Theme

The thread: One number decides the regime — page 47

Page 47 of 50, continuing through the 450 essays this motif runs through.

450 essays carry this thread — page 47 of 50.

Local transparency is the best tree only when the ends absorb. The root's reflection at the heart rate against the reflection at the tree's leaves, for the map tree, for the single exponent that is best at that leaf reflection, and for Murray's. With leaves that reflect nothing the map tree is the best of all, 0.035 against 0.0419: making every junction transparent is then the whole job. As the leaves start to reflect, the best single exponent pulls ahead, because it leaves its junctions slightly mismatched in the way that cancels the echo coming back from the ends; at a leaf reflection of 0.5 it reflects 0.0735 to the map tree's 0.13.

A tree transparent at every junction is not the quietest

The branching rule that lets a heart's pulse through one junction without reflection is set by that junction's Womersley number: area-preserving in the aorta, Murray's in the small arteries. Build a whole arterial tree that way, every junction at its own transparent rule, and it reflects more of the pulse than the best tree built to a single exponent. The single exponent wins by leaving its junctions slightly mismatched, in the way that cancels the echo from the tree's own ends — the trick an antireflection coating plays on light.

Fluids at work

The air a cavity releases is capped by the cavity

When a shut valve pulls the pressure behind it down, the air dissolved in the water starts to come out, and a little gas was already known to soften the hammer. But air leaves the water only while its pressure in the cavity is below the pressure the water was saturated at, so a cavity can fill with at most its own volume of air at that pressure — a seven-thousandth of the pipe here, however much water gives up its air and however fast. That is less than a third of what removes the hammer. Released quickly, the air does not cushion the collapse at all: it turns the vapour cavity into an air cavity, and the pulse becomes the column-separation pulse with the margin measured to the atmosphere — taller, on this main.

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

A thin interface stays unstable far past a quarter. The fastest growth rate of any disturbance against the bulk Richardson number J, on a logarithmic scale, for a density interface as thick as the shear and two, two and a half and three times thinner. The matched layer's billow dies at a quarter, as Miles' theorem requires. Two times thinner, the billow dies sooner, its last growth at J = 0.12, and nothing replaces it. Three times thinner, the billow's last is at 0.08 and a travelling wave takes over: 0.0335 at a quarter, 0.0109 at one and 0.00629 at 1.3, falling steadily with no threshold in the range solved. At 2.5 the waves are weaker and reach 0.001 at 1.3.

Transition and turbulence

A thin interface keeps its waves past a quarter

Miles' quarter rules a shear layer whose density changes over the same depth as its velocity. Make the density interface three times thinner and the stationary billow dies early, but a pair of travelling waves takes its place and is still growing at five times the quarter. No theorem is broken: at the edges of the shear, where the waves draw their energy, the local Richardson number has fallen to nothing.

Mark's criterion lets the wall gas through above Mach 6.5; the gas inside the layer never gets through. The stagnation pressure of the incident shock's boundary-layer gas in the reflected shock's frame, over the reservoir pressure p₅, against the incident shock's Mach number in air: for the gas at the wall, which is Mark's criterion, and for the lowest anywhere in the layer. Below one the gas cannot pass into the reservoir and the shock bifurcates. The wall gas is refused from Mach 1.32 to 6.5; the layer's lowest from 1.32 on, levelling near 0.69. At the tailored helium condition, Mach 3.41, both give 0.51; at hydrogen's, Mach 6, the wall gas gives 0.89 and the layer 0.64.

Compressible flow

The gas a reflected shock refuses is inside the layer

A reflected shock in a shock tube bifurcates when the gas in the wall's boundary layer cannot be pushed into the reservoir behind it: its stagnation pressure, in the shock's frame, is below the reservoir's. Mark's criterion asks that of the gas at the wall, and in air it says the bifurcation stops above an incident Mach number of 6.5. But the gas that fails at high Mach numbers is not at the wall. It is inside the layer, heated by friction until it meets the shock too slowly for its sound speed, and on that measure a reflected shock in air bifurcates at every Mach number above 1.3.

$Half the forced summit is gone in a fifth of an eddy time. The third moment of the velocity differences as a fraction of four-fifths εr, against separation in Kolmogorov lengths of the forced flow, at the instant the forcing is switched off and 0.1, 0.2, 0.5, 1 and 4 large-eddy times later. Forced, the curve peaks at 0.94 near 67 η. A tenth of an eddy time later the summit is 0.899, at a fifth 0.838, at a half 0.759, and after that it hardly moves: 0.742 at one and 0.72 at four, where the flow is simply decaying. The large separations fall first and farthest.$ Transition and turbulence

The summit forgets the forcing before the dissipation does

A flow forced at its largest scales carries the four-fifths law closer to exact than a decaying one. Switch the forcing off and the third moment has to pass from one value to the other. It does not wait for the cascade to drain: half the forced summit is gone in a fifth of an eddy turnover, while the dissipation has hardly moved, and the largest separations overshoot the decaying value before they settle on it.

Circulation and lift

No elevon reaches past a quarter chord

A wing with no tail trims itself with the back of its own section, and the back of a section is a short lever. Thin-aerofoil theory prices it exactly: the load an elevon adds acts at most a quarter of a chord behind the quarter-chord point, however small and far aft the elevon is. So every unit of trim moment costs at least four units of lift, where a tail three chords back costs a third, and a camber change shaped as a pure couple would cost nothing.

Below a quarter the cloud settles on the cell walls; above it, it crosses them. Paths of particles released with the air's velocity inside one cell of the vortex lattice, at Stokes numbers of 0.15 and 0.4, over 25 time units; the cell's walls, the separatrices joining the saddles, are drawn. At 0.15 the particles are flung outwards by the rotation and approach the walls ever more closely without crossing, so the cloud is compressed onto lines. At 0.4 they reach a wall with enough speed to overshoot it into the next cell, where they meet particles coming the other way: the cloud folds.

Flows and fields

A lattice of vortices folds a cloud at the saddles' quarter

A cloud of heavy particles folds in still air above a Stokes number of one, and in converging air above a quarter. Air that turns as well as converges might have settled between them. In a lattice of vortices it settles exactly on the quarter: the vortices fling the particles to the cell walls, but only the saddles where the walls meet can make them cross. Below the quarter nothing folds, and the cloud is gathered onto the walls without limit instead.

All themes