined by its shape and its incidence.

What decides it: It is a selection rule whose justification is a corner. Computed on a Joukowski-mapped circle with the circulation as a free parameter, a cusped section's trailing-edge speed is about U at the Kutta value and 11.3U half a Kutta circulation away, and grows without bound as the sample approaches the edge — refining the sample four times raises it four times. A section with a *rounded* trailing edge has a peak surface speed of 4.54U over the same range of circulation, unchanged under refinement, and no circulation is picked out by anything. Blowing over such an edge chooses it, and lift coefficients of five to eight follow.

Tested in The condition that can be bought, at the figure it turns on.

A yield-stress fluid below its threshold flows very slowly, so in practice the threshold is a soft one.

What decides it: In the model it is not soft at all: the flow rate is identically zero at every pressure below the threshold, for any length of time. What is soft is the model. Give the unyielded region a very large but finite viscosity — which is what every numerical code actually solves — and the fluid creeps at every pressure, so the yield stress a flow measurement infers becomes exactly proportional to the flow rate that measurement calls zero: one decade of detection threshold, one decade of apparent yield stress, with an exponent of 1.0000000000000016.

Tested in The core that does not move, at the figure it turns on.

A measured jet profile that collapses onto the similarity solution confirms the theory that produced it.

What decides it: It confirms the shape and the shape is shared. The laminar solution is exact Navier–Stokes and the turbulent one is the same formula with a fitted eddy viscosity in place of the molecular one; scaled on their own half-width and centreline speed the two are identical to seven parts in ten to the sixteenth, while their spreading rates differ by whatever the fitted constant is. A collapse of profiles tests the assumption of self-similarity, which is the cheap half; the rate is the half that carries the physics, and it is the quantity the model was fitted to.

Tested in Exactly similar, and one number short, at the figure it turns on.

A flow picture that looks right has very nearly the right energy, so energy is a good way to check it.

What decides it: It has very nearly the right energy for exactly the reason it cannot be checked that way. The excess energy of any admissible field is the energy of its own difference from the true flow, so the energy error is the square of the velocity error — a field wrong by thirty-one per cent in speed everywhere carries an energy nine and a half per cent too high, and one wrong by three per cent carries an energy wrong by one part in a thousand. The quantity that identifies the true flow is the least sensitive instrument available for finding out how far from it you are.

Tested in How much more than the least, at the figure it turns on.

A liquid poured slowly from a spout runs back under it because of the Coandă effect: the stream follows the curved surface for the same reason a jet of air does.

What decides it: An air wall jet a millimetre thick at 50 m/s on a 5 cm radius needs 60 Pa to be turned and has the atmosphere's 101 kPa to borrow, since the wall seals the side its entrainment empties — a margin of 1683. A water sheet in air has no such pressure to borrow; it is held by its own two surface tensions, 2σ against a momentum flux ρU²h, and lets go when they are equal, at √(2σ/ρh) — 0.270 m/s for a sheet two millimetres thick. The lip radius cancels from that balance, to within 0.94 per cent on a 5 mm lip, so the threshold is a speed and not a curvature.

Tested in The teapot effect is a tension, not a pressure, at the figure it turns on.

Because Newtonian theory becomes nearly exact at hypersonic speed, its rule that a surface turned away from the stream carries no pressure becomes nearly exact too.

What decides it: The leeward face of a flat plate is bounded by the vacuum coefficient −2/(γM²), and in the hypersonic limit its share of the normal force depends on K = M sin α alone: 24 per cent at K = 1, 11 per cent at K = 2 and 6 per cent at K = 3. A plate flies at small K when it flies well: with a friction coefficient of 0.001 on both faces the exact plate's best lift-to-drag ratio at Mach 10 is 7.34 at 4.2°, with 30 per cent of its normal force from the shaded face, against the Newtonian 5.07 at 7.4°.

Tested in The face Newton left in shadow, at the figure it turns on.

A jet engine is a more advanced way of producing thrust than a propeller, so it is the more efficient one.

What decides it: The efficiency with which thrust is produced — as distinct from the efficiency with which fuel is turned into power — depends on one thing only, and it is not the engine. It is the ratio of jet speed to flight speed, and the propulsive efficiency is 2/(1 + sigma) exactly. A propeller moving a great deal of air slowly sits near sigma = 1.1 and is above 95 per cent before anybody has designed anything; a turbojet throwing a thin jet at three times flight speed is at 50 per cent and cannot exceed it whatever its core does. The audit that gives the result closes to twelve decimal places here. What a jet is better at is producing power at high flight speed and high altitude, which is a different question, and the bypass ratio of every airliner since 1970 is the industry conceding this one.

Tested in A big slow push, at the figure it turns on.

A velocity profile with a point of inflection is unstable.

What decides it: Rayleigh's theorem runs the other way: an inviscid parallel flow with NO inflection point cannot be unstable. The converse is false, and Tollmien's 1935 refinement shows what has to be added — the vorticity must have an extremum and the profile must satisfy a further condition at it. The figures here find the inflection points by searching a solved profile for a sign change in its second derivative, and finding one is reported as finding one, not as a verdict.

Tested in A layer with a kink in it, at the figure it turns on.

A wing holds itself up by pushing air downwards, and the downwash behind the wing is where the lift is to be found.

What decides it: The momentum theorem evaluated face by face on a rectangle drawn in the fluid, touching the body nowhere. At eight degrees the box returns a lift of 2.9193 against ρUΓ = 2.9193 — and the downstream face, which is the one the popular account keeps, carries the least of it.

Tested in Air must be pushed down, and the usual sum is wrong, at the figure it turns on.

Momentum theory says a turbine can reach 59.3 per cent, so the difference between that and what a real machine achieves is blade losses — friction, tip vortices and imperfect aerofoils.

What decides it: Some of it is, and a computable part of it is not. A rotor extracting power applies a torque, and the reaction leaves the wake rotating; that rotational kinetic energy is taken out of the stream and never reaches the shaft, and it is present with blades of zero drag and infinite span. Glauert's optimum rotor, which has perfect blades and accounts only for the swirl, gives 0.4155 at a tip-speed ratio of one and 0.5759 at six — a shortfall against Betz of 1.7 percentage points in the band a large turbine runs in, before a single real loss is counted. The gap closes only as the tip-speed ratio goes to infinity, which is why turbines are geared the way they are.

Tested in The wake that has to spin, at the figure it turns on.

A nozzle narrows to make the flow go faster. That is what a nozzle is.

What decides it: True below Mach one and false above it. In dA/A = (Ma²−1) dV/V the bracket changes sign at Mach one, so a narrowing duct decelerates a supersonic flow. The area a stream tube needs, A/A*, has its minimum at Mach one exactly — verified to 1e-9 on a tabulated sweep — and rises on both sides of it, which is why a rocket nozzle has a throat and then flares out.

Tested in The duct that works backwards, at the figure it turns on.

A smoother pipe always carries more flow for the same pressure, so polishing a duct is always worth something.

What decides it: Only when the roughness is large enough for the flow to notice it, and whether it is depends on the speed rather than on the pipe. The roughness height in wall units, k+ = (eps/D) Re sqrt(f/8), is what decides: below about 5 the bumps are buried inside the viscous film at the wall and the pipe behaves as though polished, and a relative roughness of 10⁻⁴ at Re = 10⁴ sits at k+ = 0.06 and costs nothing measurable. Above about 70 the viscosity has stopped mattering entirely and the friction factor no longer depends on the Reynolds number — checked here at a relative roughness of 0.02, where a decade of Reynolds number moves the friction factor by 0.007 per cent. Between those two the answer is neither.

Tested in The roughness a wall cannot feel, at the figure it turns on.

The Mach number at a point in a duct is determined by the area there.

What decides it: The area–Mach relation is double-valued: A/A* = 2.5 is satisfied at Mach 0.2395 and at Mach 2.4428, both roots verified to return the area to 1e-9. Which one the flow is on is fixed by the back pressure imposed at the far end, so the answer at a station depends on a boundary condition the station cannot see.

Tested in One area, two answers, at the figure it turns on.

Water flowing over a bump in the bed has to speed up to get past it, so the surface dips over the crest.

What decides it: It dips only if the flow is subcritical. A supercritical stream over the same bump gets DEEPER and slower, and both are consequences of the same curve: raising the bed takes energy out of the flow, and on the upper branch of the specific-energy curve less energy means less depth while on the lower branch it means more. Computed on the same bump here, a 0.60 m approach at Froude 0.34 falls by 58 mm over the crest while a 0.10 m approach at Froude 5.05 rises. Which way the surface goes is a question about the Froude number, not about the bump.

Tested in The depth that costs least, at the figure it turns on.

A wing stalls when the air can no longer follow the sharp curve over the top of it, which is a matter of the shape.

What decides it: The criterion is not curvature but pressure recovery. Reading the inviscid surface velocity as a local Falkner–Skan power law and comparing with the separation value β = −0.198838, the estimate puts laminar separation at 31% of chord on this section at zero incidence, moving to 3% by six degrees — on a shape whose curvature has not changed at all.

Tested in Where the straight line stops, at the figure it turns on.

Dimensional analysis tells you what a flow depends on.

What decides it: It tells you how MANY numbers the answer can depend on, and it is silent about which quantities belong on the list. The theorem is a statement about a matrix: the drag list — force, density, speed, diameter, viscosity — gives a dimension matrix of rank three, so five minus three is two groups, and every dimensionless combination of those five is a product of powers of two independent ones. Which two is not determined either: the drag coefficient comes out as one combination of the computed basis and the Reynolds number as another, and the reciprocal of either would have served. What the theorem cannot supply is the physics of the list. Leave the viscosity out and it returns one group and a confident wrong answer, and nothing in the arithmetic will object.

Tested in Counting what matters, at the figure it turns on.

A spinning cylinder can be spun faster and faster to make as much lift as wanted, and the flow just goes round faster.

What decides it: The lift does keep rising — it is ρUΓ with nothing in the way — but the flow pattern changes character at Γ = 4πUa. Below it there are two stagnation points on the surface, at sin θ = Γ/4πUa; above it there are none, and the single stagnation point sits out in the fluid, at 1.6052 radii below the centre when Γ is 14 against a threshold of 12.5664.

Tested in How much circulation is too much, at the figure it turns on.

Viscosity is what makes a turbulent flow lose energy, so a less viscous fluid dissipates less.

What decides it: The dissipation rate in a turbulent flow is set by the large scales and is independent of the viscosity — epsilon is of order u cubed over L, with no nu in it. Lowering the viscosity does not lower the dissipation; it moves the scale at which the dissipation happens, as eta over L equals Re to the minus three-quarters. Viscosity decides WHERE the energy leaves, and the large eddies decide HOW MUCH. This is the dissipation anomaly, and it is why the inertial range widens rather than the losses falling.

Tested in Where the energy goes, at the figure it turns on.

Vortex stretching amplifies vorticity without limit, so a real fluid would develop infinite spin.

What decides it: The amplification is exact and the conclusion does not follow, because thinning a tube brings its own brake with it. Stretching a tube by a factor of lambda multiplies its vorticity by lambda and divides its radius by the square root of lambda, and viscous diffusion spreads vorticity outwards at a rate that goes as the inverse square of that radius. So the brake strengthens as lambda squared while the drive strengthens as lambda, and the two balance at a definite core radius. Burgers' vortex is the exact steady solution of that balance: the core settles at the square root of four nu over alpha, the peak vorticity settles at Gamma alpha over four pi nu, and the residual of the steady vorticity equation, differenced off the profile the figures draw, is under 10⁻⁹. At a strain rate of 2 and a viscosity of 0.005 the core is 0.100 in units where the circulation is one.

Tested in The spin that feeds itself, at the figure it turns on.

Smaller droplets are collected less efficiently than large ones, so a fine enough mist is collected inefficiently.

What decides it: Below a critical Stokes number the collection is not inefficient, it is zero — no droplet released at any offset whatever reaches the body. The reason is an eigenvalue rather than a trend. Near the front stagnation point the flow is a pure deceleration u = −Aξ and the particle equation is linear: St·ξ″ + ξ′ + Aξ = 0, whose roots are real while 4A·St < 1. Two decaying exponentials approach the wall and never reach it; past the threshold the roots go complex, the approach becomes an oscillation, and an oscillation crosses zero. A is measured off the velocity field at 2.000000 U/a, so the threshold is exactly one eighth, and a search on the stagnation streamline finds 0.1265 — above it, as any finite integration must be, because at the threshold the arrival takes forever.

Tested in Whether the droplet turns, at the figure it turns on.

Supersonic flow always goes subsonic behind a shock.

What decides it: True for a normal shock and false in general. A 10° wedge at Mach 2 carries a weak shock at β = 39.31°, across which only the normal component M sin β = 1.267 is processed, and the flow leaves at Mach 1.641 — still supersonic. The oblique relations reduce exactly to the normal ones at β = 90°, verified ratio for ratio to 1e-9.

Tested in A shock that leans, at the figure it turns on.

Making a boundary layer turbulent always increases the drag, so a smooth surface is always better than a rough one.

What decides it: It always increases the friction drag and it often decreases the pressure drag by more. On a sphere the transition from laminar to turbulent separation takes the drag coefficient from about 0.5 to about 0.1 — a fivefold reduction in TOTAL drag, bought by increasing the smaller of the two contributions. Which way the trade goes is decided by the shape: on a streamlined body at small incidence friction dominates and smoothness wins; on a bluff body pressure drag dominates and roughness wins. The dimples on a golf ball are the second case.

Tested in The cost of going turbulent, at the figure it turns on.

d'Alembert's paradox says an ideal fluid exerts no force on a body moving through it.

What decides it: The energy in the fluid around a cylinder moving at U, integrated over the exterior of the plane, comes to ½ρπa²U² — so changing U requires work, and the force needed is that of a body heavier by ρπa² per unit length. The paradox is about steady motion only, and the integral that proves it converges at all is 3.141513 against ρπa² = 3.141593.

Tested in The force of getting going, at the figure it turns on.

A siphon works because atmospheric pressure pushes the liquid up the short leg and over the hump — which is why a siphon cannot work in a vacuum, and why it cannot lift liquid more than about ten metres.

What decides it: The flow rate is the square root of 2g times the drop from the source surface to the outlet, and the height of the crown does not appear in it at all: sweeping the crown from 0.05 m to 8.95 m returns bit-identical exit velocities, because the crown height is not in the expression. So the atmosphere is not doing the driving; the level difference is. What the atmosphere DOES is keep the pressure at the crown above the liquid's vapour pressure, and that is where the ten-metre limit comes from — a limit on holding the column together rather than on pushing it. Degassed water, which sustains tension rather than boiling, has been siphoned over fifteen metres, which the atmospheric account says is impossible.

Tested in The siphon that does not need the air, at the figure it turns on.

A boundary layer's thickness is set by the geometry — the length of the plate, the size of the body, the width of the channel.

What decides it: Here is a boundary layer with no geometry in it at all. An oscillating wall is infinite in extent and has no length anywhere in the problem, and the fluid above it still organises itself into a layer of a definite thickness: √(2ν/ω), built from the viscosity and the frequency alone. In air at 100 Hz that is 0.219 mm and in water at the same frequency 0.056 mm, and neither number knows anything about the wall. What sets a layer's thickness is a COMPETITION — how far momentum diffuses against how long it has to do it in — and the plate length in Blasius' problem enters only as the time x/U that a parcel has been beside it.

Tested in The wall that shakes, at the figure it turns on.

A sharp, streamlined nose is best at supersonic speed, because it cuts through the air more cleanly.

What decides it: Sharp is best for drag and worst for heating. The detachment angle at Mach 2 is 22.974°, so a nose blunter than that carries a detached bow shock — which is nearly normal near the axis, loses most of the total pressure there, and dumps the corresponding heat into the gas rather than into the vehicle. Every re-entry vehicle ever flown is deliberately on the wrong side of that boundary.

Tested in When the wedge is too blunt, at the figure it turns on.

Chaotic systems are unpredictable, so a better measurement of the initial state does not help.

What decides it: It helps logarithmically, which is a definite and computable amount rather than none. The separation grows as e to the lambda t, so the time before a given error is reached goes as the log of the initial precision divided by lambda. On the Lorenz system at r = 28, with a measured lambda of 0.8982, improving the initial state by a factor of a million buys ln(10^6)/0.8982 = 15.4 additional time units. Halving the error buys 0.77. The exchange rate is poor and it is not zero.

Tested in A millionth is enough, at the figure it turns on.

Sucking the boundary layer away reduces drag, because it removes the slow fluid that was causing it.

What decides it: The friction on a uniformly-sucked wall is EXACTLY twice the suction coefficient — c_f = 2V/U, with nothing left over — because the momentum integral for a wall with transpiration reads c_f/2 = dθ/dx + V/U and this layer's momentum thickness does not depend on x at all. So the drag is precisely the momentum of the fluid that was swallowed, and suction cannot reduce friction below its own price. The differenced wall slope gives 1.9999997e-3 against an identity value of 2.0000000e-3 at V/U = 0.001. What suction actually buys is elsewhere: it holds the layer thin and laminar where it would otherwise have gone turbulent, and at a Reynolds number of four million the turbulent friction it avoids is 0.0028 against the 0.0020 it costs.

Tested in The layer that stops growing, at the figure it turns on.

Blood vessels and tree branches get narrower as they divide because there is less room for them — the geometry is a consequence of packing, not of anything hydrodynamic.

What decides it: Packing constrains where branches go and does not set their radii. Minimising the sum of pumping power, which goes as Q squared over r to the fourth, and the cost of owning the tissue, which goes as r squared, gives a best radius that makes Q proportional to r cubed — computed here by golden-section search and matching the closed form to eight figures. Two testable consequences follow: the cube of a parent's radius equals the sum of the cubes of its branches, and the wall shear stress is the SAME in every vessel of the network, which comes out identical to twelve decimal places across a four-to-one flow split. The second is a quantity a cell can sense, so the optimum is something a vessel can grow towards.

Tested in The radius that costs least, at the figure it turns on.

A wing put to an angle of attack has its lift immediately; the circulation appears as soon as the Kutta condition is satisfied at the trailing edge.

What decides it: The Kutta condition is satisfied at every instant here and the circulation still takes tens of chords to arrive. The reason is Kelvin's theorem: bound circulation can only be acquired by shedding its opposite, and a freshly shed vortex sits just behind the trailing edge where its induced downwash is strongest, cancelling most of the incidence the wing was set to. The lattice computes a first-step circulation of 0.14 of the settled value, half of it after 1.5 semichords, nine tenths after 14, and 99.3 per cent after 200 — approaching 2 pi alpha from below the whole way, which nothing in the model was told to do. What is instantaneous is the Kutta condition; what is slow is the wake getting out of the way.

Tested in The lift that arrives late, at the figure it turns on.

A shock tube produces a shock wave, and the rest of what happens in it is the details.

What decides it: It produces three waves of three different kinds, and only one of them is a shock. The exact solution of Sod's problem has a shock running into the low-pressure gas at Mach 1.656, an expansion fan spreading the other way as a family of rays, and between them a contact surface across which the density jumps by 61 per cent while the pressure and the velocity are continuous. The contact is not a wave at all — nothing propagates through it, it is simply the boundary between the gas that was on one side of the diaphragm and the gas that was on the other, and it moves at the fluid's own speed. Every compressible solver ever written is checked against exactly this, and the check that matters most is whether it can keep the contact sharp.

Tested in One diaphragm, every wave, at the figure it turns on.

A wing close to the ground rides on a cushion of air trapped and compressed beneath it, and that cushion is what gives it extra lift.

What decides it: Computed at the heights aircraft actually fly at, the air under the wing is not trapped and is barely compressed. At four tenths of a chord above the ground — a large airliner in the flare — the pressure under the wing reaches a coefficient of 0.35 out of a possible 1.0, and the mass flux through the gap is 81 per cent of what would pass through the same gap with no wing there at all. The lift increase at that height is 25 per cent, and it comes from the image system reducing the downwash rather than from any air being held. The cushion story does become true eventually: at a tenth of a chord the pressure coefficient reaches 0.96, the flux falls to 20 per cent, and the air genuinely is nearly stagnant — which is a ground-effect vehicle's regime and not an aeroplane's. What the story never mentions is the larger half of the effect, which is drag: at a tenth of a span the induced drag is 48 per cent lower, computed two independent ways agreeing to 10⁻¹².

Tested in The cushion that is not there, at the figure it turns on.

Because air sticks to a surface, a moving wing drags a layer of air along with it, and skin friction is the cost of hauling that air.

What decides it: Both halves are checkable and both come out badly. The amount of air genuinely travelling with the wing is the displacement thickness — the mass deficit in the layer — and at a Reynolds number of a million it is 1.72 millimetres per metre of chord against a layer 4.91 millimetres thick, so under a third of the layer is being carried in any sense. It is not the same air: only the fluid exactly at the wall shares the wing's velocity, 69 per cent of the layer's depth is moving at more than half the free-stream speed relative to the wing, and the contents are replaced continuously as the layer entrains from outside. And the drag prediction has the wrong sign: skin friction is μ ∂u/∂y at the wall, which falls as the layer thickens — at Re 10⁵ the local coefficient is 2.10 × 10⁻³ and at Re 10⁷ it is 2.10 × 10⁻⁴, a factor of ten less drag with a thicker layer, while the mass carried has grown.

Tested in The air a wing does not carry, at the figure it turns on.

A wave transports its shape and its energy but not the water itself, so a floating object rises and falls in place while the wave passes under it.

What decides it: True at first order and false at second. The Eulerian mean velocity at a fixed point is exactly zero, and the trajectory of a parcel — integrated here through the exact linear field — does not close: it advances by the same small amount every cycle, at a rate matching (ak)²c e^(2kz) to 0.16 per cent at a steepness of 0.025. For an ocean swell of one metre amplitude and eight seconds' period that is five centimetres a second, which moves a floating object four kilometres a day.

Tested in The drift in a wave that has none, at the figure it turns on.

A turbine works because the gas pushes on the blades, so the work it does is the pressure drop across it multiplied by the volume that passed through.

What decides it: The work is the change of angular momentum, and pressure need not change across a rotor at all. An impulse stage has the same static pressure either side of its rotor and does full work by turning the flow; a rotor with a pressure drop across it and no change in swirl does exactly none. Euler's equation contains no pressure term and no blade shape: the work per unit mass is U2 V-theta-2 minus U1 V-theta-1, computed here two ways that share no arithmetic and agreeing to twelve decimal places, and a stage with equal swirl on both sides returns exactly zero. What the pressure change decides is the degree of reaction, which is how the work is split between the rotor and the stator, and not how much there is.

Tested in Work out of a change of swirl, at the figure it turns on.

A porous wall can be treated as an ordinary solid wall, since the seepage velocity through it is negligible compared with the flow along the channel.

What decides it: The seepage is indeed negligible and the slip is not. Matching Brinkman's equation inside the medium to Stokes flow above it gives an interfacial velocity of √K times the shear rate plus the Darcy velocity — the first of order √K and the second of order K. For a pore scale one per cent of the channel height the flow rate is 3.03 per cent above what a no-slip wall gives, and a wall told only u = √K du/dy reproduces it to 0.058 per cent. It is the term the claim keeps that is negligible, and the one it discards that is not.

Tested in A wall that is not quite there, at the figure it turns on.

The logarithmic velocity profile near a wall is a consequence of Prandtl's mixing-length hypothesis, and stands or falls with it.

What decides it: The mixing length does imply the logarithm — integrating it gives the profile, and κ comes back out of the integration — but so does every other closure that has replaced it, and the law was known before any of them. What actually produces it is the requirement that an overlap region exist in which neither the viscous length ν/u_τ nor the layer thickness δ may appear, which leaves y du/dy with nothing to depend on and forces it to be a constant. The diagnostic function computed here is flat over two decades at Re_τ = 100,000 and over nothing at all at Re_τ = 180, which is a statement about the limit rather than about the model.

Tested in The layer with no length in it, at the figure it turns on.

A measured energy spectrum with a −5/3 slope over a decade or more is evidence that a Kolmogorov cascade is operating in the flow.

What decides it: A field synthesised with exactly those amplitudes and independent phases is built here, and its spectrum is the Kolmogorov one by construction while its energy flux is zero. Every second-order statistic of the two fields agrees — spectrum, correlation, second-order structure function, integral scale, Taylor microscale — because all of them are the same information. What distinguishes a cascade from a coincidence is the third moment, which the synthesised field has only as sampling noise: one realisation gives 0.18, and sixty-four average to 0.019 and keep falling as one over the square root of the ensemble.

Tested in The moment a spectrum cannot hold, at the figure it turns on.

Air can be treated as incompressible below Mach 0.3, so a body in a stream at Mach 0.3 or less may be analysed with the incompressible equations.

What decides it: It is a five per cent tolerance on the density at a stagnation point, and it is exact for that: the stagnation density rises by five per cent at M = 0.314. It says nothing about the body, and every body has a suction peak as well as a nose. The two exchange places at exactly cp₀ = −1, where the local speed is root two times the free stream; a section at cp₀ = −2 reaches the same five per cent at M = 0.22, and at a free-stream Mach of 0.3 its shoulder is at Mach 0.55 with the density down by nine and a half per cent.

Tested in Which speed goes in the number, at the figure it turns on.

A body's drag is the rate at which it is heating the fluid, so measuring the heat made in a region around it measures the drag.

What decides it: The equality holds only over the whole of the fluid. A creeping sphere's dissipation inside a radius R falls short of the drag power by exactly 3a/2R — a seventh at ten radii — and at Reynolds number 100 a box five diameters long behind a cylinder has not caught most of it. Two things leave a control volume that are not heat: work done by the tractions on its surface, and kinetic energy carried out through it.

Tested in Where the heat of a drag is made, at the figure it turns on.

Geostrophic balance holds when the Rossby number is small compared with one, so a weather system with Ro well below unity may be analysed by setting the Coriolis force against the pressure gradient.

What decides it: Keeping the centripetal term gives V_g/V = 1 ± Ro exactly for circular flow, so the fractional error in the geostrophic wind IS the Rossby number. "Ro is small" and "the balance is accurate" are therefore the same statement with the same number in it — which makes the folklore threshold of one a hundred per cent error rather than a boundary. And the anticyclonic branch is a quadratic whose roots vanish at a geostrophic Rossby number of exactly a quarter: past that there is no balanced wind at all, which no reading of the geostrophic approximation can suggest.

Tested in The balance that is its own error, at the figure it turns on.

A flow arranges itself to dissipate as little as it can, so of all the ways fluid could move between two walls, nature picks the cheapest.

What decides it: True for creeping flow and false as soon as there is inertia. Turbulent pipe flow at Reynolds number 10^5 dissipates roughly three times as much as laminar flow at the same flow rate between the same walls, and the laminar solution still exists there — it is merely unstable. The theorem is a property of the Stokes equations, whose solutions are unique and minimising; it says nothing whatever about which of several solutions a real flow chooses.

Tested in The cheapest shape the walls allow, at the figure it turns on.

The bumblebee story is a story about arithmetic: somebody used the forward flight speed instead of the wing speed, and doing the blade-element sum properly settles it.

What decides it: Doing the sum properly is necessary and is not sufficient. A quasi-steady sum assumes the wing has reached its steady circulation and that it flies in still air. Wagner's function averaged over each of four insects' own half-strokes gives 0.69 to 0.76, and momentum theory puts the animal's own induced velocity at 30 to 45 per cent of the wing's stroke speed. Both errors go the same way, and the mechanism that closes the gap — a leading-edge vortex that has not had time to shed — is a memory too.

Tested in A calculation with no memory in it, at the figure it turns on.

A jet pump works by suction — the fast jet has a low pressure by Bernoulli's equation, and that low pressure pulls the secondary fluid in.

What decides it: The low pressure at the nozzle is real and it is what starts the secondary flow, but Bernoulli does not explain the machine, because Bernoulli forbids what the machine does. Between the mixing tube's inlet and its outlet the pressure RISES while the flow keeps moving, which no Bernoulli argument permits: it is a momentum result, and it comes out of Cauchy-Schwarz — a single uniform stream always carries less momentum flux than the two it was made from, and the difference has to appear as pressure. The energy that pays for it is destroyed by the mixing, in exactly the Borda-Carnot expression a sudden enlargement produces, and the audit closes here to one part in ten billion. A device explained by Bernoulli alone would be a device that cannot raise the pressure of anything.

Tested in Mixing is a pump, at the figure it turns on.

A small neutrally buoyant particle released in a flow moves with the fluid, so a particle image measures the velocity at the particle's position.

What decides it: Faxén's first law says the particle moves at the surface average of the ambient flow, which differs from the value at its centre by a sixth of its radius squared times the Laplacian. On the axis of a round tube that is a lag of two-thirds of (a/R)² — 0.7 per cent for a particle a tenth of the radius, 6 per cent for one three-tenths — and the sign is always the same, so it does not average out over many particles.

Tested in A force without the flow that makes it, at the figure it turns on.

A body moving steadily through an ideal fluid carries a definite quantity of momentum with it, equal to its added mass times its speed.

What decides it: The momentum of the fluid is ∫ρu dA, which for a translating cylinder is conditionally convergent: it is exactly zero over a disc, and over rectangles of equal area it runs from −ρπa²U for a tall one to +ρπa²U for a long one, taking every value in between. The added mass times the speed is a real and unique quantity — it is the hydrodynamic impulse, a surface integral over the body — but it is not the momentum of the fluid, and the fluid's momentum has no value at all until the shape of the limit is named.

Tested in The momentum with no value, at the figure it turns on.

Streamlines drawn at equal intervals of the stream function show the speed: where they crowd together the flow is fast.

What decides it: True in a plane flow, where |∇ψ| is the speed exactly. In an axisymmetric meridional plot |∇ψ| = q·r sin θ, so the same reading is wrong by the distance from the axis — a factor running from 0.77 to 4.00 across one figure of ideal flow past a sphere, measured against the closed-form velocity to 10⁻¹⁰. The crowding of streamlines in a meridional plane is a statement about the volume flux in an annulus, and the annulus grows with radius.

Tested in The one number that runs out at three dimensions, at the figure it turns on.

Faster means a stronger shock and a bigger effect, so hypersonic flow is supersonic flow with the numbers turned up.

What decides it: The pressure ratio and the temperature ratio do grow without bound as M². The pressure coefficient at the nose, the density ratio across the shock and the shock standoff distance do not: they reach limits that are functions of γ alone. Measured, Cp at the nose is 1.8089 at Mach 5 and 1.8389 at Mach 40, against a limit of 1.8394, and the standoff changes by 8.5 per cent between Mach 8 and Mach 40 while changing by a factor of three between Mach 2 and Mach 8. Above Mach eight the question stops being about the speed and becomes a question about the molecule.

Tested in A shock that lies on the body, at the figure it turns on.

Choosing between a Pelton wheel, a Francis runner and a Kaplan propeller is an engineering judgement based on experience with similar installations — there is no principle that decides it.

What decides it: There is a principle and it is dimensional. A rotating machine's variables — shaft speed, diameter, flow, specific energy gH, density and power — have a dimension matrix of rank three, so three independent groups exist, and exactly one combination of them has a diameter exponent of zero. That combination, omega root Q over (gH) to the three quarters, is therefore a statement about a machine's SHAPE that is independent of its size, and this site checks that it survives rescaling to a machine four times the diameter at a third of the speed to nine decimal places. What IS a matter of experience is where the band boundaries sit — 0.35 or 0.4 for the Pelton-Francis crossover — and the bands are drawn here in the colour reserved for a borrowed claim.

Tested in One number picks the machine, at the figure it turns on.

Two flat plates with a film of liquid between them are held together by the surface tension of the liquid, which is why a wet glass sticks to a table.

What decides it: Surface tension supplies a force that does not depend on how fast the plates are being pulled apart, and the observed force does. A 40 mm disc separated at a millimetre a second across a 0.1 mm film of light oil resists with 75 newtons; the capillary force on the same rim is about 0.01 newtons. Pull slowly and the film gives way; pull fast and it does not, which is a viscous signature and not a capillary one.

Tested in The last of the oil, at the figure it turns on.

The Prandtl–Glauert factor corrects a subsonic result for compressibility, so it works up to the critical Mach number and stops there.

What decides it: Prandtl–Glauert and Kármán–Tsien are both linear corrections of the same order applied to the same incompressible peak. They agree to 1.5 per cent at Mach 0.3 and differ by 37 per cent at Mach 0.85 — which is inside the range a transport aeroplane cruises in and below the critical Mach number of a thin section. Where two approximations of the same order disagree by a third, the failure is not that the correction ran out but that the equation being linearised stopped being linear.

Tested in The equation that changes type inside its own answer, at the figure it turns on.

A duct's thermal entry length is x/(D·Re·Pr) = 0.05, past which the flow is thermally developed and the Nusselt number is 3.66.

What decides it: Marching the energy equation to that station gives a local Nusselt number of 3.710 against a developed 3.657 — a 1.45 per cent excess. That is a perfectly sensible tolerance and it is never stated, so nobody can say whether their own problem needs a longer duct or a shorter one. The entry length at ten per cent is x⁺ = 0.024 and at a tenth of one per cent it is 0.085, a factor of three and a half, and the quoted 0.05 is one unlabelled point on that curve.

Tested in How far before the heat arrives, at the figure it turns on.

A steady flow cannot mix, because its trajectories are its streamlines and a particle released on one stays on it.

What decides it: Both halves of that sentence are true in any number of dimensions and the conclusion follows only in the plane. A curve in a plane separates the plane; a curve in a volume does not, and can come arbitrarily close to every point of it. The ABC flow is steady, incompressible to machine precision and an exact solution of Euler's equations, and a single streamline of it covers 18 per cent of a Poincaré section and is still finding new territory after four thousand crossings.

Tested in Steady, three-dimensional, and mixing anyway, at the figure it turns on.

A real-gas correction changes everything about a shock, so a perfect-gas calculation at high Mach number is simply wrong.

What decides it: The jump conditions do not contain the heat capacity. Mass and momentum give ρ₁u₁ = ρ₂u₂ and p₁ + ρ₁u₁² = p₂ + ρ₂u₂², with no thermodynamics in them at all, and the caloric behaviour of the gas enters only through the energy equation. Measured across Mach 2 to 15, the two models' pressure ratios differ by at most 4.4 per cent while their temperature ratios differ by 18 and their density ratios by 27. A perfect-gas calculation gets the force on the body nearly right and the heat into it badly wrong.

Tested in When gamma stops being a number, at the figure it turns on.

A sonic boom carries information about the aeroplane that made it, so a quieter shape can be designed by measuring what arrives and working backwards.

What decides it: The far-field N-wave has two parameters, an overpressure and a duration, so at most two things about the aeroplane can be read from it — and they arrive mixed with the altitude and the atmosphere in between. Two bodies whose F-functions share only the maximum of their running integral, and differ by fifty-five per cent in peak and by their whole shape, produce signatures converging from seventy-three per cent apart to two and a half. Everything else the near field contained is destroyed by the propagation, which is why quiet supersonic design works on preventing the ageing rather than on making the wave smaller.

Tested in The signature that forgets the shape, at the figure it turns on.

Every dimensionless group's threshold is really a rule of thumb with a tolerance hidden in it, so the value one is never exactly where anything happens.

What decides it: It is true of every group formed by comparing two term sizes, which is most of them. It is false of a group formed from a characteristic condition. At Fr = 1 the flow speed equals the speed of a surface wave, so the upstream signal speed U − √(gh) changes sign — exactly, with no tolerance in the statement — the specific-energy curve has its minimum at the same place, and the governing equations change from elliptic to hyperbolic. Asking for the threshold at a tenth of a per cent or at ten per cent returns the same number.

Tested in The number that really is one, at the figure it turns on.

A shear-thinning fluid flows more easily, so pumping it through a pipe costs less than pumping a Newtonian fluid of the same viscosity.

What decides it: At the same pressure gradient the wall shear stress is GR/2 for every fluid, whatever its flow index — it is a momentum balance on the whole pipe and knows nothing about the fluid in it. What shear-thinning changes is where the shearing happens: at n = 0.2 the profile is flat in the middle and steep at the wall, and 96 per cent of the dissipation is made outside half the radius against 94 per cent for a Newtonian fluid. Nothing has been made easier; the cost has been moved.

Tested in A viscosity that depends on the question, at the figure it turns on.

A photograph of a Hele-Shaw cell shows what an inviscid flow looks like, so it is a demonstration of potential flow.

What decides it: The depth-averaged velocity in a cell is the gradient of a harmonic potential, so the streamline pattern is the ideal one to a part in ten billion — and the flow itself satisfies no-slip on the obstacle, which the averaged model violates at 2U, and its profile across the gap is a parabola, so a tracer at mid-gap travels at 1.5 times the average and one near a plate barely moves. The pattern is exact. Everything about the speeds is wrong.

Tested in The exact theory, drawn by viscosity, at the figure it turns on.

Dimensional analysis gives the answer to the strong-blast problem.

What decides it: It gives the exponent and cannot give the constant. Four quantities in three dimensions leave one group, so R = C(Et²/ρ₀)¹⁄⁵ with C undetermined. Getting C needs a solution: a thin-shell energy balance gives 0.907 against Sedov's exact 1.033 at γ = 1.4, and across γ the model's error runs from −32 per cent at 1.2 to +7 at 5/3, so it crosses the exact answer near γ = 1.5. Agreeing at any one γ would be a coincidence rather than a validation, which is why the comparison is a sweep.

Tested in A radius that gives the energy away, at the figure it turns on.

A chaotic flow stretches material lines at a rate given by its Lyapunov exponent, so the length of a line after time t is its initial length times e to the lambda t.

What decides it: That is the rate of the typical element and not of the average one. Measured over eight hundred elements in a steady three-dimensional flow, the mean of the logarithm of the length grows at 0.157 and the logarithm of the mean length grows at 0.273 — a ratio of 1.74 — and the rate keeps rising with the moment, reaching 0.504 at the fourth. Half the total length is carried by the longest 0.75 per cent of the elements.

Tested in The stretching rate that is not one number, at the figure it turns on.

A turbulent spectrum has a minus five thirds range, so fitting a power law to the middle of a measured spectrum measures the Kolmogorov exponent.

What decides it: The band over which the local slope is within 0.05 of −5/3 is 0.04 decades at a Taylor Reynolds number of 30, 0.33 at 100 and 0.72 at 200 — it does not reach one decade until the high hundreds. A one-decade least-squares fit at the most favourable band returns −1.82 at 30, −1.62 at 100 and −1.67 at 3,200, so its error changes sign on the way. A laboratory watching its fitted exponent improve as it raises the Reynolds number may be watching two errors cancel.

Tested in The range a real Reynolds number does not have, at the figure it turns on.

A system with no viscosity, no approximation and exactly conserved energy is deterministic, so its future can be computed for as long as anybody is willing to keep integrating.

What decides it: Four point vortices conserve their energy and both impulses to fourteen digits over two hundred time units here, and two runs from initial conditions a billionth apart separate by a factor of 140,000 in the same interval — an exponential rate of 0.033 per time unit. Three vortices, with the same integrator and the same invariants, separate by a factor of 180. Determinism is intact and prediction is not, and the difference between three and four is one degree of freedom.

Tested in Three is the most that can be predicted, at the figure it turns on.

A shock is thin but finite, and the Navier–Stokes equations describe what is inside it.

What decides it: They describe it at weak strengths and not at interesting ones. Becker's exact solution gives a thickness of 3,154 nanometres at Mach 1.05 — fifty mean free paths, comfortably a continuum — and 35 nanometres at Mach 5, which is half of one. A structure thinner than the distance a molecule travels between collisions is outside the hypothesis that produced it, and measurements give a thickness several times larger. The end states either side remain exact however thin it gets, because the jump conditions contain no transport coefficient at all.

Tested in The discontinuity that has a thickness, at the figure it turns on.

The Strouhal number of a bluff body is about 0.2, and that is a constant of the shedding process.

What decides it: Three bodies measured in the same range give 0.212, 0.200 and 0.145 — a spread of 1.462 — so it is not one number across shapes. It is not a constant across Reynolds number either: Roshko's own fit is 0.212 − 2.7/Re, which is 0.137 at Re = 60, thirty-five per cent below the asymptote, and only comes within one per cent of 0.212 above Re = 1274. What is nearly constant is the number formed with the wake's width and the speed on the streamline bounding it: 0.1628 for all three bodies, to a tenth of a per cent.

Tested in The frequency a wake chooses, at the figure it turns on.

The log-normal model of intermittency, with mu about 0.25, describes the measured scaling exponents.

What decides it: It describes the low moments and it cannot be right about the high ones, because its exponent turns over: the slope of p/3 − mu·p(p−3)/18 goes negative at p = 3/2 + 3/mu = 13.5, so beyond that it says a higher moment scales with a smaller exponent. Novikov's constraint forbids that — the velocity difference is bounded, so the exponents must be non-decreasing — and 13.5 is inside the range routinely measured, not out at some academic moment.

Tested in The exponents that stop being thirds, at the figure it turns on.

In a closed wave tank the Stokes drift is cancelled by a return current, so there is no net motion.

What decides it: The depth-integrated transport is exactly cancelled, because the water has nowhere to go. The motion is not. Three return currents with exactly the required transport — a uniform slab, a parabola with no slip at the bed, and one concentrated near the surface — put the depth at which the drift reverses at 23, 98 and 100 per cent of the water column, and disagree about the sign of the drift at the bed. The constraint fixes an integral; where the water actually goes needs the dynamics.

Tested in The drift a closed box will not allow, at the figure it turns on.

An infinite velocity in a potential-flow solution is a sign that the solution has broken down and nothing quantitative can be taken from it.

What decides it: The square-root singularity at a sharp leading edge carries a finite force, ½πρC², and rounding the edge off to a parabola of radius r returns exactly the same number for every r over four decades — the peak speed rising as r^(−1/2) while the region it acts over shrinks as r^(1/2). The force is a limit rather than an artefact, and without it a flat plate in an inviscid fluid would have drag.

Tested in A finite force from an infinite speed, at the figure it turns on.

Requiring the flow to be finite and physically sensible is enough to pick out the right solution near a corner.

What decides it: Every mode r^(mπ/α) with m a non-zero integer satisfies Laplace's equation with both walls as streamlines, and the finite-energy condition selects exactly the positive m — the energy integral converges for m > 0 and grows without limit as the cut-off falls for m < 0, measured here at 0.785 against 7.9×10⁵. It is a physical requirement imposed from outside the equations, and even after it the surviving solution at a 360° corner still has an unbounded velocity.

Tested in Nothing turns a sharp corner, at the figure it turns on.

Sound is a small disturbance that travels at the speed of sound and keeps its shape.

What decides it: Linear acoustics contains no time scale at all, which is what says it has thrown something away. Each point of a waveform travels at a₀ + ½(γ+1)u rather than at a₀, so the profile distorts and becomes vertical after a distance a₀²/(βωu₀). For a 120-decibel tone at 1 kHz that is 307 metres; for a jet engine at a metre it is 20. The distance goes exactly as the reciprocal of the amplitude, from any amplitude whatever, and the second harmonic appears in direct proportion to the distance travelled with no threshold at all.

Tested in Every compression becomes a shock in the end, at the figure it turns on.

A drop breaks when the capillary number exceeds a critical value of about a half.

What decides it: In simple shear that is true for viscosity ratios between about 0.1 and 2, where the measured critical number runs from 0.61 to 0.70. Outside that window it is false in both directions: at a ratio of 10⁻³ the measured value is 8.7, fourteen times larger, and above 4.08 there is no critical value at all because a drop in simple shear cannot be broken at any shear rate. The first-order theory cannot see this — its own critical number spans a factor of 1.14 over the entire range — and the reason is that simple shear's rotation is exactly as strong as its strain, so the drop turns out of the stretching direction before it can be pulled apart.

Tested in The number that cannot break a drop, at the figure it turns on.

Energy passes steadily from large scales to small ones through the inertial range at a constant rate equal to the dissipation.

What decides it: That is true of the average and of nothing else. In a shell model whose nonlinear term conserves energy to 2·10⁻¹⁶, the mean transfer is flat to ten per cent across seven shells and equals the dissipation — and the instantaneous transfer through those shells is negative between two and fifteen per cent of the time depending on the shell, nine per cent averaged across the band, with a standard deviation between two and five times its own mean.

Tested in A flux that runs both ways, at the figure it turns on.

A scheme that is second-order accurate in space and time makes second-order errors in every quantity, so refining it enough makes any particular error negligible.

What decides it: A second-order face rule on a moving mesh changes a uniform scalar in a fluid at rest by 2.2·10⁻⁴ over two time units at a step of 0.05 — an error in a quantity that is not changing, of a size that would be read as a physical transient. It does fall as the square of the step, 1.977 measured, so refinement removes it. What refinement cannot do is tell anybody it was there, because a smooth excursion in a conserved scalar is exactly what a real flow produces.

Tested in The mesh that makes its own mass, at the figure it turns on.

A porous medium's permeability can be estimated from its porosity and its specific surface, which is what the Kozeny–Carman relation is for.

What decides it: Forty-nine bundles of capillaries are built with exactly the same porosity and exactly the same specific surface — matched to eight figures — and nothing else in common. Their permeabilities span a factor of eighty-two. The permeability of a bundle is the fourth moment of its radius distribution over the second; its specific surface is the first over the second. They are different moments of one distribution and neither determines the other.

Tested in A permeability that is only the geometry, at the figure it turns on.

The dissipation in a two-dimensional flow is the viscosity times the integral of the squared vorticity, so a region with no vorticity in it is destroying no energy.

What decides it: The identity holds over the whole plane and not over a region. Outside a Lamb–Oseen vortex's core the fluid is irrotational and contributes nothing to the enstrophy — and it is being sheared, so it contributes plenty to the dissipation. Inside a circle of radius R the two differ by exactly μΓ²/πR², measured here and matching the closed form, and the difference is the work the fluid outside is doing on the fluid inside.

Tested in The energy a vortex cannot have, at the figure it turns on.

Opening a valve at the bottom of a tank produces Torricelli's velocity √(2gh) at the outlet.

What decides it: Torricelli's result is the steady answer, and dropping the ∂φ/∂t term from Bernoulli's equation gives it at every instant including the first — full speed before any fluid has moved. The unsteady equation gives U = U∞ tanh(U∞t/2L), a time constant of 2L/√(2gh) set by the length of the tube and by nothing else, and for a two-metre tube under a one-metre head it takes 2.4 seconds to come within one per cent.

Tested in The pressure that depends on the past, at the figure it turns on.

Two-dimensional turbulence has a k^(−3) enstrophy range, so a measured spectrum should show a slope of −3.

What decides it: Kraichnan's own correction multiplies it by the logarithm of the wavenumber ratio to the power minus one third, because a cubic spectrum makes the strain rate the same at every scale and the cascade non-local. The local slope is therefore steeper than −3 by one third of the reciprocal of that logarithm — it is −3.060 eight octaves above the forcing and −3.0075 at sixty-four, and reaching −3.01 requires forty-eight octaves. No measurement and no computation will show −3.

Tested in Where the inverse cascade stops, at the figure it turns on.

A supersonic compression is irreversible, so an isentropic inlet is an approximation that gets closer to reversible as more ramps are added.

What decides it: It is exact arithmetic rather than an approximation, and the rate is the interesting part: the entropy across a weak shock is third order in its strength, so N equal ramps cost N times a cube of one Nth, which is exactly one over N squared. Sixty-four ramps cost a two-hundred-and-fifty sixth of what one costs, and the limit is genuinely isentropic. What the limit costs instead is length — the Mach waves from a smooth turn converge on a focus, and making the turn gentler brings the focus nearer rather than pushing it away.

Tested in A compression that costs nothing in the end, at the figure it turns on.

The mean velocity of a turbulent flow is a well-defined quantity, so an experiment's mean and a simulation's mean are the same thing and can be compared directly.

What decides it: In a flow whose density varies there are two means. On an intermittent field with a density ratio of seven — a flame — the time-weighted mean is 60 metres a second and the mass-weighted one is 30. Both are correct averages of the same signal; a hot wire returns the first and a compressible model computes the second; and the two turbulence intensities that go with them differ by a third.

Tested in Two averages of one flow, at the figure it turns on.

A microscopic swimmer goes faster by beating harder, and beating harder costs more per unit distance travelled, so there is an economical stroke and a wasteful one.

What decides it: For a waving sheet the power goes as the square of the amplitude and the speed goes as the square of the amplitude, so the work per unit distance travelled is 2μkc and contains no amplitude at all. What does change the cost per metre is the wave speed, linearly. A large slow undulation and a small one of the same wave speed cost exactly the same to travel a given distance, and the small one merely takes longer.

Tested in A swimmer that cannot go backwards, at the figure it turns on.

Solving unsteady gas dynamics needs a numerical method, because the equations are nonlinear and coupled.

What decides it: In a simple wave it needs no method at all. One invariant is the same everywhere in a region reached by characteristics from a uniform state, which reduces the pair of equations to a one-parameter family of straight lines whose slopes are known from the boundary alone — so the whole solution is arithmetic. Computed for a piston-driven fan, the constant invariant holds to four parts in ten to the sixteenth, and substituting the resulting field back into Euler's equations leaves a residual of a part in ten million.

Tested in Two numbers that do not change, at the figure it turns on.

A flowmeter's discharge coefficient accounts for the approach flow, so a calibrated meter reads correctly whatever the profile upstream of it.

What decides it: The inference from pressure drop to flow rate contains the approach profile's kinetic-energy coefficient, which is exactly one for a uniform profile, 1.058 for a fully developed turbulent one and exactly two for a laminar one. At a contraction ratio of 0.6 that is an error of 0.4 per cent, 7.7 per cent and 17.6 per cent for an annular jet — and two profiles built to share both coefficients differ by 48 per cent of the mean velocity and read identically.

Tested in The profile a meter cannot see, at the figure it turns on.

Taylor–Couette flow becomes unstable when the Taylor number exceeds about 1,708.

What decides it: Two conditions have to hold first and neither is in the Taylor number. If the outer cylinder turns faster than η² times the inner, the angular momentum increases outwards everywhere and no Taylor number is large enough — Rayleigh's criterion, found here by bisection at μ = η² to a part in 10¹⁶, contains no viscosity and no Taylor number at all. And 1,708 is a minimum over a continuum of axial wavenumbers, which a column of finite height does not have: at an aspect ratio of 2.5 the threshold is 5.66 per cent higher, and the vortex count that appears is decided by the height of the apparatus.

Tested in A transition that needs a second number, at the figure it turns on.

A helicopter's forward speed is limited by the advancing blade going supersonic.

What decides it: That is a limit and it is not usually the binding one. The trimmed solve here shows the retreating side running out first: at μ = 0.5 the section at 0.7R on the advancing side carries fifty-eight times what the retreating one does, the retreating section is at −0.09° of incidence and carrying nothing, and 2.07 per cent of the disc is at its lift-coefficient cap. The advancing tip at that advance ratio is at Mach 0.9 or so on a typical machine — high, and behind the retreating blade's problem, which is that lift goes as the square of a velocity that has fallen to a half.

Tested in The side that cannot keep up, at the figure it turns on.

Whether a shear flow is unstable, and how fast, depends on the details of its velocity profile.

What decides it: Whether it is unstable does. How fast is bounded, before any calculation, by two numbers taken from the ends of the profile: every unstable mode has a complex phase speed inside a circle of radius half the velocity range centred on the mean, so the growth rate cannot exceed the wavenumber times half the range whatever the shape between the ends. Across six profiles with different inflection structures the bound holds with the closest approach at 0.915 of the radius, and the actual maxima come in at 30 to 55 per cent of what it allows.

Tested in Every unstable wave is inside one circle, at the figure it turns on.

A surge calculation is specified by the closure time: state how long the valve takes to shut and the peak pressure follows.

What decides it: Three closure laws of exactly six seconds on the same 1,200 m line give peak rises of 50.2, 84.4 and 99.2 metres of head — 97 per cent apart on the smallest. The peak is a functional of the whole opening curve, and the sensitivity to it is measured here: bumping the valve at 5.6 s moves the peak by 345 metres per unit of opening, and bumping it before 0.45 s moves it by less than a hundredth of that.

Tested in The part of the closure a pipe cannot see, at the figure it turns on.

A concentrated vortex may always be replaced by a point vortex of the same circulation, since the exterior fields agree.

What decides it: They agree exactly for a circular patch and only for a circular patch, and a patch in anything but a uniform flow does not stay circular. The exterior field of an elliptical patch of the same circulation differs by 27 per cent at two radii and falls as the inverse square thereafter, because the first moment the two do not share is the quadrupole. More seriously, the point vortex has no strain limit: a patch is torn out into a filament above a strain-to-vorticity ratio of 0.150142, and a point vortex survives any strain whatever.

Tested in The shape a vortex keeps, at the figure it turns on.

At a low enough Reynolds number, inertia may be neglected and the Stokes equations solved instead.

What decides it: Near the body, yes; far from it, never. The neglected inertial term exceeds the retained viscous one beyond a radius of the body size divided by the Reynolds number, whatever the Reynolds number is, so the limit of vanishing Reynolds number does not commute with the limit of large distance. In two dimensions the consequence is that no solution exists at all: with the far-field conditions imposed there is one constant left for two wall conditions, and the tangential velocity that survives is exactly twice the free stream, at every radius and every speed.

Tested in The flow with no solution, at the figure it turns on.

The dissipation can be computed either as twice the viscosity times the strain-rate squared or as the viscosity times the enstrophy, since the two are equivalent.

What decides it: Their averages are equal, exactly, because their difference is a divergence and a divergence integrates to nothing over a homogeneous domain. Their values are not. On six synthetic incompressible fields the two correlate between minus 0.18 and plus 0.24 — uncorrelated — with a root-mean-square difference of 150 to 185 per cent of the mean they share, and their maxima 0.46 apart on a box of side two pi. So "where is the dissipation" has two answers and only the integral is well posed.

Tested in Equal on average, and nothing else, at the figure it turns on.

Cavitation begins when the local pressure falls below the vapour pressure, so inception is decided by a pressure and the cavitation number is the number that decides it.

What decides it: A 5 micron air bubble is given a rectangular tension pulse and asked whether it grows past the radius at which its own gas can no longer hold it. At 100 microseconds it needs 1.05 bar; at 0.3 microseconds it needs 43.2, a factor of 41 for the same bubble in the same liquid. The threshold is a curve in amplitude and duration, and the duration that separates its two ends is the time the bubble needs to grow to its critical radius.

Tested in A threshold that is also a duration, at the figure it turns on.

In level flight an aeroplane's lift equals its weight.

What decides it: The *total* lift equals the weight and the wing's does not. With a quarter-chord static margin and an ordinary zero-lift pitching moment, the trim solve gives a wing load of 11,672 N against a weight of 11,000 and a tail load of −672 N: the wing is carrying 106.1 per cent of the weight, and the excess is the tail's download. The two-equation solve residual is 1.6 × 10⁻¹⁷, so this is arithmetic rather than an approximation, and the excess grows as the centre of gravity moves forward.

Tested in More lift than weight, at the figure it turns on.

Vortex breakdown occurs above a swirl number of about 1.9, so a swirl number is what decides whether a vortex will break down.

What decides it: 1.9158530 is the critical swirl of one profile — uniform axial velocity with solid-body rotation filling the pipe — and it is exact for that one. Shrink the core to a quarter of the radius and the critical value computed from the same eigenvalue problem is 4.917; measure the same family's swirl at the wall instead and it is 0.307; a q-vortex with its swirl concentrated near the axis gives 0.403. Across twelve ordinary columnar vortices the critical swirl number spans a factor of sixteen, and the radial distribution decides.

Tested in The swirl that holds a wave still, at the figure it turns on.

Matched asymptotic expansions work because there is a region in which both the inner and the outer solution are valid, and the matching is done there.

What decides it: The overlap region is measured here, as the range of x over which both approximations are within five per cent of the exact solution. At ε = 10⁻⁴ it is 1.67 decades wide. At ε = 10⁻² it does not exist at all — there is no x at which both are within five per cent — and the composite built by the same recipe is nevertheless accurate to 2.4 per cent everywhere. The construction works well before its own justification does, which is the honest state of the method rather than a criticism of it.

Tested in One formula for both ends, at the figure it turns on.

The tensile strength of water is a property of water, so a careful enough measurement will find one number for it.

What decides it: The careful measurements find two. Spinning tubes and most other methods break water near −26 to −28 MPa; water sealed in quartz inclusions holds near −140, and classical nucleation theory for a pure liquid gives −148 MPa for a cubic millimetre watched for a second — a number that twenty-one decades of volume and time move only between −175 and −126, so the sample size cannot explain the gap. Read through the Laplace threshold, −27.7 MPa is a cavity of 5.25 nm and −140 MPa one of 1.04 nm: each measurement reports the worst flaw its sample contained.

Tested in A breaking strength that is the size of a flaw, at the figure it turns on.

An inviscid, irrotational, steady, incompressible flow past a body produces no drag, so any drag a ship pays must come from viscosity or from separation.

What decides it: Every one of those four words holds for a body under a free surface, and the drag is not zero. A submerged doublet leaves a wave train behind whose amplitude is four pi times the wavenumber times the doublet moment, decaying exponentially with depth, and the resistance is one quarter of rho·g·A² — the energy the train carries away at the group velocity, which in deep water is half the phase velocity. The paradox's hypothesis includes the domain, and a free surface is a boundary through which energy can leave.

Tested in The drag that is made of waves, at the figure it turns on.

The optimum convergence ratio for a plane slider bearing is 2.1889.

What decides it: The maximum of the load bracket is at 2.1887048, found by bisecting the derivative rather than by searching the load — which is flat to parts in ten to the thirteenth over that range and cannot locate its own maximum. The difference matters not at all: 2.1889 costs eight parts in a hundred billion of load, and every convergence ratio from 1.99 to 2.43 is within one per cent of the best. Both halves are the point. The number is exact, and it is worth about as much as its neighbours.

Tested in The gap that carries the most, at the figure it turns on.

An asymptotic result becomes accurate once the small parameter is small enough.

What decides it: Only if the correction is a power. When the leading and next terms differ by a logarithm the correction is 1/ln(1/eps), whose local exponent is the correction itself — so at any scale the result looks like a power law with a small exponent, and at the next scale down it looks like a different one with a smaller exponent. Reaching one per cent takes forty-three decades of the small parameter and a tenth of a per cent takes four hundred and thirty-four; a square-root correction reaches one per cent in four.

Tested in The three that never converge, at the figure it turns on.

Point-vortex dynamics is an approximation whose error grows smoothly with the size of the vortices, so a calculation with small enough cores is always safe.

What decides it: The error is not smooth in the size and it is not an approximation while the patches are round: each component of a point vortex's velocity is harmonic, so its average over a disc is its value at the centre, and a circular patch's centroid moves at exactly the point-vortex speed — 1.3·10⁻¹³ by quadrature, at every separation tried. What breaks the model is the strain deforming the patches, and that has a threshold: at five radii the orbit is followed to 0.9 per cent and at three and a half to 14, a fifteenfold change for a thirty per cent move.

Tested in What a point vortex is not, at the figure it turns on.

Dimensional analysis tells you what the answer to a fluid-mechanics problem is, up to a constant.

What decides it: It tells you which dimensionless groups the answer may depend on, and it never produces a value. Every pure number in this collection is the answer to a problem that had to be solved — an algebraic identity, an integral over a solved field, a root of a computed function, or its stationary point — and the fourth kind is systematically the worst known: with an objective known to a part in a million, a root is located to five parts in ten million and an optimum only to one part in a thousand, an exponent of exactly one against exactly a half.

Tested in Where a pure number comes from, at the figure it turns on.

A siphon that has started and has not broken will run until its source is empty, since nothing about its heights changes as it drains.

What decides it: Nothing about its heights does, and something about its water does. Saturated with air at one atmosphere, each litre arriving at a crown at 23 kPa carries 76.7 mL more free gas than that pressure can hold. While the flow is faster than a long bubble climbs the falling leg — 0.351√(gD), 0.123 m/s in a 12.5 mm hose — the gas is carried out. The reference drain falls below that speed after 30.9 hours with 4.7 cm of level left, and an 18.4 mL pocket then fills in 2 to 74 minutes across a twenty-fold range of how much gas actually comes out.

Tested in The air that breaks a siphon nothing else can, at the figure it turns on.

The Karman momentum integral is insensitive to the assumed velocity profile, so any reasonable guess gives the drag.

What decides it: Four reasonable guesses — a straight line, a parabola, a cubic and a quarter sine — give skin-friction coefficients from thirteen per cent below Blasius to ten per cent above, a span of twenty-three. The two that also satisfy the conditions the true profile satisfies are within 1.4 and 2.7 per cent, and the two that satisfy neither are the two that are ten and thirteen per cent out. The insensitivity is a property of the family, not of the integral.

Tested in Four profiles, one drag, at the figure it turns on.

A boundary layer is parabolic, so the flow at a station cannot be affected by anything downstream of it.

What decides it: It is parabolic only when the pressure is prescribed, which is the leading-order limit and not the flow. Let the pressure respond to the layer's own displacement and the problem admits a free interaction growing exponentially in the streamwise direction, at a rate of 0.8272 times the wall shear to the five-fourths power — a disturbance appearing out of nothing upstream. At a Reynolds number of a million it decays over 27 boundary-layer thicknesses, which is what every measurement ahead of a shock has always shown.

Tested in The length the limit invents, at the figure it turns on.

A capillary viscometer measures a viscosity, since it measures a shear stress and a shear rate.

What decides it: It measures a shear stress and a flow rate. The wall stress follows from a force balance and is exact for every fluid; the wall shear rate does not follow from the flow rate at all unless the fluid is Newtonian. For a power-law fluid the true wall shear rate exceeds the apparent one by (3n+1)/4n — a factor of two at n = 0.2 — and recovering it needs the local slope of the flow curve, which takes runs at neighbouring pressure drops.

Tested in The stress a pipe knows, at the figure it turns on.

An elliptic planform is the efficient one, so a wing that is not elliptic is paying for it.

What decides it: It is paying quadratically, which at any sensible taper ratio means it is paying almost nothing. The induced-drag penalty is a sum of squares of the loading's departure from elliptic, so a loading eight per cent away from elliptic costs under two per cent in induced drag, and every straight taper ratio between 0.25 and 0.51 sits within half a per cent of the best one. The exact optimum is real; choosing it rather than its neighbour buys nothing a wing can measure, and the choice is made on root bending moment instead.

Tested in The optimum that does not matter, at the figure it turns on.

Particle image velocimetry measures the velocity field itself, so a PIV vector is the flow's velocity at the point where it is drawn.

What decides it: A PIV vector is, to first order, the average of the velocity over its interrogation window, and an average over a window of width W is a filter whose response to a wave of wavenumber k is sin(kW/2)/(kW/2). A window one core radius wide reads a Lamb–Oseen vortex's peak velocity at 92.5 per cent, places it 1.084 times too far out, and gives 76.7 per cent of the central vorticity. A wave three quarters of a window long is reported at 20.7 per cent of its amplitude with its sign reversed, and a shear layer thinner than the window is reported as exactly as thick as the window.

Tested in The window every vector is averaged over, at the figure it turns on.

Flutter is found in flight test by measuring the damping at increasing speeds and extrapolating it to zero.

What decides it: It is, and the extrapolation is unreliable in the unsafe direction for a computable reason. The damping falls towards zero with a small slope — it is still at 0.19 per cent of critical two per cent below the boundary — so a straight line fitted to three sub-critical points between sixty and eighty per cent of the flutter speed predicts 101.8 metres a second where the true boundary is 80.8. That is twenty-six per cent high, from a curve that looks perfectly straight over the range fitted.

Tested in The speed where the damping is exactly zero, at the figure it turns on.

Viscous diffusion has a time scale, so the fluid a given distance from a wall responds to what the wall did about one diffusion time ago.

What decides it: There is a most likely delay and it is y²/6ν, and it is a poor summary of the distribution around it. The kernel that carries a wall's motion into the fluid is a probability density in the delay whose tail falls as the delay to the power −3/2. At two millimetres in water the mode is 0.67 seconds, the median is 4.40 — 6.6 times later — a tenth of the response is older than 127 seconds, and the mean does not exist: computed over a window it grows as that window's square root and does not converge.

Tested in The wall the fluid is listening to, at the figure it turns on.

One photograph of an axisymmetric flow determines the flow, and a sharper photograph determines it better.

What decides it: The first half is true: Abel's integral inverts exactly, and onion peeling on 200 rings recovers a Gaussian field from its exact projection to four parts in ten thousand. The second is false for an interferogram. With noise on each ray the recovered field's noise on the axis grows exactly in proportion to the number of rings — a gain of 4.75 at 12 rings, 19.8 at 50 and 79.1 at 200 — while the same relative noise on a schlieren deflection gives a gain of 0.94 at every resolution.

Tested in One view is enough, and the axis pays for it, at the figure it turns on.

Cavitation starts wherever the flow is fastest, so it starts at the shoulder of the body.

What decides it: In an irrotational flow, yes, and the reason is stronger than the observation: the pressure of an incompressible flow satisfies a Poisson equation whose source is twice the density times Q, and Q is never positive when the vorticity is zero, so the pressure is superharmonic and attains its minimum on a boundary. Where the flow has vorticity, Q can be positive and the minimum moves off every surface — which is why cavitation in a propeller wake begins in the tip-vortex core, several diameters behind the blade that made it.

Tested in The lowest pressure is on the body, at the figure it turns on.

A wake sheds at the Strouhal frequency, so a body oscillating at a different frequency produces a wake with both frequencies in it.

What decides it: It does, outside a band. Inside the band the wake abandons its own frequency entirely and adopts the forcing's, which is lock-in, and the width of that band is proportional to how hard the body is forced — from 0.093 at a forcing amplitude of 0.05 to 0.65 at 0.4, in units of the natural frequency. Outside the band the two frequencies coexist and the amplitude beats at their difference. Near the edge the capture is slow: the settling time rises from 14 time units at small detuning to 86 near the boundary.

Tested in A wake told what to do, at the figure it turns on.

A seeding particle small enough to follow the flow measures the flow, so the only question is whether it is small enough.

What decides it: "Small enough" is a statement about a frequency, not about the flow, because a particle is a first-order filter and its response depends on which frequency it is being asked about. Its magnitude is one over the root of one plus the Stokes number squared, so at a Stokes number of one it follows 71 per cent and lags by 45 degrees. Following to within one per cent needs a Stokes number below 0.14 at every frequency of interest — and the same particle is faithful at a hundred hertz and useless at ten kilohertz.

Tested in A particle is a low-pass filter, at the figure it turns on.

Induced drag is a fixed property of the wing, so a survey plane anywhere behind it reads the same induced drag as the kinetic energy of the trailing vortices.

What decides it: The inviscid half is exact: a smoothed trailing sheet rolling up into two vortices keeps its crossflow energy to 10⁻⁹, so roll-up changes nothing. Viscosity does. Two Gaussian vortices holding the sheet's energy keep 81 per cent of it as crossflow energy at a viscous age 4νt/b₀² of 0.01, 43 per cent at 0.1 and 9 per cent at 1; the rest is still in the plane, as a total-pressure defect, and a survey that separates the two reports it as profile drag.

Tested in The drag a wake keeps however it rolls up, at the figure it turns on.

A photograph of a rolled-up shear layer shows the state of the flow at the moment it was taken.

What decides it: It shows a state, and it also shows a history, because the material points of a sheet cannot change their order: the sheet is a one-dimensional object carried by the flow, so the parameter along it is a label the fluid keeps for ever. In the computation here 200 points wind through a full turn without a single pair ever changing places, so each arm of the spiral is a known interval of the initial sheet, and the number of turns is a clock. What the picture cannot say on its own is how thin the sheet was taken to be, and three smoothing lengths give arclengths spanning 76 per cent.

Tested in A spiral is a legible record, at the figure it turns on.

A duct's development length is the distance at which its flow becomes fully developed, so it is a property of the duct and the Reynolds number.

What decides it: It is a property of the duct, the Reynolds number and a threshold nobody usually states. A disturbance to the developed profile decays mode by mode at rates going as the square of the mode number, so what is left after a short distance is one mode decaying exponentially — and the distance at which it falls below a stated fraction is the logarithm of that fraction divided by one rate. Sweeping the fraction from a tenth to a thousandth moves the length from 1.02 m to 5.69, a factor of 5.6, with nothing about the duct or the flow having changed.

Tested in A duct that forgets everything but one number, at the figure it turns on.

The dimensionless groups of fluid mechanics are a long list to be learned, each belonging to its own corner of the subject.

What decides it: A large family of them is one group. Eight produced by this collection alone — Deborah, reduced frequency, Stokes, Keulegan-Carpenter, Womersley squared, Damköhler, the inverse root Ekman and a turnover-to-strain ratio — are each a memory time divided by a process time, and their representative values span a factor of 394 while the construction is identical. All eight govern the same first-order response: unity below about a tenth, a reciprocal above about ten, and two decades in between where the answer depends on the history and nothing simpler will do.

Tested in Every memory number is one time over another, at the figure it turns on.

A flow whose equations are reversible has kept all the information about where its fluid came from, so that information can be recovered.

What decides it: The information is kept and recovering it is a different question. Reversing a point-vortex stirring exactly returns 120 tracer particles to within 1.4·10⁻⁹ of their starting places. Displacing one vortex by 10⁻⁸ at the turning point — a perturbation to the state rather than to the tracers — returns the worst of them 4.9·10⁻⁶ away, a gain of 486, and the gain grows as the 2.87 power of the window. Below a nudge of about 10⁻⁹ the round trip's own arithmetic dominates at 2.2·10⁻⁷, so there is a floor beneath which no improvement in the state is visible at all.

Tested in Reversible, and unusable, at the figure it turns on.

Langmuir circulation is driven by the waves: the rolls draw their energy from the Stokes drift.

What decides it: The rolls need two shears of the same sign — the wind-driven current's and the drift's — and do not grow with either removed. Their growth rate depends only on the product of the two, so it is identical whether the current or the waves supply more of that product. What the split decides is where the energy goes: the current's shear powers the downwind jet under each windrow and the drift's shear powers the overturning. For an 8-second, one-metre swell under a moderate wind, with an eddy viscosity of 0.01 m²/s, the current supplies 85 per cent of the fastest roll's energy.

Tested in The drift that turns a current into rolls, at the figure it turns on.

The Womersley number describes the shape of the velocity profile in a pulsating vessel, and wave propagation is a separate matter governed by the wall's stiffness.

What decides it: In an elastic tube the same function of the Womersley number that shapes the profile, F₁₀(α), multiplies the wave speed: c = c₀√(1 − F₁₀). At the aorta's α of 14.7 the wave travels at 96 per cent of the Moens–Korteweg speed and keeps 73 per cent of its amplitude per wavelength; at α = 2, 73 per cent of the speed and 3 per cent per wavelength; at small α the speed tends to α/2 of the inviscid speed and the disturbance diffuses rather than travels. The wall sets the scale of the speed. The Womersley number decides whether there is a wave.

Tested in The pulse that has to travel, at the figure it turns on.

Ideal flow is a good approximation for short times, because viscosity has not had time to act yet.

What decides it: It is a good approximation at short times for the *magnitude* of the force and a bad one for its *character*, and the difference is what the indicial kernel shows. An ideal flow delivers the whole of its response at the instant of a step and exactly nothing afterwards. The viscous response falls as the inverse square root of time — it has no time constant, so there is no moment at which it has finished, and its tail carries most of its own total. Over six crossing times the ideal kernel has delivered all of its impulse and the viscous one is still delivering.

Tested in The theory with no memory in it, at the figure it turns on.

Solving the equations of motion exactly gives the force a body really feels.

What decides it: Pressure integrated over 720 points of the exact surface solution. The drag comes out at −1.1 × 10⁻¹⁶, and the same routine on a cylinder with circulation returns a lift of 2.994 against the theorem's 3 — so the machinery finds the force that is there and reports the absence of the one that is not.

Tested in The exact theory says nothing has any drag, at the figure it turns on.

Transition on a flat plate occurs at a Reynolds number of 500,000.

What decides it: The figure is a convention, not a measurement of the fluid. Careful low-turbulence experiments delay flat-plate transition past Re_x = 3·10⁶ and a rough or noisy environment brings it below 10⁵ — a spread of thirty to one on the same fluid, the same plate and the same equations. The linear instability that begins the process sets in at Re_x of roughly 10⁵ and produces no transition for three decades afterwards. What 5·10⁵ names is an ordinary wind tunnel.

Tested in The number that is not a number, at the figure it turns on.

The exact theory and the real flow differ only in small details.

What decides it: The same body in the same stream, solved both ways side by side. The exact field is divergence-free to 3.3 × 10⁻⁸ and tangent to 7 × 10⁻¹⁵, and predicts no drag and no wake; the grid solve at the same conditions separates and carries one.

Tested in The two theories, side by side, at the figure it turns on.

The Kelvin–Helmholtz result shows that a shear layer breaks up at the shortest wavelength present.

What decides it: It shows that of the vortex sheet, which has no thickness. The growth rate sigma = k dU/2 rises without bound as k rises, so the sheet has no fastest-growing wavelength at all and the initial-value problem for it is ill-posed. Give the layer a thickness and the spectrum is cut off at k delta = 1 — the neutral wavenumber computed from Rayleigh's equation, not fitted — and a fastest-growing wavelength exists, at about seven times the layer thickness.

Tested in Every wavelength at once, at the figure it turns on.

Creeping flow is the easy limit: the equations are linear, so everything about it is straightforward.

What decides it: The Stokes equations for a cylinder in an unbounded fluid have no solution at all. Solved exactly inside a room of radius R, the drag comes to 7.058 at 16 radii and 2.119 at 1024, and 1/drag rises as ln R at 0.079464 against the predicted 1/4πμU = 0.079577 — so it never converges, and the build refuses a sweep in which it appears to.

Tested in The world with no inertia, at the figure it turns on.

The Lorenz attractor is a picture of convection.

What decides it: It is a picture of three ordinary differential equations obtained by keeping three Fourier modes of a convection problem and discarding the rest. The discarded modes are negligible only just above onset; by r of about 5 they are not, and the standard picture is drawn at r = 28. The equations remain exactly what they are — their fixed points sit at plus or minus the square root of beta times r minus one, checked here to 1e-9, and the Hopf threshold at 24.7368 comes out of the closed form — and what they are is a dynamical system rather than a fluid.

Tested in Three numbers left of a fluid, at the figure it turns on.

Pressure disturbances travel through a liquid or a gas at the speed of sound, so a numerical model of an incompressible flow is computing how they propagate.

What decides it: An incompressible flow's pressure does not propagate at all. Taking the divergence of the momentum equation kills the time derivative, because the divergence of the velocity is identically zero, and kills the viscous term for the same reason — leaving Poisson's equation, which is elliptic and has no characteristics and no signal speed. A compact source of acceleration in a quiet box produces a response everywhere in the box, at once, falling off only as the logarithm of the distance in two dimensions: the relaxation here fits a straight line in ln r to under one per cent over the outer half of the domain. That is not what a fluid does; it is what the incompressible ASSUMPTION does, and the ratio of the two speeds is exactly 1/M, so the assumption is a good one at a tenth of the speed of sound and is nonsense near it.

Tested in Pressure has no speed, at the figure it turns on.

d'Alembert's paradox: a closed body in steady, inviscid, irrotational flow experiences no drag.

What decides it: Every hypothesis but one still holds above Mach one, and the drag is not zero: a flat plate at Mach 2 and 5° has cd = 0.01768 with no viscosity anywhere in the calculation, computed face by face from shocks and fans. The hypothesis that fails is incompressibility — a compressible fluid can carry energy away to infinity along Mach waves, which an incompressible one has no mechanism to do.

Tested in Drag with nothing to rub, at the figure it turns on.

A figure showing an alternating row of vortices behind a body is a solution of the flow past that body.

What decides it: The figure in this essay is an exact ideal-flow field for two infinite rows of point vortices and contains no body, no viscosity and no mechanism that would shed anything. What it settles is the one thing a model of that kind can settle: which ratio of row spacing to along-row spacing is neutrally stable, bisected here to 0.28054993 against the exact ln(1 + root 2) over pi. Why the along-row spacing takes the value it does, and why anything sheds at all, are outside it entirely.

Tested in The street this site cannot draw, at the figure it turns on.

A swept wing's pressure coefficients are the unswept section's scaled by cos²Λ, so its critical Mach number is higher by 1/cos Λ.

What decides it: That is exact for an infinite yawed wing and it is a statement about a wing with no ends. A lattice solve of a 35°-swept wing puts the local sweep of the half-load line at 35.2° at mid-span — the theory being right — and at 14.3° at the root and 31.6° at the tip. Since the benefit goes as the *local* isobar sweep rather than the geometric one, the root of that wing is getting the benefit of a 14° sweep, which is why the shock forms there first on every swept wing ever built.

Tested in The sweep a root does not have, at the figure it turns on.