A cone intake's shock sees the whole capture

A cone finishes its turn after the shock solved Taylor and Maccoll’s equation for supersonic flow past a cone and found that only part of the cone’s turn is made at its shock. The rest is an isentropic compression in the space between the shock and the surface, which costs nothing, so a cone’s shock is much weaker than a wedge’s of the same angle and keeps far more of the free stream’s total pressure: at Mach 2 and fifteen degrees, 99.84 per cent against 95.24. That essay drew the conclusion that a conical intake spike is a continuously graded compression ramp, and named the question it left: how much of an intake’s total-pressure recovery the continuous compression actually buys.

The question is not answered by the shock alone, because an intake does not end at the shock. A supersonic intake’s job is to deliver air to the engine subsonic, and a spike cannot do it smoothly; a terminal normal shock at the cowl takes the flow the rest of the way, and at Mach 2 it is that shock, not the cone’s, that costs most. It acts on whatever Mach number the cone delivers. Behind a wedge that is one number. Behind a cone it is not: the flow keeps compressing between shock and surface, so across the captured annulus the Mach number varies from its value just behind the shock to its value at the surface, and the terminal shock loses a different amount on every ray.

The intake, and what it averages #

The model is the simplest supersonic intake that shows the effect. A cone of half-angle $\theta_c$ points into a stream at Mach $M$. Its conical shock stands at the angle $\beta$ Taylor–Maccoll gives, and the cowl lip is placed on the shock — the design condition for which the whole stream tube of the cowl’s radius is captured and nothing spills. At the cowl’s axial station every ray of the conical field, from the shock at $\beta$ to the surface at $\theta_c$, crosses the plane of the lip. A normal shock stands across that plane, and on each ray it drops the flow from the ray’s local Mach number to subsonic.

The recovery of the intake is the total pressure delivered, averaged over the captured air by mass flux, since what the engine receives is so many kilograms a second at each total pressure. On a ray at polar angle $\theta$ crossing the plane at radius $r = x\tan\theta$, the mass flux through a ring is $\rho u_x\, 2\pi r\,dr$, with the density and the axial velocity taken from the conical solution. The recovery is the conical shock’s own ratio times the mass-weighted mean of the normal shock’s.

One check comes free and is exact. Every kilogram of free stream inside the cowl’s radius has crossed the conical shock and must cross the plane of the lip between cone and shock, so the integrated mass flux there has to equal the free stream’s through a disc of the cowl’s radius. At four cones and three Mach numbers it does, to two parts in $10^9$ — a test of the Taylor–Maccoll profile’s density and velocity everywhere between shock and surface, and not only at its ends.

The wedge intake it is compared with is the two-dimensional version: one oblique shock at the wedge’s angle, uniform flow behind it, and a normal shock at the cowl on that one Mach number. Neither intake has boundary layers, bleed or a subsonic diffuser’s losses; the comparison is between the two shock systems and nothing else.

A spread of Mach numbers at the cowl #

The first figure is the flow the terminal shock meets at Mach 2. For every cone the Mach number is highest at the outer edge of the annulus, just behind the conical shock, and falls towards the surface, where the continuous compression has had longest to act. For the 28° cone — which turns out to be the best — it runs from 1.43 behind the shock to 1.32 at the surface. The best wedge, at 16.4°, delivers a uniform 1.39.

That picture already suggests why a comparison at the surface misleads. A normal shock’s loss rises steeply with its Mach number, so the terminal shock loses more on the rays by the shock than on the rays by the surface, and the question of how much the cone’s compression helps becomes a question of where the captured air is.

Most of the air arrives where the Mach number is highest #

The second figure answers it. The annulus between cone and shock at the cowl is a ring whose area grows with radius, and the flow is densest and fastest in the axial direction near the shock. So the mass flux per unit radius is largest at the outer edge — just where the Mach number is highest and the terminal shock loses most — and small near the surface, where the flow the cone has slowed furthest is a thin ring.

The ray-by-ray recovery runs from 0.926 at the surface to 0.902 at the shock. The mass-weighted average is 0.916, much nearer the shock’s end than the surface’s. The surface ray alone would say 0.926.

The geometry behind that is simple enough to state in numbers. At the cowl station of the best Mach-2 cone the annulus runs from 52 per cent of the cowl’s radius, at the cone’s surface, out to the lip; the shock stands at 45.7° from the axis. The outer half of that radial span carries 58 per cent of the captured mass, because an annulus near the lip is simply a larger ring, and the flow there is the least compressed in the field.

This is the part of the cone’s advantage that the argument for splitting a compression into weak pieces does not reach. The isentropic compression between shock and surface is real and lossless, and it acts most on the air that matters least, because it is strongest near the surface and the surface is where there is least air to act on.

The cone wins only at its own, steeper, angle #

The third figure sweeps the angle. At a given half-angle the wedge recovers more than the cone: at Mach 2 and 16.4°, the wedge’s best angle, the wedge recovers 0.903 and a cone of the same half-angle 0.842. The reason is the one the earlier essay found turned round. A wedge makes its whole turn at the shock, so at a given angle it compresses the flow further and hands the terminal shock a lower Mach number. The cone’s gentler shock is gentler because it turns the flow less, and an intake needs the turn.

The cone wins at its own best angle, which is steeper: 28.0° at Mach 2 against the wedge’s 16.4°, and 30.2° at Mach 3 against 22.6°. Both are well short of detachment — a cone at Mach 2 can carry an attached shock to about 40°, a wedge to 23°, the limit when the wedge is too blunt found — so the best intake is not pressed against the edge of the attached solutions, as it would be if the object were to turn the flow as far as possible. There its shock is strong enough to slow the flow substantially, the continuous compression slows it further, and the pair beats the wedge — at Mach 2, 0.916 against 0.903.

The gain can be split between the two shocks, and the split is instructive. The best cone’s conical shock keeps 0.950 of the total pressure and the best wedge’s oblique shock 0.939; their terminal shocks keep 0.965 and 0.961 of what reaches them. So nearly all of the cone’s advantage is at its first shock, where it is steeper and stronger than the wedge’s and still loses less, and very little is at the terminal shock, which is where the continuous compression was supposed to pay. The cone does deliver a slightly lower average Mach number to the cowl than the wedge does, and the terminal shock loses correspondingly less, but only by four-tenths of a point. The comparison the earlier essay drew, at equal angles, is the right one for the shock’s own loss and the wrong one for an intake, which chooses each centrebody’s angle for the system and not for the shock.

Why the cone has to be steeper #

The steeper angle is the earlier essay’s share of the turn, working against the intake. The best Mach-2 cone, at 28.0°, turns the flow through only 15.3° at its shock — 55 per cent of its half-angle — and the rest between the shock and the surface. A cone of the wedge’s 16.4° makes only a third of its turn at the shock, 5.6°, leaves the flow at Mach 1.80 behind it and 1.67 at its surface, and hands the terminal shock far more to do than the wedge’s uniform 1.39. To reach the same Mach number at the cowl, a cone needs a larger angle, and a larger angle means a stronger shock.

That is also why the cone’s shock still loses less than the wedge’s at the best angles, which the split above showed. A shock that leans is a normal shock at the component of the velocity across it, and its loss is set by that component alone. The cone’s shock at 45.7° from the axis meets the stream at a normal Mach number of 1.43; the wedge’s at its best angle, 1.47. The cone buys a gentler shock with a steeper body, and the terminal shock is left with nearly the same job either way.

Two shocks, and then the shape #

The fourth figure sets the whole result against flight Mach number, and it puts the two decisions in proportion. Splitting the compression into two shocks is worth a great deal: a bare normal shock at Mach 2 recovers 0.721, the best two-shock intake 0.90 to 0.92, a gain of 18 points; at Mach 3 the bare shock recovers 0.328 and the best two-shock intake 0.58 to 0.60, a gain of 25. That is what a shock costs, paid once in a strong jump or twice in weaker ones, and it is why every supersonic intake above about Mach 1.6 has a compression surface of some kind.

The bare normal shock is not only a hypothetical baseline. A pitot intake — a plain open tube facing the stream — is exactly a normal shock followed by a subsonic diffuser, and it is the intake of the airspeed indicator’s pitot probe, which in supersonic flight reads the total pressure behind a normal shock and has to be corrected for it. The loss that the probe’s reading has to undo is the same loss a fighter’s intake pays in thrust, and a shock can touch only one of the two totals: the total temperature crosses it unchanged, and it is only the total pressure that an intake loses.

Choosing a cone over a wedge is worth far less: 0.6 points at Mach 1.6, 1.3 at Mach 2, 1.7 at Mach 2.5, 1.5 at Mach 3 and 0.9 at Mach 4. It is a real gain and it is an order of magnitude smaller than the first decision. Axisymmetric intakes are chosen for other reasons as well — a round duct is light and strong under internal pressure, and a spike can translate along its axis to keep its shock on the lip as the flight Mach number changes, as the SR-71’s did — and the continuous compression is a modest bonus on top.

The usual estimate nearly doubles the gain #

The fifth figure is the one that matters for anyone checking an intake with a textbook. The quick way to estimate a cone intake’s recovery takes the cone-surface Mach number from a chart of Taylor–Maccoll solutions and applies the normal shock to it. That uses the most favourable ray in the annulus as though it were all of it. At the best cone angle the surface estimate reports a gain over the wedge of 2.3 points at Mach 2 where the mass average gives 1.3, and at every Mach number between 1.5 and 4 it reports 1.6 to 1.8 times the true gain.

So half or more of the advantage usually credited to the cone’s continuous compression is an artefact of evaluating the terminal shock on the wrong ray. The error in the recovery itself is only about a point — 0.926 against 0.916 at Mach 2 — which is why the estimate survives in use. It is the difference between two intakes that the estimate distorts, because the whole difference is about that size.

What was checked #

The ledger holds three checks. The mass crossing the cowl plane matches the free stream’s inside the cowl radius to $2\times10^{-9}$ at four cones, which tests the Taylor–Maccoll profile everywhere across the annulus. The thinnest cone the integration reaches at Mach 2.5, 3°, recovers within 0.006 of a bare normal shock, and the excess grows as the square of the angle between 3° and 4°, as a weak shock’s entropy should. And at three cones and Mach numbers the mass average lies between the recovery on the shock’s ray and the surface’s, as an average must.

What the two-shock model leaves out #

The terminal shock is not a plane. A real terminal shock stands in the cowl’s entrance where the internal flow puts it, curved, and interacting with the boundary layer on the centrebody; its strength varies across the duct in a way that one plane of normal shocks only approximates.

Boundary layers and bleed. The centrebody’s boundary layer thickens through the conical shock and again at the terminal shock, where a strong enough shock separates it. Real intakes bleed the boundary layer away through the spike’s surface, and the bleed’s cost and the separation’s are both absent here.

The subsonic diffuser. Behind the terminal shock the flow must slow further to the engine face, and its losses are counted separately. Its throat must also pass the captured mass at the reduced total pressure, and an intake whose throat is too small chokes, and stops listening to the engine behind it, pushing the terminal shock out ahead of the cowl.

Off-design. Every intake here has its shock on the lip. At lower flight Mach numbers the shock stands ahead of the lip and air spills round the cowl; at higher ones it falls inside and the cowl’s own shock appears. The comparison is at each intake’s design point.

The convention: recovery by mass #

Recovery is the total pressure delivered divided by the free stream’s, averaged over the captured air weighted by mass flux. Angles are the centrebody’s half-angle; the cone’s is the Taylor–Maccoll surface angle for the weak attached shock. Air is a perfect gas with a ratio of specific heats of 1.4. A point of recovery is one hundredth of the free stream’s total pressure.

Who worked it out #

Taylor and Maccoll gave the conical-flow equation in 1933. Oswatitsch, in the 1940s at Göttingen, showed that a train of oblique shocks ending in a normal shock recovers most when the oblique shocks are equally strong, and that axisymmetric spikes with isentropic compression surfaces could approach the ideal; Ferri and Nucci designed and tested conical-spike intakes at the NACA in the early 1950s. The mass-averaged versus area-averaged treatment of non-uniform intake flow, and the error of evaluating a terminal shock on a single representative Mach number, run through the intake literature of the 1950s onward.

Still open: a curved spike that hands the terminal shock one Mach number #

The cone hands the terminal shock a spread of Mach numbers because its compression is concentrated near its surface. A spike whose surface curves — turning the flow progressively, with each compression wave designed to reach the cowl lip — is the axisymmetric version of an isentropic ramp, and can in principle hand the terminal shock a nearly uniform Mach number lower than any cone reaches without detaching. The next calculation designs such a spike by the method of characteristics for an axisymmetric flow, with its waves focused on the lip, and asks how many points of recovery a focused spike adds over the best cone at Mach 2 and 3, and whether the whole gain survives the averaging that took half of the cone’s away.