Field

Compressible flow

What changes when density stops being a constant. Shocks, expansion fans, nozzles that work backwards, and the one place on this site where an inviscid flow has drag.

A source at Mach 0.55, and the sound it has already made. Each circle is one pressure pulse, expanding at the speed of sound from the point the source was at when it left. Below the speed of sound every circle still contains the source, so the pressure disturbance reaches every point of the fluid before the source does. Above it the circles have an envelope, and outside that envelope nothing has been heard at all.

What a signal travels at

Air finds out that an aeroplane is coming, and it finds out at a definite speed. That speed does not depend on how hard the air is squeezed or on how much of it there is — only on how hot it is — and every result in compressible flow is downstream of that one fact.

A source at Mach 2.00, and the sound it has already made. Each circle is one pressure pulse, expanding at the speed of sound from the point the source was at when it left. Below the speed of sound every circle still contains the source, so the pressure disturbance reaches every point of the fluid before the source does. Above it the circles have an envelope, and outside that envelope nothing has been heard at all.

When the warning cannot arrive

Supersonic flow is not fast flow. It is flow in which the fluid ahead has been told nothing, because the body is outrunning its own pressure signals — and that single change turns an equation of one type into an equation of another.

$The energy is all there the whole way: enthalpy spent to buy speed. At each Mach number, the fraction of the stagnation enthalpy still held as heat and the fraction converted into directed motion. The two always sum to the total, which is the compressible replacement for Bernoulli's equation — an energy statement rather than a pressure one. At Mach 2 more than four fifths of the heat has become speed.$

Energy instead of pressure

Bernoulli's equation is a statement that pressure and speed trade against each other at fixed density. Take the fixed density away and the trade is between heat and speed instead — and what survives is not a pressure at all.

A duct does the opposite thing above Mach one. The four cases of dA/A = (Ma² − 1) dV/V. Below the speed of sound a narrowing duct accelerates the flow, which is what continuity leads anyone to expect. Above it the sign of the bracket flips, and a narrowing duct decelerates: density is falling faster than the speed is rising, so the stream tube needs more room rather than less.

The duct that works backwards

Squeeze a pipe and the flow speeds up. Everyone knows this, it follows from continuity, and above the speed of sound it is false — a narrowing duct decelerates a supersonic stream, because the density is falling faster than the speed is rising.

The throat stops listening: mass flow against back pressure. Mass flow through a convergent nozzle, normalised on its choked value, as the back pressure is lowered. It rises until the throat reaches Mach one and then stops, exactly. Below that pressure the throat is sonic and nothing downstream can send a signal upstream to ask for more, so the flow does not respond however far the back pressure falls.

The throat that stops listening

Lower the pressure downstream of a nozzle and more gas flows through it. Keep lowering it and, at a definite point, the flow stops responding — not gradually, but completely, because the news that the pressure has fallen can no longer travel upstream.

One nozzle, five back pressures, five different flows. The duct above, and the static pressure along it below, computed station by station from the local area. Where a shock stands inside the divergent section its position was solved for rather than placed: the shock spends total pressure, which sets the subsonic Mach number at the exit, which has to match the imposed back pressure.

One area, two answers

The area of a duct at a station fixes the Mach number there — twice over. A ratio of 2.5 is satisfied at Mach 0.24 and again at Mach 2.44, nothing local chooses between them, and where the choice cannot be made consistently a shock appears inside the duct to join the two.

A normal shock at Mach 2.00, and what crosses it unchanged. The state in front of the shock and the state behind it. Every ratio was computed from the standard jump relations and then substituted back into mass, momentum and energy, which is an independent route — a mistyped exponent in the total-pressure expression cannot survive a momentum balance it never appeared in. The residuals are printed below because a check nobody can see is a check nobody can audit.

The jump the equations allow

A shock is a discontinuity in a fluid, which sounds like a breakdown of the description rather than a solution of it. It is a solution: mass, momentum and energy can all be satisfied across a jump, and every ratio across one follows from that alone.

The second law is the only thing that forbids the other half of this curve. Entropy change across a normal shock, against the Mach number in front of it. The solid branch is the compression shock that exists. The dashed branch below Mach one is the expansion shock, and it satisfies mass, momentum and energy exactly — the residuals are zero to machine precision. It is refused by the second law alone, the one statement in the problem that no conservation residual can show.

The only law that forbids it

The jump conditions permit a discontinuity in either direction. An expansion shock conserves mass, momentum and energy exactly — the residuals are zero to machine precision — and it does not exist. Nothing that can be drawn rules it out.

Everything a normal shock does, against the Mach number in front of it. Four quantities across a normal shock, each scaled to fit one axis. Pressure and density rise without limit and without bound as the Mach number grows; the Mach number behind falls towards a floor it never passes; and the total pressure — the flow's ability to be turned back into speed — collapses. That last curve is why a supersonic intake is designed around avoiding a single strong shock.

What a shock costs

All the heat survives a shock and none of it is lost. What is lost is the ability to turn that heat back into speed — 27.9 per cent of it at Mach 2 and 93.8 per cent at Mach 5 — and every supersonic intake ever built is a scheme for paying less.

A 10° wedge at Mach 2.00 has two shocks that solve it. The wedge turns the flow through a fixed angle, and the θ–β–M relation offers two shock angles that achieve it. The weak solution, drawn steeply forward, leaves the flow supersonic and is what a wedge in a free stream produces. The strong solution leaves it subsonic and appears where downstream pressure forces it. Nothing local to the wedge chooses between them.

A shock that leans

Tilt a shock and only the velocity component across it is changed — the component along it passes through untouched. That single observation turns every oblique shock into a normal shock in disguise, and it is why a wedge at Mach 2 leaves the flow supersonic while a blunt nose does not.

Past 23.0° at Mach 2.00 there is no attached shock. The same wedge at two half-angles. On the left the θ–β–M relation has a root and the shock sits on the nose. On the right it has none, and the solver throws rather than returning the nearest thing — which matters, because a solver that quietly clamped to the maximum would draw a neat attached shock on a body that cannot carry one. The bow shock on the right is indicative: its shape is not solved here.

When the wedge is too blunt

Every curve of shock angle against deflection has a maximum. Past it there is no attached shock at any angle — the solver has no root to return and must say so, rather than quietly handing back the nearest thing and drawing a picture that cannot exist.

The same 10° turn, taken both ways. A supersonic stream turned away from itself expands through a fan of Mach waves and keeps every bit of its total pressure. Turned into itself through the same angle it shocks, and pays. Nothing in the equations distinguishes the two cases except the sign of the angle: compression waves converge and steepen into a front, expansion waves diverge and spread.

Turning the other way is free

Compression through ten degrees at Mach 2 costs 1.54 per cent of the total pressure. Expansion through the same ten degrees costs exactly nothing — not approximately nothing, nothing — and the two are the same equations with the sign of one angle changed.

Drag with nothing to rub

d'Alembert's paradox says a closed body in a steady, inviscid flow has no drag, and four separate arguments establish it, none of them wrong. Above Mach one it is false — the flow is still inviscid, still steady, and the drag is real, finite and quadratic in incidence.

Three waves out of one discontinuity. The x–t diagram of the burst diaphragm. A shock runs right at a speed of its own, a contact surface follows it more slowly, and an expansion fan spreads left as a family of rays that opens with time — the three wave families the Euler equations possess, produced at once by an initial condition with no waves in it at all. Every straight line here is a speed the solution computed, and the fan is drawn as the rays it actually consists of.

One diaphragm, every wave

Two states of the same gas at rest, separated by nothing, is the simplest initial condition compressible flow admits — and its answer contains all three waves the equations possess at once: a shock one way, an expansion fan the other, and between them a surface across which the density jumps and the pressure does not.

Two different physics, one destination. The Fanno and Rayleigh lines, each with temperature up the page and entropy across, each referred to its own sonic state. Neither is a curve anybody drew: the entropy was swept over Mach number and its maximum located, and it sits at Mach 1 on both to within the resolution of the sweep. The subsonic branch runs up to the nose from the left and the supersonic branch runs up from below, so whatever is being done to the flow, it can only be moved towards that point.

Two ways to choke

Friction and heat are different physics acting on different conservation laws, and they drive a duct's flow to exactly the same place. Both have their entropy maximum at Mach 1, so neither can push a subsonic flow past it and both drag a supersonic one down to it — and where their two curves cross is a shock.

The wall is what cancels the waves it made. The characteristic net of a minimum-length nozzle designed for Mach 2.4, with 18 waves. The pale lines run from the sharp throat down to the axis, reflect there by symmetry, and run back up to the wall; the wall turns through exactly the angle needed to cancel each one as it arrives, so nothing reflects back into the flow and the exit is uniform at Mach 2.400 and parallel to 0.0 degrees. The area ratio decides the Mach number and this net decides the shape, and the two agree on the exit height to 0.51 per cent at this resolution.

The wall that cancels its own waves

The area ratio of a supersonic nozzle fixes its exit Mach number and says nothing whatever about its shape. What fixes the shape is a wave-by-wave construction in which the wall turns through exactly the angle needed to absorb each expansion as it arrives — and getting it wrong leaves a stream full of oblique shocks at precisely the right Mach number.

What the skin settles at, before anything is done to it. The adiabatic wall temperature against Mach number, in air at 216.7 K, with the stagnation temperature above it. The gap between the two is the recovery factor, which is 0.8417 here and stays there at every Mach number — it is a property of the Prandtl number and not of the speed. At Mach 2 the skin sits at 363 K, at Mach 3 at 545 K, and at Mach 5 at 1128 K, which is past what aluminium will do. Nothing has been burnt and nothing has been rubbed: the air was brought to rest, and this is where its kinetic energy went.

The wall that heats itself

A surface told nothing about its temperature does not settle at the air's. It settles most of the way to the stagnation temperature, and the heat flux is driven from that invented temperature rather than from the free stream's — so a wall hotter than the air can be being heated by it.

Each cancellation costs two powers of the Mach number. Radiated power against compactness for three source clusters: a single monopole, two of opposite sign, and four on a square with alternating signs. The fitted slopes are 0.00, 2.00, 4.00 — zero, two and four in (kd), measured by integrating the far field over a sphere rather than assumed. A turbulent eddy turns over in about the time sound crosses it, so kd is of order the Mach number, and those exponents become the fourth, sixth and eighth powers of speed. A flow with no moving surfaces has no monopole and no dipole available to it, which is Lighthill's whole argument, and the eighth power is what is left.

The sound that only leaves

A flow is a catastrophically bad radiator, and the reason is that it has no monopole and no dipole available to it. What is left is the eighth power of speed — and the equation is equally happy with sound converging on a jet, which is ruled out by a condition imposed at infinity.

The sound is what fails to cancel. A compact quadrupole of kd = 0.01, followed outwards. The upper curve is what one of its four sources produces on its own at each radius — every one of them is as loud as a monopole — and the lower curve is what all four produce together. Inside the source they barely cancel at all; by one radian of wavelength the sum is 9.3e-5 of what a single source is doing, and it is falling as 1/R from there on because that is what radiation does. Between the two the field falls as the cube of the distance, which is a near field rather than a sound.

The sound is what does not cancel

A quadrupole is not a weak source. Every one of the four monopoles in it is as loud as a monopole of the same strength, and what makes the assembly quiet is that they very nearly cancel — one part in ten thousand survives at a hundredth of a wavelength. The eighth-power law is a statement about how nearly, not about how little.

The Hugoniot of a gas that can burn, and the gap in the middle of it. Pressure against specific volume, both scaled on the unburnt gas. The lower curve is the ordinary shock Hugoniot, which passes through the initial state because a jump of zero strength satisfies mass, momentum and energy. Adding a heat release lifts it away, and the initial state now sits in a region no wave can reach: between the two branches a Rayleigh line would need a positive slope, and its slope is minus the square of the mass flux. A burning gas has no weak waves available to it at all.

The other branch of the same curve

Put heat into the jump conditions and the Hugoniot lifts away from the initial state, leaving a gap no wave can occupy. A burning gas has no weak waves: it must run supersonically or subsonically, and the conservation laws pick the first speed exactly and say nothing at all about the second.

The two ways a shock can meet a wall. Left: the incident shock from the wedge reaches the wall and a second shock turns the flow back parallel to it, meeting at a point. Right: at a larger wedge angle no reflected shock can turn the flow that far, and the intersection lifts off the wall into a triple point with a nearly normal Mach stem standing on the surface and a slip line trailing downstream. The two configurations are drawn at the angles the solver returns, with every shock angle computed rather than sketched.

When a shock cannot bounce

A shock reflects off a wall until the reflected shock runs out of turning, which happens at a wedge angle well below the free stream's own limit. Between the two boundaries both configurations exist, both are stable, and which one appears depends on which direction the experiment came from.

A curved shock, and the entropy each streamline picks up crossing it. A parabolic bow shock ahead of a blunt nose at Mach six, with the streamlines drawn arriving horizontally and a marker at each crossing whose size is the total pressure lost there. The streamline through the nose crosses a normal shock and keeps three per cent of its total pressure; one four nose radii out crosses at fourteen degrees and keeps ninety-four per cent. Every streamline gets a different entropy, and the stagnation enthalpy is the same on all of them.

The spin a shock leaves behind

A curved shock gives every streamline a different entropy rise and the same stagnation enthalpy. Crocco's theorem then forces vorticity into a flow with no viscosity anywhere — and it scales as the inverse of the shock's radius of curvature, exactly, so a straight shock makes none.

The density ratio and the nose pressure coefficient, against Mach number. Two quantities across a normal shock, on a logarithmic Mach axis. Both approach limits that depend on γ and on nothing else: six for the density ratio and 1.8394 for the pressure coefficient at the stagnation point. By Mach five the second is within three per cent of its limit and by Mach twenty within a tenth of a per cent. Above that the flow round a blunt body has stopped depending on how fast it is going and started depending on what the gas is.

A shock that lies on the body

Above about Mach eight the flow round a blunt body stops depending on how fast it is going. The density ratio, the nose pressure coefficient and the shock standoff all reach limits set by γ alone — and going from a perfect gas to a dissociating one halves the standoff.

The coefficient of the equation's second derivative, along a chord. The bracket multiplying the streamwise second derivative in the transonic small-disturbance equation, along a chord at Mach 0.85. Where it is positive the equation is elliptic and the flow is subsonic; where it is negative the equation is hyperbolic and the flow is supersonic. Which it is at a given point depends on the perturbation velocity there, which is the thing being solved for. Forty-two per cent of this chord is hyperbolic, and no amount of inspecting the problem beforehand could have said so.

The equation that changes type inside its own answer

Near Mach one the coefficient of the streamwise second derivative depends on the perturbation velocity, which is what is being solved for. Two solutions of the linear equation no longer add — the leftover is three times the term the linear theory keeps — and the critical Mach number approaches one as the two-thirds power of thickness.

γ for air, against temperature. The ratio of specific heats for air as a mixture of nitrogen and oxygen, with the vibrational mode filling according to the Einstein function. It is 1.400 at room temperature, where only translation and rotation are available; 1.337 at a thousand kelvin; and 1.288 at six thousand. The usual compressible-flow tables use 1.4, and that is the value for a gas that is not hot — which, behind any shock worth drawing, it is not.

When gamma stops being a number

Every perfect-gas result uses γ = 1.4, which counts the ways a nitrogen molecule can hold energy at room temperature. Behind a Mach 10 shock the gas is at 3,800 kelvin and the count is different — and the pressure barely moves while the temperature falls by fifteen per cent.

One signature, aged four times. The pressure signature of a slender body at four distances, computed by the exact Lax formula for the nonlinear propagation. Each point of the waveform moves forward in proportion to its own overpressure, so the compression at the front catches the undisturbed air and a shock forms there, while the expansion at the rear falls behind and forms a second one. What is left is an N-wave: two discontinuities and a straight line between them, spreading and weakening.

The signature that forgets the shape

The pressure field near a supersonic aeroplane depends on every part of it. What reaches the ground has two parameters. Two bodies whose near-field signatures differ by fifty-five per cent in peak and by their whole shape age into the same N-wave, to two and a half per cent.

The shock radius against time, from an equation that was given no exponent. The thin-shell energy balance integrated forward from a small initial radius, on logarithmic axes. The two-fifths power is not put in: the ordinary differential equation is Ṙ = √(E/AρR³), and the straight line is what it does. The fitted slope is 0.39983 and the fitted prefactor is 0.90721 against a closed form of 0.90702 — the small residuals being the integration's memory of where it started, which the similarity solution has no equivalent of.

A radius that gives the energy away

Four quantities, three dimensions, one group. The radius of a strong blast must be a constant times (Et²/ρ)¹⁄⁵, and nothing about the device, the chemistry or the initial size can appear. The exponent is free and the constant is not.

Four bodies of identical length and volume, and their wave drags. Each body has the same length and the same volume; only the distribution of area along it differs. The Sears–Haack body — the spindle whose area goes as the three-halves power of x(L−x) — has the least wave drag of the four, and every other shape pays between thirty-seven per cent and a hundred and seventy-five per cent more for carrying the same volume the same distance. Nothing about the cross-sections' shape enters: only the area distribution does.

The least drag a volume can have

A body's supersonic wave drag depends on nothing about it except how its cross-sectional area is distributed along its length. Minimising that for a given volume gives one shape — and the answer goes as the volume squared over the fourth power of the length.

The discontinuity that has a thickness

The jump conditions do not contain the viscosity, which is why they are exact. The thickness is entirely viscosity — 289 nanometres at Mach 1.5, 35 at Mach 5, against a mean free path of 64. At Mach five the continuum equations have produced a structure thinner than the distance between collisions.

$A sinusoid, distorting on its way to a shock. A finite-amplitude sound wave at four fractions of the distance to shock formation, computed by inverting the implicit simple-wave solution. Each point of the waveform travels at its own speed, so the compressions catch up with the rarefactions ahead of them and the profile leans forward. At σ = 1 the front is vertical. The linear theory says the first panel is the answer at every distance, for ever.$

Every compression becomes a shock in the end

Linear acoustics has no time scale in it, which is the sign that something has been thrown away. A 120-decibel tone shocks after three hundred metres and a jet engine after twenty; the distance goes exactly as the reciprocal of the amplitude, and nothing is exempt.

So splitting a turn into N ramps costs one over N squared. A twelve-degree compression at Mach 3, done in one ramp and in up to sixty-four. The entropy is N times a cube of one Nth, so it falls as exactly the inverse square of the number of ramps — the measured exponent is −2.00 — and sixty-four ramps cost a two-hundred-and-fifty-sixth of what one costs.

A compression that costs nothing in the end

Turning a supersonic stream away from itself is free and turning it into itself is not. But the price of a compression is the cube of its strength, so splitting one turn into N turns costs one over N squared — and in the limit the compression is free too.

A compression piston, and where its characteristics first cross. Sixty C+ characteristics from an accelerating piston, drawn in the distance-time plane. Each is a straight line, because the invariant makes the state along it constant; later ones are faster, because the gas ahead of them has been compressed; so they converge, and the first crossing is the shock. The envelope formula gives 2.7529 and the first actual crossing is at 2.7510.

Two numbers that do not change

One-dimensional unsteady gas flow carries two quantities that are exactly constant along two families of curves. That single fact turns a pair of coupled partial differential equations into a family of straight lines, and gives an exact speed at which a gas outruns its own expansion.

Seven shock structures, and the two states they all connect. The velocity through the shock for seven dissipation models — Prandtl numbers from a quarter to two, viscosities from constant to linear in temperature. Each curve is shifted so its midpoint sits at the origin. They start at the same speed, end at the same speed, and are nothing alike in between.

The jump does not ask what made it

Seven different dissipation mechanisms are made to smear the same shock. Their interiors are a factor of two and a third apart in thickness, their entropies overshoot the final value by between a quarter and a doubling, and the state they all arrive at agrees to seven parts in ten billion — because the end states are conservation and the interior is transport.

The sound speed of a mixture, against how much of it is gas. Wood's formula for air in water. Both ends are the pure phases at 343 and 1,481 metres a second; in between the mixture takes the water's inertia and the air's springiness and the speed collapses to twenty-four metres a second — a fourteenth of the slower constituent.

Slower than either of them

Sound travels at 343 metres a second in air and 1,481 in water. In a mixture of the two it travels at twenty-four, because the mixture takes the water's inertia and the air's springiness — and one per cent of air by volume is enough to take water down to a twelfth of its own speed.

The temperature behind a shock, which is not the jump condition's. The static temperature along the flow behind a Mach 6 normal shock, with the frozen value the jump conditions give and the equilibrium value they give a long way behind. The gas arrives at 2382 K and settles at 2059, over about four tenths of a millimetre.

A gas that has not finished being shocked

The jump conditions give the state a long way behind a shock. Immediately behind it the molecules have not started vibrating yet, so the temperature is 2,382 K where the equilibrium answer is 2,059 — and the gas takes four tenths of a millimetre to get from one to the other.

The four waves, and the one that never goes away. The shock tube in space and time: a shock running right, an expansion fan running left, and the contact surface between them. The shock and the fan are travelling disturbances that leave; the contact is made of fluid, so it is carried along and is there for ever.

A surface that remembers the diaphragm

Between the shock and the expansion in a shock tube there is a surface across which the pressure and the velocity are identical and the temperature differs by a factor of two. It is made of fluid, so it never goes away, and nothing in the pressure field says it is there.

When the sound arriving now was made. The arrival time at a fixed observer against the emission time, for a source passing at Mach 0.8. The curve is monotone and its slope is not one: while the source approaches, a long stretch of emission arrives in a short stretch of time, and while it recedes the reverse.

The sound now is the source then

Every acoustic calculation is an exercise in bookkeeping about when. The pressure arriving at a listener was emitted at an earlier time, at a place the source has since left, and the Doppler shift is not a separate effect at all — it is the slope of the curve relating the two.

The Fanno line, which has only one direction on it. The entropy relative to the sonic state, against Mach number, for both branches. Friction raises the entropy, so a duct flow moves to the right along this curve whichever branch it is on — up in Mach number from below and down from above — and it stops at the sonic point.

A duct that cannot be run backwards

Friction drives a compressible duct flow towards the speed of sound from either side, and the entropy rises the whole way. So the state of the gas at a station is an odometer: it records how much duct is behind it, and no amount of further duct can take it back.

The structure the sensitivity produces. A detonation front, schematically: a leading shock, an induction zone in which nothing measurable happens, and a reaction zone behind it. The induction zone's length is set by the shock's own strength, and because that dependence is exponential the front is unstable and breaks into cells.

A gas that has not decided to react yet

Behind a detonation's leading shock there is a zone in which nothing measurable happens. Its length is set by the temperature the shock produced, exponentially — a one per cent change in the shock shortens it by fifteen per cent — and that sensitivity is why a detonation front cannot stay flat.

Two totals across a shock. The ratio of the total temperature and the ratio of the total pressure across a normal shock, against the shock's Mach number. One of them is one at every Mach number, to the last bit of double precision; the other falls to under a hundredth by Mach eight.

Two totals, one of which a shock cannot touch

Across a normal shock the total temperature ratio is 1.000000000000000 at every Mach number, and the total pressure ratio falls to 0.0085 by Mach 8. One of the two records the energy that has been added to the gas and nothing else; the other records every irreversibility on the way.

A skin lags the air by a time its own thickness sets. A skin held at one flight condition, relaxing towards the adiabatic wall temperature. The approach is exponential with a time constant ρc τ / h — 8.37 s at 2 mm, 25.1 s at 6 mm, 50.2 s at 12 mm — so the temperature a steady recovery calculation gives is reached after several minutes rather than at once. The fluid supplies one number to this calculation, the adiabatic wall temperature, and the structure supplies everything else.

The skin that lags the flight

A wall can be told its temperature or told nothing, and both are solved problems. A real skin is told neither. It has heat capacity, so its temperature is a transient whose time constant is its own thickness divided by what the layer delivers — and the number the steady calculation returns is an upper bound a short exposure never collects.

Measuring the air temperature in flight is not measuring the air temperature. Three temperatures against Mach number in a stream at 220 K: the free stream itself, the stagnation temperature a perfect probe would read, and what a probe recovering 0.98 of the rise actually reads. At Mach 0.85 the rise is 31.79 K and the probe misses 0.64 K of it. The instrument is a small stagnation region with a thermocouple in it, so it obeys the same recovery arithmetic a wall does — a probe is a wall told nothing about its temperature, made small and put on a stalk, and its recovery factor is a calibration constant rather than a one.

The thermometer that heats itself

A total-temperature probe is a wall told nothing about its temperature, made small and put on a stalk, so it obeys the same recovery arithmetic an aircraft skin does. Its recovery factor is a calibration constant near 0.98 rather than a one — and the static temperature inferred from its reading amplifies that shortfall rather than inheriting it.

A cone sends 1.7 per cent of its jet's momentum sideways at 15°. A conical divergent section of 15° half-angle and exit area ratio 25, drawn to scale from its throat to its lip, with its virtual apex to the left. The gas leaves as a source flow: straight streamlines from the apex, each at its own angle, and a Mach number uniform on spheres centred there. On the spherical cap through the lip the flow is normal to the surface and uniform, at Mach 3.925 for a gas with γ = 1.2; its axial momentum is ρV² times the cap's projection onto the exit disc, while the mass crossing it is ρV times the cap itself. The ratio of the two areas is (1 + cos α)/2 = 0.9830, and it is the only thing the cone's shape does to the momentum. On the flat exit plane the flow is not uniform: its edge is further from the apex than its centre, and the Mach number there is higher.

The jet a cone sprays sideways

The one-dimensional nozzle sends all its gas straight out along the axis. A real divergent section is a cone, and the gas leaves it as a spray of straight lines from the cone's apex. Only the axial part of that momentum pushes, and the share that does is (1 + cos α)/2 — 98.3 per cent at fifteen degrees, 93.3 at thirty — whatever the gas, the Mach number or the area ratio, and it touches the momentum and never the pressure.

With friction the flow goes sonic after the throat, and leaves slower. The Mach number along a convergent–divergent nozzle of exit area ratio 2.5, for friction lengths 4fL/Dₜ of 0, 0.2 and 1, each solution passing smoothly through Mach 1 at the point where the sonic condition holds. Without friction that point is the throat, at x = 0.42. With friction it moves downstream — to 0.4357 and 0.4996 — and the throat itself is subsonic, at Mach 0.965 and 0.852. The exit Mach number falls from 2.443 to 2.247 and 1.747. The throat is where the area is least; the sonic point is where the widening has caught up with the friction, and the two coincide only when there is none.

Friction moves the sonic point past the throat

A choked nozzle is sonic at its throat — in a nozzle with frictionless walls. With friction the flow reaches Mach one where the section's widening rate has caught up with the friction, which is downstream of the throat, and the throat itself is subsonic. The solution through that point is a saddle that can only be found from the inside, and the mass flow it passes is less than the throat's area allows.

A 20° cone at Mach 2: the shock at 37.80°, and a flow still compressing behind it. The conical flow round a cone of 20° half-angle at Mach 2, from the Taylor–Maccoll equation. The shock sits at 37.80° from the axis and turns the flow crossing it through only 8.57°, leaving it at Mach 1.693. Between the shock and the surface every ray from the apex carries its own state: the Mach number falls from 1.693 just behind the shock to 1.568 at the surface and the pressure rises from 1.586 to 1.912 times the free stream's — the rest of the turn, made without a shock. Nothing depends on the distance from the apex, so the rays are lines of constant state.

A cone finishes its turn after the shock

A wedge turns a supersonic stream all at once, at its shock. A cone of the same angle does not: its shock turns the flow only part of the way and leaves the rest to a smooth compression between the shock and the surface. Solved from Taylor and Maccoll's equation, the cone's shock is weaker, keeps more of the total pressure, carries less than half the wedge's surface pressure, and stays attached to 40.7° at Mach 2 where the wedge gives up at 23°.

Every output of the air-data reduction, and how hard it leans on each reading. The logarithmic sensitivity of each derived quantity to each of the three readings — the total pressure, the static pressure and the indicated total temperature — at Mach 0.3 and Mach 0.85, at 11 km with a probe recovering 0.98: a one per cent error in a reading times the number is the per cent error it puts into the output. The Mach number leans on the two pressures by 8.08 and −8.08 at Mach 0.3 and by 1.13 and −1.13 at Mach 0.85, and not at all on the temperature. The static temperature leans on its probe by exactly one and on the pressures by −0.280 at Mach 0.3 and −0.281 at Mach 0.85 — almost the same. The true airspeed leans on the probe by exactly one half and the density by exactly minus one, at every speed.

Three readings, and the one each answer leans on

An aircraft works out the air it flies through from three readings — a static pressure, a total pressure and a probe's temperature — and every derived number inherits their errors through one small table of sensitivities. Written out, the table says the airspeed is afraid of the pressure sensors almost to Mach one at cruise altitude and only to Mach 0.52 at sea level, and that the density belongs to the thermometer at every speed.

The gradient the heat never hears

On a flat plate at a Prandtl number of one, a boundary layer's total enthalpy is a straight-line function of its velocity, whatever the wall's temperature. Put the same layer in a pressure gradient and the straight line fails everywhere except on an insulated wall, because the gradient enters the velocity's equation and not the enthalpy's — and a favourable gradient can leave a band of gas colder than the free stream above a wall three times hotter than it.

Below Mach 1.153 the boom turns back before it reaches the ground. The ray leaving the Mach cone straight down from an aeroplane at 11 km, at Mach 1.1, 1.15, 1.2, 1.5, 2, traced through a standard atmosphere whose sound speed rises from 295.1 m/s at the aeroplane to 340.3 m/s at the ground. A ray bends back upward where the local sound speed equals the aeroplane's speed, so Mach 1.1: turns at 4.00 km; Mach 1.15: turns at 0.25 km; Mach 1.2: lands 24.0 km on; Mach 1.5: lands 11.4 km on; Mach 2: lands 7.1 km on. The dividing speed, Mach 1.1533, is the ratio of the two sound speeds.

The boom that turns back before the ground

Sound is faster in the warm air near the ground, so a sonic boom's rays bend back upward on the way down. Whether any of them arrive is one comparison — the aeroplane's speed against the fastest sound beneath it — and the ray that just grazes the ground sets the edge of the carpet, which the uniform air of the ageing calculation cannot give it.

The arrival map, and the place where it goes backwards. Where each boom ray lands on the ground, against the Mach number the aeroplane was doing when it launched it, for four accelerations from 15 km. In level flight this would be a straight line rising at the aeroplane's own speed. Here it falls before it rises: a ray launched at a higher Mach number is shorter and steeper, and near the cut-off it shortens faster than the aeroplane advances. Every minimum in these curves is a fold — two emission times delivering to one place — and on the fold itself neighbouring rays converge onto a single line.

The carpet an accelerating aeroplane folds

Level flight launches every boom ray with the same invariant, so they run parallel and each place hears one boom. Accelerate, and each successive ray is shorter than the last — shorter, near the cut-off, than the aeroplane's own advance — so later rays overtake earlier ones and the arrival map folds onto a line.

What the rays say the edge is: nearly as loud, and then nothing. The overpressure across the carpet relative to the value under the track, from ray-tube spreading alone, at three Mach numbers from 15 km. It falls gently and then stops: at Mach 1.8 it is 0.92 ten kilometres out, 0.77 at twenty-three and 0.65 on the last ray, 36 km out, with silence beyond. Geometrical acoustics says the carpet ends at a cliff, and the boom at the edge of a real carpet fades as a rumble. The cliff is what the model says; it is also where the model stops.

The edge is a rumble, not a quieter bang

The rays that reach the outer half of a sonic-boom carpet arrive nearly horizontally, having travelled almost three times as far as the one under the track. Ray theory says they still carry two-thirds of the overpressure, right up to a line beyond which there is nothing. Neither half of that is what is heard — which is the useful result, because it says the edge's loudness is not a ray quantity at all.

A boom is aged in the thin air it starts in

The rays that reach the edge of a sonic-boom carpet travel two and a half times as far as the one under the track, and it is natural to expect their signatures to have aged accordingly. They have not. A pressure wave distorts thin air far faster than dense air, so two-thirds of a boom's ageing is done in the stratosphere near the aeroplane, and the extra kilometres near the ground add little — which decides how far out a boom shaped to be quiet stays quiet.

Three different flows, drawn on top of one another. The surface pressure of three parabolic arcs of different thickness, each at the Mach number that gives them the same transonic similarity parameter, divided by the scale the similarity rule prescribes. They lie on one curve. The three solutions were computed separately from an equation with different coefficients in each case, and nothing in the solver knows what the parameter is — the scaled peaks differ by 0.01 per cent.

One curve for every thickness

Near Mach one there is no linear theory, so there is no superposition, no scaling of a solution by the thing that caused it, and no shortcut of any kind. There is exactly one thing left — a change of variables in which the thickness and the Mach number vanish and only their combination survives — and it is worth more than the theory that was lost.

A drag that is nothing until there is something to be discontinuous about. The scaled wave drag of a parabolic arc against the similarity parameter. Above about 1.4 it is not small but zero to five decimal places — d'Alembert's paradox, which the small-disturbance equation inherits. Below it the drag rises by four orders of magnitude over a range of the parameter that corresponds to a few hundredths in Mach number, and the only thing that changed is that the flow now contains a shock.

A drag that needs a discontinuity

A closed body in a steady inviscid flow has no drag, and the transonic equation obeys that as faithfully as any other — until a shock appears inside its own answer. Then the same pressure integral that was zero to five decimal places becomes four decades larger, with no viscosity added and nothing about the body changed.

A frequency set almost entirely by how nearly sonic the flow behind the shock is. The Strouhal number of the loop against the local Mach number behind the shock. The band drawn across it is what is measured on aerofoils in buffet, 0.06 to 0.08, and the arithmetic meets it at a post-shock Mach number of 0.971. Two per cent either side of that number changes the frequency by a third, which is why buffet onset is predicted by correlation rather than by calculation.

A shock that will not stand still

A converged transonic solution puts the shock in a definite place. Over a band of conditions the real one sweeps back and forth across a fifth of the chord at a fixed frequency, and the period is set almost entirely by how nearly sonic the flow just behind it is — a quantity that is small, hard to compute, and multiplies everything.

Where the reflected shock meets the driver gas. Distance against time in a helium-driven air tube, the incident shock in red running from the diaphragm to the end wall, the contact surface behind it in green, and the reflected shock coming back. Where it meets the contact, an under-tailored tube sends an expansion back to the wall — in ochre — which lowers the reservoir's pressure, and the contact drifts back towards the diaphragm; an over-tailored one sends a shock, which raises it. At the tailored Mach number, 3.41, nothing is sent back, the contact stops dead, and the reservoir at the end wall holds until something slower arrives.

The shock that passes without an echo

A shock tunnel's reservoir lasts until a wave comes back from the driver gas to spoil it. For each pair of gases there is one incident shock strength at which nothing comes back at all, and it is found by asking the reflected shock to stop two different gases at the same pressure.

$A pipe at the gas's temperature does not keep the gas at it. Left, the static temperature of the gas along a pipe, as a fraction of the wall's, against the local Mach number, entered at Mach 0.1: with no heat transfer, with the heat transfer Reynolds' analogy gives, and with ten thousand times that. Right, the stagnation temperature. The isothermal model holds the static temperature at the wall's and needs the stagnation temperature to climb by 14 per cent; every real case keeps it within two per cent, cools, and runs on to Mach one.$

A pipe cannot hold its gas at the wall's temperature

The textbook model of a long gas pipe in contact with the ground holds the gas at the ground's temperature and has it choke at 0.845 of the speed of sound. No pipe does either. A wall at the gas's temperature draws heat out of it rather than putting heat in, and with any strength of heat transfer at all the flow runs on to Mach one, within a tenth of a per cent of the length a perfectly insulated pipe would need.

$Inside a strong blast: a shell of gas around a hot, nearly empty core. Left, the density, velocity and pressure inside a spherical strong blast wave at γ = 1.4, each as a fraction of its value just behind the shock, against the distance from the centre as a fraction of the shock radius. The density falls to nothing well inside the shock while the pressure levels off at 0.37 of its post-shock value. Right, the temperature, which is the pressure over the density and so rises without limit towards the centre.$

The fireball is hollow

Inside a strong blast wave there is almost nothing. Half the air the shock has swept up lies in the outer four per cent of its radius; at half the radius the density is a hundredth of the air's; the centre is empty and, formally, infinitely hot. Integrating Sedov's equations to see this also shows that a published table of his constants was wrong at three of its four entries.

Dry air makes a slow bang. The 10-to-90 per cent rise time of a steady shock against relative humidity, for jumps of 25, 50 and 90 pascals, all below the strength at which a discontinuity returns. Each falls roughly as the nitrogen relaxation time does, and doubles when the jump is halved. The rule marks the thermoviscous rise time of the 50-pascal shock, a few microseconds — two to three orders of magnitude below any of the curves.

Oxygen makes a boom's crack, nitrogen its rise time

The shock at the front of a sonic boom is not the viscous shock of a textbook, a few microseconds thick. It is spread over a large fraction of a millisecond by the vibrational relaxation of the air's molecules, and the two gases do different jobs. Oxygen, fast, removes the discontinuity for any boom weaker than about ninety pascals and decides how much of the front is left at the frequencies the ear weighs most; nitrogen, slow, sets the rise time that is measured. Water vapour speeds both, so a dry day makes a softer bang.

A bow shock that grows as a line explosion does. The radius of the bow shock around a hemisphere-nosed cylinder, in body diameters, against distance behind the nose: the blast-wave analogy, R/d = 0.795 C^¼ (x/d)^½ with C, the drag coefficient, 0.919, which contains no Mach number, and Billig's correlation of measured bow shocks on spheres at Mach 5, 10 and 20, anchored at the nose and continued as a hyperbola to the Mach cone. Both grow as the square root of the distance; at Mach 20 the analogy's shock is a steady 0.71 of Billig's.

A hypersonic body leaves a line explosion behind it

A blunt body at hypersonic speed does work on the air at a rate equal to its drag, and each slice of air it passes through is struck once and left to expand. Seen from the ground, that is a line explosion, and Sedov's cylindrical blast wave gives the bow shock's width and the pressure on the afterbody without a Mach number in either. The analogy's classical constants come straight out of the blast solution. So do its limits: it holds only while its own shock stays strong, over a length that grows as the square of the flight Mach number, and it puts the body inside a core hotter than anything the flow can reach.

A long line empties from the break backwards. The pressure along a gas line three hundred friction lengths long — fL/D = 300 — at five times after it breaks at its far end into still air at a twentieth of its pressure, the break on the right. Half a transit time in, the rarefaction has crossed half the line and the gas beyond it has not moved. Friction then holds the pressure in a slope that lengthens back towards the closed end; at six transit times the closed end is at 0.55 of its starting pressure, at fifteen 0.21, and the break has long since stopped being choked.

A broken line empties at the pace of its friction

When a gas line breaks, its open end chokes and the textbook stops there: a choked outlet passes gas at its own speed of sound and nothing downstream can change that. On a long line the choke is the least of it. The break stays choked only while friction lets enough gas reach it, which on a line a thousand friction lengths long is the first eighth of the time to half pressure. The pressure falls at a pace friction sets, the half-time grows as the square root of the line's length in friction lengths, and the wall's heat — negligible in the steady line — makes the blowdown a third slower.

Next to the body the gas remembers the nose. The gas temperature, over the free stream's, against distance from the axis of a hemisphere-nosed cylinder at Mach 15, at 2, 20 and 200 diameters behind the nose; the body's surface is at a half. The gas next to the body crossed the nearly normal part of the bow shock and carries its entropy: 16.7 times the free stream's temperature at 2 diameters, 9.29 at 200, falling outwards to the gas that crossed the weaker, oblique shock. The blast-wave analogy's core at 2 diameters, dashed, runs off the top of the frame on its way to infinity.

The gas beside a hypersonic body remembers its nose

The blast-wave analogy gets a blunt body's bow shock and afterbody pressure right and its temperature absurdly wrong: it puts the body in the empty core of an explosion, where the temperature has no bound. The real gas beside the body crossed the nearly normal shock at the nose and carries that crossing's entropy all the way down. Expanded to the afterbody's pressure, it is ten to twenty times the free stream's temperature at Mach 15, it never falls below a floor set by the nose alone, and the sheath that carries it is wider than the body.

$The wall lets go long before the textbook's shock arrives. Where the flow leaves the nozzle's wall, as a fraction of the divergent section's length, against the ratio of chamber to ambient pressure. The inviscid picture runs full to the exit until the pressure ratio falls to 24.5, where a normal shock stands at the exit, and then moves the shock inside. The wall flow separates instead, when its pressure falls to Summerfield's 0.4 of ambient — at a pressure ratio of 192 — or to Schmucker's Mach-dependent value, at 138: six to eight times higher. Across that whole band the textbook's nozzle is full and the real one is not.$

An overexpanded nozzle lets go before its shock arrives

The textbook's overexpanded nozzle runs full until its back pressure is high enough to hold a normal shock at the exit, and then draws the shock walking inside. A real nozzle never shows that sequence. Its wall's boundary layer cannot climb the pressure rise a normal shock imposes; it separates at a wall pressure of about four-tenths of ambient, which for a rocket nozzle happens at six to eight times the pressure ratio the textbook's shock needs. The separation is not a failure: it is what keeps the nozzle's thrust, and it is why a sea-level engine can be built twice the size it expands to.

Tailored, the reservoir waits for the driver's own expansion. Distance against time in a helium-driven air tube at the tailored Mach number, 3.41, with a driver half as long as the driven tube: the incident shock, the contact surface, the reflected shock and the shock it transmits into the driver gas, and the driver's expansion — its head running back to the driver's closed end, and the head reflected from there. Nothing returns from the contact, and the reservoir at the end wall holds from 0.293 until the reflected head arrives at 0.555 L/a₁: a test time of 0.261.

A tailored tube buys its test time with its driver

Tailoring a shock tube removes the wave the contact surface would send back to the reservoir. What ends the reservoir then is slower: the driver's own expansion, which runs back to the driver's closed end, reflects, and has to cross the whole tube to reach the end wall. Its arrival is exact in one dimension, because the reflected head crosses the incident fan as a simple wave, and the answer is that test time is bought with driver length — about seven-tenths of a driven-tube crossing time per driver length for helium — and that tailoring is worth nothing with a driver shorter than a quarter of the tube.

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.

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 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.

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.

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.

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.

A cold wall holds its layer on longer

A compressible boundary layer's pressure gradient pushes on its density, and its density is its temperature. Solve the velocity and the enthalpy together and the wall's temperature reaches the velocity: a cold wall holds its layer on to an adverse gradient two-thirds steeper, a hot one lets it go a third sooner, the stagnation point's heat-to-friction ratio more than doubles between them, and the cold band an uncoupled layer put above a cooled nose disappears.

$Below the bubbles' resonance the mixture is slow; above its stop band it is faster than water. The phase speed of sound in water carrying millimetre air bubbles, against frequency, at void fractions of 10⁻⁴, 10⁻³ and 10⁻². At low frequency each is Wood's mixture speed with the gas isothermal — 312 m/s at 10⁻³, not the 366 an adiabatic gas would give. Approaching the bubbles' resonance near 3.2 kHz the speed falls further, then jumps through a stop band in which the wave hardly propagates, and above it the phase speed exceeds water's — at 10⁻³ 1640 m/s at 30 kHz — before returning to 1481 m/s from above.$

Above their resonance, bubbles make water faster

Wood's formula says a pinch of air makes water's sound slower than air's, and it is right at low frequency. Each bubble, though, is a spring with water for its mass, and the wave drives it. Below the bubbles' resonance the mixture is slow; through a band above it no sound propagates at all; and above that band the same bubbly water carries sound faster than pure water does. Even the slow end is not quite Wood's, because a slowly squeezed bubble keeps its heat.

Zero wave drag at one Mach number, inside a loop it cannot enter from below. The wave drag of the pair, per unit of one element's chord, against Mach number, beside a single diamond of the pair's combined thickness. The biplane's drag is zero at its design Mach number, 1.944, and rises either side. Shaded: below 1.6 the channel cannot pass the flow at all; between 1.6 and 2.15 it can, but only if it was started above the band. Accelerating from subsonic, the biplane stays choked until Mach 2.15, past its design point.

Busemann's biplane has to be flown past its design point

Two half-diamonds facing across a channel cancel each other's waves at one Mach number and carry their thickness at no wave drag. The cancellation needs the channel narrowed at its middle, and a narrowed supersonic channel will not swallow the shock that forms in it on the way up. A biplane 3 per cent thick per element, designed for Mach 1.94, stays choked until Mach 2.15, and only one that is thinner than about 2.6 per cent can start by itself at any speed.

$A spread of sizes smears the stop band and pushes the fast band up. The phase speed of sound against frequency in water carrying air bubbles at a void fraction of 10⁻³: all of one millimetre radius, and spread from 10 µm to 1 mm as n ∝ R^(−q) with q = 0, 3/2 and 10/3. Every cloud starts at Wood's speed, 312 m/s, because at low frequency every bubble is driven below its resonance. The single size jumps through its stop band near 3 to 13 kHz into a fast band. The distributions climb through a band as wide as their resonances, 3.2 kHz for the largest to 301 kHz for the smallest, and the more small bubbles they carry the later they emerge faster than water.$

The more small bubbles a cloud holds, the later it turns fast

Bubbles of one size give water three sound speeds: slow below their resonance, nothing in a stop band, faster than water above it. A cloud of many sizes has every resonance at once, so at any frequency its small bubbles soften it and its large ones stiffen it. The stop band becomes a wide band of attenuation, the fast band survives above it, and where the fast band starts is set by how much of the gas is in the smallest bubbles.

Each oblique shock added before the normal one buys less than the last. The best total-pressure recovery of a train of oblique shocks ending in a normal shock, against the number of oblique shocks, at Mach 2, 2.5, 3 and 4. At Mach 3 one oblique shock recovers 58.1%, two 74.4%, four 88.9% and eight 96.4%; a normal shock alone keeps 32.8%.

The last shock should be the weakest

An intake slows a supersonic stream through a train of oblique shocks and one normal shock, and the rule for laying out the train is that its shocks should be equally strong. Solved exactly rather than quoted, the rule is right for the oblique shocks to four figures and wrong for the last one: the terminal normal shock should be about five per cent weaker than the others, and the turns that give equal strengths are not equal turns but grow along the train.

At Mach 3 the cowl caps external compression at 88 per cent. Best recovery at Mach 3 against the number of oblique shocks, free and with the total turn capped at the free stream's detachment angle, 34.07°, which is as far as the cowl's outer surface can turn. Free, six shocks reach 94.2%; capped, 86.7%, and no number of shocks, nor a continuous isentropic ramp turning the same angle, can pass 88.5%.

The cowl caps what the ramps can buy

Every ramp in front of a supersonic intake turns the flow towards the cowl, and the cowl's outer surface has to turn the free stream at least as far. Past the free stream's own detachment angle it cannot, and that one angle caps every external intake: at Mach 3 no number of ramps, nor a smooth ramp, recovers more than 88 per cent, and the shock that is left at the throat is the strongest in the train. Above Mach 3 the cap falls fast, which is why fast intakes compress inside.

Behind the lip the ceiling is set by starting, not by the cowl. The best total-pressure recovery with every compression isentropic but the terminal normal shock, against free-stream Mach number: an external ramp turning to the free stream's detachment angle; the same ramp followed by the largest internal contraction a fixed duct can start through, Kantrowitz's limit; and the same ramp followed by a contraction to a throat Mach number of 1.2, which only a duct that opened its throat to start can hold. At Mach 3 the three are 88.5%, 95.1% and 99.3%; at Mach 4, 71%, 82.5% and 99.3%.

Compression behind the lip is bought by opening the throat

The cowl caps how far an intake's ramps can turn the flow, and moving some of the compression inside the duct escapes the cap. It does not escape the duct's other limit: a duct that contracts a supersonic stream has to swallow the shock that stands at its lip while it starts. With fixed walls that allows a contraction of about a tenth, and at Mach 3 six shocks with three inside recover exactly what six outside do. A duct that opens its throat to start and closes it after recovers 94 per cent — the internal compression is paid for by the moving wall, not by the walls being inside.

A cone swallows its sheath as the cube root of the Reynolds number; a cylinder in proportion to it. The swallowing distance in nose radii against the Reynolds number on the nose radius at Mach 15: a 7° sphere-cone, laminar and turbulent, and the sphere-cylinder of the same nose, laminar. At Reₙ = 10⁵ the cone's laminar layer swallows the sheath at 160 nose radii and its turbulent one at 28.1; the cylinder's laminar layer at 1.36·10⁵.

A blunted cone eats its nose's sheath in a few hundred radii

A sphere-cylinder's boundary layer takes a third of a Reynolds number of diameters to get through the hot sheath its nose leaves, which on any real body is never. Put the same nose on a slender cone and the answer collapses: the cone's widening perimeter feeds the boundary layer so fast that it swallows the sheath in tens to hundreds of nose radii, growing only as the cube root of the Reynolds number. On a two-metre cone at Mach 15 that makes a nose of about two centimetres the line between a base still bathed in the nose's gas and one that has forgotten it.

The body's dipole outshouts the flow's quadrupoles by one over the Mach number squared. The power of a wire's lift tone over the generous bound on its wake's quadrupole noise, against the wind's Mach number, for rms lift coefficients 0.2, 0.45 and 0.6 — the spread of the measurements. At Mach 0.03, a 10 m/s wind, the ratio is 9.4·10⁴; it falls as one over the Mach number squared and is still 104 at Mach 0.9, where a real cylinder's shedding has long since changed.

A wire sings with the force on it, not the flow round it

A free flow can only radiate as a quadrupole, four sources nearly cancelling, which is why a jet's noise grows as the eighth power of its speed and so little of its power becomes sound. Put a body in the flow and the cancellation is broken: the fluctuating force the body exerts is a dipole, its sound grows as the sixth power, and at the speed of any wind it outshouts the turbulence that made it by a factor of tens of thousands. The whistle of a wire is the force of its own wake, radiated.

Bluntness delays a cone's transition by a factor that peaks, and the correlation's Mach exponent sets how high. The distance at which a blunted 8° cone's layer transitions, over the sharp cone's 0.3 m, against the nose Reynolds number, at Mach 6 and 2.6·10⁷ per metre, for three correlations of the form Reθ/Mₑⁿ = C. Reθ alone: a largest delay of 4.96 at a nose of 43.9 mm, nose Reynolds number 1.14·10⁶; Reθ / √Mₑ: a largest delay of 2.57 at a nose of 20.1 mm, nose Reynolds number 5.23·10⁵; Reθ / Mₑ: a largest delay of 1.41 at a nose of 7.29 mm, nose Reynolds number 1.9·10⁵. Past a nose Reynolds number of 8.81·10⁵ the Mach-ratio criterion transitions within ten nose radii of the tip, where the model's edge is least trustworthy.

A blunted nose delays its cone's transition only so far, and the correlation decides how far

A blunted tip leaves a sheath of hot, slow gas along a hypersonic cone, and the boundary layer growing inside it sees a lower Reynolds number and a lower Mach number than a sharp cone's. That is why bluntness is said to push transition back. Read the layer's edge through the sheath and the push has a limit: the delay grows with the nose, peaks, and falls again once the transition point lies inside the sheath. How large the peak is depends on whether the transition correlation divides by the edge Mach number — five times the sharp distance at Mach 6 for the momentum-thickness Reynolds number alone, 1.4 for its ratio to the Mach number — and every peak sits forty to sixty-five nose radii from the tip.

A steady force carried round a circle sings, and each harmonic rises steeply with the tip's Mach number. The loudest level, 30 m away, of the first four blade-passing harmonics of a two-bladed 1.88 m propeller giving 2 kN of thrust from 127 kW, against its tip Mach number. At a tip Mach number of 0.4 the fundamental is 89.8 dB and the fourth harmonic 51; at 0.78, the climb, 98.2 and 88.6; at 0.95, 101 and 98.9.

More blades hush a propeller only while its tips are slow

A propeller's thrust and torque are steady in the blade's own frame, and a steady force held still makes no sound. Carried round a circle it sings at the blade-passing frequency, and Gutin's formula says how loudly: through a Bessel function whose order is the harmonic's and whose argument is that order times the tip's speed towards the listener over the speed of sound. Such a function is exponentially small until its argument nears its order. So spreading the same thrust over more blades, which raises the order of the fundamental, silences it by twenty-five decibels at a tip Mach number of 0.4 and by one at 0.95.

A flap a fifth of the chord long, turned in three and a half degrees, starts the biplane at Mach 1.72 instead of 2.15. The Mach number at which the channel between the elements swallows the normal shock at its mouth, against the leading-edge flap's deflection, for flaps of 0.1, 0.2, 0.3 and 0.4 of the chord; the deflection runs from shut to the ramp's own angle, 3.43°, where the flap lies flat. Shut, the channel starts at 2.15; flat, at 1.93 (0.1), 1.72 (0.2), 1.53 (0.3), 1.32 (0.4). The design Mach number is 1.944.

A flap that closes the mouth starts Busemann's biplane below its design point

Busemann's biplane cancels its own waves at one Mach number and cannot get there: its channel narrows to a throat that will not swallow the shock standing in its mouth until the aeroplane is flying faster than the design point. Hinge the front of each element's inner ramp and turn it into the channel, and the mouth closes towards the throat while the throat stays put. A flap a fifth of the chord long, turned in three and a half degrees, starts the channel at Mach 1.72 instead of 2.15; once started, the flaps can shut at any speed above 1.60 and the zero at the design point returns exactly.

Behind a strong shock the layer is an exact rotational flow, squeezed into a tenth of the shock's radius. Streamlines of the constant-density shock layer between a spherical shock (radius 1) and a sphere at density ratio ε = 0.167, from the solution ψ = sin²χ(ar⁴ + br² + c/r): the body sits at 0.8944 of the shock's radius, a standoff of 0.118 body radii. The gas enters along the shock, turns, and runs round the body in a layer whose vorticity over r sin χ is the same everywhere.

The shock layer is Hill's vortex, and it heats the nose a tenth harder than Newton says

Behind a strong bow shock the gas is squeezed six-fold or more and turned round the nose in a thin layer, and if its density is taken as constant there the layer has an exact solution: the same family of rotational flows as Hill's spherical vortex, fitted to the shock's three conditions. It stands off further than a mass balance says, and its gas accelerates away from the stagnation point a fifth faster than Newton's pressure law implies. Since a nose's heating goes as the square root of that acceleration, the layer heats the nose about ten per cent harder than the estimate every first calculation of re-entry uses.

At its design Mach number the biplane's channel lifts nothing: all its lift is on the two outer faces. The lift coefficient of the pair at 2° incidence against Mach number, per unit of one element's chord, beside what the two outer faces give alone, 4α/β, and two isolated flat plates, 8α/β. At the design Mach number, 1.944, the pair's lift is 0.08378 and the outer faces' 0.08378: the channel's net lift is zero to rounding. At Mach 1.7 the channel adds 13.9 per cent, at 2.2 21.1.

Busemann's biplane lifts only on its outer faces, and its channel turns the lift into a quarter-chord load

Tilt Busemann's biplane to make it lift and the same cancellation that removed its wave drag removes the lift of its channel: at the design Mach number the waves the incidence sends across the gap meet and cancel just as the thickness's did, and every bit of lift is carried by the two outer faces — half what two separate plates would give. The channel does not go idle. It carries a couple, lift on the front half and download on the rear, which moves the centre of pressure from mid-chord, where every supersonic plate has it, to the quarter chord. The pair pays for its lift two and a half times as steeply as flat plates do, and still beats a diamond fourfold.

A five per cent distortion gives six blades a tone that no longer falls with the tip's speed, and two blades hardly notice. The sound power of the blade-passing fundamental against tip Mach number, thrust and shaft power held, for two and six blades: from the steady load, and from the thrust's fluctuation as each blade cuts an inflow distortion of 5 per cent rms. At tip Mach 0.4 six blades give 100 dB from the steady load and 113 from the distortion; two blades 127 and 109. The six-bladed steady tone overtakes its distortion tone at tip Mach 0.593.

A distorted inflow gives back the tone more blades took away

A propeller's steady thrust sings through a Bessel function whose order is the number of blades times the harmonic, and that function is exponentially small at low tip speeds — which is why six blades are quieter than two. But a blade that cuts a non-uniform inflow carries a thrust that changes round the circle, and its s-th harmonic radiates through a Bessel function of order six minus s. The sixth needs no tip speed at all. A distortion of five per cent turns six blades' tone into one that hardly falls as the tips slow, fills the axis where the steady tone is silent, and cuts the six-blade advantage over two from twenty decibels to thirteen. Two blades never notice.

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