The last shock should be the weakest

What a shock costs made the case that every supersonic intake is a device for taking a compression in small bites, because the entropy a shock makes rises as the cube of its strength: at Mach 2 a fifteen-degree turn taken as one oblique shock keeps 95.9 per cent of the total pressure and taken as four keeps 99.4. It quoted the design rule that follows, which goes back to Oswatitsch in 1944 — a train of shocks recovers most when its shocks are equally strong — and said that equal cubes minimise their sum. A cone intake’s shock sees the whole capture built an intake on one conical shock and a terminal normal shock and quoted the rule again.

Neither essay solved the optimum it quoted, and the cube-law argument has a gap in it. An intake’s train is not a set of interchangeable shocks: every oblique shock is followed by more compression, but the last one, the normal shock that makes the flow subsonic for the engine, is followed by none, and the Mach number it acts at is whatever the train leaves it. Solved without any assumption about the shape of the answer, the rule turns out to be exact for the oblique shocks and wrong for the last one.

The problem, stated so it can be solved #

An intake at free-stream Mach number $M$ turns the flow through $n$ planar oblique shocks and then slows it through a normal shock. Each oblique shock is fixed by its normal Mach number $M_{n,i} = M_i\sin\beta_i$, which sets its total-pressure ratio, its turning angle and the Mach number it hands on; the terminal shock acts at whatever Mach number is left. The recovery is the product of all $n + 1$ total-pressure ratios, and it is maximised over the $n$ strengths. The search is coordinate by coordinate, each coordinate by golden section, repeated until nothing moves, with each strength written as a fraction of the way from a Mach wave to the strongest shock the local Mach number allows, so that every trial is a train that exists.

Nothing in the search knows Oswatitsch’s answer. One oblique shock’s best strength, found this way, agrees with a direct four-thousand-point sweep over its turning angle to $3\times10^{-11}$ in recovery.

What the trains recover #

At Mach 3 a normal shock alone keeps 32.8 per cent of the total pressure. One well-chosen oblique shock in front of it raises that to 58.1, two to 74.4, four to 88.9 and eight to 96.4. Each shock added buys less than the last, and the curves at higher Mach numbers start lower and climb more slowly: at Mach 4 eight oblique shocks reach the recovery four reach at Mach 3. At Mach 2 the gains are smaller because there is less to lose — one oblique shock already recovers 90.3 per cent and two 95.7.

These are the inviscid numbers a designer starts from, and they are the reason the first supersonic fighters’ intakes had one ramp, Mach 2 airliners’ two or three, and Mach 3 aircraft’s a whole moving spike.

Equal oblique shocks, and a weaker last one #

Read the strengths of the best train and the rule appears — almost. The four oblique shocks of the best train at Mach 3 come out with the same normal Mach number to four figures, 1.326, though the search started them unequal and was never told to equalise them. The terminal normal shock does not join them: it acts at 1.261, 4.9 per cent weaker.

The reason is the asymmetry the cube-law argument leaves out. Strengthening an oblique shock costs entropy at that shock and lowers the Mach number for every shock after it, including the terminal one, so the cost and the benefit are both spread down the train, and at the optimum the trade between any two oblique shocks balances only when their strengths are equal. Strengthening the last oblique shock relative to the terminal one moves compression from a shock whose loss rises as the cube of its normal Mach number to a shock that also turns the flow, and a shock that turns the flow has less of its strength in loss per unit of compression than one that does not. The optimum therefore hands the terminal shock a slightly lower Mach number than an equal share would. The rule as it is usually quoted — equal shocks throughout — keeps the first half of that and loses the second.

The ratio, at every Mach number #

Across the range intakes are designed for, Mach 1.6 to 4, the ratio of the terminal shock’s normal Mach number to the oblique shocks’ stays between 0.942 and 0.98, and it is closest to one for long trains at low Mach numbers, where every shock is weak and the difference between turning and not turning matters least. The correction is small in Mach number and smaller in recovery — the optimum is flat, as every smooth optimum is — but it is systematic, and a train designed on the equal-strength rule throughout is not the best train.

Equal strengths are not equal turns #

The other half of the rule that tends to be lost is the geometry. Equal strengths mean equal normal Mach numbers, and the turn a shock of given normal Mach number makes depends on the Mach number in front of it: it grows as that Mach number falls, at first, and shrinks again as it approaches the shock’s own normal Mach number, where the shock becomes normal and turns nothing. The best train of four at Mach 3 is on the growing side throughout and turns the flow by 8.7°, 10.0°, 11.4° and 12.6°, each ramp steeper than the last by more than a degree; the best train of three at Mach 2 turns 7.5°, 8.1° and 8.0°, already past the turnover at its last shock. Two shocks at Mach 3 turning a total of 33.8° should split it 15.0° then 18.8°, the first taking 44 per cent.

Dividing the same total equally — two ramps of 16.9° — recovers 0.739 against the best 0.744. The figure shows the whole curve, and the flatness is the practical news: half a point for a split that is wrong by four degrees is a small price, and it is why ramp angles in practice are chosen as much by the structure, the boundary layers and the off-design behaviour as by the optimum.

Three rules, one Mach number #

Set the equal-turn rule against the best train at every train length and the gap closes as the shocks multiply: 0.45 points at two shocks, 0.2 at eight. That is the cube law doing its work from the other side. A long train’s shocks are all weak, the loss at each goes as the cube of a small number, and whether the turns are distributed exactly right matters less and less. In the limit the train becomes a compression that costs nothing in the end — an isentropic ramp, with no loss at all until its waves meet — and only the terminal shock is left to pay. Forty weak shocks of normal Mach number 1.004 at Mach 2.5 leave the flow within $2\times10^{-6}$ of the Mach number an isentropic ramp of the same turn leaves, and lose three parts in a million of total pressure between them.

Flat in every direction but one #

The optimum is flat, and it is worth saying how flat and in which directions. Take the best train of four at Mach 3 and move any one shock’s normal Mach number by one per cent either way: the worst of those eight changes costs 0.018 points of recovery. Three per cent costs 0.16, five per cent 0.46. Errors of a single ramp angle of the size a fixed-geometry intake makes off its design point are cheap.

Moving all four together is not. Strengthen every oblique shock by one per cent and the recovery falls by 0.21 points, ten times the cost of moving one of them by the same amount, and the terminal shock now meets the flow at Mach 1.19 instead of 1.26. Strengthen them by three per cent and the train hands the terminal shock Mach 1.03, having turned the flow so far that the compression is nearly all oblique, and the recovery has fallen by 1.7 points — the obliques’ own losses rising faster than the terminal shock’s fall. At five per cent the train no longer exists in this form: its later shocks ask for a normal Mach number the flow in front of them no longer has. A ramp schedule can be loose in its parts and must be exact in its sum. The one direction in which the train is sensitive is the total, which is the direction the cowl constrains. Turning the other way is free states the asymmetry this rests on: an expansion can turn the flow any amount at no cost, and a compression cannot, so a train’s total turn is a resource that runs out.

What the shorthand costs, and where it matters #

Design the train the shorthand’s way instead — every shock, the terminal one included, at one common normal Mach number — and the price can be stated exactly. At Mach 3 it costs 0.09 points of recovery with one oblique shock, 0.15 with two and 0.12 with four; at Mach 2, 0.12, 0.09 and 0.03; at Mach 4, 0.04, 0.10 and 0.15. A tenth of a point is below what a wind tunnel resolves on an intake model, and nobody should redesign a ramp for it.

The geometry is where it matters. The best train, letting its oblique shocks be stronger than its terminal one, turns the flow further: at Mach 3 with two oblique shocks, 33.8° against the shorthand’s 32.3°; with four, 42.8° against 41.4°; at Mach 4 with four, 52.2° against 50.5°. A degree and a half of extra turn buys a tenth of a point inside the intake and must be paid for outside it, on the cowl, and the next section says why that bill can be the larger one. The exact optimum is a better intake only while the cowl can follow it.

A train at Mach 2, in degrees #

For an aircraft cruising at Mach 2 the numbers are concrete enough to draw. One ramp, at the best angle, turns the flow 16.4° and recovers 90.3 per cent with its terminal shock at Mach 1.39. Two ramps turn 10.4° and then 11.1° — the second steeper, because the flow reaching it is at Mach 1.63 — and recover 95.7 per cent with the terminal shock at 1.22. Three ramps, at 7.5°, 8.1° and 8.0°, recover 97.7 per cent with the terminal shock at 1.15. The step from one ramp to two is worth five and a half points and from two to three two points; the third ramp also adds a hinge, a seal and a bleed slot. That arithmetic is why two-ramp intakes were the common choice for Mach 2 cruise, and it is the same arithmetic the spot a local theory cannot see used for two shocks crossing at hypersonic speed, where the gain from splitting one shock into two is far larger because the loss being avoided is.

Every one of those recoveries is the inviscid upper bound. The entropy each shock leaves is carried downstream as a layer of lower total pressure — the spin a shock leaves behind traces its vorticity — and an engine face sees an average over streamlines that crossed different numbers of shocks at different places, which the planar train here does not.

The turn grows faster than the recovery #

What the best trains also do, steadily, is turn the flow further. At Mach 3 one oblique shock turns it 22.6°, two 33.8°, three 39.6°, four 42.8° and eight 47.3°, approaching the 49.8° of the Prandtl–Meyer angle that would take the flow isentropically all the way to Mach one. At Mach 2 the turns approach 26.4°. Every degree of that turn points the flow towards the intake’s cowl, and the cowl has to receive it.

That is where the free optimum stops being buildable. The cowl’s inner lip lies along the turned flow, so its outer surface meets the free stream at an angle at least as large as the train’s total turn, and an oblique shock on that outer surface detaches once the angle passes the free stream’s own detachment angle — 23.0° at Mach 2, 34.1° at Mach 3. The best trains pass that limit at three shocks at both Mach numbers. When the wedge is too blunt is the cowl’s problem too.

How the search was checked #

The single-shock search agrees with a direct sweep over deflection to $3\times10^{-11}$ in recovery. The equal oblique strengths are a result, not an input: the search started its four shocks at unequal strengths and the spread of their normal Mach numbers at convergence is below $10^{-4}$. A train of forty weak shocks reaches the Mach number of the isentropic ramp of the same turn to $2\times10^{-6}$. The normal shock alone at Mach 3 keeps 0.32834 of the total pressure, the tables’ value, and every oblique shock is computed from the same θ–β–M relation a shock that leans set out.

The convention: recovery as a product of totals #

Recovery is the ratio of the total pressure after the terminal shock to the free stream’s, the product of each shock’s ratio; it is what an engine face sees before any subsonic diffuser loss. The ramps are planar and the gas perfect with $\gamma = 1.4$. A shock’s strength is its normal Mach number, $M\sin\beta$, which alone fixes its total-pressure ratio. The turns are on the weak branch of every oblique shock.

What the inviscid train leaves out #

The boundary layer. Every ramp shock lands on the boundary layer growing along the ramps, and the terminal shock on the layer at the throat, and the interaction thickens or separates it; real intakes bleed that layer away through slots, which costs mass flow the recovery numbers above do not count. The cowl. It is the subject of the limit above and of the next calculation, and none of the free trains here respects it. Off design. A train laid out for one Mach number puts its shocks in the wrong places at every other, which is why intakes have moving ramps and spikes. And three dimensions: the ramps here are planar; a cone’s compression continues behind its shock and its train is not this one.

Oswatitsch, and what survived of him #

Klaus Oswatitsch worked out the best distribution of compression among a train of shocks at Göttingen in 1944, in a report on supersonic intakes that is still the basis of external-compression design, and the equal-strength condition for the oblique shocks carries his name. The shorthand that reached later accounts — equal shocks throughout, justified by the cube law — treats the terminal shock as one more member of the train. The search here, which assumes neither, finds the oblique shocks equal and the terminal one about five per cent weaker, and prices the shorthand at a tenth of a point.

Still open: what the cowl allows #

The best trains turn the flow further than any cowl can follow once they have three shocks. The next calculation puts the cowl’s limit into the optimisation — the train’s total turn held at or below the free stream’s detachment angle — and asks what the best recovery then is at each number of shocks, how the strengths redistribute when the turn is capped, and what the ceiling on external compression becomes at each Mach number: the recovery of an isentropic ramp turned exactly as far as the cowl allows, which no number of shocks can beat.