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The thread: A smooth picture proves nothing — page 10

Page 10 of 20, continuing through the 172 essays this motif runs through.

172 essays carry this thread — page 10 of 20.

Model spectra at four Reynolds numbers, compensated. The spectrum multiplied by k^(5/3) and divided by eps^(2/3), so that a true inertial range is a horizontal line at the Kolmogorov constant. What a finite Reynolds number has instead is a single maximum: it reaches 1.4996 at the highest and 1.49 at the lowest, and the band over which it is flat to one per cent goes from a third of a decade to two.

The range a real Reynolds number does not have

Kolmogorov's minus five thirds is a statement about a band of scales that has forgotten the forcing and does not feel the viscosity. Both conditions are about separation, and separation is exactly what a finite Reynolds number does not have much of.

A patch of dye, folded. A circle of marker particles carried by the flow, at four times. It is stretched into a filament and folded through itself, and its area does not change at any point of that — which is not visible in the picture, and is the whole difficulty. A scheme that lost eight per cent an orbit would produce a picture indistinguishable from this one.

Flows and fields

The area that must not move

A patch of dye in an incompressible two-dimensional flow keeps exactly the area it started with, for ever. Two respectable integrators are put on the same flow: one respects that identically at any step size, the other does not, and the pictures they draw are the same picture.

One sign change, and both of a stall's surprises follow from it. A lift curve with a peak, and the rolling moment a wing makes against its own roll at the same incidence. Below the peak the slope of the lift curve is positive, the down-going wing makes more lift, and the roll is opposed. Past the peak the slope is negative, the down-going wing makes less, and the roll is reinforced. The stalling angle here is 16.46° and the damping changes sign at 17.07°. Nothing in this picture is a spin yet — a spin needs yaw as well — but the engine that drives one is the crossing of that line.

What is taught wrongly

A roll that feeds itself

A spin is routinely described as a stall that got worse, and it is not a stall at all in the sense of an angle rather than a speed. It is autorotation — a roll that sustains itself because past the peak of the lift curve the down-going wing makes less lift rather than more — and the arithmetic says it begins a little past the stalling angle rather than at it.

Four vortices, and the end of prediction. The same three vortices as before with a fourth, weaker one added near the middle. Three point vortices have three independent invariants for three degrees of freedom and cannot be chaotic; four have the same three invariants and one more degree of freedom, and generically are. The energy and the impulses are conserved here to fourteen digits over the whole run, which is what makes the tangle a property of the system rather than of the arithmetic.

Ideal flow

Three is the most that can be predicted

Point vortices are the simplest dynamical system fluid mechanics has — no cores, no viscosity, no approximations, four exactly conserved quantities. Three of them are integrable and cannot be chaotic. Add a fourth and the same equations, conserving the same quantities to fourteen digits, stop being predictable at all.

Three bluff bodies, whose Strouhal numbers agree once the wake's width is used. The measured Strouhal number of each body, and Roshko's universal number formed with the wake's width and the speed on the free streamline that bounds it. The raw numbers span a factor of 1.462; the collapsed ones span 1.0011, with a mean of 0.16281. The shedding was never body-dependent — the length in the number was.

Regimes and numbers

The frequency a wake chooses

Bluff bodies shed at Strouhal numbers from 0.145 to 0.212, and the spread is not a fact about shedding — it is a fact about which length went into the number. Change the length to the wake's own width and three bodies agree to a tenth of a per cent. Then let the body move, and the number stops deciding anything at all.

Four sets of scaling exponents, all of them exact at the third moment. zeta_p against p for K41, the beta-model, the log-normal model and She–Leveque. Every one of them passes through zeta_3 = 1 exactly, because the four-fifths law is a consequence of the equations and a model that missed it would be wrong about the one thing that is known. What they disagree about is every other moment.

Transition and turbulence

The exponents that stop being thirds

Kolmogorov's 1941 theory says every moment of the velocity difference scales with the same exponent, p over three, so the distribution keeps its shape at every scale. It does not. The exponents fall below the line, by more the higher the moment, and what the departure measures is a dimension.

And the net drift, which is where they disagree. The Stokes drift plus each return flow. All three drift forward at the surface, because the Stokes drift there swamps any return current of the right size. Below that they part company completely: the uniform current has most of the column moving upstream, and the two that satisfy no slip have almost none of it. The reversal depths span 77 per cent of the water column.

Flows and fields

The drift a closed box will not allow

A wave in a wave tank carries mass forward, and the tank has nowhere to put it. So a return current appears carrying exactly the opposite transport — exactly, from mass conservation and nothing else. Which fixes a total and leaves the answer anybody wants entirely open.

Fluids at work

The section that decides the river

A reach of open channel has a normal depth and a critical depth, and the water has neither. What it has is a profile obeying a first-order equation, which needs exactly one condition — and whether that condition belongs at the upstream end or the downstream end is not a choice, because the equation is stable in one direction and unstable in the other.

A high tail buys a second trim point, at 31.52 degrees. The pitching moment of two layouts against incidence, with the stable trim points marked. Both cross zero with a negative slope near 0.76°, which is the ordinary cruise trim. Past the stall the conventional layout's tail is caught only glancingly by the wake and its moment stays nose-down, so it has no second crossing; the T-tail's tail is swept into the wake and sees 12 per cent of the dynamic pressure, its download collapses, and the wing's own nose-up moment carries the curve back across zero at 31.52°. That second crossing is stable — the slope there is negative too — which means an aircraft that reaches it stays there.

What is taught wrongly

A stall that is a place

A stall is an event, and the usual accounts treat it as one — a boundary reached, a damping lost. A high tailplane makes it something else. Swept into the wing's wake, the tail loses the download that held the nose up, and the aircraft finds a second stable trim point thirty degrees past the stall from which the elevator cannot bring it back.

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