A hypersonic body leaves a line explosion behind it

The fireball is hollow integrates Sedov’s blast wave and finds that an explosion piles nearly all its gas into a thin shell behind the shock, leaving a hot, nearly empty core. It ends on an analogy that aerodynamicists have used since the 1950s. A blunt body flying through still air at hypersonic speed does work on the air at a rate equal to its drag. Each slice of air, a thin disc across the flight path, is struck once as the body passes and then left to itself. Seen from the ground, the slice has received an impulse and a quantity of energy along a line, and it expands as a cylindrical explosion would, with the drag per unit length as its energy and the distance behind the nose, divided by the flight speed, as its time.

This essay takes the analogy at its word. It reads the bow shock’s width and the pressure on the body’s sides off the cylindrical blast wave, compares the shock with a measured one, and asks the analogy’s own solution where it must stop being right.

The body as a line of charges #

The energy argument is exact, and it is short. In the frame of the still air, a body moving at speed $U$ against a drag $D$ does work $DU$ per second, and it covers a length $U$ in that second. So it leaves energy $D$ in every unit length of its path. For a body of diameter $d$ with drag coefficient $C_D$ on its frontal area,

$$E = D = C_D\cdot\tfrac12\rho U^2\cdot\frac{\pi d^2}{4}\quad\text{per unit length.}$$

The analogy then makes two approximations. It treats each slice as independent of its neighbours, which is the hypersonic small-disturbance idea that a slender flow at very high speed is a sequence of two-dimensional unsteady flows in the planes across it. And it treats the energy as released instantly on the axis, as though the body were a line of charges fired in sequence. Sedov’s cylindrical solution then gives the shock radius at a time $t = x/U$ after the nose passes as $R = \xi_0\,(Et^2/\rho)^{1/4}$, which is

$$\frac{R}{d} = \xi_0\left(\frac{\pi C_D}{8}\right)^{1/4}\left(\frac{x}{d}\right)^{1/2}.$$

The flight speed has cancelled. The bow shock behind a blunt nose has a width set by the drag and not by the Mach number. That is the hypersonic independence principle arriving by a different route, the one a shock lying on the body reaches for the standoff distance at the nose: above a Mach number of five or so, the flow’s shape stops depending on how fast it is going.

The constant $\xi_0$ is the one a radius that gives the energy away corrected for the spherical case. Integrated here for the cylindrical case at $\gamma = 1.4$ it is 1.00402, against the 1.004 that is usually quoted, and the analogy’s coefficient $\xi_0(\pi/8)^{1/4}$ comes out as 0.7948 — the 0.795 of the textbooks, derived rather than copied.

The shock against a measured one #

The first figure puts the analogy against a measured shock. Frederick Billig’s correlation, published in 1967, fits the bow shocks on spheres with three pieces: the standoff distance at the nose, $\Delta/R = 0.143\,e^{3.24/M^2}$; the shock’s radius of curvature at its vertex, $R_c/R = 1.143\,e^{0.54/(M-1)^{1.2}}$; and a hyperbola from that vertex to the Mach cone. It is a fit to measurement, anchored at the nose, and the hyperbola is what carries it downstream. For the analogy the drag coefficient is the modified-Newtonian value for a hemisphere, 0.919 at Mach 20, from the theory Newton would have recognised.

The two curves are parallel on logarithmic axes from a third of a diameter to twenty: both are parabolas, radius growing as the square root of distance. At Mach 20 the analogy’s shock is 0.708 of Billig’s at two diameters, 0.713 at five, 0.712 at ten and 0.706 at twenty — constant to within one per cent. The constant is not mysterious. Near its vertex Billig’s shock is the parabola $y^2 = 2R_c x$ set by the nose’s own curvature, and the ratio of the two parabolas’ coefficients is 0.722. The analogy knows the nose only through its drag; Billig’s correlation knows it through its shape. Neither can say which is right twenty diameters back without a computed flow, because the correlation’s hyperbola is an extrapolation there and the analogy’s line explosion is an idealisation everywhere. What the comparison does settle is the exponent, and the near-absence of the Mach number from it: Billig’s curves for Mach 10 and 20 lie within a few per cent of each other over the first ten diameters, and only the Mach 5 curve departs, as its Mach cone opens wide enough to matter.

One slice, from the inside #

The second figure is one slice seen from the inside. It is the cylindrical version of what the earlier essay drew for a sphere: the gas piled into a thin shell against the shock, with half of it in the outer 6 per cent of the radius, and a core that is nearly empty. The pressure does not follow the density down. It falls from its post-shock value to 37 per cent of it and then stays level across the whole core, because the core’s gas is too light to need a pressure gradient to hold it.

Then the analogy has to put the body back, and the figure marks where the body’s surface falls: at 29 per cent of the shock radius five diameters behind the nose, at 14 per cent twenty diameters back. Both are deep in the core. So the body’s sides are bathed in the blast’s hollow — at a pressure that is the core’s nearly uniform pressure, and at a density that is a small fraction of anything the shock produced. That is the second half of the analogy, and it is its most useful: the pressure on a hypersonic afterbody is the blast’s core pressure.

The pressure on the afterbody #

The core pressure is $p_c = \pi_c\,\rho\dot R^2$, with $\pi_c = 0.3108$ the cylindrical solution’s central value and $\dot R$ the shock’s speed outwards. With $\dot R = UR/2x$ and the radius above,

$$\frac{p_c}{p_\infty} = \gamma M^2\cdot\frac{\pi_c\,\xi_0^2}{4}\sqrt{\frac{\pi C_D}{8}}\cdot\frac{d}{x} = 0.0687\,\frac{M^2\sqrt{C_D}}{x/d}.$$

The constant 0.0687 is, again, a number the literature quotes — as 0.067 — and it follows here from $\pi_c$ and $\xi_0$, which the blast integration computes. The Mach number has come back, but only because the pressure is referred to the ambient one: as a fraction of the dynamic pressure it is independent of speed like the shock’s radius.

The third figure draws it. The pressure on the cylinder falls as one over the distance behind the nose, from several times the ambient pressure a diameter or two back to the ambient value at 6.6 diameters at Mach 10 and 26 at Mach 20, and below it thereafter. That last part cannot be right: a body’s sides do not sit in a partial vacuum behind a bow shock in still air. The analogy has neglected the ambient pressure, as every strong-blast solution does, and the prediction has passed the point where neglecting it is harmless.

The analogy’s own limit #

A strong-blast solution has a strong shock by assumption, and the analogy can be asked how strong its own shock is. The shock moves outwards at $\dot R$, so its Mach number relative to the still air is $M\,dR/dx$, which the square-root law makes fall as one over the square root of the distance.

The fourth figure follows it. At Mach 5 the analogy’s shock is already below Mach 2 one diameter behind the nose — the analogy is not a statement about Mach 5 flight at all. At Mach 10 it falls to three at 1.7 diameters, at Mach 20 at 6.7, at Mach 40 at 27. The reach grows as the square of the flight Mach number, and as the square root of the drag coefficient, and that is the whole of the analogy’s domain: from the nose, where the flow is a detached shock over a body and nothing like a line explosion, to a few diameters or a few tens back, where the shock has weakened and the ambient pressure has taken over. The pressure’s failure in the third figure comes later than the shock’s, which is the usual order: a solution first becomes inaccurate and only later absurd.

Beyond the reach, the flow does what a weak shock does. The shock bends back towards the Mach cone, as Billig’s hyperbola eventually does, and far enough downstream the disturbance is a weak cylindrical wave spreading from a line — the far field that a sonic boom’s signature settles into, in which the body is remembered only through a few numbers.

What a flat nose buys #

The analogy knows the nose only through its drag coefficient, and it says exactly how much the drag is worth: the shock’s width goes as its quarter power and the afterbody pressure as its square root. A flat-faced cylinder at Mach 20 has, in modified-Newtonian theory, the full stagnation pressure coefficient over its whole face, a drag coefficient of 1.84 against the hemisphere’s 0.919. Doubling the drag widens the shock behind the body by $2^{1/4}$, 19 per cent, and raises the pressure on its sides by $\sqrt{2}$, 41 per cent. Near the nose the two bodies look very different — the flat face’s shock stands off far further than the thin layer over a sphere, and the face’s pressure is the part of the flow that Newton’s impact theory is built on — and a few diameters back the analogy says the difference has been reduced to those two powers of one number.

That is the analogy’s practical content for the designers who used it. A blunt re-entry capsule is blunt on purpose: a large drag slows it high in the atmosphere, and a detached shock keeps the hottest gas away from its surface. The analogy adds that the price of that drag is paid again behind the capsule, in a wider disturbance and a higher pressure on anything that follows it through the air — a trailing parachute, a second body, or the capsule’s own afterbody — in proportion to the square root of what it bought in front.

How much room the body takes #

The analogy puts all the energy on a line and then asks where the body is. It is consistent only if the body is small against the blast it sits in, and the second figure’s rules say how small. At Mach 20 the body’s radius is 29 per cent of the shock’s five diameters behind the nose and 14 per cent twenty diameters back, so its cross-section takes 8 per cent and then 2 per cent of the disturbed area. Far enough back, the body is a thin stick in a wide blast and neglecting its volume costs little. Near the nose it is not: one diameter back the body’s radius is two-thirds of the shock’s, the flow is a shock layer wrapped round a solid, and nothing in the line explosion describes it.

That fixes the other end of the analogy’s domain. It needs the body small against the shock, which holds from a few diameters back, and it needs its own shock to be strong, which holds up to the reach in the fourth figure. At Mach 20 those two conditions overlap between about three diameters and seven, and at Mach 40 between three and twenty-seven. The window is real only for very fast flight — which is where the analogy was invented, for re-entry, and where Newtonian ideas about hypersonic flow are also at their best.

A core hotter than anything that flowed #

Sedov’s core is hot, and in the analogy the body’s surface lies in it. The temperature in the blast rises towards the centre without limit, because the gas there was struck when the shock was strongest and has expanded ever since with the entropy it was given. Reading it at the body’s surface, at Mach 20, gives 580 times the post-shock temperature five diameters back, 3,300 times at ten and 18,600 at twenty.

No gas behind a bow shock can be hotter than the stagnation temperature, and the post-shock temperature at Mach 20 is already most of that: the entropy a shock creates is what a shock costs, and it cannot be spent again as heat. So the analogy’s core is not a flow the body could have made. What it caricatures is real: the gas that passed through the nearly normal part of the bow shock, close to the axis, carries a far higher entropy than gas that crossed the shock further out, and it wraps the body in a layer that is hotter and much less dense than the flow outside it. That is the entropy layer, and the analogy predicts it qualitatively — hot, light gas next to the body, carrying the nose’s history downstream. It overstates its temperature without bound because it has put all the energy on an infinitely thin line, where the real nose spreads it over its own diameter.

A meteor and a lightning stroke #

The same solution describes two things that have nothing to do with aircraft. A meteor entering the atmosphere at tens of kilometres per second is a hypersonic body with a drag and a path, and the pressure wave it leaves behind — heard on the ground, and recorded by infrasound arrays — is a cylindrical blast with the meteor’s drag per unit length as its energy. A lightning stroke deposits its energy along its channel in microseconds, and the channel expands as a line explosion whose weakening shock, far out, becomes the sound of thunder. Neither is moving through the air in the way a re-entry vehicle is, but both are energy released along a line, and Sedov’s cylindrical solution is the first thing each is compared with.

What was checked #

The fifth figure lists the checks. The cylindrical constant, integrated from Sedov’s equations without being told the answer, reproduces the quoted 1.004 to two parts in $10^5$; the analogy’s two coefficients reproduce the quoted 0.795 and 0.067, which is the check that the analogy has been assembled correctly rather than that it is right. Billig’s hyperbola was checked against its own vertex curvature and its own Mach-cone asymptote, so that the comparison is against the correlation as published. And the analogy against Billig’s shock was checked for the one thing the two share, the square-root growth, whose ratio stays within one per cent of 0.71 between two and twenty diameters.

What the picture cannot show #

A computed bow shock. The comparison is with a correlation of measurements anchored at the nose and extrapolated as a hyperbola, not with a solution of the Euler equations round the body. Which of the two is closer twenty diameters back is not settled here.

The ambient pressure. The analogy is a strong-blast solution, and the ambient pressure enters only through the check on its own shock’s strength. Sakurai’s correction for counter-pressure, which extends the blast to weaker shocks, is not computed.

The nose’s shape. The analogy knows the nose through one number, its drag coefficient. Two noses with the same drag and different shapes have the same shock in the analogy and different shocks near the body.

A perfect gas. At the speeds where the analogy’s reach is long, the gas behind the nose is dissociating, and the ratio of specific heats is not 1.4.

The convention the numbers depend on #

$x$ is measured downstream from the nose and $d$ is the body’s diameter. The drag coefficient is based on the frontal area and taken, for the hemisphere, from the modified-Newtonian pressure coefficient, $C_D = C_{p,\max}/2$, which is 0.904, 0.916 and 0.919 at Mach 5, 10 and 20. “No longer strong” is a shock Mach number of three, where the density ratio across it is 3.9 against the strong-shock limit of six. Temperatures in the core are ratios to the temperature just behind the blast’s shock.

Who found it, and when #

The idea that a slender body’s hypersonic flow is a sequence of unsteady two-dimensional flows is Wallace Hayes’s equivalence principle of 1947. Shao-Chi Lin solved the cylindrical blast wave in 1954 with this application in mind, and Harold Cheng and Adrian Pallone, and Lester Lees and Toshi Kubota, turned it into the analogy for blunt-nosed bodies in 1956 and 1957, including the pressure law used here. Billig’s correlation of shock shapes dates from 1967. Taylor and Sedov had solved the point explosion in the 1940s, for a different purpose.

Still open: the entropy layer’s real temperature #

The analogy places the body in a core whose temperature has no bound, and the real entropy layer has one. The next calculation replaces the line of charges with the nose itself: it follows each streamline from where it crosses the curved bow shock, with the entropy that crossing gives it, and expands it to the afterbody pressure the analogy predicts. That gives the entropy layer’s temperature and thickness along the body — a computation the analogy cannot do and the nose’s shock can — and it says whether the layer, swallowed downstream by the growing boundary layer, is thick enough at the afterbody to change the heating a hypersonic vehicle’s sides receive.