What is refuted here, and by what — page 2

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 supposes 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 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 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 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.