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 calculation requires the two ratios to 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 calculation 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 calculation 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 tests 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 step — 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; 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 calculation 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 the solved cylinder flow 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 both sides are computed 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.

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 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.

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.

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.

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.

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: 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 conventional 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 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.

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 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.

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.