What is refuted here, and by what — page 5

The deeper a protection cuts the motor when the bearing runs hot, the better it protects the oil film — a harder cut means a cooler film.

What decides it: Under a trip-and-reset protection the film's peak mean excess is 3.493, the unprotected value, at every cut, because each reset restores full drive and the film is hot again at once. What the cut's depth decides is the trough: a cut shallower than the film's window, 0.350 of full drive, keeps the film on its hot branch, while every deeper cut drops it to the cool branch and throws it back twice a cycle, with the period fixed near 0.76 housing times by the band rather than by the cut.

Tested in A trip deeper than the film's window flips the bearing twice a cycle, at the figure it turns on.

Sweeping a wing back spreads a gust's arrival along the span and so smooths the loads it causes, cutting the root's fatigue cycles towards the lift's.

What decides it: It smooths both and leaves the gap. With each station of an elliptic wing of aspect ratio nine meeting frozen turbulence late by its distance aft, 35° of sweep cuts the lift's crossing rate to 0.899 of the straight wing's and one root's to 0.887, so the root still crosses its mean 1.17 times as often as the lift, against 1.19 straight; the variance changes by a third of a per cent.

Tested in Sweep slows a wing root's gust cycles and leaves their excess over the lift, at the figure it turns on.

Below a Reynolds number of a few hundred a branching tube carries Poiseuille flow, so the lung's conducting airways have the resistance Poiseuille's law gives them and follow Murray's cube law at their cheapest.

What decides it: Every daughter airway restarts its boundary layer at the flow divider, and along Weibel's symmetric tree Pedley's measured factor puts the dissipation above Poiseuille's down to generation 11 at 1 L/s — 9.8 times it in generation 1 and 5.2 in generation 5 — so generations 1 to 16 have 4.1 times Poiseuille's resistance, and a resistance that grows as the square root of the flow. The factor depends on the flow and the length and not on the radius, so the cheapest segment follows Q ∝ r^12/5 at fixed length or r^13/5 at fixed shape rather than r³.

Tested in The layer every airway restarts bends Murray's cube towards the turbulent pipe's, at the figure it turns on.

A squeeze-film levitator filled with a gas that cannot shed its heat of compression in time is stiffer and so lifts more: heavy, slowly conducting gases make better levitators.

What decides it: Solved to every order in the amplitude with the heat conducted across the gap, a trapped film squeezed by a quarter of its gap at the thermal numbers of a 20 kHz, 20 µm levitator at one atmosphere lifts within 0.5 per cent of the isothermal film for helium, air, argon, carbon dioxide and sulphur hexafluoride and 4.2 per cent above it for xenon, whose loss is twenty times helium's thermal loss. Even far from the walls' reach, at a thermal number of 300, a monatomic film lifts only 1.21 times the isothermal film, not the 1.69 of an adiabat at the walls' temperature, because its middle settles 1.2 per cent cooler.

Tested in A levitator lifts by its amplitude, not its gas, at the figure it turns on.

A gas at lower pressure has fewer molecules to carry momentum, so its viscosity falls as it is pumped down: a thinner gas is a less viscous one.

What decides it: Fewer molecules each carry momentum further, and the two cancel: sheared between plates a millimetre apart, air transmits its full atmospheric viscous stress to within one per cent at every pressure from an atmosphere down to 1.7 kPa, and half of it only at 15 Pa, where the free path has grown to a few hundredths of the gap. The fall below that is not a smaller viscosity but a gas whose molecules cross the gap without colliding, so that the stress becomes proportional to the pressure.

Tested in A gas's viscosity ignores its pressure until the free path reaches the walls, at the figure it turns on.

In thin-aerofoil theory a section's thickness carries no load, so the lift and the chordwise loading of a section are fixed by its camber and incidence alone and are the same for a thick section as for its camber line.

What decides it: On an exact Joukowski section 12 per cent thick with 4 per cent camber at 4°, the loading at a tenth of the chord is 1.59 against 1.26 for the camber line and flat plate added, and at nine-tenths 0.40 against 0.51: thickness moves the load forward. The section lifts 1.026 against 0.941 from thin-aerofoil theory, 9.0 per cent more — 1 + 0.75 t/c, against Joukowski's 1 + 0.77 t/c — although a symmetric section of the same thickness lifts nothing at zero incidence.

Tested in Thickness carries no load of its own and still moves the load forward, at the figure it turns on.

Knudsen's minimum is a dip in the flow rate below what both the continuum-with-slip formula and the free-molecular flux predict, so the truth in the transition regime lies under both closed forms.

What decides it: Between plates the linearised BGK equation gives a least reduced flow rate of 1.538 at δ = 1.07, and everywhere in the transition regime the solve lies above both closed forms: where the slip formula δ/6 + 1.016 and the uncollided flux cross, at δ = 0.335, both read 1.07 and the solve reads 1.68. The minimum is where a transport rising with density (viscous momentum exchange) crosses one falling with it (molecules skimming the walls), not a gap under the asymptotes.

Tested in A thinning gas flows least at one free path across, and then more without end, at the figure it turns on.

A golfer reaches the best launch conditions for a drive by choosing the right driver loft: more loft for a higher launch, less for less spin.

What decides it: Loft raises launch and spin together, because a strike's backspin is set by the angle between the head's path and its face and its launch by the face. At a level strike with a 45 m/s head, no loft reaches the best drive: the best loft, 9.1°, gives 2,129 rpm at 7.4° and 241.4 m. Striking 9° upward with 13.9° of loft gives 1,123 rpm at 13.0° and 246.0 m, and near their best the total changes by under half a metre per degree of loft and by about a metre per degree of attack.

Tested in A driver reaches the best drive only by striking upward, at the figure it turns on.

A wing's drag due to lift falls as one over its aspect ratio, so a longer span always buys a proportionally cleaner wing.

What decides it: That is true only of the induced drag, which is all the drag due to lift a wing has when it collects its leading-edge suction. A flat wing that loses the suction has a drag due to lift of Cₗ²/Cₗα, and its lift slope stops rising at the section's 2π, so with a zero-lift drag of 0.008 its best lift-to-drag ratio rises only from 6.75 at aspect ratio 1 to 12.5 at 12, and can never pass the square root of π over twice the profile drag coefficient, 14.0. With the suction the same wings go from 9.9 to 33.4.

Tested in A wing that cannot hold its suction gains little from its span, at the figure it turns on.

A supersonic channel that will not start can be helped to start by opening its inlet wider, so that it captures the flow more easily.

What decides it: Starting is decided by the throat's area against the mouth's: a channel with a normal shock at its mouth swallows it when the throat can pass the flow behind that shock, which at Mach 1.944 needs a throat of 0.831 of the mouth. Widening the mouth lowers the ratio and delays the start. Busemann's pair, 0.8 shut, starts only at Mach 2.15; turning the front fifth of each ramp in by 3.4° narrows the mouth from 0.300 to 0.276 chords, raises the ratio to 0.870, and starts it at Mach 1.72.

Tested in A flap that closes the mouth starts Busemann's biplane below its design point, at the figure it turns on.

A jet held further from the surface it strikes arrives slower and so does less to the surface: the wall flow it drives weakens with the standoff.

What decides it: Schlichting's round jet arrives at a plate with a centreline speed falling as one over the standoff, but the radial wall jet it leaves keeps Glauert's invariant F = Kνh/(2π), which grows in proportion to the standoff because the jet's gathered volume flux 8πνh does. A water jet from a 2 mm nozzle at 15 cm/s held 40 cm from a plate arrives at a quarter of its speed at 10 cm, hands on four times the invariant, and its wall jet 5 cm from the stagnation point runs twice as fast.

Tested in A round jet hands its wall jet a strength proportional to its throw, at the figure it turns on.

A forward-facing step in a stream is pushed back by the high pressure that builds up against its face, so even in an ideal flow it must have drag.

What decides it: Solved exactly by a Schwarz–Christoffel map, the face of a right-angled step is pushed back over its lower 82 per cent with a force of 2/π of a dynamic pressure times its height, and pulled forward by the suction round its lip with exactly the same force: the net is zero to rounding at every ramp angle from 5° to 90°, because the reflection w → −w of the mapping exchanges the stagnation at the foot with the infinite speed at the lip. The drag a real step has is the lip's suction, lost when the flow separates there.

Tested in A step's face is pushed at its foot and pulled at its lip, equally, at the figure it turns on.

A large Deborah number says a liquid will behave elastically, so any liquid whose relaxation time is long against the period of a forcing will show the same elastic response to it.

What decides it: Oscillated along a tube 1 cm in radius, a single-mode Maxwell liquid with λ = 1 s and η₀ = 30 Pa·s resonates at 6.63 Hz, De = 41.7, and carries 1661 times the flow of a Newtonian liquid of the same viscosity. A 40-mode Rouse spectrum with the same longest time and the same viscosity, at the same elasticity number, peaks at 20.3: its loss tangent stays near one across the band where the resonance falls, and a shear wave in it keeps 0.11 of its amplitude per wavelength against the Maxwell liquid's 0.97 at De = 100.

Tested in A liquid with one relaxation time rings in a tube, and a spectrum of times silences it, at the figure it turns on.

Prandtl's bell-shaped loading is the least-drag answer only to an artificial question — a wing held to a fixed bending moment — so a real aeroplane, whose span is set by its structure's weight, should load its wing elliptically.

What decides it: Count the spar's weight as lift it must carry and minimise the drag over span and loading at a fixed payload. With the spar's depth held, the best wing is the bell exactly — √(3/2) of the best elliptic wing's span for 8/9 of its drag, its spar a third of the lift. With the wing's area held, so that the spar thins as the wing lengthens, the best loading lies between: a third harmonic of −0.1356 of the first, five per cent more span and 4.3 per cent less drag than the best ellipse, its spar a quarter.

Tested in A spar that must lift itself stops the span at a quarter of the weight, at the figure it turns on.

Newton's pressure law is exact enough at hypersonic speeds that the velocity gradient it gives at a blunt nose's stagnation point — and the heating that follows from it — needs no correction for the flow in the shock layer.

What decides it: The constant-density shock layer, an exact rotational solution satisfying the strong shock's conditions, gives a stagnation velocity gradient of 0.667 times the flight speed over the body radius at a perfect gas's density ratio of 1/6, against Newton's 0.553: 1.206 times as large, and √(4/3) = 1.155 times in the thin-layer limit. Its pressure falls round the nose faster than cos²χ — 0.83 of the stagnation value at 20° against 0.883 — and the Fay–Riddell heating, which goes as the gradient's square root, is about ten per cent above the Newtonian estimate.

Tested in The shock layer is Hill's vortex, and it heats the nose a tenth harder than Newton says, at the figure it turns on.

Two particles settling side by side or one above the other in a viscous liquid soon separate or rearrange, because each disturbs the flow round the other unevenly.

What decides it: In Stokes flow two equal spheres under equal weights always move with one velocity, so a pair keeps its separation and its slant exactly: a pair five radii apart falls 1.286 times as fast as one sphere in line and 1.154 times side by side, and at a slant of 46.5° drifts 3.1° sideways for as long as it falls. Rearrangement needs a third sphere: three released ten radii apart in a row close to within 2.5 radii of one another within 520 radii of fall.

Tested in Two spheres settling at a slant keep their slant for ever, and three do not, at the figure it turns on.

An attached sheet cavity, being the larger and longer-lived cavity, is the first to feel the cooling its own boiling causes; travelling bubbles are too small and too brief for the thermal effect to reach them.

What decides it: Under a tension of 10 kPa, a bubble growing for a millisecond makes about 140 times its cooled liquid's volume in vapour, while a 2 cm sheet at 20 m/s, which keeps renewing the liquid it cools, makes about seven times. Against the critical value that uses up the whole tension, the bubble is suppressed in water above 60.5 °C and the sheet only above 104 °C.

Tested in A bubble is throttled by its age, and a sheet by its length, at the figure it turns on.

Thin-aerofoil theory's error on a thick section is concentrated at the leading edge, where its surface speed is infinite, so a leading-edge correction such as Riegels's recovers the section's lift.

What decides it: On a NACA 2412 at 4°, solved by panels, the section lifts 0.743 against thin-aerofoil theory's 0.666. Linear theory with Riegels's factor makes the nose speed finite and gives a suction peak of 1.11 against the section's 1.44, but its integrated load is 0.666 — thin theory's lift to four figures. The missing eleven per cent is spread over the whole chord, not concentrated at the nose.

Tested in A NACA section multiplies its lift by nearly one plus its thickness, and a nose correction cannot supply it, at the figure it turns on.

An oil-flow picture of the skin-friction lines on a wall shows where the flow separates: a line onto which the skin-friction lines converge, running out of a saddle of the pattern, is a separation line.

What decides it: In a wall flow whose skin friction is (x² − μ, y), the pattern is the same for every μ above zero: a saddle and a node, with the lines converging on the saddle's outgoing separatrix. For μ above a quarter a stagnation point hangs above the saddle and the flow rises off the wall there; for μ below, the point has landed and the flow falls onto the wall. The same converging line is a separation in one case and an attachment in the other.

Tested in A stagnation point off the axis can land only on a saddle or a node, and turns it inside out, at the figure it turns on.

A turbulent boundary layer's resistance to an adverse pressure gradient is fixed by its shape factor where the gradient begins; how it came to have that shape does not matter.

What decides it: In Green's lag-entrainment model, a layer released for twelve thicknesses after one gradient and decelerated again separates 5.07 of its thicknesses into the second gradient, against 4.81 for a layer of exactly the same shape and thickness whose entrainment starts at its shape's equilibrium; released for none, 2.91 against 3.16. The history moves the separation by up to eight per cent either way.

Tested in A layer's memory of one gradient helps it through the next, unless the next comes too soon, at the figure it turns on.

When a larger particle sinks past a smaller one in a viscous liquid, the encounter pushes the two apart sideways, and repeated encounters of this kind are what spread a cloud of settling particles.

What decides it: For two smooth spheres of one material in Stokes flow, one 2% larger, the larger sphere overtakes round a path that is its own mirror image and leaves at its entry offset to a few parts in ten thousand at every offset from 0.85 to 5 radii. A net sideways displacement appears only when the spheres touch: with a roughness of a hundredth of a radius every touching path leaves 0.534 radii out, whatever offset it came in at.

Tested in A larger sphere overtakes its neighbour and leaves on the line it came in on, at the figure it turns on.

A fast enough throttle with enough gain on the plenum pressure can always damp surge, so stall is the only obstacle to running a compressor left of its peak.

What decides it: In the Moore–Greitzer model the pressure gain that damps the surge oscillation grows as B², but the throttle may not open with the pressure faster than the compressor's slope allows, or the operating point becomes a saddle; that ceiling does not contain B. At 90 per cent of the peak flow the floor is 7.3 at B = 2 and the ceiling 2.76, so no pressure gain works there. Adding a throttle that opens as the flow falls, with a gain of −3.8, makes every pressure gain from 7.3 to 12.5 stable.

Tested in A compressor needs one actuator for each way it can fail, at the figure it turns on.

Because unloading the tips makes the spar lighter, a structurally efficient wing should always carry a loading closer to Prandtl's bell than to the ellipse, whatever sizes its outer spar.

What decides it: With the spar's weight charged as lift and a strength-sized spar, the best loading sheds tip load to a third-harmonic ratio of −0.136. Put a stiffness floor under the spar's weight per unit span, governing the outer 65 per cent of the semi-span at a floor of 0.8, and the best ratio is −0.075; at 3.2 it is −0.028, almost the ellipse, and the saving over the ellipse has fallen from 4.25 per cent to 0.23. Outboard of where the floor governs, unloading the tips saves no spar.

Tested in A stiffness floor keeps the span and gives the tips their load back, at the figure it turns on.

Tilted to an angle of attack, Busemann's biplane lifts like two wings, each of its elements making the lift a flat plate would, with the channel between them adding its own.

What decides it: In linear theory at the design Mach number the waves the incidence sends across the channel cancel exactly as the thickness's do: the channel's net lift is zero to rounding at every incidence, and the pair lifts 4α/β per unit chord — the outer faces' share, half of two flat plates' 8α/β. Off design the channel lifts again, adding 14 per cent at Mach 1.7 and 21 at 2.2.

Tested in Busemann's biplane lifts only on its outer faces, and its channel turns the lift into a quarter-chord load, at the figure it turns on.

In ideal flow the front face of a bluff block standing in a stream is pushed downstream by the stagnation pressure on it, and its back is sucked; the block has no drag only because those two happen to balance.

What decides it: By an exact conformal map, the front face of a block as long as it is tall is pushed by 0.594 of the dynamic pressure times its height near its foot and pulled by 0.895 near its lip: net, the front face is pulled upstream by 0.300, and the back face, by its own lip, is pulled downstream by the same 0.300. The ideal block's zero drag is a balance of two faces each pulled by its lip, not of a pushed front against a sucked back.

Tested in A block's front face is pulled upstream by its own lip, at the figure it turns on.

A wake vortex pair that loses fluid as it sinks through stable air is weakened by the loss, so it stops sooner and higher than one that keeps its fluid.

What decides it: In Saffman's model with Lamb's oval detraining at a rate set by its buoyancy jump, in air of buoyancy frequency 0.22 times the pair's speed over its spacing, a pair that keeps its fluid stops 6.75 spacings down after 10.6 spacing-times; one detraining with α = 0.05 stops at 6.93 after 12.4, and one with α = 0.08 at 7.31 after 16.0. Above α = 0.085 the oval empties before the buoyancy can stop the pair at all. The loss of fluid removes the brake faster than it removes the momentum.

Tested in A wake that sheds its fluid sinks further, and past a point does not stop, at the figure it turns on.

Eddies that can hold a particle up in a steady flow keep doing so in a turbulent one, which is why fine dust and droplets stay suspended in turbulent air far longer than their settling speed would allow.

What decides it: In a lattice of vortices whose strength flickers at random with a memory of three turnovers, the band of inertial particles a steady lattice holds up — 22 per cent of a cloud at Stokes number 3 falling at a tenth of the cells' speed — falls to 5.5 per cent at a flicker of five per cent of the cells' strength and to nothing at a half. The same flicker leaves the particles falling 4.75 and 2.2 times faster than in still air: flickering eddies speed settling far more reliably than they suspend.

Tested in Flickering vortices cannot hold a settling particle up, and still make it fall faster, at the figure it turns on.

The Moody chart gives the friction factor of any rough pipe of a stated equivalent sand roughness, across the whole range from smooth to fully rough, because the equivalent sand roughness is defined to make it do so.

What decides it: The equivalent sand roughness matches a pipe to Nikuradse's sand in the fully rough limit only. Through the transitional range the friction depends on how the roughness function rises, which differs between sand and commercial roughness: with Nikuradse's function a pipe of relative roughness 1/120 has 21.9 per cent less friction than the Colebrook curve the Moody chart draws, at a Reynolds number of 11,000, where the grains are six wall units high.

Tested in Sand and a commercial pipe part company at six wall units, at the figure it turns on.

The spacing of roll waves on a chute records how far the flow has travelled since the disturbance began: surges keep merging, so a longer chute always shows longer waves.

What decides it: Marched from noise down a periodic chute at Froude number 3, the train's 71 surges at time 100 merge to 55 at 200 and 45 at 400, and then the count stays at 45 until time 1000: the merging stops and the spacing settles at 4.4 units. A chute short enough that the train is still merging shows a spacing that grows with distance; past the settling distance it does not.

Tested in Roll waves stop merging at a spacing their fronts choose, at the figure it turns on.

Pressing a scraper or doctor blade harder onto a moving surface simply squeezes the film under it thinner, and a sharp edge with no flat on it leaves the thinnest and most even film.

What decides it: In Reynolds's equation for a blade carrying 100 N/m on ink of 0.1 Pa·s at 5 m/s, a 30° blade with no flat leaves 0.050 µm, but ten per cent more load takes the film to 0.38 of itself, and above 44° no gap a liquid can hold carries the load at all, so the edge is in contact. A 0.1 mm flat land leaves 1.69 µm at 30° and makes the same ten per cent change only a factor of 0.83: the land is what makes the film a steady function of the load.

Tested in A pressed blade floats on its land and leaves half the gap behind, at the figure it turns on.

A strut between the wings of a biplane or box wing works by carrying the lift of one wing to the other in tension or compression, as a truss member does.

What decides it: In Prandtl's best box both wings lift equally, so they deflect together and a vertical strut between them carries an axial force of a hundred-thousandth of the lift. Joined rigidly and as stiff as a spar, at two fifths of the semi-span, it still halves the spars' flange material — by its joints, as the post of a frame — while a strut with three thousandths of a spar's bending stiffness saves only one per cent.

Tested in A strut inside a box wing is invisible to the wake, and halves the spar, at the figure it turns on.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The pressure coefficient is the pressure, in convenient units.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Tested in Turbulent some of the time, at the figure it turns on.

A viscoelastic material has a relaxation time, and the Deborah number formed from it says whether the material behaves as a solid or as a liquid.

What decides it: The statement is exactly true of a Maxwell fluid, which has one relaxation time by construction, and the crossover it predicts sits at λω = 1 with no coefficient in it. A real polymer has a spectrum, and summing the same two formulas over a Rouse chain gives moduli that never separate: the loss tangent settles at 1.04 instead of falling as 1/λω, so the ratio of the two moduli at λ₁ω = 100 is 1.0 rather than 100. The crossover stops being a point and becomes a graze 2.98 decades wide, thirty-one times wider than the single time's 0.095, and where it sits is decided by the number of modes rather than by any property called "the" relaxation time.

Tested in A solid, if it is not given time, at the figure it turns on.

The hydraulic diameter lets one friction correlation cover any duct shape.

What decides it: In turbulent flow it is good to about ten per cent, and it is the reason the idea is worth having. In laminar flow it is not a correlation with scatter, it is a different number for every shape, known exactly: 64 for a circle, 96 for parallel plates, 160/3 for an equilateral triangle, 56.908 for a square. Using the circular value on parallel plates under-predicts the pressure drop by a third, and the error is not scatter — it is a factor of 1.8 across the shapes, computed to four figures.

Tested in A number that is only the shape of the hole, at the figure it turns on.

Fitting Morison's equation to a measured force record determines the drag and inertia coefficients of the body.

What decides it: Fitting the peak does not. At KC = 10 there is a one-parameter family of (C_D, C_M) pairs whose maximum force is identical to fifteen digits, and the energy they dissipate spans a factor of exactly 4 across it. Worse, over the lower half of that family the inertia coefficient does not move at all: when the peak is the inertia peak it occurs where the velocity is zero, so it carries no information whatever about the drag coefficient. The phase of the record separates them; its maximum cannot.

Tested in Long enough to make a wake, at the figure it turns on.

A thicker aerofoil has a higher lift-curve slope, because potential-flow theory says the slope rises with thickness.

What decides it: Potential flow does say that, and the Joukowski map gives it exactly: 2π at zero thickness and 7.098 per radian at 16.9 per cent, which is 2π(1 + 0.766 t/c). Measurements say the opposite — the slope of a real section falls with thickness, by roughly seven per cent at fifteen per cent thickness — because the boundary layer thickens towards the trailing edge and decambers the section. Thin-aerofoil theory, which has no thickness term at all and returns 2π for every section, is closer to the measurement than the exact inviscid answer is.

Tested in The half that carries nothing, at the figure it turns on.

The Kutta condition is a law of fluid mechanics, so a real aerofoil's circulation is determined by its shape and its incidence.

What decides it: It is a selection rule whose justification is a corner. Computed on a Joukowski-mapped circle with the circulation as a free parameter, a cusped section's trailing-edge speed is about U at the Kutta value and 11.3U half a Kutta circulation away, and grows without bound as the sample approaches the edge — refining the sample four times raises it four times. A section with a *rounded* trailing edge has a peak surface speed of 4.54U over the same range of circulation, unchanged under refinement, and no circulation is picked out by anything. Blowing over such an edge chooses it, and lift coefficients of five to eight follow.

Tested in The condition that can be bought, at the figure it turns on.

A yield-stress fluid below its threshold flows very slowly, so in practice the threshold is a soft one.

What decides it: In the model it is not soft at all: the flow rate is identically zero at every pressure below the threshold, for any length of time. What is soft is the model. Give the unyielded region a very large but finite viscosity — which is what every numerical code actually solves — and the fluid creeps at every pressure, so the yield stress a flow measurement infers becomes exactly proportional to the flow rate that measurement calls zero: one decade of detection threshold, one decade of apparent yield stress, with an exponent of 1.0000000000000016.

Tested in The core that does not move, at the figure it turns on.

A measured jet profile that collapses onto the similarity solution confirms the theory that produced it.

What decides it: It confirms the shape and the shape is shared. The laminar solution is exact Navier–Stokes and the turbulent one is the same formula with a fitted eddy viscosity in place of the molecular one; scaled on their own half-width and centreline speed the two are identical to seven parts in ten to the sixteenth, while their spreading rates differ by whatever the fitted constant is. A collapse of profiles tests the assumption of self-similarity, which is the cheap half; the rate is the half that carries the physics, and it is the quantity the model was fitted to.

Tested in Exactly similar, and one number short, at the figure it turns on.

A flow picture that looks right has very nearly the right energy, so energy is a good way to check it.

What decides it: It has very nearly the right energy for exactly the reason it cannot be checked that way. The excess energy of any admissible field is the energy of its own difference from the true flow, so the energy error is the square of the velocity error — a field wrong by thirty-one per cent in speed everywhere carries an energy nine and a half per cent too high, and one wrong by three per cent carries an energy wrong by one part in a thousand. The quantity that identifies the true flow is the least sensitive instrument available for finding out how far from it you are.

Tested in How much more than the least, at the figure it turns on.

A liquid poured slowly from a spout runs back under it because of the Coandă effect: the stream follows the curved surface for the same reason a jet of air does.

What decides it: An air wall jet a millimetre thick at 50 m/s on a 5 cm radius needs 60 Pa to be turned and has the atmosphere's 101 kPa to borrow, since the wall seals the side its entrainment empties — a margin of 1683. A water sheet in air has no such pressure to borrow; it is held by its own two surface tensions, 2σ against a momentum flux ρU²h, and lets go when they are equal, at √(2σ/ρh) — 0.270 m/s for a sheet two millimetres thick. The lip radius cancels from that balance, to within 0.94 per cent on a 5 mm lip, so the threshold is a speed and not a curvature.

Tested in The teapot effect is a tension, not a pressure, at the figure it turns on.

Because Newtonian theory becomes nearly exact at hypersonic speed, its rule that a surface turned away from the stream carries no pressure becomes nearly exact too.

What decides it: The leeward face of a flat plate is bounded by the vacuum coefficient −2/(γM²), and in the hypersonic limit its share of the normal force depends on K = M sin α alone: 24 per cent at K = 1, 11 per cent at K = 2 and 6 per cent at K = 3. A plate flies at small K when it flies well: with a friction coefficient of 0.001 on both faces the exact plate's best lift-to-drag ratio at Mach 10 is 7.34 at 4.2°, with 30 per cent of its normal force from the shaded face, against the Newtonian 5.07 at 7.4°.

Tested in The face Newton left in shadow, at the figure it turns on.

Rivers in the northern hemisphere erode their right banks because the Coriolis force pushes their water to the right.

What decides it: The force is real and so is its direction: on a straight river it drives a transverse circulation throwing surface water right and bed water left, exactly as a bend of radius 2V/f turning left would — 21 km at 50°N for a river running at 1.2 m/s. But a river's own bends drive the same helix at their own radius, and the Earth's share is fR/2V: 0.07 per cent on a mountain stream, 0.45 on a gravel-bed river and 3 on a lowland river 150 m wide. Only the wide, slow, gently curving rivers of high latitudes, 22 per cent on the middle Volga and 39 on the lower Ob, give rotation a share that could show.

Tested in The Earth bends every river a little, at the figure it turns on.

Fitting an exponential to the start of a measured autocorrelation, or reading the time scale off its first lag, gives the integral scale more accurately than integrating the autocorrelation.

What decides it: Only if the correlation really is exponential. On an exponentially correlated record three hundred integral scales long the lag-one estimate is within 9 per cent and the integral to the first zero within 27. On a record of the same length and integral scale whose correlation is smooth at zero lag, as a real velocity's is, the exponential fit settles at 1.18 of the truth however long the record, and the lag-one estimate reports 3.1 times the truth at five samples per integral scale and 10.1 times at twenty.

Tested in The best estimate of a scale assumes its shape, at the figure it turns on.

A carburettor ices because the air cools as it expands through the pressure drop in the venturi.

What decides it: Expansion to a hundred metres a second cools the moving air by 4.98 K, but the metal the ice forms on sits at the air's recovery temperature and is cooled by only 0.55 K. Petrol evaporating at an air–fuel ratio of 12.5 cools the mixture by 27.9 K if all of it evaporates and 13.9 if half does — five times the expansion's effect on the air and fifty times its effect on the metal. Fuel-injected engines, whose fuel evaporates downstream of the throttle, keep the venturi's expansion and lose the fuel's cooling, and rarely suffer carburettor-type icing.

Tested in Carburettor ice is made by the fuel, not the venturi, at the figure it turns on.

Carburettor icing is decided by the temperature of the fuel–air mixture, so it needs the mixture below freezing, which the fuel's evaporation can manage only on days up to about 15 to 20 °C.

What decides it: Ice grows on the fuel-wetted throttle plate, which settles at a wet-bulb temperature for petrol rather than at the mixture's: on a 20 °C, 80 per cent day a plate wet with fresh petrol is at −10.5 °C while the stream over it is at +10.4. The warmest day on which it ices is 33.7 °C at 80 per cent humidity and 40.4 °C at 50, against 15.6 and 20.4 for the well-mixed mixture. A plate wet with fuel that has lost its light ends ices only up to 14.2 °C at 80 per cent.

Tested in The throttle plate is a wet bulb for petrol, at the figure it turns on.

A melt's slip can be described equally well by a slip length or by a slip velocity law, and a slip length can be read off a measured slip law at whatever rate is convenient.

What decides it: The two agree only for a Newtonian liquid. For a power-law melt with n = 0.4, a constant slip length forces the slip law's exponent to be exactly 1/n = 2.5; a measured exponent of 2 therefore implies a slip length that falls from 0.22 to 0.08 mm across an ordinary range of rates. The descriptions also predict different temperature dependence at fixed stress — a tenfold change in viscosity moves the slip velocity 316-fold if the length is the property and tenfold if the law is.

Tested in A melt's slip length has an exponent it cannot choose, at the figure it turns on.

A diffuser-augmented wind turbine beats the Betz limit, so a duct lets a turbine take more than 16/27 of the wind's power.

What decides it: The coefficient that beats 16/27 is measured on the rotor's area, and it scales with the diffuser's flare. On the diffuser's exit area, momentum theory gives a best of (2/3√3)(1 + c)^(3/2) for an exit pressure coefficient −c: 0.385 with the exit at ambient, below Betz, and 16/27 only at c = 1/3, the pressure behind Betz's own disc. Beating the limit on the device's own frontal area needs a suction behind the exit larger than that, bought with at most a few per cent more frontal area.

Tested in A duct beats Betz only on the area it chooses, at the figure it turns on.

A slender fuselage disturbs the air so little — a couple of per cent of overspeed — that it can be ignored in estimating when the wing reaches the speed of sound.

What decides it: Alone, the fuselage's overspeed is small; added to a wing section's at the root, it is not. A fuselage of fineness ten moves a 12-per-cent root section's critical Mach number from 0.809 to 0.796, the same as making the isolated wing 13.5 per cent thick, and its effect is still 0.005 five fuselage radii along the span, because the middle of a long body spreads its overspeed like a line of sources.

Tested in A fuselage lowers its wing root's critical Mach number, at the figure it turns on.

A little air in a pipeline cushions water hammer, so gas coming out of solution when a valve is shut can only lower the pressures.

What decides it: A gas pocket at the valve is a spring with a time of its own, the gas volume over the margin to vapour pressure. In a frictionless 600 m main it lowers pulses shorter than a few of those times — a hundred-thousandth of the pipe's volume takes a 1.94-rise pulse to 1.62 — but a ten-thousandth re-times the cavity's collapse and raises the main's pulse from 1.30 to 1.34 rises, and on a line that never separates it reflects each step as a spike to 1.56. The largest head falls below the Joukowsky rise only past about half a thousandth of the pipe's volume.

Tested in Air at a valve softens the hammer only in quantity, at the figure it turns on.

If the Earth's rotation biased river erosion towards the right bank, the bias would show in individual bends of the great northern rivers, so the absence of an obvious right-bank preference bend by bend is evidence against Baer's law.

What decides it: In the linear migration law the Earth adds a constant to the curvature, which leaves every meander mode unchanged and moves the whole river right at the bend migration rate times fR/2V — 0.39 widths a century on the lower Ob. A single bend moves about 0.86 widths in the same century on its own, in a direction set by its phase, so one bend shows nothing; the mean over about twenty independent points along the reach shows the drift at two standard errors.

Tested in A river drifts right, and only a reach can show it, at the figure it turns on.

A three-surface aircraft is more efficient than a conventional one because its canard lets the tail stop pushing down, so the wing no longer carries a download as well as the aircraft.

What decides it: With the neutral point held, using the canard lets the trimmed split slide along the trim line to the point nearest the least-drag foot, and the induced drag saved is (d/2)(fₜ₀ − 2f*)², the square of the tail aircraft's distance from twice the foot share. At a margin of a tenth of a chord that is 0.19 per cent of an elliptic wing's induced drag with an uncambered wing and 0.06 per cent with a cruise camber of Cₘ₀ = −0.08, against a skin friction for a canard of a tenth of the wing's area worth 8 per cent at Cₗ = 0.5. The saving exceeds that skin friction only for wing moments beyond about −0.33.

Tested in A third surface is worth a square, at the figure it turns on.

A bulbous bow works because the bow is where a ship makes its biggest wave, so a bulb placed anywhere else — at the stern, say — would do nothing.

What decides it: In Michell's linear theory, which is the theory in which a bulb's wave cancellation is usually explained, reversing a hull conjugates its Kochin function and leaves the wave resistance unchanged. A spherical bulb that cuts the Wigley hull's wave resistance to 0.53 of its bare value at Fr 0.30 when placed at the bow cuts it to exactly 0.53 at the stern too. The theory's reason for a bulb is a phase, not a location, and what makes a bow bulb better than a stern one in practice is outside it: the stern's boundary layer and wake, which the integral does not contain.

Tested in The hull Michell's integral prefers, at the figure it turns on.

A conical intake spike recovers much more total pressure than a two-dimensional ramp, because a cone's shock loses far less than a wedge's — 99.8 per cent kept against 95.2 at fifteen degrees and Mach 2.

What decides it: The comparison at equal angle is the wrong one for an intake, which chooses each centrebody's angle to suit the terminal shock. At the best angle for each, a cone at Mach 2 recovers 0.916 and a wedge 0.903, a gain of 1.3 points; the gain peaks at 1.7 points near Mach 2.5. The cone must be steeper — 28° against 16° — to deliver a low enough Mach number to the terminal shock, and most of its captured air arrives near the shock at the highest Mach number in the annulus. Estimated from the surface Mach number alone, the gain would look 1.75 times larger.

Tested in A cone intake's shock sees the whole capture, at the figure it turns on.

When a hammer pulls a line's pressure below atmospheric, the dissolved air that comes out forms a cushion, so a line carrying air-saturated water is protected from the worst column-separation pulses by its own air.

What decides it: Air comes out only while its pressure in the cavity is below the saturation pressure, so one cavity can hold at most its own volume of air at that pressure — 1.42·10⁻⁴ of this main's volume, a third of what removes the hammer. Released slowly, over ten round trips, it lowers the first pulse from 1.30 to 1.16 rises; released fast, it holds the cavity at atmospheric pressure rather than vapour pressure, and the first pulse is the column-separation pulse at the reduced margin, 1.45.

Tested in The air a cavity releases is capped by the cavity, at the figure it turns on.

Adding blades to a propeller while keeping its thrust and power makes it quieter, because each blade carries less load.

What decides it: The load per blade is not the reason, and the gain depends on the tip speed. The total force round the circle is unchanged; what more blades change is the order of the tone's Bessel function. That function is exponentially small at a low tip Mach number and not near one. For a 1.88 m propeller giving 2 kN from 127 kW, six blades make a fundamental quieter than two by 25 dB at a tip Mach number of 0.4, by 5.8 at 0.78, and by 1.1 at 0.95.

Tested in More blades hush a propeller only while its tips are slow, at the figure it turns on.

A jet engine is a more advanced way of producing thrust than a propeller, so it is the more efficient one.

What decides it: The efficiency with which thrust is produced — as distinct from the efficiency with which fuel is turned into power — depends on one thing only, and it is not the engine. It is the ratio of jet speed to flight speed, and the propulsive efficiency is 2/(1 + sigma) exactly. A propeller moving a great deal of air slowly sits near sigma = 1.1 and is above 95 per cent before anybody has designed anything; a turbojet throwing a thin jet at three times flight speed is at 50 per cent and cannot exceed it whatever its core does. The audit that gives the result closes to twelve decimal places here. What a jet is better at is producing power at high flight speed and high altitude, which is a different question, and the bypass ratio of every airliner since 1970 is the industry conceding this one.

Tested in A big slow push, at the figure it turns on.

A velocity profile with a point of inflection is unstable.

What decides it: Rayleigh's theorem runs the other way: an inviscid parallel flow with NO inflection point cannot be unstable. The converse is false, and Tollmien's 1935 refinement shows what has to be added — the vorticity must have an extremum and the profile must satisfy a further condition at it. The figures here find the inflection points by searching a solved profile for a sign change in its second derivative, and finding one is reported as finding one, not as a verdict.

Tested in A layer with a kink in it, at the figure it turns on.

A wing holds itself up by pushing air downwards, and the downwash behind the wing is where the lift is to be found.

What decides it: The momentum theorem evaluated face by face on a rectangle drawn in the fluid, touching the body nowhere. At eight degrees the box returns a lift of 2.9193 against ρUΓ = 2.9193 — and the downstream face, which is the one the popular account keeps, carries the least of it.

Tested in Air must be pushed down, and the usual sum is wrong, at the figure it turns on.

Momentum theory says a turbine can reach 59.3 per cent, so the difference between that and what a real machine achieves is blade losses — friction, tip vortices and imperfect aerofoils.

What decides it: Some of it is, and a computable part of it is not. A rotor extracting power applies a torque, and the reaction leaves the wake rotating; that rotational kinetic energy is taken out of the stream and never reaches the shaft, and it is present with blades of zero drag and infinite span. Glauert's optimum rotor, which has perfect blades and accounts only for the swirl, gives 0.4155 at a tip-speed ratio of one and 0.5759 at six — a shortfall against Betz of 1.7 percentage points in the band a large turbine runs in, before a single real loss is counted. The gap closes only as the tip-speed ratio goes to infinity, which is why turbines are geared the way they are.

Tested in The wake that has to spin, at the figure it turns on.

A nozzle narrows to make the flow go faster. That is what a nozzle is.

What decides it: True below Mach one and false above it. In dA/A = (Ma²−1) dV/V the bracket changes sign at Mach one, so a narrowing duct decelerates a supersonic flow. The area a stream tube needs, A/A*, has its minimum at Mach one exactly — verified to 1e-9 on a tabulated sweep — and rises on both sides of it, which is why a rocket nozzle has a throat and then flares out.

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

A smoother pipe always carries more flow for the same pressure, so polishing a duct is always worth something.

What decides it: Only when the roughness is large enough for the flow to notice it, and whether it is depends on the speed rather than on the pipe. The roughness height in wall units, k+ = (eps/D) Re sqrt(f/8), is what decides: below about 5 the bumps are buried inside the viscous film at the wall and the pipe behaves as though polished, and a relative roughness of 10⁻⁴ at Re = 10⁴ sits at k+ = 0.06 and costs nothing measurable. Above about 70 the viscosity has stopped mattering entirely and the friction factor no longer depends on the Reynolds number — checked here at a relative roughness of 0.02, where a decade of Reynolds number moves the friction factor by 0.007 per cent. Between those two the answer is neither.

Tested in The roughness a wall cannot feel, at the figure it turns on.

The Mach number at a point in a duct is determined by the area there.

What decides it: The area–Mach relation is double-valued: A/A* = 2.5 is satisfied at Mach 0.2395 and at Mach 2.4428, both roots verified to return the area to 1e-9. Which one the flow is on is fixed by the back pressure imposed at the far end, so the answer at a station depends on a boundary condition the station cannot see.

Tested in One area, two answers, at the figure it turns on.

Water flowing over a bump in the bed has to speed up to get past it, so the surface dips over the crest.

What decides it: It dips only if the flow is subcritical. A supercritical stream over the same bump gets DEEPER and slower, and both are consequences of the same curve: raising the bed takes energy out of the flow, and on the upper branch of the specific-energy curve less energy means less depth while on the lower branch it means more. Computed on the same bump here, a 0.60 m approach at Froude 0.34 falls by 58 mm over the crest while a 0.10 m approach at Froude 5.05 rises. Which way the surface goes is a question about the Froude number, not about the bump.

Tested in The depth that costs least, at the figure it turns on.

A wing stalls when the air can no longer follow the sharp curve over the top of it, which is a matter of the shape.

What decides it: The criterion is not curvature but pressure recovery. Reading the inviscid surface velocity as a local Falkner–Skan power law and comparing with the separation value β = −0.198838, the estimate puts laminar separation at 31% of chord on this section at zero incidence, moving to 3% by six degrees — on a shape whose curvature has not changed at all.

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

Dimensional analysis tells you what a flow depends on.

What decides it: It tells you how MANY numbers the answer can depend on, and it is silent about which quantities belong on the list. The theorem is a statement about a matrix: the drag list — force, density, speed, diameter, viscosity — gives a dimension matrix of rank three, so five minus three is two groups, and every dimensionless combination of those five is a product of powers of two independent ones. Which two is not determined either: the drag coefficient comes out as one combination of the computed basis and the Reynolds number as another, and the reciprocal of either would have served. What the theorem cannot supply is the physics of the list. Leave the viscosity out and it returns one group and a confident wrong answer, and nothing in the arithmetic will object.

Tested in Counting what matters, at the figure it turns on.

A spinning cylinder can be spun faster and faster to make as much lift as wanted, and the flow just goes round faster.

What decides it: The lift does keep rising — it is ρUΓ with nothing in the way — but the flow pattern changes character at Γ = 4πUa. Below it there are two stagnation points on the surface, at sin θ = Γ/4πUa; above it there are none, and the single stagnation point sits out in the fluid, at 1.6052 radii below the centre when Γ is 14 against a threshold of 12.5664.

Tested in How much circulation is too much, at the figure it turns on.

Viscosity is what makes a turbulent flow lose energy, so a less viscous fluid dissipates less.

What decides it: The dissipation rate in a turbulent flow is set by the large scales and is independent of the viscosity — epsilon is of order u cubed over L, with no nu in it. Lowering the viscosity does not lower the dissipation; it moves the scale at which the dissipation happens, as eta over L equals Re to the minus three-quarters. Viscosity decides WHERE the energy leaves, and the large eddies decide HOW MUCH. This is the dissipation anomaly, and it is why the inertial range widens rather than the losses falling.

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

Vortex stretching amplifies vorticity without limit, so a real fluid would develop infinite spin.

What decides it: The amplification is exact and the conclusion does not follow, because thinning a tube brings its own brake with it. Stretching a tube by a factor of lambda multiplies its vorticity by lambda and divides its radius by the square root of lambda, and viscous diffusion spreads vorticity outwards at a rate that goes as the inverse square of that radius. So the brake strengthens as lambda squared while the drive strengthens as lambda, and the two balance at a definite core radius. Burgers' vortex is the exact steady solution of that balance: the core settles at the square root of four nu over alpha, the peak vorticity settles at Gamma alpha over four pi nu, and the residual of the steady vorticity equation, differenced off the profile the figures draw, is under 10⁻⁹. At a strain rate of 2 and a viscosity of 0.005 the core is 0.100 in units where the circulation is one.

Tested in The spin that feeds itself, at the figure it turns on.

Smaller droplets are collected less efficiently than large ones, so a fine enough mist is collected inefficiently.

What decides it: Below a critical Stokes number the collection is not inefficient, it is zero — no droplet released at any offset whatever reaches the body. The reason is an eigenvalue rather than a trend. Near the front stagnation point the flow is a pure deceleration u = −Aξ and the particle equation is linear: St·ξ″ + ξ′ + Aξ = 0, whose roots are real while 4A·St < 1. Two decaying exponentials approach the wall and never reach it; past the threshold the roots go complex, the approach becomes an oscillation, and an oscillation crosses zero. A is measured off the velocity field at 2.000000 U/a, so the threshold is exactly one eighth, and a search on the stagnation streamline finds 0.1265 — above it, as any finite integration must be, because at the threshold the arrival takes forever.

Tested in Whether the droplet turns, at the figure it turns on.

Supersonic flow always goes subsonic behind a shock.

What decides it: True for a normal shock and false in general. A 10° wedge at Mach 2 carries a weak shock at β = 39.31°, across which only the normal component M sin β = 1.267 is processed, and the flow leaves at Mach 1.641 — still supersonic. The oblique relations reduce exactly to the normal ones at β = 90°, verified ratio for ratio to 1e-9.

Tested in A shock that leans, at the figure it turns on.

Making a boundary layer turbulent always increases the drag, so a smooth surface is always better than a rough one.

What decides it: It always increases the friction drag and it often decreases the pressure drag by more. On a sphere the transition from laminar to turbulent separation takes the drag coefficient from about 0.5 to about 0.1 — a fivefold reduction in TOTAL drag, bought by increasing the smaller of the two contributions. Which way the trade goes is decided by the shape: on a streamlined body at small incidence friction dominates and smoothness wins; on a bluff body pressure drag dominates and roughness wins. The dimples on a golf ball are the second case.

Tested in The cost of going turbulent, at the figure it turns on.

d'Alembert's paradox says an ideal fluid exerts no force on a body moving through it.

What decides it: The energy in the fluid around a cylinder moving at U, integrated over the exterior of the plane, comes to ½ρπa²U² — so changing U requires work, and the force needed is that of a body heavier by ρπa² per unit length. The paradox is about steady motion only, and the integral that proves it converges at all is 3.141513 against ρπa² = 3.141593.

Tested in The force of getting going, at the figure it turns on.

A siphon works because atmospheric pressure pushes the liquid up the short leg and over the hump — which is why a siphon cannot work in a vacuum, and why it cannot lift liquid more than about ten metres.

What decides it: The flow rate is the square root of 2g times the drop from the source surface to the outlet, and the height of the crown does not appear in it at all: sweeping the crown from 0.05 m to 8.95 m returns bit-identical exit velocities, because the crown height is not in the expression. So the atmosphere is not doing the driving; the level difference is. What the atmosphere DOES is keep the pressure at the crown above the liquid's vapour pressure, and that is where the ten-metre limit comes from — a limit on holding the column together rather than on pushing it. Degassed water, which sustains tension rather than boiling, has been siphoned over fifteen metres, which the atmospheric account says is impossible.

Tested in The siphon that does not need the air, at the figure it turns on.

A boundary layer's thickness is set by the geometry — the length of the plate, the size of the body, the width of the channel.

What decides it: Here is a boundary layer with no geometry in it at all. An oscillating wall is infinite in extent and has no length anywhere in the problem, and the fluid above it still organises itself into a layer of a definite thickness: √(2ν/ω), built from the viscosity and the frequency alone. In air at 100 Hz that is 0.219 mm and in water at the same frequency 0.056 mm, and neither number knows anything about the wall. What sets a layer's thickness is a COMPETITION — how far momentum diffuses against how long it has to do it in — and the plate length in Blasius' problem enters only as the time x/U that a parcel has been beside it.

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

A sharp, streamlined nose is best at supersonic speed, because it cuts through the air more cleanly.

What decides it: Sharp is best for drag and worst for heating. The detachment angle at Mach 2 is 22.974°, so a nose blunter than that carries a detached bow shock — which is nearly normal near the axis, loses most of the total pressure there, and dumps the corresponding heat into the gas rather than into the vehicle. Every re-entry vehicle ever flown is deliberately on the wrong side of that boundary.

Tested in When the wedge is too blunt, at the figure it turns on.

Chaotic systems are unpredictable, so a better measurement of the initial state does not help.

What decides it: It helps logarithmically, which is a definite and computable amount rather than none. The separation grows as e to the lambda t, so the time before a given error is reached goes as the log of the initial precision divided by lambda. On the Lorenz system at r = 28, with a measured lambda of 0.8982, improving the initial state by a factor of a million buys ln(10^6)/0.8982 = 15.4 additional time units. Halving the error buys 0.77. The exchange rate is poor and it is not zero.

Tested in A millionth is enough, at the figure it turns on.

Sucking the boundary layer away reduces drag, because it removes the slow fluid that was causing it.

What decides it: The friction on a uniformly-sucked wall is EXACTLY twice the suction coefficient — c_f = 2V/U, with nothing left over — because the momentum integral for a wall with transpiration reads c_f/2 = dθ/dx + V/U and this layer's momentum thickness does not depend on x at all. So the drag is precisely the momentum of the fluid that was swallowed, and suction cannot reduce friction below its own price. The differenced wall slope gives 1.9999997e-3 against an identity value of 2.0000000e-3 at V/U = 0.001. What suction actually buys is elsewhere: it holds the layer thin and laminar where it would otherwise have gone turbulent, and at a Reynolds number of four million the turbulent friction it avoids is 0.0028 against the 0.0020 it costs.

Tested in The layer that stops growing, at the figure it turns on.

Blood vessels and tree branches get narrower as they divide because there is less room for them — the geometry is a consequence of packing, not of anything hydrodynamic.

What decides it: Packing constrains where branches go and does not set their radii. Minimising the sum of pumping power, which goes as Q squared over r to the fourth, and the cost of owning the tissue, which goes as r squared, gives a best radius that makes Q proportional to r cubed — computed here by golden-section search and matching the closed form to eight figures. Two testable consequences follow: the cube of a parent's radius equals the sum of the cubes of its branches, and the wall shear stress is the SAME in every vessel of the network, which comes out identical to twelve decimal places across a four-to-one flow split. The second is a quantity a cell can sense, so the optimum is something a vessel can grow towards.

Tested in The radius that costs least, at the figure it turns on.

A wing put to an angle of attack has its lift immediately; the circulation appears as soon as the Kutta condition is satisfied at the trailing edge.

What decides it: The Kutta condition is satisfied at every instant here and the circulation still takes tens of chords to arrive. The reason is Kelvin's theorem: bound circulation can only be acquired by shedding its opposite, and a freshly shed vortex sits just behind the trailing edge where its induced downwash is strongest, cancelling most of the incidence the wing was set to. The lattice computes a first-step circulation of 0.14 of the settled value, half of it after 1.5 semichords, nine tenths after 14, and 99.3 per cent after 200 — approaching 2 pi alpha from below the whole way, which nothing in the model was told to do. What is instantaneous is the Kutta condition; what is slow is the wake getting out of the way.

Tested in The lift that arrives late, at the figure it turns on.

A shock tube produces a shock wave, and the rest of what happens in it is the details.

What decides it: It produces three waves of three different kinds, and only one of them is a shock. The exact solution of Sod's problem has a shock running into the low-pressure gas at Mach 1.656, an expansion fan spreading the other way as a family of rays, and between them a contact surface across which the density jumps by 61 per cent while the pressure and the velocity are continuous. The contact is not a wave at all — nothing propagates through it, it is simply the boundary between the gas that was on one side of the diaphragm and the gas that was on the other, and it moves at the fluid's own speed. Every compressible solver ever written is checked against exactly this, and the check that matters most is whether it can keep the contact sharp.

Tested in One diaphragm, every wave, at the figure it turns on.

A wing close to the ground rides on a cushion of air trapped and compressed beneath it, and that cushion is what gives it extra lift.

What decides it: Computed at the heights aircraft actually fly at, the air under the wing is not trapped and is barely compressed. At four tenths of a chord above the ground — a large airliner in the flare — the pressure under the wing reaches a coefficient of 0.35 out of a possible 1.0, and the mass flux through the gap is 81 per cent of what would pass through the same gap with no wing there at all. The lift increase at that height is 25 per cent, and it comes from the image system reducing the downwash rather than from any air being held. The cushion story does become true eventually: at a tenth of a chord the pressure coefficient reaches 0.96, the flux falls to 20 per cent, and the air genuinely is nearly stagnant — which is a ground-effect vehicle's regime and not an aeroplane's. What the story never mentions is the larger half of the effect, which is drag: at a tenth of a span the induced drag is 48 per cent lower, computed two independent ways agreeing to 10⁻¹².

Tested in The cushion that is not there, at the figure it turns on.

Because air sticks to a surface, a moving wing drags a layer of air along with it, and skin friction is the cost of hauling that air.

What decides it: Both halves are checkable and both come out badly. The amount of air genuinely travelling with the wing is the displacement thickness — the mass deficit in the layer — and at a Reynolds number of a million it is 1.72 millimetres per metre of chord against a layer 4.91 millimetres thick, so under a third of the layer is being carried in any sense. It is not the same air: only the fluid exactly at the wall shares the wing's velocity, 69 per cent of the layer's depth is moving at more than half the free-stream speed relative to the wing, and the contents are replaced continuously as the layer entrains from outside. And the drag prediction has the wrong sign: skin friction is μ ∂u/∂y at the wall, which falls as the layer thickens — at Re 10⁵ the local coefficient is 2.10 × 10⁻³ and at Re 10⁷ it is 2.10 × 10⁻⁴, a factor of ten less drag with a thicker layer, while the mass carried has grown.

Tested in The air a wing does not carry, at the figure it turns on.

A wave transports its shape and its energy but not the water itself, so a floating object rises and falls in place while the wave passes under it.

What decides it: True at first order and false at second. The Eulerian mean velocity at a fixed point is exactly zero, and the trajectory of a parcel — integrated here through the exact linear field — does not close: it advances by the same small amount every cycle, at a rate matching (ak)²c e^(2kz) to 0.16 per cent at a steepness of 0.025. For an ocean swell of one metre amplitude and eight seconds' period that is five centimetres a second, which moves a floating object four kilometres a day.

Tested in The drift in a wave that has none, at the figure it turns on.

A turbine works because the gas pushes on the blades, so the work it does is the pressure drop across it multiplied by the volume that passed through.

What decides it: The work is the change of angular momentum, and pressure need not change across a rotor at all. An impulse stage has the same static pressure either side of its rotor and does full work by turning the flow; a rotor with a pressure drop across it and no change in swirl does exactly none. Euler's equation contains no pressure term and no blade shape: the work per unit mass is U2 V-theta-2 minus U1 V-theta-1, computed here two ways that share no arithmetic and agreeing to twelve decimal places, and a stage with equal swirl on both sides returns exactly zero. What the pressure change decides is the degree of reaction, which is how the work is split between the rotor and the stator, and not how much there is.

Tested in Work out of a change of swirl, at the figure it turns on.

A porous wall can be treated as an ordinary solid wall, since the seepage velocity through it is negligible compared with the flow along the channel.

What decides it: The seepage is indeed negligible and the slip is not. Matching Brinkman's equation inside the medium to Stokes flow above it gives an interfacial velocity of √K times the shear rate plus the Darcy velocity — the first of order √K and the second of order K. For a pore scale one per cent of the channel height the flow rate is 3.03 per cent above what a no-slip wall gives, and a wall told only u = √K du/dy reproduces it to 0.058 per cent. It is the term the claim keeps that is negligible, and the one it discards that is not.

Tested in A wall that is not quite there, at the figure it turns on.