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The thread: What is conserved — page 41

Page 41 of 43, continuing through the 384 essays this motif runs through.

384 essays carry this thread — page 41 of 43.

Moving air needs a quarter of the inertia still air does. The first crossing against Stokes number for a cloud thrown into still air — time in units of the cloud's fastest convergence rate — and for a cloud carried by air converging on a line, in units of the air's strain rate. Still air only brakes: its threshold is one. Converging air keeps pushing: its threshold is a quarter, and heavy particles in it cross in about one strain time. Dots: particles integrated in the converging sine flow; the curve is the linearised oscillator.

Drag decides whether a cloud folds, and the air decides at what

A cloud of free particles keeps its velocities, and wherever it converges its paths cross in finite time: a caustic, the density infinite along a sheet. Give each particle a drag on the air and the answer depends on what the air is doing. In still air the drag only brakes, and a cloud folds only if its Stokes number is above one. In air that is itself converging the drag keeps pushing, and a quarter is enough — the same quarter that decides whether a droplet hits a wing.

Dissolved gas takes height from the largest nuclei first. The tallest crown a siphon runs over, against the radius of its largest nucleus at the reservoir, when a nucleus at the crown must neither break the water at once nor grow to break it by diffusion, for water at 0, 10, 30, 60 and 100 per cent of saturation. Fully degassed water gives the static limit. For a one-micron nucleus, gas below about a fifth of saturation changes nothing — 14.41 m against 14.41 — and saturated water brings it to 10.7 m. A five-micron nucleus in saturated water breaks the siphon at 2.4 m.

What is taught wrongly

A crown dissolves its nuclei or breaks on them, and fast

A siphon of degassed water runs until its crown's tension reaches the breaking threshold of its largest nucleus. A nucleus lodged at the crown does not stay the size it arrived: it gains or loses gas by diffusion, and across a micron diffusion takes milliseconds, so the crown gives its verdict almost at once. Every stable nucleus loses gas unless the water holds more than half the crown's tension, and past a knee in gas content, dissolved air takes height from a siphon's crown that no amount of further degassing gives back.

A cone hands the terminal shock a spread of Mach numbers. At Mach 2, the Mach number across the annulus a cowl on the conical shock captures, from the cone's surface to the shock, for cones of 15°, 22° and 28°, the last the best for recovery. Behind a wedge the flow is uniform; behind a cone it is fastest just behind the shock and slowest at the surface, 1.43 to 1.32 for the best cone. The terminal shock acts on all of it.

Compressible flow

A cone intake's shock sees the whole capture

A cone's shock keeps far more total pressure than a wedge's at the same angle, because the cone finishes its turn in an isentropic compression after the shock. An intake then needs a terminal normal shock, and that shock meets not one Mach number but the whole spread the cone leaves across the captured annulus — fastest by the shock, where most of the air is, and slowest at the surface, where little is. Averaged over the mass, the cone's advantage over a wedge at the best angle for each is under two points of recovery, and the usual estimate from the surface Mach number nearly doubles it.

A frozen run counts the area it sweeps. Independent values in a run of frames while a frozen pattern is swept 100 integral lengths past a window 12.8 integral lengths square, against the distance the pattern moves between frames. For a scalar the run holds 353 values with a frame every half integral length, 362 with one every four and 388 with one every window width: the swept area in integral areas, whatever the frame rate. Only past a window width, when gaps open between frames, does the count fall. Treating frames an integral scale or two apart as independent, as a record would allow, counts 2075 at two integral lengths — 5.8 times too many.

Transition and turbulence

Frames that share their eddies count as one

A particle-image run is a sequence of snapshots, and snapshots closer together than an integral time photograph the same eddies. When a mean flow carries a frozen pattern past the window, a run holds exactly the area it sweeps, counted in integral areas, and a faster camera adds nothing until frames stop overlapping — a threshold set by the window, not by the turbulence. In a flat flow one component escapes the rule: its run mean is fixed by the pattern at the two ends.

Ideal flow

Drag makes every free pair meet

Two cylinders set free in an ideal stream swing towards side by side like a pendulum, and whether they collide or part is fixed, far apart, by forty-five degrees. Give each a drag on its motion relative to the stream and the pendulum stops swinging — but it does not stop the pair. A drag coefficient of order one turns the swing into a creep, lets no pair get far from where it started, and brings every one of them round to meet near side by side.

$At a coast the swell's water goes back, or goes along. The Lagrangian mean current — the Stokes drift plus the Eulerian current — under an 8-second swell arriving straight at a coast, in seas 8, 25 and 60 metres deep, each as a fraction of its own surface drift against depth as a fraction of the water's. Left, the part across the shore, positive onshore; right, the part along it. The cross-shore parts carry no net water. At 8 metres, inside the Ekman layer, the whole exchange is across the shore; at 25 the return is shared with a current along the coast; at 60 the open shelf's own rotation has already taken the transport back, and the coast has little left to do.$ Flows and fields

A coast sends the drift back, or sends it along

A shelf sea under a swell keeps part of the Stokes drift's transport, turned to the right by the rotating Earth, and near a coast some of it points at the land. The sea answers with a slope in its surface. In water shallower than an Ekman depth the slope drives the water straight back as an undertow, as it would without rotation; a few Ekman depths down the coast turns the transport into a current along the shore instead, and there the sea can slope down towards the land it faces.

$The free waist is flat-bottomed, and follows the sweep aft. The cross-section each waist removes along the body, as a fraction of the fuselage's greatest, for the root strip unswept, at 30° and at 45°. The linear programme's waists have flat bottoms and steep sides: depth is spent only where the strip needs it. With sweep they lengthen aft rather than deepen — 24.8, 24.3 and 26.9 per cent. The best single dents, dashed, must widen and deepen instead: 39.8, 45.7 and 52.6 per cent.$ Ideal flow

A free waist needs half the depth, and the sweep costs it nothing

A cosine-squared dent in a fuselage cancels its overspeed at an unswept wing's root for about a third of the fuselage's cross-section, and nothing can remove the overspeed, only move it. Let a linear programme choose the waist's shape instead, and half of that third turns out to have been the dent's fault. The free waist is flat-bottomed and steep-sided, and for a swept root it simply runs aft with the chord, where a single dent has to widen and deepen.

$Holding the neutral point with a canard ahead takes a tail two and a half times the size. The tail area, as a fraction of the wing's, that puts the neutral point where the tail aircraft with a tail of 0.2 has it, against the tail's height above the wing's plane in spans: with no canard, with the canard, and with the canard but its wake at the tail removed. At the reference height the tail aircraft needs 0.2; with the canard it needs 0.496, of which 0.22 is for the canard's own lift ahead of the centre of gravity and 0.077 for its wake at the tail. A tail 0.25 span up needs 0.354 with the canard and 0.154 without.$ Circulation and lift

The tail a canard needs costs more than the canard saves

A canard added to an aircraft as a second trimming surface saves induced drag only if the neutral point is held where it was, and holding it is not free. The canard's own lift ahead of the centre of gravity pulls the neutral point forward, its wake reaches the tail and weakens it, and the tail has to grow to put the neutral point back. For a canard of a tenth of the wing's area the tail grows two and a half times, and its skin friction is twenty to thirty counts against a saving of one to twenty. A T-tail escapes most of the wake and a third of the bill, and still only a flapped wing comes near to paying.

A laminar boundary layer eats through the sheath a Reynolds number of diameters back. How far out the boundary layer has eaten into the entropy layer, as the upstream radius of the streamline at its edge in nose diameters, against distance behind the nose, at Mach 15 for Reynolds numbers on the diameter of 10⁴ to 10⁷, laminar, and 10⁶ turbulent. The sheath's edge is the streamline whose entropy is half the axis's, 1.22 diameters out. The laminar layer reaches it at 3.3·10³ diameters for the smallest Reynolds number and 3.4·10⁶ for the largest; turbulent at 10⁶, at 536.

Compressible flow

The sheath outlasts the body

A blunt hypersonic body wraps itself in a sheath of hot, thin gas from its nose's shock, and its boundary layer grows by eating that gas from the inside. The usual picture has the boundary layer through the sheath within a few nose diameters. It is not: a laminar boundary layer takes about a third of a Reynolds number of diameters to swallow the sheath at Mach 15, which is thousands to millions of diameters, and a turbulent one hundreds. On any body of ordinary length the boundary layer never sees the cooler gas outside, and it is heated by the sheath's gas all the way down.

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