What is refuted here, and by what — page 3

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The no-slip condition holds for all fluids at all walls, and apparent slip in a rheometer is an artefact of a badly prepared surface or a poorly wetted one.

What decides it: An entangled polymer melt has a slip length of microns to millimetres on an ordinary clean wall, which follows from the ratio of the melt's viscosity to its own monomer's and needs no surface treatment at all. It is measurable without assuming it: at a fixed wall shear stress the apparent shear rate is a straight line in the reciprocal of the die radius, whose slope is four times the slip velocity, so three dies of different bore separate the sliding from the shearing with nothing assumed about either.

Tested in The wall a melt really does slip on, at the figure it turns on.

Turbulence stretches vortices into tubes, so the typical local straining in a turbulent flow is one direction pulled out and two squeezed in.

What decides it: The local dynamics favour the opposite sign. Under the restricted Euler equation an axisymmetric strain with one stretching and two squashing directions decays for ever, while the same strain with two stretching and one squashing reaches infinity at exactly √6 of its own time units, and three hundred random starting gradients all end on the sheet-forming branch with the vorticity lying along the middle strain axis. Measured turbulent gradients show the same preference — two positive strain rates more often than one — and the tubes seen in simulations are what the non-local pressure makes of it, not what the local stretching asks for.

Tested in A gradient left to itself, at the figure it turns on.

A drop hanging from a tube falls when its weight exceeds the surface tension acting round the tube's rim, 2πRσ.

What decides it: Every static pendant drop satisfies an exact force balance at the rim with two terms Tate's law leaves out — the angle the surface meets the rim at, and the pressure inside the drop acting on the mouth's area — and it holds for each of 475 computed drops to 3·10⁻⁷. The drop falls at the largest volume the family of static shapes contains, which on a tube 0.4 capillary lengths in radius is 0.758 of 2πRσ/ρg, and on a tube wider than 1.83 is more than it.

Tested in The drop falls at a fold, at the figure it turns on.

A gust of a given speed produces a given load on an aeroplane — the lift-curve slope times the angle the gust adds, times the dynamic pressure.

What decides it: That is the load on a wing held still, and only after the lift has had time to build. An aeroplane free to rise escapes part of it by rising: computed through a one-minus-cosine gust with the unsteady lift functions, the peak load is 0.43 of that figure at a mass parameter of 5 and 0.88 at 200 — a light single at sea level keeps 0.68 of it and an airliner at eleven kilometres 0.87 — and the gust length that does the most harm is eight chords for the one and twenty-three for the other.

Tested in The aeroplane rises out of its own gust, at the figure it turns on.

A body's added mass is the mass of fluid it carries with it — a fixed property of its shape and of the boundaries around it.

What decides it: Under a surface that can make waves it depends on the frequency. A circle with one radius of water above it borrows 1.13458 times its own volume of water when shaken slowly, 0.883294 when shaken fast, 1.258 at a wavenumber of 0.134 per radius and 0.707 at 0.98. A circle with a tenth of a radius of water above it borrows 2.37 at one frequency and −0.182 at another. No quantity of carried water can be negative.

Tested in The borrowed mass that goes negative, at the figure it turns on.

A puff of smoke spreads like a diffusing substance, with a width growing as the square root of the time, just faster than molecular diffusion because the eddy diffusivity is larger.

What decides it: Inside the inertial range the diffusivity for the puff's own size grows as that size to the four-thirds power, so the mean square size grows as the cube of the time, not linearly — three pairs released one, thirty and a thousand Kolmogorov lengths apart converge onto one curve g ε t³ and lose every trace of their start. The puff's cloud of separations is not Gaussian either: its kurtosis is 3.76, more than twice a diffusing cloud's 5/3. A constant eddy diffusivity describes the puff only once it is larger than the largest eddies.

Tested in The further apart, the faster they part, at the figure it turns on.

In a Hele-Shaw cell the obstacle feels no pressure drag, because the flow is the ideal flow and its pressure recovers behind the body as d'Alembert's paradox requires.

What decides it: The cell's pressure is proportional to the velocity potential, not to the square of the speed, so it falls steadily from the front of the obstacle to the back. A cylinder in a cell carries Darcy's drag, 24πμUa²/h, and once no-slip on its side is included the drag is exactly Darcy's drag on a cylinder one Brinkman length larger — checked by integrating pressure and wall shear round the real cylinder, to 5·10⁻⁷ at four sizes.

Tested in The cell draws a larger cylinder, at the figure it turns on.

Jurin's law is a narrow-tube approximation, so in a wide tube the liquid rises less than it says.

What decides it: The weight of all the liquid lifted above the outside level is exactly the wall's tension times the perimeter, whatever the tube, because the plane that level defines is at atmospheric pressure. Dividing by the tube's area gives Jurin's height as the exact mean: computed lifted volumes match 2πR cos θ to a part in a million from R = 0.05 to 8 capillary lengths at three contact angles. What fails in a wide tube is only how the lifted liquid is spread — a rim near √2 capillary lengths at the wall and a centre that at twelve capillary lengths rises by five hundred-thousandths.

Tested in A law that is exact as an average, at the figure it turns on.

A slender body of revolution disturbs the flow about as much as a wing section of the same thickness-to-length ratio; thickness is thickness.

What decides it: A plane ellipse speeds the flow over its middle by exactly its thickness ratio, 0.1 at a tenth. A prolate spheroid of the same ratio does it by Lamb's coefficient, 0.0207 — a fifth — and the ratio falls as the thickness does, because the body of revolution's excess goes as the thickness squared with a logarithm. Compressibility widens the gap: by Göthert's rule the spheroid's peak suction grows 23 per cent between Mach 0 and 0.8 where the ellipse's grows 67, and the spheroid first meets sound on its surface at Mach 0.96 against the ellipse's 0.83.

Tested in Thickness costs a fuselage far less than a wing, at the figure it turns on.

Once the reflected shock has brought the test gas to rest against the end wall, the reservoir holds its pressure until the driver gas itself arrives.

What decides it: It holds only if the tube is tailored. When the reflected shock reaches the contact surface it meets a second gas, and in general it sends a wave back. With helium driving air at an incident Mach number of 2.6 the wave is an expansion, it reaches the end wall 0.154 L/a₁ after the reflection, and it lowers the reservoir's pressure by 17%. At Mach 3.4085 the returning wave has zero strength and nothing arrives.

Tested in The shock that passes without an echo, at the figure it turns on.

Weather helm is drag: a rudder angled to carry side force slows the boat, so a boat is fastest balanced with its rudder unloaded.

What decides it: Unloading the rudder takes 1.22° of lee helm and costs 0.97 per cent of the foils' drag against the best division, which puts 12.4 per cent of the side force on the rudder at 0.78° of weather helm. The rudder does feel a disproportionate drag there — a quarter of the induced drag for an eighth of the load — but that is the keel's sidewash billed to it, and the pair's total is lower.

Tested in The rudder pays for the keel's wake, at the figure it turns on.

An approximate solution of a flow problem can only be judged against an exact one or an experiment; on its own it says nothing about how wrong it is.

What decides it: For viscous duct flow a trial velocity that vanishes on the walls always underestimates the flow rate, and a trial stress in balance with the pressure gradient always overestimates it. Computed on one mesh for an L-shaped duct with no closed form, the two give 0.213991 and 0.214154 — the answer to within 0.08 per cent, certified by the calculation itself.

Tested in A flow pinned between two guesses, at the figure it turns on.

A gas moving far below the speed of sound can be treated as incompressible: its density changes by about the square of its Mach number, and its velocity field has no divergence to speak of.

What decides it: That estimate counts only what the pressure does to the density. A hot spot of a millimetre radius in air, three times the surrounding temperature, moves at Mach 1.3 × 10⁻⁴ and changes its volume threefold as it cools. The gas flows inward everywhere at exactly (γ − 1)/(γp) times the heat flux, reversed — computed from the gas's own motion and from the heat flux separately, agreeing to 7 × 10⁻⁵ of the peak.

Tested in The air flows in against the heat, at the figure it turns on.

A turbulence statistic becomes more accurate the more samples it is computed from, so a higher sampling rate buys precision.

What decides it: A record's precision is set by how many integral time scales it spans, not by how many samples it holds. For a record a hundred integral scales long the scatter of the mean falls as one over the square root of the samples only while they are sparser than one per integral scale; beyond that it stops at √(2Tᵢ/T) = 0.141 of the standard deviation — 0.142 at five samples per scale and 0.142 at twenty, across five hundred records each.

Tested in A record is as long as its integral scales, at the figure it turns on.

A liquid's contact angle is fixed by the liquid and the solid, so on a given surface a drop of a given volume can take only one shape.

What decides it: On a smooth face it is fixed. On an edge the line stops and the angle is free across a band. Water with a Young angle of 60° on a coin the size of a US cent sits on the rim at every apparent angle from 60° to 150°, holding anywhere from 0.52 to 1.44 mL without the line moving — nearly three times what the chemistry alone lets the same surface hold.

Tested in An edge holds any angle it is given, at the figure it turns on.

A falling raindrop oscillates because the vortices its wake sheds strike it at its own natural frequency.

What decides it: At the textbook Strouhal number of a sphere's wake, 0.2, and the terminal speed a sphere-drag calculation gives, the wake of every raindrop large enough to shed at all sheds between 2.3 and 18 times faster than the drop's fundamental — 661 Hz against 122 at two millimetres. The shedding passes through the fifth and sixth modes instead, which damp thirteen times faster. A drop that rings at its fundamental is not being driven at it; being lightly damped, with a quality factor near 76, it picks its own pitch out of whatever broadband buffeting reaches it.

Tested in A drop rings like a bell, at the figure it turns on.

An aeroplane's wake vortices fade away as the air's turbulence slowly wears them down, so in very still air a wake lasts almost indefinitely.

What decides it: In still air the pair ends itself. A mirror-image bend about eight and a half spacings long grows by 0.827 e-folds each time the pair sinks one spacing, because each vortex sits in its partner's strain. For an airliner's wake, whose pair sinks one spacing in 27 seconds, a bend starting at one per cent of the spacing brings the two cores together in about 143 seconds, and a bend a hundred times smaller only doubles that, to 296 seconds. Stiller air buys logarithmically little time.

Tested in A wake ends by bending, not by fading, at the figure it turns on.

A dilute suspension has one viscosity, fixed by the particles' shape and volume fraction, as a dilute suspension of spheres does.

What decides it: Rods in a shear flow each follow one of a family of closed Jeffery orbits, and slow viscous flow, being reversible, never moves a rod from its orbit. The orientation part of the intrinsic viscosity for rods of aspect ratio five is zero if every rod spins on the vorticity axis, 1.24 if they start isotropic, and 2.16 if every rod tumbles in the plane of shear — all permitted, all permanent. Only a weak effect outside the equations, such as rotary diffusion, selects one: 1.22, the same at two diffusivities threefold apart.

Tested in A viscosity the flow cannot decide, at the figure it turns on.

A wet ceiling drips at a spacing fixed by the liquid's surface tension: the fastest-growing wave of the Rayleigh–Taylor instability, 2π√3 capillary lengths, about 30 mm for water.

What decides it: That is the spacing for a deep, inviscid layer. A film thinner than a capillary length, which is what condenses on a ceiling, grows fastest at 2π√2 capillary lengths — 24.1 mm for water, with the fastest wavenumber recovered at 0.99944 of 1/√2 — because the film must be squeezed sideways to thicken. A deep viscous layer grows fastest at a wavelength set by its depth: 88 mm under 3 cm of honey. The depth of the layer chooses between the three.

Tested in How far apart a ceiling drips, at the figure it turns on.

A long gas pipe in contact with the ground flows isothermally, and so it chokes at a Mach number of 1/√γ, 0.845 for air, rather than at Mach one.

What decides it: Holding the gas at the wall's temperature needs heat put in, as its kinetic energy grows — 11.5 units per unit friction length at Mach 0.84 — while a wall at the gas's temperature takes heat out, because friction raises the gas beside it to its recovery temperature. Marched with heat transfer from a tenth to a thousand times what Reynolds' analogy gives, a pipe entered at Mach 0.1 chokes at Mach one in every case, at 0.835 to 0.848 of the wall's temperature, within 0.1 per cent of the length an insulated pipe needs.

Tested in A pipe cannot hold its gas at the wall's temperature, at the figure it turns on.

A flat, steady flame is a stable state: a small wrinkle in it is smoothed out as the flame burns on, because the parts that stick out are exposed to more fresh gas.

What decides it: A thin flame that expands its gas sevenfold grows every wrinkle at 1.578 times its burning speed times the wrinkle's wavenumber, whatever the wrinkle's size — the jump-condition determinant reproduces Landau's closed form to 3·10⁻¹⁶. Only the flame's response to its own curvature, a Markstein length of a few tenths of a millimetre for methane and air, smooths anything: wrinkles shorter than 4πσ/(σ − 1) Markstein lengths, about 3 mm. Everything longer grows.

Tested in A flat flame is unstable at every size, at the figure it turns on.

The drag on a bluff body comes from the eddies trapped behind it: the low pressure inside the standing eddies sucks the body backwards.

What decides it: A pair of vortices standing behind a cylinder in ideal flow, on Föppl's curve, exerts a drag on the cylinder of exactly zero — 5·10⁻¹⁶ of ρU²a by Blasius' theorem integrated round the surface — at every position on the curve. The same pair held at a strength a fifth too weak or too strong would pull on the cylinder with a force of order ρU²a, but free standing eddies do not. Drag needs vorticity carried away, not vorticity held in place.

Tested in Two eddies can stand behind a cylinder, but not for long, at the figure it turns on.

A duct round a rotor is a guard that at best stops air leaking round the blade tips; the rotor still makes all the thrust.

What decides it: In hover, momentum theory puts exactly half the thrust of a rotor in a straight duct on the duct itself — the suction on its inlet lip — and the rotor carries the other half, 1/2σ for an exit σ times the disc's area. By holding the slipstream at the disc's full area instead of letting it contract to half, the duct also cuts the induced power for a given thrust by a factor of √2: a 1 m rotor lifting 100 N needs 721 W open and 510 W in a straight duct, checked against the closed forms to 3·10⁻¹⁶.

Tested in A duct is worth the square root of two, at the figure it turns on.

Birds fly in a V because the V shape itself is what saves energy; a flock flying in a line abreast would save less.

What decides it: Nine wings one span apart across the stream, each carrying the same elliptic loading, pay 0.4244 of their lone induced drag between them whether they fly in a line abreast or in a V swept at any angle up to seventy degrees, because the total depends only on the wakes' positions across the stream. The sweep changes only the division: abreast, the two end birds pay 0.657 and the centre 0.340; in a V swept 45°, the leader pays 0.894 and every other member 0.39 or less. The V's advantages are that it can be flown and that it moves the saving backwards, not that it is larger.

Tested in The bird at the point pays for the V, at the figure it turns on.

A film pushed past its thermal runaway limit runs away straight away, so an overload that crosses the limit, however briefly, is lost.

What decides it: A film one per cent past the fold climbs to the temperature the vanished steady state would have had and lingers there, running away only after 16.3 thermal diffusion times — 31 milliseconds for a 25-micron oil film. The delay grows as one over the square root of the overload, with a measured exponent of −0.492. And if the stress is dropped back below the fold at any time up to 16.05 of those 16.3, the film recovers.

Tested in A lost steady state still holds the film, at the figure it turns on.

An explosion fills its fireball with compressed air, pushed outward and packed densely behind the shock.

What decides it: The air the shock sweeps up is packed against the shock itself and nowhere else. In the exact spherical solution at γ = 1.4, half the swept-up mass lies in the outer 4.2 per cent of the radius and nine-tenths outside 0.844 of it; at half the radius the density is 0.17 per cent of its post-shock value, a hundredth of the undisturbed air's. The interior is a hot, nearly empty core at a nearly uniform pressure, 0.37 of the post-shock value, and its temperature rises without limit towards the centre.

Tested in The fireball is hollow, at the figure it turns on.

A region where the velocity gradient has complex eigenvalues — where Q is positive, rotation beating strain — at every instant is a vortex that holds its fluid.

What decides it: In Haller's rotating saddle the velocity gradient has eigenvalues ±i√3 and Q = 3 at every point and every instant, to 10⁻¹⁵, in the frame the flow is measured in. Every particle leaves: the flow map over one period has multipliers of −4.810 and −0.208, minus e to the ±π/2, matched to 7·10⁻¹⁴, and neighbouring particles separate as e to the time, a Lyapunov exponent of exactly one. The strain's axes are turning at twice the strain rate, and in the frame that turns with them the flow is a pure saddle.

Tested in Every snapshot says vortex, and every particle leaves, at the figure it turns on.

A vortex is a rigid spinning column of fluid, and anything disturbing it is simply carried round with the flow.

What decides it: A vortex core is a waveguide. A helical bend of wavelength 2π core radii on a Rankine vortex rotates against the flow at a third of the core's rotation rate, from Kelvin's exact dispersion relation — not with it — and axisymmetric waves travel along the core, the long ones at 0.832 of the swirl speed at the core's edge, 2Ωa/j₀,₁, checked to a part in ten thousand. For every pattern round the axis there are infinitely many such waves, one for each number of radial nodes.

Tested in A vortex is a waveguide, at the figure it turns on.

A filter clogs because its cloth fills up, so a stronger pump or a coarser cloth keeps the flow going.

What decides it: Within the first three hundredths of a cubic metre of filtrate per square metre of cloth, the cake of caught particles has more resistance than the cloth, and it keeps growing: the time per litre rises in a straight line with the litres already filtered, Ruth's line, matched by direct integration to 7·10⁻⁹, and the filtrate grows as the square root of both time and pressure — four times the pressure for twice the output. For a compressible cake even that fails: with a compressibility index of 1.3 a hundredfold more pressure buys 1.7 times the flow.

Tested in A filter is slowed by what it has caught, at the figure it turns on.

D'Alembert's paradox means that in ideal flow no body in a steady stream feels any force, so an inviscid calculation has nothing to say about the forces between neighbouring bodies.

What decides it: The paradox is a statement about a surface drawn round the whole flow, and with two bodies it proves only that the pair feels nothing. Each cylinder does: side by side a fifth of a radius apart, each is pulled towards the other with a force of 3.60 ρU²a; in tandem, the front one is pushed upstream and the rear one downstream with 0.61. The sum is zero to 6·10⁻¹⁵ in every arrangement computed.

Tested in The paradox is for the pair, not for each body, at the figure it turns on.

A vortex is where the velocity gradient says rotation beats strain, so criteria such as Q, λ₂ or the Okubo–Weiss parameter find every vortex in a flow and nothing that is not one.

What decides it: Round two co-rotating vortices Q is positive only inside the two cores, an area of 0.25 half-separations squared. In the frame turning with the pair, two eddies of total area 17.5 circulate round the points that make equilateral triangles with the vortices, and fluid released there goes round with the pair: a blob at an eddy's centre is still within 0.32 of it after three orbits. The eddies carry no vorticity, so every snapshot criterion calls them strain.

Tested in A vortex pair carries eddies no snapshot can see, at the figure it turns on.

A self-heating film has a thermal runaway limit, and a drive either holds the speed, which removes it, or holds the force, which keeps it; a real motor falls somewhere between and delays the runaway in proportion.

What decides it: Any droop at all removes the runaway: along the hot branch the film's resistance falls faster than any straight torque–speed line, so the line always crosses it again. A drive with droop below 0.0738 of the cold film's resistance jumps instead — at k = 0.04 and five per cent past its fold, a film marched from rest settles at 4.987 e-folds against the hot operating point's 4.988 — and a slow ramp up and down leaves a loop four e-folds wide. Above 0.0738 there is no jump and no loop, and the film tracks one curve both ways.

Tested in A motor turns the runaway into a jump, at the figure it turns on.

A wet ceiling that slopes drips just as a level one does, at the same spacing and in the same places; the slope only makes the film thinner.

What decides it: Tilting the ceiling carries every disturbance downhill at V = 3 tanθ ℓ/h₀ in the film's own units, and once V passes 1.62208 — the speed at which a level ceiling's disturbance spreads — the growth at the point disturbed turns negative: 0.042 at 0.9 of that speed, −0.044 at 1.1, from the exact impulse response. For water that is 1.1° of slope for a 0.1 mm film and 5.7° for 0.5 mm. The drips then form downhill of where they began, at a distance that rises with the slope — half a metre for a 0.2 mm film at 10°.

Tested in A sloping ceiling drips downhill, or not at all, at the figure it turns on.

Two identical pumps in parallel deliver twice the flow of one, so a station can meet half its design flow by switching one of two pumps off.

What decides it: Against a system whose head rises with flow, one pump of two runs further out along its own curve, where it makes less head, and delivers 76 per cent of the design flow with no static lift, 71 per cent with 30 per cent lift and 65 per cent with 60. The second pump adds a third, two-fifths and a half, not a doubling — and the pump left alone runs at up to 1.5 times its best flow coefficient, drawing up to 17 per cent more than its own design power.

Tested in One fast pump or two slow ones, at the figure it turns on.

A shock wave is a discontinuity a few mean free paths thick, so the crack at the front of a sonic boom is as sharp as the air can make it.

What decides it: For a 50-pascal boom in air at 20 °C and 50 per cent humidity, viscosity and heat conduction alone would make a rise of 3.3 microseconds. The vibrational relaxation of oxygen and nitrogen spreads it to 0.62 milliseconds, nearly two hundred times longer, and for any boom weaker than 94 pascals there is no discontinuity in the steady profile at all. At 10 per cent humidity the rise takes 2.95 milliseconds.

Tested in Oxygen makes a boom's crack, nitrogen its rise time, at the figure it turns on.

A vortex in a strain is torn apart before the strain reaches Moore and Saffman's limit, because the rounder steady ellipse becomes unstable to lobed disturbances first; the existence limit overestimates how much strain a vortex survives.

What decides it: On the rounder branch, contour dynamics with disturbances of two, three, four and five lobes finds none growing exponentially at aspect ratios of 1.5, 2.2 and 2.7 — the last at a strain within a thousandth of the limit of 0.150; Kida's equations make the two-lobed disturbance exactly neutral all the way to the limit. The rounder shape holds until it no longer exists.

Tested in A strained vortex holds until it has no shape to hold, at the figure it turns on.

Leapfrogging smoke rings stop after a few passes because viscosity drains the energy that holds them together, so the rings drift apart.

What decides it: With cores diffusing at a vortex Reynolds number of 3,000 the pair's binding margin — its energy less the most two free rings with its impulse could hold — falls by only 3 per cent and never changes sign: the rings cannot part. The leapfrogging ends when the cores, fattened by diffusion, pass within 0.24 core separations of each other and merge, after about one pass at Re 3,000 and in proportion to Re above it.

Tested in Leapfrogging rings end by merging, not by parting, at the figure it turns on.

Because a premixed flame is unstable at every wavelength longer than its cut-off, a wider flame develops more wrinkles and burns faster in proportion to its width.

What decides it: In the Michelson–Sivashinsky equation every front settles, whatever its width, into a single arc and a single cusp per period; the steady front with N pole pairs has a speed excess of exactly 2Nν(1 − Nν), and with the most pairs a domain can hold that never exceeds one half. A flame marched from random noise in a domain 12.5 neutral wavelengths wide settles at 0.49986. Beyond about five wavelengths a wider flame is the same arc at a larger size, with the same slopes, and no faster.

Tested in A wrinkled flame has one cusp and a speed limit, at the figure it turns on.

Where a squeeze film being pulled apart cavitates, it stops resisting, so a sphere rebounding through a film that tears at a tenth of a millimetre escapes the decades of gap below the tear and its rebound threshold falls by about a third.

What decides it: With a pressure floor, the force on the way out is 2πRΔp(h − x) + 6πμR²u/h, with h the gap at which the film tears: the vapour disc holds the full tension over its area and the liquid round it still pulls as a film at the tear gap would. The saving over the outward leg is ln r − 3/2 + 2/r − 1/2r², not ln r — 0.49 rather than 1.61 for a tear at five times a 20 μm contact gap — and the threshold moves from 10.18 to 9.63, by five per cent.

Tested in A torn film still pulls, at the figure it turns on.

A hydrofoil near the surface is a wing in ground effect reflected in the surface: its lift moves between a gain at low speed and a loss at high speed, and is never worse than the high-speed value.

What decides it: The lift dips below both images between them. Half a chord down, the foil keeps 0.84 of its deep-water lift when fast and 1.22 times it when slow, and 0.43 at a chord Froude number near 1.2; one chord down, 0.94 and 1.07 at the ends and 0.61 at the worst speed. The dip sits where the wave drag peaks, near a depth Froude number of √2, and for a 0.3 m chord at 0.3 m depth that is 2.7 metres a second.

Tested in A hydrofoil loses most lift on the way up, at the figure it turns on.

Which relation a relaxed two-dimensional flow obeys — sinh, tanh or linear — can be read off its vorticity–streamfunction scatter plot, or failing that from its vorticity statistics at the measured ratio of enstrophy to energy.

What decides it: The scatter plots differ only in curvature: a straight line fitted to the sinh or the tanh state at Z/E = 1.1 misses by 0.27 of the rms vorticity, the size of the scatter unrelaxed eddies make. Matching by Z/E fails too, because small-scale vorticity at 40 per cent of the rms raises Z/E by sixteen per cent. The streamfunction's flatness is unmoved by the same contamination, and lies above the linear state's 9/4 for every sinh dipole and below it for every tanh dipole at every amplitude computed.

Tested in The streamfunction says which relaxed state, at the figure it turns on.

Surge and rotating stall are two degrees of the same failure: a compressor that stalls badly enough surges, so what surges is a machine with worse blades or a higher loading.

What decides it: With the compressor, its characteristic and its throttle held identical, the Moore–Greitzer model settles into steady rotating stall at B = 0.5 and goes into deep surge, with the flow reversing to −0.23, at B = 2. The only difference between the two runs is the plenum's volume against the duct's inertia. The blades decide where the peak of the characteristic is; the system decides what happens past it.

Tested in The plenum decides whether a compressor surges, at the figure it turns on.

Water drops stick to a surface because it is wettable: the more readily water spreads on it, the larger the drop a tilted pane can hold.

What decides it: What holds a drop on a tilted plate is the difference between the cosines of its receding and advancing angles, not how well it wets. A ridge of water on clean glass (advancing 30°, receding 10°) can have a cross-section of at most 0.24 square capillary lengths on a plate tilted 30°; on a plastic that water wets far less (90°, 70°), 0.68 — nearly three times as much — because the same 20 degrees of hysteresis make a larger difference of cosines at larger angles.

Tested in The force a contact line holds is a range, at the figure it turns on.

Two bodies drawn together in an ideal fluid cannot actually touch, because the fluid trapped between them has to be squeezed out through a vanishing gap and resists without limit.

What decides it: The resistance is the growth of each body's added mass as the gap closes, and for two circular cylinders it is only logarithmic — 1.47 displaced masses at a gap of four-tenths of a radius, 2.03 at a fiftieth, 2.20 at a five-hundredth, by successive images agreeing with a panel method to 4·10⁻⁵. A neutrally buoyant pair released side by side four radii apart in a stream collides after 6.5 radii of stream, arriving at about the stream's speed.

Tested in A pair set free in a stream collides or parts, at the figure it turns on.

A soft motor drive whose rotor has an inertia comparable with the oil film's thermal diffusion time can make a self-heating bearing oscillate between its cool and hot states — a thermal relaxation oscillator built from a bearing and a motor.

What decides it: The linearised film and rotor, with the rotor's speed rescaled by √(h/4M), are a symmetric matrix at every steady state, so every eigenvalue is real at every inertia: none of the three branches can ring, and none changes its stability as M runs from 0.01 to 100. Computed on all three, the largest imaginary part is zero. Only a drive that lowers its torque as the film warms — a link of the opposite sign — produces an oscillatory pair.

Tested in A rotor's inertia slows the jump and cannot make it ring, at the figure it turns on.

A siphon cannot run over a crown more than about ten metres above its upper reservoir, because the pressure at the crown cannot fall below zero.

What decides it: The crown's pressure can fall below zero: water holds tension until its largest gas nucleus reaches Blake's breaking point. With the energy equation for the running siphon and Blake's threshold, a slow siphon whose largest nucleus is a micron runs over a 14.4 m crown, and one whose largest is three-tenths of a micron over 27.3 m; ten metres is the limit only for water carrying nuclei of ten microns and more, as ordinary water does.

Tested in A degassed siphon is as tall as its largest nucleus allows, at the figure it turns on.

The shortfall of the third-order structure function below the four-fifths law at laboratory Reynolds numbers is a finite-Reynolds-number effect, so any two flows at the same Taylor-scale Reynolds number fall short by the same amount.

What decides it: On one model spectrum at a Taylor-scale Reynolds number of 200, a flow forced in a band at its integral scale peaks at 0.965 of four-fifths and a decaying flow at 0.747. A decaying flow needs a Reynolds number near 4,800 to peak as high as the forced one does at 200, and the factor between the two grows with the Reynolds number, because the forced shortfall closes as the six-fifths power and the decaying one as the two-thirds.

Tested in Forcing at the integral scale leaves the cascade alone, at the figure it turns on.