What is refuted here, and by what — page 4

A fibre drawn out of a bath carries the Landau–Levich film, 0.94 ℓ_c Ca^(2/3), the same as a plate, because the film is set by the capillary length and not by the object being coated.

What decides it: Only on a fibre thicker than the capillary length. A thinner one pulls up a meniscus only a few of its own radii high, and the curvature that meniscus asks of the film is set by the radius: the coat becomes 1.3376 b Ca^(2/3). A 125 µm fibre drawn at 1 cm/s from a silicone oil of 20 mPa·s carries 7.6 µm, where the plate's law gives 64 µm; at a tenth of the capillary length the coat is 0.137 of the plate's.

Tested in A thin fibre coats by its own radius, and beads by it, at the figure it turns on.

A canard aircraft is more efficient than one with a tail, because both of a canard's surfaces lift upwards while a tail pushes down and the wing must carry that load as well.

What decides it: By Munk's stagger theorem the least induced drag of the two surfaces depends only on how the lift is split between them. Trimmed at the same static margin of a tenth of a chord, a tail carries 6.7% of the weight upwards and the aircraft pays 1.009 of an elliptic wing's induced drag; a canard must carry 19% on 0.35 of the span and pays 1.128. With a cambered wing the tail's small download costs 1.012 at a margin of 0.3, the canard 1.332.

Tested in A canard pays for its stability in induced drag, at the figure it turns on.

A roller pump or a peristaltic gut pumps best when it occludes the tube completely, which a hard enough squeeze guarantees.

What decides it: With the fluid's pressure acting back on an elastic wall, no squeeze closes the channel: a squeeze that would push an empty tube's wall 1.5 half-widths inward leaves a gap of 0.012 to 0.074 of the half-width, growing as about the square root of the wall's compliance. The free-pumping flow peaks at a squeeze near 1.2 and falls beyond it, and the best efficiency is 0.96 for the stiffest wall computed and 0.81 for a wall ten times as compliant, where a prescribed-shape pump approaches one as its wave closes the channel.

Tested in A squeezed tube is held open by its own fluid, at the figure it turns on.

A drop on a tilted pane behaves like a block on a ramp: once it breaks free it keeps sliding at a smaller tilt than the one that started it, because kinetic friction is lower than static.

What decides it: A contact line's resistance to sliding is the difference of the cosines of its receding and advancing angles, and Cox's law moves both angles away from their static values as the drop speeds up. On a surface whose static angles are 90° and 70°, the resistance per unit length rises from 0.342 of the surface tension at rest to 1.217 at the speed at which the receding angle reaches zero, and it rises at every speed in between. A drop that has started to slide is held at least as hard as it was held still, and it stops at the tilt it started at.

Tested in A sliding drop is held harder the faster it goes, at the figure it turns on.

Thermal protection that lowers a motor's torque as the bearing's housing warms can only make a self-heating oil film settle: a slow sensor delays the correction, and a delayed negative feedback on a stable film rings down to rest.

What decides it: With the housing a separate state relaxing to the film's mean temperature over Tₕ film diffusion times, and the stall group derated as (1 − 0.22H)², the film has no steady state the housing can rest on: the hot branch always warms the housing and the cool branch always cools it. Marched from rest at Tₕ = 300 it settles onto a cycle of 259 diffusion times, the film's centre swinging between 0.64 and 4.87 in its own temperature units, and the cycle does not decay.

Tested in A slow sensor makes the bearing cycle, at the figure it turns on.

As a supersonic nozzle's back pressure rises, the flow runs full until a normal shock stands at the exit, and then the shock moves inside the nozzle to where the subsonic flow behind it can leave at the back pressure.

What decides it: That is the inviscid picture, and for a rocket nozzle of area ratio 40 it puts the shock at the exit at a pressure ratio of 24.5. The wall's boundary layer separates when the wall pressure falls to about 0.4 of ambient — at a pressure ratio of 192 by Summerfield's criterion, 138 by Schmucker's — and below that the flow leaves the wall as a free jet instead. The inviscid shock's thrust at a pressure ratio of 20 would be 0.21; the separated nozzle's is 1.34.

Tested in An overexpanded nozzle lets go before its shock arrives, at the figure it turns on.

Losses in a steam ejector's mixing tube and diffuser cut its capacity, so an ejector that entrains less than its rating has a poor diffuser or a rough mixing section.

What decides it: In the critical mode the entrained flow is fixed where the two streams first share a pressure, upstream of the mixing. Lowering the mixing efficiency from one to 0.8 in the one-dimensional model leaves the entrainment of a 60-throat machine at 0.5465 to every digit and lowers its critical compression ratio from 3.42 to 2.62; the diffuser's efficiency at 0.8 leaves it again and lowers the ratio to 3.32. Only the primary nozzle's loss reaches the entrainment in the ordinary range, 8 per cent of it at a nozzle efficiency of 0.9, and the mixing loss does so only once it is large enough, below 0.76, to choke the mixing tube itself. A downstream loss shows as a lower break pressure, not as a lower flow.

Tested in An ejector's nozzle exit is a condition, not a choice, at the figure it turns on.

An air squeeze film turns adiabatic near where it turns from a damper into a spring, because its thermal squeeze number differs from the viscous one only by the Prandtl number, 0.71 — so every stiff gas film is about γ = 1.4 stiffer than the isothermal theory says.

What decides it: The heat of compression is conducted across the gap, so the thermal number is ωh²/κ and contains no disc radius; its ratio to the squeeze number is (h/a)² p h²/(12 μ κ). For six real air films that ratio runs from 0.035 down to 4·10⁻¹², and five of the six sit more than two decades below the thermal crossing at ωh²/κ ≈ 18. The one that comes near it, a levitator with a twenty-micron gap, gains less than one per cent of stiffness and 27% of damping.

Tested in The heat of a squeeze leaves across the gap, at the figure it turns on.

A snapshot of n × n samples holds as many independent values for estimating a mean as a record of n² samples at the same spacing, because both are counted in integral scales.

What decides it: A record counts integral lengths and a snapshot counts integral areas. For fields with one integral length sampled five times per integral length, a 64 × 64 snapshot of a scalar holds 45 independent values for its mean, a plane cut through a three-dimensional flow 86, and a record of 4,096 samples 409. A two-dimensional incompressible flow's snapshot holds 855, because its window mean is a streamfunction difference round the edge and converges as the inverse cube of the window.

Tested in A snapshot counts areas, and a flow counts its edge, at the figure it turns on.

Pinning at an edge is a property of a perfectly sharp edge; a real edge, rounded at some scale, lets the contact line slide, and so holds liquid no better than a flat surface of the same material.

What decides it: On a quarter-round of radius r the line slides and meets the surface at the Young angle throughout, but the apparent angle still rises by the angle it has gone round. Water with a Young angle of 60° keeps 96 per cent of a sharp edge's extra height when the rounding is a tenth of a millimetre, 62 per cent at a millimetre and about 30 at three; only for roundings many capillary lengths across does the edge's advantage fall towards nothing, as cos²(θ/2)/2r.

Tested in A rounded edge spills before its line goes round, at the figure it turns on.

Stratification suppresses vertical motion and so shields the ocean from its floor, making a current less likely to be blocked by a seamount than it would be in a homogeneous sea.

What decides it: Stratification confines the hill's anticyclone near the floor, where it is stronger, so the current stops sooner: the threshold (h₀/H)/Ro falls from 3.134 in a homogeneous layer to 1.92 when NH/fa = 1 and as 2.243/B beyond, which in deep water is a blocking condition N h₀/U > 2.243 that holds for currents far faster than a homogeneous sea would block. What stratification does shield is the water above: the trapped cap reaches only about 0.7 fa/N above the floor at three times the threshold.

Tested in A stratified sea blocks a current sooner and traps less, at the figure it turns on.

A parcel follows the streamline it is on, so where the streamlines curve the parcels must be turning.

What decides it: In a uniform stream carrying a transverse velocity pattern with it, v = 0.6 sin(x − t), the streamlines at every instant are waves, and every parcel moves in a straight line at the velocity it started with: integrated through the changing field its path strays from a straight line by 10⁻¹³. The parcel's local acceleration, ∂v/∂t, and its convective acceleration, u∂v/∂x, are each as large as 0.6 and cancel to 6·10⁻¹², so nothing turns it; the curved streamlines are a snapshot of a pattern sliding past.

Tested in A parcel goes straight while the streamlines curve, at the figure it turns on.

Two bodies free to move in an ideal stream end side by side, since side by side is the arrangement the stream pulls them towards.

What decides it: Side by side is where the stream's torque on the pair vanishes, but nothing in an ideal fluid removes the energy the pair gains swinging towards it, so a released pair swings through side by side and back. Released from rest six radii apart at 50° from the stream, a neutrally buoyant pair meets at 77° after swinging past side by side; from four radii at 47° it meets at 48°. Released within 45° of tandem from far apart, it turns and flies apart.

Tested in A free pair turns, and forty-five degrees decides, at the figure it turns on.

A hydrofoil running just under the surface is a wing in ground effect turned upside down, so near the surface its induced drag falls, as a wing's does over a runway.

What decides it: At a speed where the surface cannot hold a pressure, its image of the foil's trailing vortices keeps their sense instead of reversing it, so the image is an equally loaded biplane partner rather than a cancelling one. A vortex-lattice foil of aspect ratio 9 one chord below the surface pays 1.43 times its deep-water induced drag at the same lift, a quarter-chord down 1.74, and the elliptic biplane's 1 + σ runs to 2 as the depth closes. The same foil near a wall at a chord's height pays 0.56.

Tested in A foil under the surface flies with a phantom, at the figure it turns on.

A gas's bulk viscosity is a transport coefficient like its shear viscosity — a property of the gas at a given temperature and pressure — so a measured value can be used in the Navier–Stokes equations at any frequency.

What decides it: A bulk viscosity is produced by an internal mode that lags a compression by a relaxation time τ, and equals that time multiplied by the pressure and the lagging share of heat capacity. The quoted fifteen hundred times μ for carbon dioxide implies τ = 2.75 µs at one atmosphere. Computed exactly, the absorption it produces departs from the constant-ζ answer by ten per cent at 24 kHz, by a factor of two near 76 kHz and by a factor of 181 at a megahertz.

Tested in The largest bulk viscosity is the first to expire, at the figure it turns on.

A cloud of particles falls into dense sheets wherever it converges, as a pressure-free cloud does, and drag only delays the moment.

What decides it: In still air a particle of response time τ free-streams at most a distance τ times its starting speed, so the cloud's map is the free one with time capped at τ. Its Jacobian reaches zero only if the Stokes number τ|λₘᵢₙ| exceeds one. At 0.8 the one-dimensional cloud bunches and freezes with its paths parallel; at 0.99 its densest point stops at a hundred times its starting density and never folds.

Tested in Drag decides whether a cloud folds, and the air decides at what, at the figure it turns on.

A siphon of degassed water running near its height limit is on borrowed time: its largest nucleus slowly gathers the gas still dissolved in the water, and the siphon breaks when enough has collected.

What decides it: Across a micron diffusion takes milliseconds, not hours, and it runs the other way unless the water holds more gas than half the crown's tension. At a 14-metre crown, −36.7 kPa, a nucleus lodged there decides in 7 to 60 ms whether it dissolves or breaks the water; in water under a fifth saturated with air, a one-micron nucleus dissolves, and the siphon's limit is exactly the static one.

Tested in A crown dissolves its nuclei or breaks on them, and fast, at the figure it turns on.

Frames of a particle-image run separated by more than an integral time scale are independent, so a run's count of independent values is the number of such frames times the count of one frame.

What decides it: For a pattern swept past the window by a mean flow faster than it forgets itself, frames overlap in space until the flow has carried the pattern a whole window width between them. A scalar field swept 100 integral lengths past a window 12.8 integral lengths wide holds 357 independent values with a frame every two integral lengths; counting those 51 frames as independent gives 2,075, 5.8 times too many.

Tested in Frames that share their eddies count as one, at the figure it turns on.

In a seaway a foiling boat should keep its foil at the depth where its drag is least on flat water, following the surface up and down the waves, since any other depth costs drag.

What decides it: Following the surface keeps the depth but makes the foil carry the boat's heave, and the induced drag rises as 1 + ½(aωₑ²/g)², where ωₑ is the frequency the waves are met at. For a 110 kg dinghy at 8 m/s in a sea 0.4 m high, heading into waves 12 m long, that costs 71.6 N against 68.7 N for flying level at a deeper mean depth; the two cost the same only at 15.6 m, and into waves 8 m long following would need 1.67 g, which a foil that can only push up cannot give.

Tested in A foil flies level through a short sea and follows a long one, at the figure it turns on.

A bush that warms with the bearing housing adds a positive feedback — a warmer bush makes a warmer film, which warms the bush — so it can only make a hunting bearing hunt harder, or run away.

What decides it: A film on walls at βH is exactly the fixed-wall film under a drive whose stall group and droop are both multiplied by exp(βH). The droop's growth pushes the drive towards its cusp, where the film's two states merge. At a derating of 0.22 the cycle's housing window narrows from 1.237–1.593 to 1.833–1.846 at β = 0.3 and closes at β = 0.324; past it the bearing settles on its one remaining state, and no derating of any strength makes it cycle.

Tested in A warm bush stops the bearing hunting, at the figure it turns on.

An arterial tree that passes the heart's pulse best is one in which every junction is sized to reflect nothing: since each junction's transparent rule is known from its Womersley number, building every junction to its own rule gives the least reflecting tree.

What decides it: Built that way — each junction at the exponent where it alone reflects least at the heart rate — a six-generation asymmetric tree with leaves reflecting half the pulse reflects 0.130 at its root. The same tree built to the single best exponent, 2.354, reflects 0.0735: its junctions are individually mismatched in the way that cancels the echo from the leaves. Only with leaves that absorb is the junction-by-junction tree the best, 0.035 against 0.042.

Tested in A tree transparent at every junction is not the quietest, at the figure it turns on.

A canard added to a conventional aircraft as a second trimming surface is sized for trim alone; the tail behind it is unaffected, so the induced drag the third surface saves is the whole of the account.

What decides it: Holding the neutral point with a canard of a tenth of the wing's area 3.5 chords ahead takes a tail of 0.496 of the wing's area instead of 0.2 — 0.22 for the canard's own lift ahead of the centre of gravity and 0.077 for its wake at the tail — and the extra skin friction, with the canard's own, is 31.7 counts against an induced saving of 0.8 to 5.1. With the tail 0.3 span above the wing the bill falls to 22.9 and the saving rises to 20.4 for a flapped wing, which is still short.

Tested in The tail a canard needs costs more than the canard saves, at the figure it turns on.

The entropy layer of a blunt hypersonic body is thin, so the body's boundary layer grows through it within a few nose diameters, and beyond that the afterbody's boundary layer sees edge conditions set by the free stream rather than by the nose.

What decides it: A laminar boundary layer grown through the entropy layer on its own edge conditions reaches the sheath's edge — the streamline carrying half the axis's entropy, 1.22 diameters out upstream — at 0.34 times the Reynolds number on the diameter: 3,400 diameters at a Reynolds number of 10⁴ and 340,000 at 10⁶. A turbulent one needs 160 to 960 diameters over the same range. Ten diameters back at 10⁶ the boundary layer's edge is at 11.4 times the free stream's temperature, against 1.4 outside the sheath.

Tested in The sheath outlasts the body, at the figure it turns on.

A stratified shear layer whose bulk Richardson number is well above a quarter is stable, because Miles' theorem forbids instability above a quarter.

What decides it: Miles' theorem bounds the smallest gradient Richardson number, not the bulk one. With the density interface three times thinner than the shear, the bulk number can be 1.3 — more than five times the quarter — while the gradient number at the shear's edges falls towards zero, and the Taylor–Goldstein equation then has travelling waves growing at 0.0063 per unit time, their critical levels sitting where the local number is below 0.232.

Tested in A thin interface keeps its waves past a quarter, at the figure it turns on.

A flying wing's elevons work like a tail mounted at the trailing edge: being at the back of the wing, they act through a lever arm of most of the chord.

What decides it: Thin-aerofoil theory's exact flap solution puts the load an elevon adds at 0.185 chords behind the quarter chord for an elevon of a fifth of the chord, and at a quarter chord in the limit of a vanishing tab at the very trailing edge. Deflecting a flap changes the circulation of the whole section, and most of that added load sits near the leading edge.

Tested in No elevon reaches past a quarter chord, at the figure it turns on.

Turbulent air folds a cloud of heavy particles at a Stokes number somewhere between the strain's quarter and the still air's one, because each particle spends part of its time in vortices, which fling rather than converge.

What decides it: In a lattice of vortices joined at saddles of unit strain the threshold is the strain's quarter itself, 0.2501 over runs of 80 and 160 time units. The vortices bring the particles to the cell walls; only at the saddles, where a particle approaching along the converging direction overshoots exactly when τ > ¼, do paths cross.

Tested in A lattice of vortices folds a cloud at the saddles' quarter, at the figure it turns on.

A gas squeeze film's stiffness and loss are properties of its geometry and frequency; the gas in the gap matters only through its viscosity.

What decides it: At one atmosphere six gases give the same levitator stiffnesses within eight per cent, 26 to 28 N per unit relative amplitude, and loss tangents from 0.19 to 0.36. The difference is thermal: the heat of compression crosses the gap at a rate set by the gas's diffusivity, xenon's film loses 53 per cent of its damping that way and helium's 5, and sulphur hexafluoride, on the thermal loss peak, loses only 16 because its ratio of specific heats is 1.098.

Tested in The gas sets a squeeze film's loss, not its stiffness, at the figure it turns on.

Because a liquid metal carries its heat across a turbulent pipe by conduction rather than by eddies, its thermal entrance is many times longer than an ordinary fluid's in the same pipe.

What decides it: The entrance length grows as the Péclet number while conduction dominates, as 0.029·Pe diameters, and a liquid metal's Péclet number is a hundredth of a gas's at the same Reynolds number. At Reτ = 1000 a fluid of Prandtl number 0.005 settles to within 5 per cent of its developed Nusselt number in 4.1 diameters from a wall-temperature step, against 9.6 for air and 3.5 for water. Only at Péclet numbers of several hundred does the metal's entrance overtake air's, and the longest of all belongs to fluids of Prandtl number near 0.1.

Tested in The longest thermal entrance belongs to neither metal nor gas, at the figure it turns on.

Two bodies falling or towed side by side are held apart by their wakes: each is pushed towards the faster fluid beside the other's wake, and that repulsion balances the ideal stream's attraction at a spacing set by the wake's width.

What decides it: A wake trails behind its body, so a pair side by side is outside both wakes and feels no push from either. Given wakes that carry their drag and a shear lift away from the wake's centre, a damped pair of cylinders released at any angle still touches near side by side, and near tandem it touches sooner: at Re = 200, released two degrees from tandem six radii apart, after 76 radii of travel instead of 2006. The wake's push throws the follower out of the wake towards side by side, which is the way the stream already turns the pair.

Tested in A trailing wake hurries a pair together, at the figure it turns on.

A laminar boundary layer separates at the same pressure gradient whatever the wall temperature, β = −0.1988 in the Falkner–Skan family, because the momentum equation of a similar layer does not contain the temperature.

What decides it: It contains it once the density is allowed to vary: the pressure gradient acts on the local density, which is the total enthalpy's. Solved together with the enthalpy equation, the separating gradient is −0.3264 for a wall at absolute zero and −0.1295 for a wall at twice the edge's total enthalpy. The −0.1988 belongs to the single wall temperature at which the two equations decouple.

Tested in A cold wall holds its layer on longer, at the figure it turns on.

Since a foiling boat's choice between flying level and following the surface hardly depends on the height of a regular wave, the crossover frequency of a wand's filter can be set once for a given speed and heading and left alone whatever the sea is doing.

What decides it: In a random sea the best crossover moves with the roughness. For a 110 kg dinghy at 8 m/s heading into a Pierson–Moskowitz sea of peak period 3 s, the shortest wave worth following rises from 1.0 times the peak frequency in a sea 0.2 m high to 1.17 in one 0.6 m high, and the cost of getting it wrong is not symmetric: set for a calm sea, the control flies level through waves tall enough to bring the foil within centimetres of the surface, or through it.

Tested in A foil in a random sea follows it up to its peak, at the figure it turns on.

A ship's afterbody is fuller than its forebody because the stern sits in the wake at the propeller, and the strength of that wake fixes how far aft the least-resistance hull should carry its volume.

What decides it: In Michell's integral, a linear wake falling to 0.7 of the ship's speed at the stern moves the least-resistance hull's centre of volume 2.8 per cent of the half-length aft at Fr 0.30. A wake with the same 0.3 averaged over a propeller disc but confined to the last twentieth of the hull moves it 0.86 per cent if uniform in depth and 0.11 per cent if deepest at the keel. The lean returns only as the wake is spread along the run: 2.3 per cent with the wake spread over the whole half-length.

Tested in A wake held at the stern keeps the bulb and loses the lean, at the figure it turns on.

A bubble behaves like a rigid sphere only when its surface is contaminated — coated with surfactant — and a trace of impurity, well short of a monolayer, leaves it nearly as mobile as a clean one.

What decides it: The surfactant gathers into a cap at the rear, and what holds the cap still is a surface pressure of the order of the viscous stress μU, not of the surface tension. For an air bubble of radius 50 µm rising in water, a cap of 120° — enough to bring the drag within two per cent of a rigid sphere's — needs a mean coverage of 7·10⁻⁴ of a monolayer.

Tested in A thousandth of a monolayer holds a bubble still, at the figure it turns on.

A vortex whose vorticity falls off outward is stripped layer by layer in a strain: each layer goes when the strain reaches the uniform patch's limit, 0.150 times that layer's own vorticity.

What decides it: A ring of unit vorticity round a core of twice that, the core six-tenths of the radius, is stripped by contour dynamics at a strain of 0.227 — 1.56 times the 0.146 the same method gives for its own vorticity, and 1.15 times the limit of the mean vorticity inside its edge. Across seven such vortices the own-vorticity reading underrates the ring by 1.3 to 3.35 times, and the enclosed-mean reading by 1.11 to 1.16.

Tested in A strained vortex loses its ring at the limit of what it holds, at the figure it turns on.

In a flapping V each bird times its beat to the bird ahead, so timing errors accumulate along the arm and the birds at the ends, furthest from the leader, lose the most.

What decides it: With every bird wandering independently and re-timing through a first-order lag, each bird's phase error has exactly the same variance, 0.632 rad² for half a span of wander, whatever its place: the single lag is power-complementary, |L|² + |1 − L|² = 1 at every frequency, and the sum telescopes. Only a re-timing that is not complementary — two lags in series — accumulates, and then by a third before it saturates.

Tested in A wandering flock passes no error down the V, at the figure it turns on.

A ball thrown or struck for distance should leave at forty-five degrees; air resistance shortens the flight but leaves the best angle where it is.

What decides it: With quadratic drag the best angle depends on β, the square of the launch speed over the terminal speed. It is 40.98° at β = 1, where a struck baseball and a kicked football sit, 38.1° for a golf ball hit without spin and 30.2° for a shuttlecock cleared at 30 m/s; only the shot put, at β = 0.009, is still at 45°.

Tested in The terminal speed sets the best launch angle, at the figure it turns on.

A hull slamming into the sea is wetted where it crosses the calm water surface, so the slamming load follows from the added mass of that wetted width.

What decides it: The water pushed aside rises up the hull before the hull reaches it. For a wedge the wetted half-width is π/2 times the geometric one, which a time-marched free surface confirms to 7·10⁻⁴ of the penetration, and the slamming force, which carries the width and its rate of growth, is π²/4 = 2.47 times the geometric estimate.

Tested in Water climbs a falling wedge, and the load comes from the climbing, at the figure it turns on.

The quarter at which particles fold in a lattice of vortices is a property of its steady saddles; let the lattice move and particles will fold at smaller Stokes numbers, as they do in turbulence.

What decides it: In the lattice ψ = sin(x + ε sin ωt) sin y the folding threshold within sixty time units never falls below a quarter at any amplitude up to one and any frequency from one to ten times the strain rate, and rises to 0.32; a cloud at St = 0.24 in a lattice wobbling at ε = 1 does not fold in three hundred time units, though most of it has crossed into other cells.

Tested in Wobbling saddles keep the quarter, and raise it, at the figure it turns on.

A viscous liquid jet does not break up because its viscosity holds it together; that is why a stream of honey falls as an unbroken thread.

What decides it: Viscosity enters the growth rate as damping, and cannot change its sign: every ripple longer than the circumference still grows. At an Ohnesorge number of 10 the fastest ripple grows in 88 capillary times rather than 2.9 and is 25.5 diameters long rather than 4.5, so the thread breaks far further from the nozzle and into drops 3.4 times its diameter — but it breaks.

Tested in Viscosity lets a jet break, later and into bigger drops, at the figure it turns on.

A turbulent boundary layer's velocity profile is the logarithmic law of the wall, with small corrections near the edge that matter little for the drag.

What decides it: Above about a fifth of the layer the profile departs from the logarithm by Coles's wake, which at Π = 0.55 is a tenth of the edge velocity but carries a third of the displacement thickness and decides the edge velocity and so the friction coefficient: with it the friction matches Ludwieg and Tillmann's measured correlation within 0.26% at a momentum-thickness Reynolds number of 10⁴.

Tested in The wake is a tenth of the velocity and a third of the displacement, at the figure it turns on.

A round laminar jet from a small source is Schlichting's slender boundary-layer jet, whatever its strength; the boundary-layer approximation is only an approximation in the details.

What decides it: The exact point-force solution reaches Schlichting's jet only as the force grows. At a jet Reynolds number of 1 it is Stokes's point force, pushing fluid outwards over 89° either side of the axis; at 10 its half-width is 26° against Schlichting's 28° and its axis speed 7% higher than his; its outflow volume reaches 92% of Schlichting's only at a jet Reynolds number of 100.

Tested in A point force makes a jet only when it is strong, at the figure it turns on.

A ground-effect craft rides on a cushion of air that presses a trough into the sea beneath it, and the waves it makes are a large part of the drag it has to overcome.

What decides it: The wing does press its whole weight on the water, but the water's response is set by the ratio of air's density to water's. For a 5 m chord at a lift coefficient of 0.8, one metre up, the wave drag is at most 4.4·10⁻⁴ of the lift, at 4.43 m/s, and 9.3·10⁻⁶ at 50 m/s, where the sea under the wing is lifted 3.7 mm rather than dented.

Tested in A wing over the sea barely touches it, at the figure it turns on.

To hit a ball further, launch it higher: the best launch angle with spin is the same as without it, and spin only adds a little lift on top.

What decides it: Giving a ball a lift a fraction ℓ of its drag takes the best angle down by 2.2° for every 0.1 of ℓ at a golf drive's β of 2.39. A 70 m/s drive with 3,000 rpm of backspin carries 262 m at its own best angle of 17.1° and 213 m at the unspun ball's 38.1°, where it climbs 98 m.

Tested in A spinning ball should leave low, at the figure it turns on.

The ideal load for a section is a uniform one, the same lift on every part of the chord, and thin-aerofoil theory will design a camber line to deliver it.

What decides it: The line that delivers a uniform load, z/c = −(design lift coefficient/4π)·[(1−x)ln(1−x) + x ln x], has a slope that grows without bound as +ln(1 − x) at the trailing edge, because a load that is finite there violates the Kutta condition; run forwards, Glauert's series cannot reproduce the load at the edge with any finite number of terms.

Tested in A load carried to the trailing edge needs a hook, at the figure it turns on.

Busemann's biplane flies with no wave drag at its design Mach number; its only drawback is that the cancellation is lost at other speeds.

What decides it: Its channel's throat, 0.8 of the inlet for elements 3% thick at a gap of 0.3 chords, is wide enough to run started down to Mach 1.60 and too narrow to swallow the shock that forms on the way up below Mach 2.15; the design point, 1.944, is between the two, so accelerating from subsonic the biplane is still choked when it passes it.

Tested in Busemann's biplane has to be flown past its design point, at the figure it turns on.

The fingers that form when a less viscous fluid displaces a more viscous one in a narrow cell are random; their spacing is set by imperfections in the plates and in how the flow was started.

What decides it: Linearising the cell's averaged law about a flat interface gives a growth rate that rises with wavenumber until surface tension stops it, with a single fastest wavelength, π b/√Ca, exactly: 44.4 mm for air into 0.1 Pa·s oil at 1 mm/s in a 1 mm gap. Imperfections supply the seed; the cell selects the spacing that grows fastest.

Tested in The fastest finger is set by the gap and one number, at the figure it turns on.

A wall creates vorticity pointing across the flow — the vorticity of a boundary layer is spanwise — and streamwise vorticity appears only when something tilts or stretches it afterwards.

What decides it: The wall's source lies along the isobars, not across the flow. On a cylinder swept 35° the isobars run along the span while the flow outside the layer turns from spanwise to chordwise, so the share of the source pointing along the local flow is one at the attachment line, a third at the suction peak and half when integrated over the accelerating front: streamwise vorticity created at the wall, with no tilting.

Tested in A swept wall makes vorticity along its isobars, not across its flow, at the figure it turns on.

Transit time over a wing cannot be defined, because a parcel near the stagnation streamline can take as long as one likes to pass the nose; so nothing quantitative can be said about which parcel arrives first or by how much.

What decides it: The difference between the transit times of two parcels released either side of the dividing streamline converges as they are released closer to it: 1.1383, 1.1270 and 1.1268 time units at offsets of 10⁻², 10⁻⁴ and 10⁻⁵ for an 11.8% Joukowski section at 4°, while each transit time grows without limit. On a thin plate it is the circulation divided by the square of the speed.

Tested in The upper parcel leads by the circulation, at the figure it turns on.

A liquid jet breaks up because of surface tension alone, so its drops are set by its diameter: about twice the jet's width, at any speed.

What decides it: Adding the air's pressure over the rippled surface to the instability shortens the fastest ripple by 7.4% at a gas Weber number of 0.4 and makes it shorter than the jet's circumference past 1.87; a water jet's drops fall from 1.89 diameters to 1.04 at a gas Weber number of 13, and its break-up length stops growing with speed and falls past 10.6 m/s.

Tested in The air shortens a jet's fastest ripple, and the drops follow, at the figure it turns on.

A turbulent boundary layer in a rising pressure separates when its friction falls to zero, and any pressure rise steep enough will do that; until then the layer is simply thicker.

What decides it: In Clauser's equilibrium layers the friction falls with the pressure-gradient parameter β — by 2.8 times at β = 10 and 27.5 at β = 100 — without reaching zero at any finite β, while the profile changes shape: the wake strength grows from 0.48 to 4.7 and 15, and the shape factor from 1.32 to 1.91 and 3.1. And the momentum integral puts a limit on the pressure rise such a layer can follow, a free stream falling no faster than about the inverse fourth root of distance.

Tested in An adverse gradient grows the wake, not the log law, at the figure it turns on.

A bubble that moves through a still liquid is being pushed — by buoyancy, by an applied force — and the liquid round it carries the signature of that push to large distances, as it does round any body dragged through a viscous fluid.

What decides it: A bubble migrating in a temperature gradient carries no net force: the stress its own surface puts on the liquid sums to zero. Its outer flow is a pure potential dipole, every lab-frame streamline a closed loop from front to back, and the liquid a hundred radii ahead moves at a millionth of the bubble's speed, against a hundredth for the same bubble rising under its weight.

Tested in The warm side pulls a bubble with no force on it, at the figure it turns on.

Murray's law, flow proportional to the cube of the radius, is the rule for any vessel or pipe built to carry its flow at least cost, and the wall shear stress it fixes is the same in every vessel of an optimal network.

What decides it: Minimising the same cost with a turbulent friction factor gives Q ∝ r^(27/11) for Blasius's smooth-pipe law and Q ∝ r^(7/3) for constant friction, recovered by the optimisation to five decimal places; with Colebrook's equation the smooth-walled optimum sits at 2.40 to 2.47 from Re = 6,000 to 10⁷, and its wall shear grows with the radius.

Tested in The cheapest turbulent pipe follows two and a half, not three, at the figure it turns on.

Washout always costs induced drag, because elliptic loading is optimal and twisting the tips down moves the loading away from it.

What decides it: A rectangular wing of aspect ratio 8 has a span efficiency of 0.937 untwisted; with four degrees of washout it reaches 0.991 at a lift coefficient of 0.51 and is better than untwisted at every lift coefficient above 0.25. Washout costs drag only on a wing whose untwisted loading is already near elliptic, such as a 0.4 taper, whose best linear twist is a tenth of a degree.

Tested in Washout is right at one lift coefficient, at the figure it turns on.

A bubbly liquid is acoustically slow: its sound speed is Wood's mixture value, far below the liquid's, at every frequency that matters.

What decides it: Summed over bubbles from 10 µm to 1 mm at a void fraction of 10⁻³, the dispersion relation gives Wood's 312 m/s only below a few kilohertz; above the band where the bubbles resonate the cloud carries sound faster than water with little loss, from 27.8 kHz when the gas is spread evenly in radius and 54.1 kHz when n ∝ R^(−3/2).

Tested in The more small bubbles a cloud holds, the later it turns fast, at the figure it turns on.

Contamination affects a bubble in a temperature gradient as it affects one rising under its weight: at worst the surface is held still, the bubble behaves as a solid sphere, and it loses about a third of its speed.

What decides it: A solid sphere in a temperature gradient does not migrate at all, because the pull acts only on a surface that can move. Solving the capped bubble's migration shows the speed falls to a fifth of the clean value with a cap over the cold hemisphere, and to four per cent with a cap of 120°, where the same caps leave a rising bubble 74 and 68 per cent.

Tested in A cap that slows a rising bubble stops a migrating one, at the figure it turns on.

A poorly conducting pipe wall makes a liquid metal's thermal entrance much longer, because the heat must cross the wall as well as the fluid and the wall is the slower path.

What decides it: A resistive wall slows the overall heating, but the entrance — how far the fluid's temperature profile takes to settle — is set by how much faster the profile's second mode decays than its first, and a resistive wall slows both. For a slug the entrance grows by at most 1.50 times, at a wall Biot number of 2.29, and by 1.46 times for a wall that fixes the flux entirely.

Tested in A steel wall lengthens a sodium entrance by half, at the figure it turns on.

Diffusion always destroys a sharp density interface's Holmboe waves, because it thickens the interface until its thickness ratio to the shear falls below two.

What decides it: Viscosity thickens the shear layer too, and faster than conduction thickens the interface whenever the Prandtl number exceeds one, so a free interface's thickness ratio tends to √Pr: 2.64 for heat in water and 25.8 for salt. Only when the shear is held at its thickness does the ratio fall through two — after 50 s for heat and 4724 s for salt in a centimetre-scale layer.

Tested in What holds the shear decides whether the waves last, at the figure it turns on.

Heavy particles fold onto a flow's walls whenever its strain exceeds a quarter of their inverse response time, so a flow whose strain is sometimes stronger folds them at the quarter of its strongest strain.

What decides it: In a lattice whose strain pulses as 1 + 0.3 sin ωt the threshold is the peak strain's quarter, 0.192, only for slow pulses (0.194 at ω = 0.05); at ω = 5 it is the mean strain's quarter, 0.249. The particle responds to the strain averaged over about four strain times, so a stronger strain that does not last does not fold it.

Tested in A stronger strain folds a cloud only if it lasts, at the figure it turns on.

Inside the boundary layer, where the air is slowed by friction, parcels on the two sides of a wing travel at similar crawling speeds, so the upper parcel's lead shrinks towards nothing.

What decides it: At a fixed fraction of the layer's thickness the upper parcel's lead grows with depth: on a thin cambered section at zero incidence it is 0.183 chord-over-speed units outside the layer at mid-chord and 0.695 at a quarter of the thickness, because the suction side's layer has a fuller profile and a faster edge than the pressure side's.

Tested in Inside the layer the upper parcel leads by more, at the figure it turns on.

A heavier hull slams harder in proportion, and a lighter one proportionally less, whatever the shape of its bottom.

What decides it: A wedge's peak force grows only as the square root of the hull's mass, peaking when the recruited water weighs a fifth of the hull; a round bottom's peak is at first contact and is the same at every mass — 314 kN/m at 5 m/s for a 2 m radius, whether the hull weighs a third or three times the water under its width.

Tested in A light hull slams softly, and a round one does not, at the figure it turns on.

An aeroplane's washout should be chosen for its average lift coefficient, or for the condition it spends most of its time at.

What decides it: A training sortie that is half cruise at Cₗ 0.33 and half climb, slow flight and circuits has a time-mean lift coefficient of 0.67, whose best washout is 5.28°. The sortie's induced energy is least at 3.43°, the optimum for Cₗ 0.436 — the distance-weighted harmonic mean — and 5.28° loses 1.2% of the sortie's span efficiency against it.

Tested in One twist is chosen for the speed flown per metre, at the figure it turns on.

A bush that lags its housing will reopen the hunting band a well-coupled bush closed, because a delayed feedback is what makes a thermal system oscillate.

What decides it: The hunting band is where the bearing's equilibrium lies on the middle branch of the film's S, and the equilibrium has the bush at βH whatever its time; marched bearings with bush times from zero to ten housing times hunt between deratings of 0.18 and 0.28 at β = 0.2 and settle outside it. The settled states' growth rates stay real and negative at every bush time, because the lagged link is the positive one.

Tested in A slow bush quickens the hunt it cannot start, at the figure it turns on.

The best supersonic intake divides its compression equally among all its shocks, the terminal normal shock included, because entropy rises as the cube of shock strength and equal cubes minimise the sum.

What decides it: Optimised exactly over every shock's strength, the oblique shocks come out equal to four figures but the terminal normal shock is weaker by about five per cent — 1.261 against 1.326 in normal Mach number for four oblique shocks at Mach 3 — and the ratio stays between 0.94 and 0.98 from Mach 1.6 to 4.

Tested in The last shock should be the weakest, at the figure it turns on.

Correcting the Saffman–Taylor calculation for the meniscus in the gap changes the predicted finger width substantially, because the effective surface tension is only π/4 of the true one.

What decides it: The π/4 tension shortens the fastest wavelength by 11.4 per cent, but the same meniscus makes the pressure jump rise with the interface's speed, which damps short waves and lengthens it; the two cancel at a capillary number of 0.0131, and at the bench's 0.022 the fastest finger moves only from 21.2 to 21.4 mm. What the meniscus does change is the growth rate, by 23 per cent there, and the onset, by π/4.

Tested in The meniscus slows the finger and hardly moves it, at the figure it turns on.

An arterial tree built to one compromise branching exponent reflects less of the heart's pulse than one whose every junction is sized to pass the pulse, and Murray's tree takes the least of the heart's oscillatory work — so a pulse-transparent tree is neither the quiet design nor the economical one.

What decides it: Both findings came from trees grown for a fixed six generations, which ended in leaves of different sizes. Grown from the same 12 mm aorta to the same 0.5 mm vessels, the junction-by-junction tree reflects 0.028 at the heart rate against the single exponent's 0.035 and Murray's 0.083, and less than 0.005 at every harmonic; Murray's tree then takes 1.14 times a matched root's oscillatory power, the most of the four.

Tested in Between fixed ends the transparent tree is the quiet one, at the figure it turns on.

The finite-Reynolds-number steepening that makes the second-order exponent read 0.70 is a bias towards apparent intermittency, so the anomalous exponents measured at higher orders in laboratory flows overstate the intermittency the flow really carries.

What decides it: The steepening multiplies every order's local slope by the same factor, while intermittency lowers the higher orders. In a log-normal field whose distribution carries a sixth-order departure of 0.333, the local slope at a Taylor Reynolds number of 200 reads 1.795, an anomaly of only 0.205. At fourth order it shows 0.026 of 0.111. So at higher orders the bias hides intermittency rather than inventing it.

Tested in At higher orders the bottleneck hides intermittency, at the figure it turns on.

A body only makes streamwise vorticity where it is swept or where its flow separates; on an unswept nose the wall's vorticity points across the flow.

What decides it: On a prolate spheroid at 10° of incidence in attached potential flow, 55% of the wall's vorticity source over the accelerating region points along the local flow at a fineness of 4 and 81% at 8, because the crossflow's pressure gradient runs round the body; only at zero incidence, or on a sphere, is it all across the flow.

Tested in A nose at incidence makes its vorticity along the flow, at the figure it turns on.

A pump sized for half the flow at the same head and speed has half the specific speed and is a scaled-down copy of the larger pump.

What decides it: The specific speed goes as the square root of the flow, so half the flow is 1/√2 of it — 0.385 against 0.545 — and the head fixes the tip speed, so the smaller pump keeps the larger one's impeller diameter, 369 mm here. It is not a scaled copy but a narrower one: the same two discs, an eye smaller by a fifth, an outlet half as wide, and twice the disc friction per unit of flow.

Tested in Two small pumps carry two big discs, at the figure it turns on.

A bending-moment constraint fixes the optimum span: holding the lift and the root moment, there is a best span, and the bell at a quarter more span is it.

What decides it: With lift and either bending-moment measure held, the least induced drag falls without limit as the span grows — to 0.79 of the elliptic drag at 1.7 times the span for the moment along the span — with the tips carrying downloads. A best span appears only when no section is allowed to push down, and it is where the free loading's tip slope reaches zero: √(3/2) for the moment along the span and 4/3, not 5/4, for the root moment.

Tested in A bending moment never chooses a span, at the figure it turns on.

A model pump's efficiency can be stepped up to full size with a fixed power of the size ratio, Moody's fifth root, because a pump's losses scale with the Reynolds number.

What decides it: Only part of a pump's loss is friction. Leakage past a clearance that cannot be made smaller falls as the inverse size, and mixing, incidence and diffusion losses do not fall at all, so the local exponent of the loss runs from 0.33 at a 25 mm impeller to 0.044 at 1.6 m and equals Moody's 1/5 only near 72 mm. Stepped up tenfold from a 50 mm model, Moody over-predicts by 1.5 points.

Tested in Moody's fifth root is true at one size, at the figure it turns on.

A jet blown along a wall behaves like half of a free jet: it conserves its momentum, as every jet does, and spreads and slows at the free jet's rates.

What decides it: The wall's shear removes the jet's momentum, which falls as x^(−1/4) at exactly the rate the wall drag accumulates. The conserved quantity is Glauert's ∫u(∫_y^∞u²)dy, and it makes the wall jet thicken as x^(3/4) against the free jet's x^(2/3), slow as x^(−1/2) against x^(−1/3), and gather fluid as x^(1/4) against x^(1/3).

Tested in A jet along a wall keeps the flux of its momentum flux, at the figure it turns on.

An external-compression intake can approach a perfect recovery at any Mach number by adding enough ramps, or by replacing them with a smooth isentropic ramp.

What decides it: The ramps' total turn must be followed by the cowl's outer surface, which detaches the free stream's shock past 34.07° at Mach 3. Held to that turn, six optimised ramps recover 86.7% and an isentropic ramp 88.5%, the ceiling, leaving a terminal shock at Mach 1.63; at Mach 4 the ceiling is 71.0%.

Tested in The cowl caps what the ramps can buy, at the figure it turns on.

A noisy flame's speed grows with its width only until the small cusps noise makes are as large as an arc can hold; beyond a few tens of neutral wavelengths the speed saturates at a new, higher ceiling.

What decides it: Marched at widths of 12.5 to 100 neutral wavelengths, the speed keeps rising as about the 0.43 power of the width at noise strengths of 10⁻³ and 10⁻⁶. The two fits match the six widths to within 4 per cent, and the local power between neighbouring widths scatters about that value without falling. What grows alongside is the front's slope, out of the range the equation assumes.

Tested in A noisy flame is still speeding up a hundred wavelengths wide, at the figure it turns on.

Below a Prandtl number of about 0.1 the integral method for a warm vertical plate over-predicts the heat transfer badly, by more than a quarter at Pr = 0.01, because it gives the two layers one thickness.

What decides it: Against the exact similarity solution computed by the box scheme, the integral method is 0.900 of the exact heat transfer at Pr = 0.01 and 0.858 of it as Pr → 0: it under-predicts. The over-prediction came from a tabulated value divided by √2 before it was copied and again in the comparison.

Tested in Over a plate in mercury the warm layer flows as if inviscid, at the figure it turns on.

A simulation forced by a random process with a long correlation time keeps flipping between more and less forced states, so its measured four-fifths ratio depends on the forcing's correlation time, which most simulations choose for convenience.

What decides it: In Leith's closure at Reλ 200 the third moment averaged and divided by the averaged dissipation has its summit between 0.937 and 0.944 for correlation times from 0.03 to 10 eddy times, against the steady forcing's 0.940: the averaged balance is linear, and stationarity makes its averaged source the mean forcing whatever its history. Only the average of the instantaneous ratio depends on the correlation time, reaching 0.99.

Tested in A wandering forcing leaves the averaged four-fifths law alone, at the figure it turns on.

A wing holds itself up by pushing air downwards, and that downward-moving air carries the aeroplane's weight away behind it for as long as anyone cares to follow the wake.

What decides it: In neutral air the trailing pair keeps its impulse, but the atmosphere is usually stably stratified, and the column the pair carries down is then lighter than its surroundings. Buoyancy returns the impulse in π/(2ω) with ω = 0.6735 N — 233 seconds at N = 0.01 per second for every aeroplane — after a descent of a few hundred metres behind an airliner; beyond about 58 kilometres behind it, the face across the wake carries none of the weight.

Tested in Stable air takes back the momentum a wing pushes down, at the figure it turns on.

The bottom plating of a planing hull should be sized for the peak slamming pressure, since that is the largest load the water can put on it, and anything designed to carry it statically is safe.

What decides it: The peak pressure is a ridge a few centimetres wide that crosses a panel in a few milliseconds. For a 10 mm aluminium strip 0.3 m wide on a 10° wedge entering at 5 m/s, a uniform pressure equal to Wagner's inner peak, 0.99 MPa, gives 670 MPa of bending stress; the panel's actual response is 361 MPa. Sized to stay under a 215 MPa proof stress, the peak-pressure rule asks for 17.6 mm of plating where 12.2 mm suffices — and the static stress of Wagner's own pressure, 287 MPa, errs the other way, because the panel overshoots it.

Tested in A hull bottom feels the slam it has time to answer, at the figure it turns on.

A horizontal warm plate transfers heat like a vertical one with a different constant: the Nusselt number goes as the fourth root of the Grashof or Rayleigh number in both, which is why the standard correlation for a plate facing up is 0.54·Ra^¼.

What decides it: The laminar layer on a flat plate is driven by the pressure deficit of its own warm fluid, not by buoyancy along it, and its local Nusselt number is 0.357·Gr^⅕ in air. The fourth-root correlation sits 1.3 to 2.1 times above that layer over its range, so whatever it describes, it is not the laminar edge layer.

Tested in Laid flat, a warm plate is pushed by the pressure of its own layer, at the figure it turns on.

Mixed compression lets a supersonic intake beat the external-compression ceiling because the duct's own walls can turn the flow back and the cowl's detachment limit no longer applies to that part of the turn.

What decides it: The cowl's limit no longer applies inside, but the duct must start: with fixed walls its contraction is capped by the normal shock it has to swallow at start-up, and at Mach 3 the best six-shock design with three inside recovers 86.7 per cent, exactly the all-external design's. Only a duct whose throat opens for the start and closes afterwards reaches 94.1.

Tested in Compression behind the lip is bought by opening the throat, at the figure it turns on.

The frame rate past which a faster particle-image camera adds no independent values for the mean is one frame per integral time, and the integral time is what a single probe in the flow measures.

What decides it: When large scales remember longer than small ones the relevant time is the window mean's own memory, which a probe's integral time underestimates: 1.26 times it for a sweeping memory saturating at the field's scale and 1.9 times if it saturates at twice that scale. With no sweep the frame interval that stops buying moves from 1.17 to 1.47 probe times; computed from the window's own memory the single-memory rule gives it to within 2.4 per cent.

Tested in The camera needs the window's memory, not the probe's, at the figure it turns on.

A jet that strikes a wall loses strength with every centimetre it travels before it gets there, so the wall jets it makes are strongest when the nozzle is held close to the plate.

What decides it: The wall jets keep Glauert's invariant, and each inherits F = 3KQ/32 from the arriving jet: its momentum flux K, which the free jet keeps, times its volume flux Q, which grows as the cube root of the distance travelled. A nozzle twice as far from the plate leaves wall jets with an invariant 1.26 times as large and a speed far along the plate 1.12 times as large, though the jet arrives 0.79 times as fast.

Tested in A jet striking a plate leaves a stronger wall jet the further it travelled, at the figure it turns on.

More backspin always makes a golf ball fly further, because backspin makes lift and lift keeps the ball in the air; the only reason to limit spin is that the ball balloons or stops dead on landing.

What decides it: Only with lift in proportion to spin and drag independent of it. With the coefficients measured on spinning golf balls — lift growing as the spin ratio to the 0.4 power, drag rising with it — the carry of a 70 m/s drive peaks at 234.6 m at 2,273 rpm and falls either side, before any landing is considered.

Tested in A drive's best spin falls where players hit it only because lift tires and drag grows, at the figure it turns on.

A sea breeze penetrates further inland on a longer, hotter afternoon, since the front keeps moving for as long as the land is warmer than the sea.

What decides it: Only near the equator. In a shallow-water model of the cool marine layer, 600 m deep, with a land–sea contrast rising to 3 K, lengthening the afternoon from 10 to 14 hours carries the equatorial front from 154 to 216 km inland but moves the front at 45° from 97.5 to 98.9 km. There the Earth's rotation stops the head near hour nine at about 1.33 times the Rossby radius of the cool layer, whatever the afternoon does afterwards; only the contrast and the layer's depth, through the Rossby radius, can move it.

Tested in The sea-breeze front runs till evening only near the equator, at the figure it turns on.

A ground-effect craft rides on a cushion of air that smooths out the waves below it, so over a moderate swell it flies as steadily as over a calm sea at the same mean height.

What decides it: The cushion is the sea's reflection of the wing, and it depends steeply on the gap. A 5 m-chord wing at 4° and 50 m/s with its leading edge 1.2 m above the mean sea has a calm lift coefficient of 0.684; a 50 m swell of amplitude 0.5 m swings it between 0.534 and 1.00 every 0.85 s. Holding the swing to a tenth of the lift needs a cruise height of 2.6 m, where the calm lift is only 16 per cent above its free-air value.

Tested in A swell lifts a ground-effect wing more at its crests than it takes in its troughs, at the figure it turns on.

A wing's root bending moment follows its lift: the moment is the lift times an arm, so the gust load spectrum and the rate of load cycles a structural designer should count at the root are the lift's.

What decides it: The gust weight of one root's moment is the loading of a wing washed in towards that root's tip only, which is not symmetric and passes the rolling half of the gust the lift cancels. On an elliptic wing of aspect ratio nine one root's moment crosses its mean 1.19 times as often as the lift at a span of a hundredth of the turbulence scale and 1.15 at one scale; only the two roots added together follow the lift.

Tested in One wing root feels the rolling gust its lift cannot, at the figure it turns on.

If each saddle of a flow fluctuates on its own, heavy particles should fold at least as often as when every saddle fluctuates together, since independent saddles give more strong excursions somewhere in the flow at any moment.

What decides it: In a lattice of 4 × 4 saddles whose strains are random signals of mean one, deviation 0.3 and correlation time 2, particles at a Stokes number of 0.14 fold at 2.7·10⁻⁴ per unit time when each saddle has its own signal and at 1.3·10⁻³ when all share one, five times as often; at 0.16, 2.8·10⁻³ against 5.7·10⁻³. A particle folds only if it is at a saddle while that saddle is strong, and with independent saddles most of the strong excursions happen where the particle is not.

Tested in A cloud folds where a saddle is strong, not when, at the figure it turns on.

A plume rising from a heat source slows down as it rises, because it spreads, mixes in the surrounding air and loses its buoyancy by dilution.

What decides it: Above a line source in laminar flow the centreline speed grows as the fifth root of height: the heat flux is fixed, the width grows only by diffusion, as √(νx/u), and buoyancy keeps adding momentum faster than viscosity spreads it. Above a 10 W/m source in air the speed is 11 cm/s a centimetre up and 20 cm/s at twenty. Only the turbulent plume above a point source slows down, as height to the minus one third.

Tested in Above a warm wire the plume speeds up as it rises, at the figure it turns on.

A blunt nose wraps a hypersonic body in a sheath of hot gas that its boundary layer takes millions of nose diameters to get through, so every afterbody behind a blunt nose is heated by the nose's gas along its whole length.

What decides it: That holds for a sphere-cylinder, whose boundary layer grows round a fixed perimeter. On a slender cone the perimeter grows with distance, the layer's mass grows as the cube of it, and at Mach 15 a 7° cone's laminar layer swallows the sheath at 160 nose radii for a nose Reynolds number of 10⁵, against 1.4·10⁵ on the cylinder: the swallowing distance grows as the cube root of the Reynolds number instead of in proportion to it.

Tested in A blunted cone eats its nose's sheath in a few hundred radii, at the figure it turns on.

A wing that is flexible in torsion adapts its twist to the flight condition, washing out more where it needs to, so a little flexibility helps a fixed-twist wing approach the best twist at every speed.

What decides it: The washout a lift coefficient wants rises with the lift coefficient, so slow flight wants the most. The twist a cambered wing's pitching moment gives rises with dynamic pressure, so it adds washout at cruise, where the least is wanted: on a trainer's sortie one degree of it at cruise raises the induced energy above an elliptic wing's from 2.31 to 2.68 per cent. The best twist driven by dynamic pressure washes in.

Tested in A wing that twists with its speed washes out the wrong way, at the figure it turns on.

A second-order filter cuts more sharply than a first-order one, so a foiling boat whose height control divides the sea with a second-order complementary pair comes closer to the clean division between following and flying level, and does better in a rough sea where the division matters most.

What decides it: For a 110 kg dinghy at 8 m/s heading into a Pierson–Moskowitz sea 0.6 m high with a peak period of 3 s, averaged over three seeded seas, the second-order pair lets the foil breach at every crossover up to five times the peak frequency, and where it first keeps the foil under it asks the foil to pull the boat down more than a fifth of the time; the first-order pair does the same at 11 per cent and the ideal cut at none. Near its crossover the second-order pair flies through up to 1.27 times the height of the waves it meets.

Tested in A wand that lags its sea makes flying level the better bet, at the figure it turns on.

Since a turbulent boundary layer mixes fast fluid down towards the wall, the parcels inside a wing's turbulent layers move almost at the outer speed on both surfaces, so the upper parcel's lead deep in the layer comes back to its value outside it.

What decides it: It does so only when both layers turn turbulent together. On a thin cambered section at a chord Reynolds number of a million with transition where Michel's criterion puts it, the lag at 95% chord a tenth of the way into the layer is 12.5 times the outside lag at 4°, where the upper layer turns at 11% chord and the lower stays laminar, and −1.95 times it at 0°, where the lower turns first; with both layers tripped at the leading edge it is 1.3 and 1.8. The turbulent profile is full, but the parcels remember where on each side the layer turned.

Tested in The side that turns turbulent first decides who arrives first near the wall, at the figure it turns on.

A turbulent boundary layer that meets an adverse pressure gradient suddenly is safer than one that has been in it all along, since it carries the full, healthy profile of the flat plate into the gradient and takes some distance to lose it.

What decides it: The flat plate's eddies draw in too little fluid for the new gradient, and the deficit piles momentum loss into the layer faster, not slower. For a 23 mm layer in air at 30 m/s meeting an outer speed falling as the power −0.4 of distance, Green's lag-entrainment method separates it 35 layer thicknesses downstream of the step; with the entrainment fixed by the layer's shape, which assumes the eddies keep up, it separates 60 thicknesses on.

Tested in A layer whose stress lags the gradient separates sooner, at the figure it turns on.

The launch that carries a drive furthest is also the one that goes furthest in all; the run only adds a little to whatever the carry was.

What decides it: Not on an ordinary fairway. A 70 m/s drive carries furthest, 234.6 m, at 2,273 rpm and 16.8°, and then runs 14.6 m. With the turf yielding into a crater at each bounce and the roll decelerated at 4 m/s², the drive that goes furthest in all is 1,359 rpm at 11.6°: it carries 13 m less, lands at 35.5° instead of 46.8°, runs 35 m, and finishes 7 m further.

Tested in The run moves the best drive below the ridge that carries furthest, at the figure it turns on.

A blade held flat against a moving surface sits on it more firmly than one held steeply, so a scraper or a doctor blade laid low lets less of the liquid through.

What decides it: In Taylor's creeping corner flow the liquid dragged into the wedge pushes the still blade away from the moving wall with a force μU ln(R/ε) times 196 at 10°, 8.56 at 45° and 1.36 at 90° per unit span — within a per cent of lubrication theory's 6/α² at small angles. The flatter the blade, the harder the liquid lifts it, so a blade pressed with a given force rides up on a thicker film the lower it is laid.

Tested in A scraper held flat is pushed off the wall it scrapes, at the figure it turns on.

The tone a wire makes in the wind is the wire vibrating, like a plucked string, so its pitch is set by the wire's tension and length and a wire held rigid would be silent.

What decides it: A rigid wire sings. Its tone is the reaction of the fluctuating lift its own vortex shedding puts on it, at a frequency of 0.2 times the wind speed over the diameter whatever the tension — 667 Hz for a 3 mm wire in a 10 m/s wind and twice that in twice the wind — and its sound power grows as the sixth power of the wind. Vibration, where it happens, matters only through what it does to the shedding: it lengthens the span over which the force is coherent, and the power grows with that length.

Tested in A wire sings with the force on it, not the flow round it, at the figure it turns on.

A junction's own loss is a small correction to a network that has been sized for its pipes; it can be added afterwards without changing the network's shape.

What decides it: For turbulent smooth pipes fifty parent diameters long, pricing the split as Borda's loss of the velocity difference costs 1.3 per cent of the bill and moves the cheapest half-angle from 28.4° to 21.4°, because the pipes' bill is flat at its best angle while the split's loss rises with the angle at first order.

Tested in The cheapest fork loses only its sideways speed, and pricing it closes the fork, at the figure it turns on.

A wingtip free to pivot on a hinge near its own aerodynamic centre weathervanes into the local flow, so it adapts its twist to every flight condition automatically, whatever its section.

What decides it: Only a symmetric panel does. On a trainer's sortie a free outer panel with no camber moment, held by a constant torque, reaches every segment's own best twist, 1.34% above an elliptic wing; the same panel with a cambered section's moment of −0.05 spends 9.31%, far worse than a fixed twist's 2.65%, because the camber moment about the hinge grows with dynamic pressure and adds washout at cruise.

Tested in A pivoted wingtip follows the lift only with no spring stiffness and no camber, at the figure it turns on.

Blunting a hypersonic cone's tip moves transition aft, and a blunter tip moves it further, because its entropy layer lowers the Reynolds number at the boundary layer's edge.

What decides it: Only up to a point. With the layer's edge read through the sheath of Billig's shock, the delay of a Mach 6, 8° cone's transition peaks and falls again, whichever correlation is used: at 4.96 times the sharp cone's distance for the momentum-thickness Reynolds number alone, at 1.41 for its ratio to the edge Mach number, at noses of 44 and 7.3 mm. Past the peak the transition point lies inside the sheath, where a bigger nose only raises the layer's Reynolds number.

Tested in A blunted nose delays its cone's transition only so far, and the correlation decides how far, at the figure it turns on.

Once a turbulent layer leaves an adverse pressure gradient it recovers gradually towards a flat-plate layer, never getting ahead of a layer that adjusts instantly to the new conditions.

What decides it: With the shear stress allowed to lag, as Green's lag-entrainment method allows it, a layer released from a deceleration of power −0.3 arrives with the worse shape factor, 1.66 against 1.57, then overtakes the instantly adjusting layer 11.8 thicknesses past the release and falls below the flat plate's own shape factor, 1.287 against 1.293 at 40 thicknesses, while its friction runs up to 4.1 per cent above the instantly adjusting layer's.

Tested in A layer released from a gradient overshoots the shape it left, at the figure it turns on.

A ground-effect craft is at its most dangerous over a swell because it holds its height while the crests rise towards it, so every crest closes the gap by the swell's whole amplitude.

What decides it: Free in heave, the craft does not hold its height. Its own cushion is a spring, so over long swells it rises and falls with the sea, and over short ones the convex gap law lifts it above its calm height. Over a half-metre swell the least gap under the wing is 0.42 to 0.74 m, depending on the swell's length, against 0.36 m for the same wing held at its height; the closest approach is over swells met near the craft's own natural frequency, not over the shortest.

Tested in Free to rise and fall, a ground-effect craft rides its own cushion over the swell, at the figure it turns on.