The boom that turns back before the ground

The signature that forgets the shape propagated a boom straight down through air of one temperature, and in that air every signature arrives. It then named, in words, three things the real atmosphere does instead — a carpet with an edge, a Mach number below which nothing lands, and rays that converge into superbooms — and said plainly that none of them was in its figures. This essay computes the first two. They turn out to be one calculation, and a short one: a boom’s ray obeys Snell’s law in air whose sound speed rises towards the ground, and the quantity Snell’s law conserves is fixed by the aeroplane’s speed.

What comes out is a single comparison. A boom reaches the ground only if the aeroplane is moving faster than sound travels anywhere between it and the ground, and on an ordinary day the fastest sound in that column is at the bottom of it. Everything else here — the cut-off Mach number, the width of the carpet, what a wind aloft does, and why a cold morning spreads a boom farther than a hot afternoon — is that comparison evaluated in another direction or on another day.

The aeroplane’s speed is the invariant the rays conserve #

Four conventions first. The atmosphere is the standard one: the temperature falls linearly from 15 °C at the surface to −56.5 °C at 11 km and is constant from there to 20 km, and where a different surface temperature is quoted the tropopause is held where it is and the lapse rate changes to meet it. The flight Mach number is measured against the air at flight height, which is the only Mach number an aeroplane’s instruments can report, as What the airspeed indicator believes worked through. A ray’s azimuth φ is its angle around the flight path, zero straight down. And distances on the ground are measured from the point beneath the aeroplane at the moment the boom arrives, so a footprint is drawn as it would be seen from the aeroplane.

The derivation takes a paragraph. The Mach cone is steady in the aeroplane’s frame, and each ray leaves it along the cone’s normal. In air whose properties change only with height, the horizontal part of a ray’s slowness — its wave normal divided by the local sound speed — cannot change along it; that is Snell’s law, and it holds for sound for the same reason it holds for light. The along-track part of the slowness is the reciprocal of the aeroplane’s speed, because the whole pattern sweeps along at that speed and every piece of it must keep pace. So the vertical slowness at height z is $q = \sqrt{1/a(z)^2 - 1/V^2 - s_y^2}$, where $s_y$ is the sideways part set by the azimuth. Straight down, $s_y = 0$, and $q$ vanishes where $a(z) = V$. There the ray is travelling horizontally, and below that height it cannot go: it turns and climbs.

The sound speed depends on the temperature and on nothing else, which What a signal travels at established for exactly this use. It is 340.3 m/s at a 15 °C surface and 295.1 m/s at the tropopause. An aeroplane at 11 km at Mach 1.1 is moving at 324.6 m/s, which is the sound speed at 4.0 km, and so that is where its boom stops descending.

The first figure shows how gradual the turning is. At Mach 1.05 the ray bottoms out at 7.58 km, at 1.1 at 4.00 km, at 1.13 at 1.77 km and at 1.15 at 250 m, and each climbs back symmetrically — the Mach 1.1 ray regains flight height 65.2 km farther on, 213.6 seconds after it left. At Mach 1.16 the ray lands, 33.8 km on and 111 seconds later; at 1.2 it lands 24.0 km on after 82.9 s; at Mach 1.5 11.4 km on after 49.9 s; and at Mach 2 7.1 km on after 41.1 s. Between a ray that passes 250 m above the ground and one that strikes it lies a hundredth of a Mach number, which is what a threshold looks like when the quantity approaching it is smooth.

The cut-off is the ratio of two sound speeds #

Straight down, landing requires the aeroplane’s speed to exceed the sound speed at every height below it. On a day whose temperature falls with height the largest of those is at the ground, so the condition is simply $M > a_{\text{ground}}/a_{\text{flight}}$.

On a standard day the cut-off is 1.036 for an aeroplane at 3 km, 1.062 at 5 km and 1.105 at 8 km, and from 11 km upward it is 1.1533 at every height, because the air above the tropopause has one temperature and the ratio no longer changes. A 0 °C surface lowers that plateau to 1.1228 and a 30 °C surface raises it to 1.1829. The figure of “about Mach one and a sixth” that the ageing essay quoted is right in size: seven-sixths exactly is the cut-off above a 21.7 °C surface.

The consequence is easy to state and odd to hold in mind. Above the tropopause, whether a town hears an aeroplane flying at Mach 1.15 is decided not by anything at the aeroplane’s height but by the thermometer in the town. On a −10 °C winter morning the cut-off is 1.102 and the boom arrives; on a 30 °C summer afternoon it is 1.183 and the same flight, on the same track, is silent below. Raised ground works the other way: an airfield 1,500 m up sits in air at 5.25 °C on a standard day, its sound speed is 334.5 m/s, and the cut-off over it is 1.134.

The geometry is one most people have already seen, in light. Over a hot road on a summer day, light is slightly faster in the thin hot layer at the surface than in the air above it, so rays coming down at shallow angles bend back up before they reach the tarmac, and a driver sees the sky where the road should be — an inferior mirage. It is the same bending of light by a gradient that An instrument that takes a derivative turns into a measurement. A boom below its cut-off is that mirage seen from the other end: the ground lies in the shadow the bending makes, and the rays that would have reached it are the ones a driver sees as sky.

The carpet ends at the ray that grazes the ground #

Off the track the sideways slowness adds to the invariant, so a lateral ray needs a larger margin: it lands only if $1/a_{\text{ground}}^2 > 1/V^2 + s_y^2$. The ray for which the two sides are equal arrives travelling horizontally, grazing the ground, and it is the edge of the carpet. Its azimuth has a closed form, $\sin\varphi_c = (a_{\text{flight}}/b)\sqrt{1/a_{\text{ground}}^2 - 1/V^2}$ with $b = \sqrt{1 - 1/M^2}$.

That azimuth is 25.7° at Mach 1.2, 38.8° at 1.3, 48.1° at 1.5, 54.9° at Mach 2 and 58.1° at Mach 3, the same from any height above the tropopause. As the Mach number grows without limit it tends to $\arcsin(a_{\text{flight}}/a_{\text{ground}})$, which is 60.12°; at Mach 50 it is already 60.12°. No ray leaving the cone more than sixty degrees from straight down ever reaches the ground, however fast the aeroplane, because a ray launched that far to the side is already nearly horizontal before it has come down through any warm air at all.

In uniform air the footprint is the cone’s hyperbola, and it runs out sideways without end: every ray below the horizontal lands somewhere. When the warning cannot arrive measured that hyperbola’s vertex for an aeroplane at 11 km at Mach 2 and put it 19.1 km behind; traced through the standard atmosphere the same boom arrives under the track 17.2 km behind, because the ray has steepened on its way down. The traced footprints also end. From 15 km at Mach 1.3 the boom arrives under the track 10.7 km behind the aeroplane and at the carpet’s edge 22.7 km to the side and 18.9 km behind; at Mach 1.6 those are 17.1 km behind, and 34.1 km to the side and 41.5 km behind; at Mach 2, 24.1 km behind, and 40.2 km to the side and 67.9 km behind. At the same sideways distances the uniform-air hyperbola would have the boom 22.6, 46.6 and 74.4 km behind, so refraction also pulls the outer footprint forward, towards the aeroplane, by four to seven kilometres — the lateral rays have bent away from the ground and come down steeper than a straight line would.

The edge ray is also the one that has travelled farthest through the air. At Mach 2 from 17 km the ray under the track covers 20.0 km of path in 64.6 seconds and lands 27.6 km behind the aeroplane; the grazing ray covers 56.2 km in 176.7 seconds and lands 43.1 km to the side and 73.9 km behind. It has come nearly three times as far, most of it through the lowest and densest air, and Every compression becomes a shock in the end is the reminder that distance travelled is what a wave’s nonlinear distortion accumulates against. What that does to the edge’s loudness is a separate calculation, and the last two sections come back to it.

A mile of carpet per thousand feet is a Mach 2 rule #

The rule quoted for carpet width — a mile of width for every thousand feet of height — can now be tested rather than repeated.

From 11 km, 36,100 ft, the rule gives 58.1 km; the traced carpet is 23.4 km wide at Mach 1.2, 39.0 at 1.3, 54.1 at 1.5, 69.1 at Mach 2 and 78.0 at Mach 3. From 15 km, 49,200 ft, the rule gives 79.2 km against 45.4 at Mach 1.3, 80.4 at Mach 2 and 90.9 at Mach 3. From 18.3 km, 60,000 ft, it gives 96.6 km against 50.7 at 1.3, 89.8 at Mach 2 and 101.5 at Mach 3. The widths level off as the edge azimuth approaches its limit: from 15 km the carpet is 95.8 km wide at Mach 5, 97.8 at Mach 10 and 98.4 at Mach 50.

So the rule is a Mach 2 rule. For an aeroplane cruising at Mach 2 near 18 km — which is what a supersonic transport did — it lands within seven per cent, and the ageing essay’s band “some fifty miles across” is right. Applied to a business jet at Mach 1.3, it overstates the carpet by a factor approaching two. Above the tropopause the height dependence is simple, because rays are straight in isothermal air: every extra kilometre of height widens the carpet by $2\tan\varphi_c$, which is 2.84 km at Mach 2. Below the tropopause the rays curve and the dependence is not linear. The ground swept by the boom along a track is therefore mainly a question of Mach number close to the cut-off, and mainly a question of height well above it.

A wind that changes with height moves the cut-off; a steady one does not #

Wind enters in exactly one place. With a wind $u(z)$ along the track, the boom pattern moves over the ground at the aeroplane’s ground speed, and the ray in the track plane turns where the sound speed plus the local wind equals it. Since the Mach number is measured against the air at flight height, the cut-off becomes the largest value of $a + u - u_{\text{flight}}$ in the column, divided by the sound speed at the aeroplane. A wind the same at every height cancels out of that expression entirely. Only the difference between the wind aloft and the wind below counts. A layer of air moving faster than the one beneath it bends sound exactly as a warmer layer does, and it is the same refraction by shear and temperature that The sound that only leaves names as the reason a jet’s noise leaves in a cone its scaling law does not predict — there the shear is the jet’s own, here it is the weather’s.

From 15 km over calm ground, a 40 m/s headwind aloft raises the cut-off from 1.153 to 1.289 and a 40 m/s tailwind lowers it to 1.018. At a tailwind of 45.2 m/s — the difference between the sound speeds at the ground and at the tropopause — the cut-off reaches Mach 1, and above that every supersonic ray under the track lands. From 8 km the calm cut-off is 1.105, a 40 m/s headwind raises it to 1.234, and a tailwind of 32.2 m/s brings it to Mach 1. Winds of forty and fifty metres a second are ordinary near the tropopause in winter.

The operational reading is sharp. Flying just below the cut-off has been proposed and flight-tested as a way to cross land supersonically without a boom reaching the ground, and this figure is what makes it a forecasting problem rather than a piloting one. The Mach number in the cockpit is computed from two pressures at the aeroplane and says nothing about the wind or the temperature in the ten kilometres of air beneath it — the reduction that Three readings, and the one each answer leans on laid out has no term for either. The same flight at Mach 1.15 is silent below with a headwind aloft and audible with a tailwind, and which one it has is known only from the column.

A cold morning lays a wider carpet than a hot afternoon #

The surface temperature moves the edge as well as the cut-off, and in the direction that is least convenient.

From 15 km at Mach 1.3 the carpet is 73.6 km wide over a −20 °C surface, 45.4 km on a standard day and 33.1 km over a 40 °C surface. At Mach 1.6 those are 97.7, 68.2 and 57.4 km, and at Mach 2 111.5, 80.4 and 69.5 km. The cut-off from that height runs from 1.081 on the cold day to 1.202 on the hot one. A cold surface has a slower sound speed, the column’s sound-speed gradient is weaker, the lateral rays bend less, and more of them reach the ground — so the day on which a flight is most likely to be heard is also the day on which it is heard across the widest band.

The figure’s cold end is also where the model runs out. A real winter surface is often colder than the air a few hundred metres above it — an inversion — and there the gradient near the ground reverses, sound speed rises with height for a while, and rays in that layer bend down rather than up. This atmosphere refuses a surface colder than the tropopause outright and has no inversion in it at all, which is the first thing a cut-off forecast for a real morning would have to add.

Two routes to one ray #

Every width and cut-off above rests on a quadrature of the conserved slownesses, and the quadrature has to cope with the one awkward place in the problem: at a turning point or a grazing arrival the vertical slowness goes to zero and the integrands for distance and time blow up. They blow up only as the inverse square root of the height above that point, and substituting the square of a new variable for that height removes the singularity exactly, so a plain midpoint rule then converges as the square of its step. That is a claim about a method, and it deserves a test that shares none of its algebra.

The test is to march the same rays forward in time through Hamilton’s equations for sound in a stratified medium, using the sound-speed gradient at each step and never the conserved quantities’ closed form. The landing distance, sideways reach and travel time agree to 5.4 × 10⁻⁶ for a ray at Mach 1.5 from 15 km launched 40° off the track, to 6.0 × 10⁻⁶ under the track at Mach 2 from 17 km, to 2.3 × 10⁻⁵ at Mach 1.25 from 11 km and 17°, and to 1.3 × 10⁻⁸ at Mach 2.5 from 9 km and 57°. The turning ray at Mach 1.1 from 11 km regains flight height at 65,207.9 m by quadrature and 65,210.8 m by the march, 4.3 × 10⁻⁵ apart — the loosest of the five, because the march must detect its return to flight height between two steps. In uniform air the quadrature reproduces the straight cone to 5 × 10⁻¹⁶, and the carpet edge at Mach 2 from 15 km moves by 2.5 × 10⁻⁸ when the quadrature points are doubled.

The remaining claims are checked the same way. The calm cut-off equals the ratio of the two sound speeds to 10⁻¹² at 8, 11 and 17 km, and a ray a tenth of a per cent above it lands while one a tenth of a per cent below it turns above the ground. A wind equal at every height moves the cut-off by less than 10⁻¹², and a tailwind aloft moves it by exactly the wind divided by the sound speed at the aeroplane. There is no carpet below the cut-off, and above it the width grows at every Mach number sampled from 1.2 to 3. And the calculation refuses what it has no answer for: a subsonic aeroplane, a ray launched horizontally, a flight above the 20 km column, a flight at the ground, and an inversion.

What the rays cannot show #

How loud any of it is. A ray says where energy goes, not how much arrives. The overpressure at a point on the carpet needs the spreading of neighbouring rays, the change of air density along the path, the ageing of the signature along a curved path of the length computed above, absorption, and the doubling a hard ground gives a wave arriving on it. None of that is here. In particular the path lengths show that the edge ray travels nearly three times as far, but whether the edge is quieter because of that distance or because of how the rays spread is not something the geometry alone decides.

The shadow is not silent. Geometrical acoustics draws a sharp boundary at the grazing ray and below the turning height, and the real field does not have one. Low frequencies diffract into the shadow and the pressure there decays rather than vanishing, which is the rounded rumble the ageing essay described at the carpet’s edge. The rays can say where the shadow begins; they cannot say how quiet it is.

The day is an idealisation. The atmosphere here has one lapse rate, no inversion, no humidity, and a wind only along the track and only linear in height. A crosswind shifts and skews the carpet sideways, and a real sounding has layers that this profile smooths away. The ground is flat: over the 40 km out to the edge of a Mach 2 carpet the Earth’s surface falls about 130 m below a flat plane, which is small against a 15 km flight height but not small against a ray that is grazing. And the aeroplane flies straight and level at constant speed, which rules out the case the ageing essay named third.

What it does show without idealisation is the shape of the answer: a threshold set by a ratio of sound speeds, an edge set by the same inequality applied off the track, and a sensitivity to wind shear and surface temperature large enough that the question “will this flight be heard” is a question about the weather.

Still open: where an accelerating aeroplane focuses its boom #

Level flight at constant speed launches every ray with the same invariant, so neighbouring rays run parallel and never cross. An aeroplane accelerating through the transonic launches each successive ray at a slightly higher Mach number, and those rays converge: their envelope is a caustic, where geometrical acoustics predicts an infinite pressure and the real signature is several times the carpet’s. The next calculation traces rays from an accelerating aeroplane with this same invariant, finds where on the ground the caustic lands for a given acceleration and height, and asks how that line moves with the cut-off — since an aeroplane that accelerates while still below its cut-off Mach number can focus a boom that turns back before it lands. A turning aeroplane does the same thing sideways.

Beside it is the amplitude across the carpet: the ageing of the signature that forgets the shape carried along each refracted path, combined with the spreading of the rays, to say how the overpressure falls from the track to the edge. The quiet-supersonic design that the least-drag body is related to aims at a slower rise under the track; whether the same shaping helps at the edge, where the path is three times as long, is the question that calculation would answer.