What is refuted here, and by what — page 7

Narrowing a fuselage where the wing joins it removes the fuselage's overspeed at the wing root, so a waist is a way of making the junction as good as the wing alone.

What decides it: A waist is a line of sources and sinks with zero net strength, so the axial velocity it adds along any line parallel to the fuselage integrates to zero: it lowers the overspeed over its middle only by raising it at its shoulders. A narrow waist 6 per cent deep cancels the fuselage's 0.025 at the root's mid-chord and raises the chord's ends to 0.034. Clearing the whole root chord needs a waist wider than the chord and 31 per cent deep at Mach 0.8, and it lifts the fuselage's own peak overspeed from 0.025 to 0.048 ahead of and behind the wing.

Tested in A waist moves the overspeed and cannot remove it, at the figure it turns on.

Once a cloud of particle pairs has forgotten its release separation, its shape settles to a constant kurtosis set by the pairs' memory and the velocity's statistics, so a shape measured at any time after release is the same number.

What decides it: Let the velocity's flatness follow the measured growth towards small scales, F = 3(L/r)^(4μ/9) with μ = 0.25. A cloud released at a ten-thousandth of the integral scale peaks at a kurtosis of 6.85 at 0.002 L and then falls without settling, to 3.95 at 0.13 L and 2.9 at 0.57 L. Only clouds whose velocity has the same flatness at every separation keep one shape, and measured velocities do not.

Tested in A cloud grows out of its bursts and keeps their shape, at the figure it turns on.

A gas with a large bulk viscosity has thick shocks, and a computation that adds the tabulated bulk viscosity to the Navier–Stokes equations gets their structure right.

What decides it: For carbon dioxide the tabulated value reproduces a relaxing shock's thickness to 0.2 per cent at M = 1.002, and to within eight per cent up to M = 1.02. Past M = 1.0411, a 9.5 per cent pressure rise, the real shock splits: at M = 1.3 it jumps 84 per cent of the way in a few mean free paths and relaxes over 0.72 mm, where the constant coefficient draws a smooth shock 0.34 mm thick.

Tested in A bulk viscosity holds a shock together until it splits, at the figure it turns on.

A pair of cylinders released from rest in a stream parts if it starts within about 45 degrees of tandem, because far apart its energy is positive and the Lagrange–Jacobi identity then makes its spacing grow for ever.

What decides it: The identity holds only while energy is conserved. With a drag on each cylinder's motion relative to the stream, linear in that motion and set by a drag coefficient of one, pairs released six radii apart at 10° and 20° drift out to 9.6 and 8.1 radii, are turned by the stream's torque past 45°, and come back to meet near side by side. Every release does.

Tested in Drag makes every free pair meet, at the figure it turns on.

A swell running at a coast piles water up against it, so the sea's surface always slopes up towards the shore, and the water the waves carry in returns seaward along the bottom.

What decides it: That is the balance in water shallower than an Ekman depth, where the set-up is the non-rotating channel's 3νMₛ/H³. On a rotating shelf deeper than about two Ekman depths the transport the sea keeps has turned right far enough to point partly offshore, and the coast sets down instead, by up to 0.044 mm/km for a one-metre swell; the return there is an alongshore current carrying up to 0.36 of the Stokes transport.

Tested in A coast sends the drift back, or sends it along, at the figure it turns on.

A drainage experiment that finds a slip length of a nanometre has measured slip at the wall: the liquid slides over the solid by that much.

What decides it: A layer of liquid 1 nm thick at each wall, three times as viscous as the bulk and with no slip at all, changes the drainage force by the same amount as a slip length of −0.667 nm to first order. The two forces differ by 0.17 per cent at a 30 nm gap and pass one per cent only at 13 nm; a layer that does not move at all is exactly a wall moved in by its thickness, and no drainage force can tell it from an error in where the surfaces are.

Tested in A viscous nanometre reads as slip until the gap is ten of it, at the figure it turns on.

A bulbous bow works because the bow is where a ship makes its biggest wave, and linear wave theory explains it: the bulb's wave cancels the bow's.

What decides it: In Michell's linear theory without a wake the same bulb at the stern cancels exactly as much as at the bow, because reversing a hull leaves its wave resistance unchanged. Weighting the stern's sources by the wake it moves through breaks that: with a wake fraction of 0.25 at Fr 0.30 a bow bulb leaves 0.44 of the Wigley hull's wave resistance and the same bulb at the stern 0.62. The bow is preferred because the stern's waves are made by slowed water.

Tested in The stern's wake puts the bulb at the bow, at the figure it turns on.

A bird in a flapping formation that cannot hold its fore-and-aft place exactly loses the formation's benefit, since the right phase changes by a whole beat for every wavelength it drifts.

What decides it: Holding the phase that suits its average place, a follower whose tip swings a fifth of a span keeps half of its 0.837 saving until its fore-and-aft wander reaches 0.80 spans — a third of the wake's wavelength — and one swinging it four-tenths until 0.37, against 0.185 spans of sideways wander. A follower that re-times its beat within one beat of its drift, against drift correlated over four beats, keeps 0.666 of a half-span wander's saving instead of 0.506.

Tested in A flapping follower can drift fore and aft, but not sideways, at the figure it turns on.

Clearing a slender fuselage's overspeed from a wing root takes about a third of the fuselage's cross-section at the wing, because a waist can only move overspeed, never remove it.

What decides it: The third belongs to the cosine-squared shape, not to the rule. With the waist's shape chosen by a linear programme among sums of small dents — least peak depth, no positive perturbation on the root chord — the waist needs 16.2 per cent of the greatest cross-section against the single dent's 31.0, and it still integrates to zero along the body.

Tested in A free waist needs half the depth, and the sweep costs it nothing, at the figure it turns on.

A surface of posts is always more slippery than a surface of stripes with the same solid fraction, because a post holds liquid back far less than a line does.

What decides it: That holds only for sparse posts. Solving Stokes flow over a square array of no-slip discs, posts a hundredth solid slip 2.87 periods against lengthwise stripes' 1.32, but past about a tenth solid the stripes win: at half solid the posts slip 0.066 periods against 0.110 along the stripes, whose gas lanes run unbroken, though still above the 0.055 across them.

Tested in A post holds liquid back by its radius, a stripe by its length, at the figure it turns on.

A reconstruction built on a wrong prior is sharp, confident and incorrect, and the pictures cannot say so: whatever a prior assumes about the part of a flow the views did not see, the data are silent about it.

What decides it: That is true of priors that choose only among fields meeting the data, such as the smoothest field: applied to a sharp front from eight views it is 13 per cent wrong and misses its rays by a part in a million. It is not true of priors that exclude fields the data allow. Non-negativity applied to a field with a trough cannot meet its rays and misses them by 77 per cent; a dictionary of puffs applied to a streak misses by 4.5.

Tested in A prior the pictures can refute is the one worth having, at the figure it turns on.

By Mark's criterion a reflected shock in air stops bifurcating above an incident Mach number of about 6.5, so a tailored tube driven by hydrogen, at Mach 6, is close to free of the bifurcation that feeds driver gas along the wall.

What decides it: Mark's criterion tests the gas at the wall, which is cold and at rest. Inside the boundary layer, friction has heated the gas to three times the wall's temperature at Mach 6, and it meets the reflected shock at a lower Mach number: its stagnation pressure is 0.64 of the reservoir's where the wall gas's is 0.89, and 26 per cent of the layer's mass cannot enter the reservoir. Across the layer, gas is refused at every incident Mach number above 1.32.

Tested in The gas a reflected shock refuses is inside the layer, at the figure it turns on.

A decaying simulation started from a forced field keeps the forced flow's clean third moment for about a cascade time, because the energy already in the inertial range was delivered from above and takes that long to drain.

What decides it: In a spectral closure forced at a Taylor-scale Reynolds number of 199 and then left alone, the summit of the third moment completes half of its fall in 0.22 large-eddy times, while the dissipation is still at 95 per cent of its forced value. The energy in transit keeps the dissipation up; it does not keep the summit up, because the summit measures what the source term leaves for the cascade, and the source moves into the inertial range at once.

Tested in The summit forgets the forcing before the dissipation does, at the figure it turns on.

A fast throttle with enough gain can hold a compressor at the peak of its characteristic, removing both surge and the rotating-stall hysteresis.

What decides it: In the Moore–Greitzer model a throttle driven by the plenum pressure damps surge, and one driven by the stall cell's amplitude above a critical gain removes the stall's jump. At B = 0.2 the hysteresis loop closes. At B = 0.5 the same stall feedback makes the machine cycle, the cell swinging between zero and 3.3, and near the peak the controlled stalled equilibria are stable only below B of about a third, whatever pressure gain is added.

Tested in A throttle can hold the peak only in a small plenum, at the figure it turns on.

A drainage damping below Taylor's value measured against a smooth surface means the liquid slips there, by the slip length Vinogradova's formula gives.

What decides it: A surface of plane-strain modulus 1 GPa under water at 100 Hz, with no slip at all, lowers the damping against a 3 mm sphere enough to be read as 0.1 nm of slip at a 30 nm gap and 9.5 nm at 3 nm. It is told apart by two things slip does not do: the reading depends on the frequency, as its square, and the force acquires an in-phase part, linear in the softness, before the damping changes at all.

Tested in A soft wall is read as slip only at second order, at the figure it turns on.

In strip theory a wing responds to the gust averaged uniformly across its span, so the uniform average gives the span's effect on the gust load.

What decides it: The reverse-flow theorem of lifting-line theory weights the upwash at each station by the span loading a uniform incidence makes there. For an elliptic wing that weight is elliptic, its filter is the Airy pattern rather than the slit's sinc², and the load's crossing rate comes out 2.2 to 4.1 per cent higher than the uniform average gives over two decades of span — the uniform strip is the lower bound.

Tested in A wing averages a gust through its own loading, at the figure it turns on.

A two-dimensional flow that begins as patches of two vorticity levels relaxes to the tanh state of the maximum-entropy theory, because that theory keeps the area at each vorticity level, which the flow conserves.

What decides it: Decayed from two-level patches with hyperviscosity, the flow ends with a streamfunction flatness of 2.92 and 2.97, above the linear dipole's 9/4 where every tanh dipole lies below it, on the sinh branch to within 0.06 at its own Z/E; a sinh relation fits its vorticity against its streamfunction to 0.14 of the rms vorticity and a tanh can do no better than 0.48. The areas the tanh theory conserves are not conserved by a flow whose small scales are dissipated.

Tested in Decaying flows end on the sinh side, at the figure it turns on.

The current a swell drives along a coast is strongest in the shallows, where the waves are steepest and the water carries them most strongly.

What decides it: Near the beach what moves along the shore is the swell's own Stokes drift, divided by a small depth. The Eulerian current the sea makes in answer to the swell's transport runs fastest where the depth is about three Ekman depths — 37.6 m on a 1-in-500 shelf at 55°N, 19 km out — and the other way, weakly, in the shallows.

Tested in A sloping shelf puts the swell's current three Ekman depths down, at the figure it turns on.

A siphon of degassed water is limited by the largest gas bubble in it, which the tension at the crown makes grow.

What decides it: A free bubble at the crown dissolves unless the water holds more gas than half the crown's tension, so in degassed water there is soon no such bubble. The nucleus that remains is a pocket in a crevice of the wall, whose escape threshold is set by the crevice's mouth — −142.7 kPa for a micrometre mouth when empty, −138 kPa with gas on a non-wetting wall — and which, on a wall the water does not wet, keeps drawing gas from the water until its threshold rises towards the crown's own pressure.

Tested in A crevice keeps the nucleus a free bubble loses, at the figure it turns on.

Air bubbles in water always lower its speed of sound, because the mixture's compressibility is dominated by the gas — which is what Wood's formula computes.

What decides it: Wood's formula is the low-frequency end of a dispersion relation. Above the bubbles' resonance each bubble moves out of phase with the wave and stiffens the mixture, and above a stop band the phase speed exceeds water's: for millimetre bubbles at a void fraction of 10⁻³, 1640 m/s at 30 kHz against water's 1481.

Tested in Above their resonance, bubbles make water faster, at the figure it turns on.

The lift is not carried by downward momentum leaving behind the wing: in a control-volume account the face behind a wing carries a few per cent of it, and the faces above and below carry the rest.

What decides it: That is true of a two-dimensional section, whose face behind empties as it moves back: in a square box four spans across its share falls to 0.025 fifty spans' distance behind. A finite wing's trailing vortices keep carrying momentum past any face. Far behind, the face takes (2/π)·arctan(H/W) of the weight — half for a square box, nearly all for a tall one, almost none for a wide one — and only the total is the wing's.

Tested in Behind a finite wing the wake face keeps its share, at the figure it turns on.

Flexible bottom plating always softens a slam, because a panel that deflects absorbs part of the impact.

What decides it: The relief comes from the wetting, not from absorption, and it depends on the deadrise. Feeding the plating's deflection back into Wagner's condition, at impacts that load a 10 mm aluminium panel to 200 MPa, lowers its peak stress by 15 per cent on a 4° bottom, by 3 per cent at 10° and by nothing at 15°, because what slows the water is the deflection measured against the bottom's own rise across the panel.

Tested in Plating that gives way slows the water that slams it, but only on a flat bottom, at the figure it turns on.

Pressing a liquid against a superhydrophobic surface squeezes its slip away steadily, so the slip a textured wall delivers falls off smoothly with pressure long before the texture fails.

What decides it: Solving shear along stripes whose menisci bow into the grooves as pinned arcs, the slip falls with pressure but by little: a surface 0.8 gas keeps 0.913 of its slip at the flooding pressure of a 110° groove wall, 0.879 at a 120° wall's and 0.804 at a 150° wall's, and no gas fraction from 0.5 to 0.95 loses more than a quarter before its groove floods. The slip is lost at the flood, all at once, not before it.

Tested in A bowed meniscus keeps most of its slip until the groove floods, at the figure it turns on.

The Keulegan–Carpenter number decides how a cylinder's oscillatory force divides between drag and inertia: the smaller it is, the less of the force is drag.

What decides it: Only once the flow separates. While it stays attached, below Hall's onset of the Honji instability, the drag is the Stokes layer's and its coefficient grows as 1/KC, so the drag's peak over the inertia's — the drag coefficient times KC over π² times the inertia coefficient — does not depend on KC at all: by Wang's expansion it is 0.041 at a frequency parameter β of 10³ and 0.013 at 10⁴, at every KC below the onset.

Tested in Below separation a cylinder's drag is a Stokes layer, and KC stops deciding the force, at the figure it turns on.

When a wing stalls its lift collapses because the flow over it has separated, so a stalled flat plate carries little lift and a centre of pressure wherever the remaining flow happens to put it.

What decides it: A flat plate whose flow leaves both edges along free streamlines — the cleanest model of a leading-edge stall — still carries a normal force 2π sin α/(4 + π sin α), a quarter of the attached plate's lift slope near zero incidence, with its centre fixed at (3/4) cos α/(4 + π sin α) chords ahead of mid-chord: 0.315 of the chord from the leading edge at 1°, behind the attached quarter chord.

Tested in A stalled plate keeps a quarter of its lift slope and moves its centre aft, at the figure it turns on.

A vortex boundary built from the vorticity's deviation along fluid paths is free of every arbitrary choice: it is objective, it needs no threshold, and so it is simply where the vortex ends.

What decides it: For a co-rotating Lamb–Oseen pair the boundary is objective — the laboratory and the turning frame agree to 8·10⁻⁸ — and nearly free of its convexity tolerance and its window, at a radius of 1.00 round each core, where no fluid within 1.03 strays in an orbit. But the deviation is from a mean taken over some region, and discs of radius 4, 6, 10 and 20 put the edge at 0.896, 1.00, 1.13 and 1.13; at 1.13 a tenth of the fluid inside it leaves its core within the first orbit.

Tested in An edge no observer can move still depends on what it is averaged against, at the figure it turns on.

A camera's frames stop adding information once they are spaced by more than about the integral time of the quantity it averages, so a window with the same memory as a probe needs the same frame rate as the probe's integral time suggests.

What decides it: Not when the window sits inside the inertial range. A window a hundredth of an integral scale across keeps 0.987 of the variance and has a memory of 1.01 probe times under a sweeping memory law, yet its run of frames keeps nine tenths of its information only at intervals up to 0.803 probe times — a third shorter than the 1.18 the integral-time rule gives — because the small eddies inside it forget in a fraction of that time.

Tested in A camera zoomed into the cascade needs frames faster than its window remembers, at the figure it turns on.

Smoke released near a wing's leading edge shows the upper parcel arriving at the trailing edge far ahead of the lower one, and deeper in the boundary layer the upper parcel's lead is larger still.

What decides it: Only for parcels held at a fixed depth. Smoke follows streamlines, which near the attachment point lie outside the thin layer and are swallowed by it later, the upper one deeper. At 4° and a chord Reynolds number of 10⁶ the particles of a streamline a twentieth of the way into the upper layer's trailing-edge flux arrive at 95% chord with a lag −0.16 of the outside one — nearly together, the lower slightly first — where parcels held at the same final depth give 10.4.

Tested in Followed rather than held, the parcels near a wing nearly tie, at the figure it turns on.

A small particle that settles slower than the eddies around it turn is held up by them, and its average fall is slower than in still air.

What decides it: Only without inertia. In a steady lattice of vortices a particle with no inertia settling at 0.3 of the vortices' speed is held in about half of each cell for ever; with a Stokes number as small as 0.02 it leaves in about a hundred turnover times, and the cloud then falls 2.4 times faster than in still air, along the cells' downflowing edges.

Tested in Inertia empties the vortices that would hold a settling particle, and then speeds its fall, at the figure it turns on.

The noise in a particle-image run can be measured from the run itself, by extrapolating the autocorrelation of its frames back to zero lag: whatever the correlation loses between zero lag and the first frame is the noise.

What decides it: Only when the frames come much faster than the smallest eddies the window holds forget. For a velocity window a twentieth of an integral scale across, frames a tenth of a probe integral time apart read a true noise share of 0.1 as 0.2, and for a velocity gradient in the same window frames a hundredth of a probe time apart read it as 0.18 — the cascade's fast early fall in the correlation is indistinguishable from noise at any frame interval that the run's own information calls for.

Tested in Noise in every frame is a memory of zero, and the run cannot measure it at its own pace, at the figure it turns on.

A stiff mica sheet on its glue shields the glue's compliance at the nanometre gaps where slip is measured, so a drainage measurement against mica sees a rigid wall, and the glue's elasticity can be left out of the slip lengths reported there.

What decides it: The sheet shields only where the pressed region √(2RD) is narrower than its bending length on the glue, about 9 µm for 3 µm of mica on a 1 GPa glue, and then only by about the square of the ratio. Against a 3 mm sphere at 100 Hz the false slip at 3 nm falls from 9.5 nm on bare glue to 2.7 nm under the sheet — still the size of the slip lengths reported — and at 30 nm from 0.10 to 0.067.

Tested in A mica sheet hides its glue only at the smallest gaps, and changes how the false slip falls, at the figure it turns on.

The stagnation points of a three-dimensional flow can only appear and disappear in pairs of opposite index, so a point that vanishes from the fluid must have met a partner there.

What decides it: Not when the flow has a wall. In an exact swirling stagnation flow onto a plate, the interior point of index +1 descends to the wall as the skin friction's spreading rate σ falls through zero and vanishes alone; the wall's skin-friction focus changes from an attachment, of index −1, to a separation, of index +1, and with the wall's points counted as halves the total stays at −½.

Tested in A stagnation point can leave the flow through the wall, and the count is kept by halves, at the figure it turns on.

A prior the data can refute stays refutable when the data are noisy: a constraint that is wrong misses its own rays, and the miss is there to be seen whatever the noise.

What decides it: Only while the miss stands clear of the noise. A dictionary of puffs applied to a field with a streak cannot fit its exact rays better than 4.2 per cent from eight views, and is refused at one and three per cent noise; above 6.2 per cent it fits the noisy rays as well as any prior and its error is within a point of theirs. Non-negativity, which misses a field with a trough by 77 per cent, stays refused at any noise a measurement would tolerate.

Tested in A prior chosen from noisy pictures keeps most of its gain, and a wrong one loses its alibi, at the figure it turns on.

Spreading a propeller's thrust over more blades always makes its blade-passing tone quieter, so a many-bladed propeller is the quiet choice wherever it flies.

What decides it: In a clean stream six blades are quieter than two by 19.6 dB at tip Mach 0.5. In an inflow distortion of 5 per cent rms each blade's fluctuating thrust radiates through Bessel functions of orders down to zero, which need no tip speed, and six blades are quieter than two by only 13.2 dB; at 10 per cent, by 8.2. Six blades' tone is then set by the distortion, not the steady load, up to tip Mach 0.59.

Tested in A distorted inflow gives back the tone more blades took away, at the figure it turns on.

Solving the equations of motion exactly gives the force a body really feels.

What decides it: Pressure integrated over 720 points of the exact surface solution. The drag comes out at −1.1 × 10⁻¹⁶, and the same routine on a cylinder with circulation returns a lift of 2.994 against the theorem's 3 — so the method finds the force that is there and reports the absence of the one that is not.

Tested in The exact theory says nothing has any drag, at the figure it turns on.

Transition on a flat plate occurs at a Reynolds number of 500,000.

What decides it: The figure is a convention, not a measurement of the fluid. Careful low-turbulence experiments delay flat-plate transition past Re_x = 3·10⁶ and a rough or noisy environment brings it below 10⁵ — a spread of thirty to one on the same fluid, the same plate and the same equations. The linear instability that begins the process sets in at Re_x of roughly 10⁵ and produces no transition for three decades afterwards. What 5·10⁵ names is an ordinary wind tunnel.

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

The exact theory and the real flow differ only in small details.

What decides it: The same body in the same stream, solved both ways side by side. The exact field is divergence-free to 3.3 × 10⁻⁸ and tangent to 7 × 10⁻¹⁵, and predicts no drag and no wake; the grid solve at the same conditions separates and carries one.

Tested in The two theories, side by side, at the figure it turns on.

The Kelvin–Helmholtz result shows that a shear layer breaks up at the shortest wavelength present.

What decides it: It shows that of the vortex sheet, which has no thickness. The growth rate sigma = k dU/2 rises without bound as k rises, so the sheet has no fastest-growing wavelength at all and the initial-value problem for it is ill-posed. Give the layer a thickness and the spectrum is cut off at k delta = 1 — the neutral wavenumber computed from Rayleigh's equation, not fitted — and a fastest-growing wavelength exists, at about seven times the layer thickness.

Tested in Every wavelength at once, at the figure it turns on.

Creeping flow is the easy limit: the equations are linear, so everything about it is straightforward.

What decides it: The Stokes equations for a cylinder in an unbounded fluid have no solution at all. Solved exactly inside a room of radius R, the drag comes to 7.058 at 16 radii and 2.119 at 1024, and 1/drag rises as ln R at 0.079464 against the predicted 1/4πμU = 0.079577 — so it never converges, and the calculation refuses a sweep in which it appears to.

Tested in The world with no inertia, at the figure it turns on.

The Lorenz attractor is a picture of convection.

What decides it: It is a picture of three ordinary differential equations obtained by keeping three Fourier modes of a convection problem and discarding the rest. The discarded modes are negligible only just above onset; by r of about 5 they are not, and the standard picture is drawn at r = 28. The equations remain exactly what they are — their fixed points sit at plus or minus the square root of beta times r minus one, checked here to 1e-9, and the Hopf threshold at 24.7368 comes out of the closed form — and what they are is a dynamical system rather than a fluid.

Tested in Three numbers left of a fluid, at the figure it turns on.

Pressure disturbances travel through a liquid or a gas at the speed of sound, so a numerical model of an incompressible flow is computing how they propagate.

What decides it: An incompressible flow's pressure does not propagate at all. Taking the divergence of the momentum equation kills the time derivative, because the divergence of the velocity is identically zero, and kills the viscous term for the same reason — leaving Poisson's equation, which is elliptic and has no characteristics and no signal speed. A compact source of acceleration in a quiet box produces a response everywhere in the box, at once, falling off only as the logarithm of the distance in two dimensions: the relaxation here fits a straight line in ln r to under one per cent over the outer half of the domain. That is not what a fluid does; it is what the incompressible ASSUMPTION does, and the ratio of the two speeds is exactly 1/M, so the assumption is a good one at a tenth of the speed of sound and is nonsense near it.

Tested in Pressure has no speed, at the figure it turns on.

d'Alembert's paradox: a closed body in steady, inviscid, irrotational flow experiences no drag.

What decides it: Every hypothesis but one still holds above Mach one, and the drag is not zero: a flat plate at Mach 2 and 5° has cd = 0.01768 with no viscosity anywhere in the calculation, computed face by face from shocks and fans. The hypothesis that fails is incompressibility — a compressible fluid can carry energy away to infinity along Mach waves, which an incompressible one has no mechanism to do.

Tested in Drag with nothing to rub, at the figure it turns on.

A figure showing an alternating row of vortices behind a body is a solution of the flow past that body.

What decides it: The figure in this essay is an exact ideal-flow field for two infinite rows of point vortices and contains no body, no viscosity and no mechanism that would shed anything. What it settles is the one thing a model of that kind can settle: which ratio of row spacing to along-row spacing is neutrally stable, bisected here to 0.28054993 against the exact ln(1 + root 2) over pi. Why the along-row spacing takes the value it does, and why anything sheds at all, are outside it entirely.

Tested in The street a steady calculation cannot draw, at the figure it turns on.

A swept wing's pressure coefficients are the unswept section's scaled by cos²Λ, so its critical Mach number is higher by 1/cos Λ.

What decides it: That is exact for an infinite yawed wing and it is a statement about a wing with no ends. A lattice solve of a 35°-swept wing puts the local sweep of the half-load line at 35.2° at mid-span — the theory being right — and at 14.3° at the root and 31.6° at the tip. Since the benefit goes as the *local* isobar sweep rather than the geometric one, the root of that wing is getting the benefit of a 14° sweep, which is why the shock forms there first on every swept wing ever built.

Tested in The sweep a root does not have, at the figure it turns on.

A wrinkled premixed flame has a maximum speed: once it is wider than a few neutral wavelengths it settles into a single cusp and gets no faster however wide it grows.

What decides it: That is true of the Michelson–Sivashinsky equation's steady fronts and of nothing noisier. Forced by white noise in its growing band, a front 12.5 neutral wavelengths wide runs at 0.502 with noise of 10⁻¹², 0.668 with 10⁻⁶ and 1.47 with 10⁻², against the quiet 0.499; and a noisy front keeps speeding up with width — from 0.77 at eight neutral wavelengths to 1.36 at twenty with noise of 10⁻³ — where the quiet front never passes a half.

Tested in A flame's speed limit holds only in silence, at the figure it turns on.

By the blast-wave analogy, the gas next to a blunt hypersonic body's afterbody is the hot, empty core of a line explosion, far hotter than anything that flowed past the nose.

What decides it: The analogy's core temperature at the body's radius is 1,400 times the free stream's two diameters behind the nose at Mach 15 and rises without limit downstream. The gas actually there crossed the nearly normal part of the bow shock at the nose, and expanded to the afterbody pressure it is 16.7 times the free stream's temperature at two diameters, 10.7 at twenty, and never less than its floor e^(Δs₀/cₚ) = 9.1 however far back.

Tested in The gas beside a hypersonic body remembers its nose, at the figure it turns on.