The collection

Every essay — page 24

Page 24 of 39, continuing through the fields in the same order.

Flows and fields Ideal flow Circulation and lift Viscosity Regimes and numbers Compressible flow Transition and turbulence Fluids at work What is taught wrongly Series Concepts Regimes Refutations Search

What is taught wrongly

Equal transit time, Bernoulli misapplied, and the rest. Each stated fairly, then tested against a solved flow and found false.

One sheared stream, and the total pressure across it. A parallel shear flow is an exact steady solution of the Euler equations, and the momentum equation requires its static pressure to be uniform. So the total pressure is entirely the dynamic pressure, which varies with the speed — by three and a fifth dynamic heads across this layer, on a flow where the static pressure does not vary at all.

Four Bernoullis and one name

"Bernoulli's equation" names at least four statements with four different constants, three domains of validity and one shared reputation for being misapplied. A single sheared stream separates the first three: its total pressure is constant along every streamline, varies by three dynamic heads across them, and its static pressure never moves at all.

10 figures

Four sections, and what the shape story says about each. A flat plate, a symmetric section twelve per cent thick, a cambered one, and a section with a wavy upper skin. The first two have upper and lower surfaces of exactly equal length; the last has an upper surface two and a half per cent longer than its lower one, which is twice the cambered section's excess.

The wing that is flat, and flies

Refuting the equal-transit story by computing the parcels leaves its premise standing, and the premise is the part most readers believe: that the shape is what makes the lift. It is a claim about shapes, so it is tested with shapes — a flat plate, a symmetric section, and a cambered one flown upside down.

9 figures $Where four insects have to turn round. Wagner's function — the fraction of its eventual circulation a wing has built after a stated distance of travel — with the half-stroke of each of four insects marked on it. Every one of them reverses while the curve is still climbing, so no insect wing ever reaches the circulation a steady calculation assigns it.$

A calculation with no memory in it

The bee calculation is famous for using the wrong velocity. Done with the right one it still falls short, and the reason is structural: a quasi-steady sum is a statement about a wing that has always been going, and an insect's wing travels between two and five chord lengths before it turns round.

8 figures

The exposure lengths that give the true mean, and the ones that do not. How far a finite exposure of a shedding wake lands from the long-exposure mean, against how many shedding periods the shutter was open for. It is exactly zero at every whole number of periods — an average over a complete cycle is the mean, with no error at all — and between the zeros it falls as one over the exposure. Nothing about the picture tells the reader which of these they are looking at.

The shutter is part of the answer

Flow photographs are compared as though the exposure were a detail of the camera. It is a term in the measurement: an average over exactly one shedding period returns the true mean to fourteen figures, any other length carries a residue that falls only as one over the exposure, and a two-pulse velocity reading is short by exactly the sinc of the swept angle.

8 figures $What the phase reaches, and what it does not. The difference between the two records, as a fraction, for five quantities. The variance and the autocorrelation are the same to machine precision because they are the spectrum. A narrow-band linear oscillator answers its own frequency and almost nothing else, so it is nearly phase-blind too. Everything extremal — the crest, the peak drag load, the range of the running integral — is not.$

The same statistics, and a different load

A wind or wave specification is written as a spectrum, and a spectrum discards the phases. Two records built from one spectrum agree in variance to thirteen figures and in peak drag load by twenty per cent — and with the phases lined up, the same spectrum is a single impulse thirty-one times worse.

8 figures

The number that settles the argument. How many turns the earth's contribution gets through in the time the vessel takes to drain: the draining time divided by the rotation period at the drain. A bathtub manages a tenth of one and there is nothing to see. The apparatus that settled it manages nearly nineteen hundred, and that is the whole of what its enormous area-to-drain ratio was for.

The bath that was only ever a wait

The Coriolis force is far too weak to steer a draining bath, and it does steer a large enough tank that has been left alone long enough. Both are true, and the number that separates them is not a force ratio — it is how many turns the earth's contribution gets through before the vessel is empty.

8 figures

The one calculation in which the atmosphere really does push. How high a partial vacuum inside the tube will raise the liquid, against the absolute pressure achieved inside it. The lift is (p_atm − p_inside)/ρg and it is capped at the barometric height of 10.11 m, because below the vapour pressure the liquid boils and pulling harder buys nothing. The horizontal lines are five crown heights: a crown below a line's intersection with the curve can be primed by suction and one above it cannot, at any pump. This is the process the running siphon's argument explicitly excluded, and it is the one where the atmospheric account is the mechanism rather than a limit.

The one place the atmosphere pushes

The height of a siphon's hump is not in its flow rate, and the atmosphere holds the column together rather than driving it. That account excludes one process by name — starting. That is the process the atmospheric account describes correctly, and it is the only one in the whole device.

6 figures

The source falls 6 m and the crown pressure does not move. The pressure at the crown of a draining siphon, against time, as the source level falls from the top of the tank to the end of the run, a drop of 6 m. It is flat to 1.5e-11 pascals — not nearly flat, exactly flat, because the two effects of a falling source cancel identically. Losing a metre of level shrinks the drop, which slows the flow and raises the crown pressure by half a velocity head; and it grows the rise, which lowers the crown pressure by ρg per metre. Those are the same number. So a siphon that starts will not break as it drains, however far the level falls, and the run ended because the level reached the outlet instead.

The siphon that does not break

A draining reservoir shrinks the drop and grows the rise at the same time, and the siphon's own coupling — a metre of extra drop costs a metre of hump — says a siphon should break as it empties. It does not. The two effects cancel exactly, and the crown pressure of a draining siphon is a constant that does not contain the source level at all.

5 figures

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.

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.

5 figures

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.

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.

5 figures

Identical lift at every altitude, and the skin load nearly doubles. The same wing at the same dynamic pressure at seven altitudes. The lift is identical at every one of them — the flat line, computed from the absolute pressures rather than assumed, because a net force cannot depend on where the pressure datum is set. The load on a vented panel is identical too, at 4.86 kPa, because both sides of it moved together. The load on a sealed panel with 75.26 kPa inside is not: it is 28.27 kPa at sea level and 60.84 kPa at twelve kilometres. The datum cancelled in the force and it is one of the two numbers in the stress.

A force forgets the datum, a stress cannot

Nothing sucks, because the pressure datum cancels — the normals of a closed body sum to nothing, so the lift is the same in gauge or absolute pressure. That identity is about a resultant, and it is routinely carried one step too far. The load on a panel of skin has the ambient in it as one of two numbers, not as a datum.

5 figures

Four bodies, four drag coefficients, and no flow was solved. Four bodies with their Newtonian drag coefficients, each computed as a quadrature over its own surface with no flow solution anywhere. The cone's answer is exactly 2sin²δ, checked against the closed form to nine decimal places; the flat disc's is exactly 2, since every element of it faces the stream; and the sphere's is 1 against the classical Newtonian value of 1. Every one of those is an integral of one expression over a shape, and none of them required knowing what the air was doing anywhere.

The only theory simple enough to optimise

Whether Newton's sine-squared law is right has two answers — hopeless at the speeds he argued about, nearly exact behind a strong shock. This asks a different question about the same formula. Its pressure depends only on the local surface angle, so a shape's drag is a quadrature rather than a solution, and the best shape can be found by calculus.

5 figures