A cone finishes its turn after the shock

A shock that leans turned a supersonic stream with a wedge and found the whole answer in one line: tilt a normal shock, keep the velocity along it, and every ratio across it is the normal-shock ratio at $M\sin\beta$. When the wedge is too blunt followed that relation to its maximum, 22.97° at Mach 2, past which no attached shock exists. That essay also said, in a section of its own, that a cone of the same half-angle stays attached to nearly twice the angle, explained why in words — the streamlines can spread round the axis, so only part of the turn has to happen at the shock — and then said that nothing on its page computed it.

This essay computes it. The equation is Taylor and Maccoll’s of 1933, and solving it turns the explanation into numbers that are larger than the words suggested. The share of a cone’s turn that its shock makes is not a little less than all of it: at 10° it is fifteen per cent. The surface pressure is not somewhat below a wedge’s: it is under half. And the thing that decides both is a compression that happens after the shock, in a region the wedge does not have.

Why a wedge’s flow is uniform and a cone’s is not #

Behind a wedge’s shock the flow is parallel to the surface, and in two dimensions there is nothing to change that: every streamline crosses the same shock at the same angle, turns through the same deflection, and runs on parallel to its neighbours and to the wall. The whole turn is made at the shock, the flow behind is uniform, and the answer is an algebraic relation between three numbers.

Behind a cone’s shock the flow is heading towards the surface at an angle, and it cannot simply run parallel to it, because a streamline moving inward round a cone is moving into a region of smaller circumference. The cross-section available to the flow between two neighbouring streamlines changes as they travel aft, and so their speed and pressure change too. The flow behind a conical shock cannot be uniform, and there is no jump relation that gives it.

What it does have is a symmetry that nearly makes up for the loss. Nothing in the problem has a length: the cone is infinite, its shock is straight, and the free stream is uniform. So the flow at a point can depend only on its direction from the apex, not on its distance. Every ray from the apex is a line of constant state, and the whole field is a function of one angle.

One equation in one angle #

With every property a function of the polar angle $\theta$ alone, the steady Euler equations reduce to one ordinary differential equation for the radial velocity. Written in the velocity made dimensionless on the flow’s limiting speed, $\sqrt{2h_0}$, with $V_\theta = \mathrm{d}V_r/\mathrm{d}\theta$ because the flow is irrotational,

$$V_r'' = \frac{\tfrac{\gamma-1}{2}\,(1 - V_r^2 - V_\theta^2)\,(2V_r + V_\theta\cot\theta) - V_r V_\theta^2} {V_\theta^2 - \tfrac{\gamma-1}{2}\,(1 - V_r^2 - V_\theta^2)}.$$

The irrotationality is not an approximation. A straight shock gives every streamline the same entropy rise, so the flow behind it has one entropy everywhere and Crocco’s relation leaves it no vorticity; a curved bow shock would not.

The equation is solved backwards. Choose a shock angle $\beta$; the oblique-shock relation gives the velocity just behind it, with its radial and polar components; integrate inward from $\theta = \beta$ with fourth-order Runge–Kutta until the polar component $V_\theta$ reaches zero. Where it does, the flow is moving along a ray, which is the statement that it is tangent to a cone of that half-angle. Asking for a particular cone then means searching for the shock angle whose integration stops at it, and the search here is refined between rows of a table built once per Mach number, every integration at the same 12,000 steps.

The profile has a shape worth reading. The change is gentle just behind the shock and steepens towards the surface, because the flow nearest the shock has barely started to turn and the flow nearest the surface must finish turning to follow it. The whole of that change is isentropic: the total pressure the shock left, 99.0 per cent of the free stream’s here, is carried unchanged from the shock to the surface. The pressure on the cone is 1.912 free-stream pressures where the shock alone produced 1.586 — just over a third of the cone’s surface pressure rise above the free stream is made after the shock rather than at it.

The share of the turn a shock makes #

The number the earlier essay’s argument turns on is how much of the turn happens at the shock, and the integration gives it directly: the deflection the shock gives the flow crossing it, as a share of the cone’s half-angle.

At Mach 2 a 5° cone’s shock turns the flow through 0.12° — two per cent of the cone’s angle. A 10° cone’s shock turns it through 1.48°, fifteen per cent; a 15° cone’s, 4.58°, thirty per cent; a 20° cone’s, 8.57°, forty-three per cent. For a slender cone the shock is nearly a Mach wave, and almost the whole turn is an isentropic compression behind it.

The share rises with Mach number at a fixed half-angle, and the reason is that the region between shock and surface narrows. At Mach 5 a 20° cone’s shock lies close to its surface, the flow has little room to go on turning, and the shock must do 78 per cent of the work — the same thinning of the region behind a shock that makes the shock layer on a blunt body stop depending on speed at high Mach number. As the Mach number rises further the cone’s behaviour approaches a wedge’s for exactly this reason — the thin shock layer that makes Newton’s impact theory nearly right at high speed is the same thin layer that leaves no room for the compression behind the shock.

This is where the earlier essay’s phrase “three-dimensional relief” becomes something with a size. Relief is not a small correction to the wedge; for the cones a supersonic missile or an intake centrebody actually uses, it is most of the turn.

Two shock angles, two detachment angles #

The weaker work the cone asks of its shock shows first in where the shock sits.

Both curves leave the Mach angle, 30°, together — a body of vanishing angle makes a Mach wave either way — and part at once. A 10° wedge needs a shock at 39.31° to turn the flow through its whole ten degrees; a 10° cone’s shock lies at 31.21°, barely off the Mach angle, because it is turning the flow through a degree and a half.

The cone’s curve runs on long after the wedge’s has stopped. The wedge’s shock detaches at a half-angle of 22.974°; the cone’s at 40.688°, when its shock stands at 69.42°. That confirms the earlier essay’s “about 41°” and gives the figure it lacked a source. The cone’s detachment is not a weaker version of the wedge’s: it is the same event — the equation for the shock angle has no attached root — reached much later, because a cone of any angle up to that point asks its shock for less than a wedge of 23° does.

Subsonic at the surface, and still attached #

The shock angle climbs steeply as a cone approaches its detachment angle, and that climb changes what the shock does in two ways a wedge’s does not.

The first is visible in the share of the turn. At Mach 2 the share rises to a maximum of 59.1 per cent at a half-angle of 36.6°, then falls: just short of detachment, at 40.6°, the shock is making only 55.9 per cent of the turn. The same happens at Mach 3, where the share peaks at 74.7 per cent near 39.4° and falls to 67.8 per cent, and at Mach 5, peaking at 83.5 per cent near 36.8° and falling to 73.9. Close to detachment a small increase in half-angle needs a large increase in shock angle, and a shock that has swung round towards the normal is deflecting the flow through less, not more, for each degree it moves.

The second is a band the earlier essay found only a sliver of. For a wedge at Mach 2, detachment is at 22.97° and the flow behind the shock goes sonic at about 22.8°, so between them lies a fifth of a degree of attached shocks with subsonic flow behind. For a cone at Mach 2 the surface goes sonic at a half-angle of 36.60° and the shock detaches at 40.69°: a band of more than four degrees in which the shock is attached and straight and the flow reaching the cone’s surface is subsonic. At 36.60° the flow just behind the shock is still at Mach 1.106 and slows to exactly one at the surface; at 40.64° it is subsonic immediately behind the shock, at Mach 0.845, and reaches the surface at 0.731.

That band is the isentropic compression doing more than finishing the turn. Behind a fat cone’s shock it slows the flow through the speed of sound without a second shock, which is something no compression behind a wedge can do, because behind a wedge there is no compression at all. At Mach 2 the share of the turn peaks at 36.55°, within a twentieth of a degree of the half-angle where the surface goes sonic; whether the two coincide at other Mach numbers is not something this calculation has checked.

The band is a result of the conical solution, which assumes the shock stays straight and attached right up to the detachment angle. In a real flow a cone in it has a subsonic region near its surface that can feel conditions downstream — the cone’s base, a support, the flow round a body behind it — and the attached conical flow is then fragile in a way the wedge’s supersonic flow is not.

Less than half the pressure, and a check from outside #

What a body’s shape does to its drag and its load is decided by the pressure on its surface, and here the difference is large.

A 10° cone carries a surface pressure coefficient of 0.1045 at Mach 2 against a wedge’s 0.252: 41 per cent of it. At 20° it is 0.3256 against 0.6582, forty-nine per cent. A missile’s nose, an intake’s spike or a projectile’s ogive is designed against the first of these, and taking it from the second would roughly double the estimated wave drag of the forebody.

A calculation this size needs a check that does not go through its own arithmetic, and one exists. For a slender cone, linearised supersonic theory gives the surface pressure coefficient in closed form,

$$C_p = \theta_c^2\left[2\ln\frac{2}{\sqrt{M^2-1}\,\theta_c} - 1\right],$$

from a line of sources along the axis, with no shock and no integration — the same linearised theory that finds the body of least wave drag for a given volume. It should be wrong for a fat cone and right for a thin one, and the integration agrees with it in exactly that way: 15.80 per cent apart at 8°, 9.63 per cent at 6°, 4.83 per cent at 4°, and 3.86 per cent at 3.5°. The gap closes monotonically as the cone thins, which is what both methods say should happen and is the evidence that the integration is right where an independent answer exists.

The comparison also marks where the integration stops being trustworthy, and it is at the thin end rather than the fat one. A shock within a few hundred-thousandths of a radian of the Mach angle is almost a Mach wave, and the flow just behind it begins on the equation’s singular line, where the velocity normal to the ray equals the local speed of sound. Started there, the half-angle the integration arrives at depends on the size of its steps rather than on the flow. So a shock that close is not integrated, every cone quoted here is one reached at a resolution where doubling the steps changes nothing in the sixth decimal place, and a cone thinner than that is refused rather than approximated by the thinnest one available.

The total pressure a cone keeps #

A weaker shock destroys less, and the cone’s shock is weaker at every angle.

At 15° a cone’s shock leaves 99.84 per cent of the free stream’s total pressure and a wedge’s 95.24 per cent; at 20°, 99.01 against 89.29. The difference is the cube law for a weak shock’s entropy at work: a shock that makes a fraction of the turn costs far less than that fraction of the loss, and the isentropic compression that finishes the turn costs nothing.

It is the same argument the accounting of a shock’s cost makes for several weak shocks against one strong one, arriving by a different route. A multi-ramp intake chooses to split a turn into steps; a cone has the split imposed on it by its geometry, with the second step replaced by a continuous, lossless compression. A conical intake spike is, in this sense, a continuously graded compression ramp, and it is why the centrebodies of axisymmetric supersonic intakes are cones rather than wedges of revolution.

Every Mach number #

The comparison at Mach 2 generalises, and the table of detachment angles is worth having whole.

At low supersonic speed the difference is proportionally enormous: at Mach 1.15 a wedge detaches at 2.7° and a cone at 16.8°, more than six times as blunt. At Mach 1.4 it is 9.4° against 27.5°. The ratio falls as the speed rises — 1.77 at Mach 2, 1.45 at Mach 3, 1.28 at Mach 10 — because the shock layer thins and the cone’s relief shrinks, but the cone’s allowance is larger at every Mach number computed.

That low-speed end is the one with a practical sting. An aircraft flying just above Mach one with a wing leading edge or an intake lip of a few degrees is on the wedge’s curve, where the allowance is tiny, while a fuselage nose or a probe of the same angle is on the cone’s, where it is large. The same angle is attached on the body and detached on the wing, and the two parts of one aircraft meet the transonic difficulty at different speeds.

What the calculation assumes #

The cone is infinite and sharp. A real nose has a finite length and a tip radius, and a blunted tip carries a small detached shock whose curvature puts entropy gradients and vorticity into a layer near the surface. Far from the tip the conical solution is recovered; near it, it is not.

No incidence. A cone at an angle of attack has a flow that is still conical but no longer axisymmetric, and the windward and leeward rays carry different states. The equation here is the zero-incidence case only.

No boundary layer. The surface pressure is inviscid. A boundary layer thickens the effective cone slightly, and on a slender cone at high Mach number — where the shock layer is thin — that thickening is not negligible.

A perfect gas. $\gamma = 1.4$ throughout. At the Mach numbers where the cone’s shock layer is thin enough to approach a wedge’s, the gas behind a strong shock is hot enough for $\gamma$ to fall, which moves the shock closer to the surface still.

Shocks close to a Mach wave are refused. The integration cannot resolve a shock within a hundred-thousandth of a radian of the Mach angle, so the thinnest cone this essay computes is 3.27° at Mach 2 and 1.78° at Mach 5, and every thinner cone is refused rather than extrapolated. Slender-body theory, which is exact in that limit, covers it.

Taylor and Maccoll, 1933 #

G. I. Taylor and J. W. Maccoll derived the equation and integrated it numerically in 1933, for the pressure on a projectile’s conical head, at a time when integrating an ordinary differential equation meant hand computation or a mechanical calculator. Their tables of shock angle and surface pressure against cone angle and Mach number were used for decades, and their method — reduce the problem to one variable by a similarity argument, then integrate — is the pattern that the blast wave and many later self-similar flows followed.

The problem’s other name for itself is conical flow, and it is the first case in supersonic aerodynamics where three-dimensional geometry changed an answer qualitatively rather than by a correction. Kopal’s extensive tables for cones at incidence followed in the 1940s at MIT, from the same equation with the symmetry broken.

Still open: a cone at incidence, and a cone in an intake #

The equation here has one angle in it because the cone meets the stream head on. Tilt it and the flow is still conical — nothing has a length — but it depends on two angles, the polar angle and the angle round the axis, and the windward side carries a stronger shock than the leeward. Computing that field, and in particular the incidence at which the leeward flow separates and rolls into a pair of vortices above the cone, is the next calculation: it is the conical version of the wing that makes most of its lift from a vortex, and the point at which an inviscid similarity solution starts to need a viscous event it cannot compute.

Beside it is the cone doing a job: the centrebody of an axisymmetric intake, where the conical shock must fall exactly on the cowl lip at the design Mach number and the isentropic compression behind it is part of what the intake is for. How much of such an intake’s total-pressure recovery the continuous compression buys over a single oblique shock, at the Mach numbers intakes are designed for, is the practical form of the argument above.