The spot a local theory cannot see

Newtonian impact theory survives at hypersonic speed because it is almost right about the thing it claims. A stream at Mach 8 or 10 hits an inclined panel much as a hail of particles would, the pressure is very nearly $2\sin^2\theta$ times the dynamic pressure — or, in the modified form, the stagnation value behind a normal shock times $\sin^2\theta$ — and a whole vehicle’s forces can be added up panel by panel. The optimisation essay built a body on that locality, and the face Newton left in shadow found the first place it fails: a surface turned away from the stream carries a pressure the theory calls zero.

That failure costs a few per cent of a force. This essay is about the other failure of locality, which costs a structure.

A local theory says the pressure on a panel depends only on the panel’s own angle to the free stream, and so it says something else implicitly: that no spot on a vehicle can ever see more than the stagnation pressure behind the bow shock, because no panel faces the stream more squarely than normal. The implicit statement is false. It is false in a precise, computable way, and the reason is that a stream need not reach a surface through the bow shock at all.

A stream loses total pressure by the shocks it crosses #

The currency that decides stagnation pressure is total pressure, and what a shock costs put the price of a shock in it. A normal shock at Mach 8 keeps 0.85 per cent of the free stream’s total pressure; everything behind the bow shock’s normal part starts from that. The modified Newtonian ceiling is the stagnation pressure that 0.85 per cent produces.

A weaker shock costs much less, and a sequence of weak shocks turning the same stream costs far less than one strong one — which is the whole principle of a supersonic intake. So the question a local theory cannot ask is: could any stream on this vehicle reach the surface having crossed weaker shocks than the bow shock’s normal part? If one can, it stagnates higher.

The obvious candidate is a stream that crosses a shock from somewhere else first — a wing’s leading edge, an intake ramp, a pylon — and then reaches the bow shock of a second part of the vehicle already turned and compressed.

The simplest estimate, and the one the essay on the shaded face named as the calculation to do, is a stream crossing one oblique shock of turning angle θ and then a normal shock at the surface. Its advantage over the bow shock alone is the total pressure the oblique shock saves by pre-compressing the stream at a lower cost. At Mach 8 that path reaches 3.58 times the undisturbed stagnation pressure at a turn of 21.6°, and less at any other angle.

That number is a real amplification and the wrong answer, and the next section says why.

Where the two shocks meet #

When an incident shock crosses a bow shock near the bow shock’s normal part — the geometry Barry Edney labelled type IV — the flow around the crossing must settle into states that agree with each other. Behind the bow shock, the free stream has been turned sharply and compressed strongly. Behind the incident shock, the stream has been turned by θ and compressed a little, and where it meets the bow shock it crosses a second shock — the transmitted shock — that turns and compresses it again. The two resulting streams lie side by side with only a shear layer between them, so they must have the same pressure and the same direction.

That condition is a pair of equations, and the standard way to solve it is graphically, with pressure–deflection polars: for each upstream state, the locus of pressure against flow direction over every shock that state can cross.

The crossing is found by bisection on the difference between the two branches’ pressures at a common direction, and at Mach 8 with a 10° incident shock the two agree to $7\times10^{-14}$ at a direction of 36.85° and a pressure 71.6 times the free stream’s. The strong branch on one side leaves the free stream subsonic. The weak branch on the other leaves the stream behind the incident shock supersonic — at Mach 2.59.

That supersonic stream is the jet. It runs between two shear layers towards the surface and ends in a shock of its own just in front of it. It has crossed two turning shocks, of 10° and 26.85°, and it has kept 6.7 per cent of the free stream’s total pressure, against 0.85 per cent through the bow shock. Taking its terminal shock as normal at Mach 2.59, it stagnates at 7.88 times the undisturbed pressure — more than twice the one-turning-shock estimate at the same incident angle.

The single-shock estimate missed the transmitted shock. The jet’s path is a two-stage compression followed by a normal shock, which is a better intake than a one-stage one, and it is the geometry of the crossing, not anybody’s design, that builds it.

The worst angle, at every Mach number #

Sweeping the incident turn locates the worst case for each Mach number. At Mach 8 it is a 13.9° incident shock, and the jet then stagnates at 8.6 times the undisturbed pressure. At Mach 4 the worst is 3.5 times; at Mach 6, 6.2; at Mach 10, 10.7.

Each curve has a reason at both ends. A very weak incident shock barely changes the stream and the jet is barely better than the bow shock. A strong one compresses the stream so much that the transmitted shock has little room to turn it, and past a limit the polars stop crossing in the right arrangement and the interaction changes type — to Edney’s types III and V, where the jet does not form or does not reach the surface. Between them is a worst angle, and it lies in the range a wing’s leading edge, a ramp or a pylon produces without trying.

Close to the Mach number, and far from the ceiling #

The worst type IV amplification is close to the free-stream Mach number: 3.5 at Mach 4, 6.2 at Mach 6, 8.6 at Mach 8, 10.7 at Mach 10 and 12.4 at Mach 12. That is not a law this calculation proves — it is what the numbers do over this range, a little below the Mach number at the low end and a little above at the high end. But it is a useful rule of thumb with a clear reason under it. The lossless ceiling, one over the normal-shock recovery, grows roughly as the Mach number to the power three and a half, because the normal shock’s losses grow that fast; a two-stage compression recovers a growing share of what the normal shock throws away, but a fixed number of stages can only recover a fixed kind of share, and the result grows much more slowly than the ceiling.

The single-turning-shock estimate falls further behind as the Mach number rises: 2.3 against the jet’s 3.5 at Mach 4, 4.2 against 12.4 at Mach 12. A designer who had done the estimate the obvious way would have been out by a factor of nearly two at Mach 5 and three at Mach 12.

What that does to a pressure coefficient #

In the language of Newtonian theory the jet is an absurdity. The modified theory’s maximum pressure coefficient at Mach 8 is 1.827, and every panel on the vehicle is supposed to lie below it. The spot where the jet lands has a coefficient of 15.9.

No inclination produces that. The theory’s inputs at that spot are the panel’s angle and the free-stream Mach number, and neither knows that the stream arriving there was pre-compressed by a shock from a part of the vehicle several metres away. This is the sense in which the failure is one of locality and not of accuracy: the theory is not a little wrong at that spot, it is asking the wrong question, because the pressure there is a property of the whole vehicle’s shock pattern.

The spot is small. The jet is narrow — a fraction of the bow shock’s standoff distance — and it moves with the incident shock’s position, so the peak pressure acts on an area that a force calculation can ignore. That is exactly why the failure matters so much: the forces are right and a spot on the structure is not, and a local theory’s force being right is the evidence people take for its being right everywhere.

What a designer can do about it, and what sweep buys #

The interaction cannot be designed out of a vehicle that has more than one part: wherever a shock from a forebody, an intake ramp or a wing root reaches a leading edge downstream, some flight condition puts the crossing on the bow shock’s normal part. What can be controlled is how bad the worst case is, and the calculation above says which variable controls it.

Since the worst amplification follows the Mach number, anything that lowers the Mach number the leading edge’s normal part sees lowers the peak. Sweep does that. A leading edge swept back by Λ sees, at the first approximation of hypersonic independence, a normal Mach number of $M\cos\Lambda$, and its bow shock’s normal part is set by that component. Running the polars at the normal Mach numbers of an edge swept at Mach 8 gives the consequence directly: 8.6 times unswept, 7.3 at 30° of sweep, 5.7 at 45° and 3.5 at 60°. Sweeping a pylon, a fin or a cowl lip by 60 degrees cuts the worst pressure amplification by more than half.

Bluntness helps in a different way. It moves the bow shock away from the surface and widens the subsonic region the jet must cross, so the jet arrives later, bent and spread; the peak falls without the crossing itself changing. And moving the crossing — placing the incident shock so that it strikes the bow shock above or below its normal part, where the interaction becomes Edney’s type III or V — changes the pattern into one in which the pre-compressed stream never reaches the surface as a jet at all. That is the design logic of scramjet cowl lips, which must sit near the shock from the forebody by construction and are shaped and positioned to keep the crossing out of type IV over the flight envelope.

None of these is visible to a panel method that knows only each panel’s inclination. They are all statements about where one part’s shock goes relative to another part’s leading edge.

Why the heating is worse than the pressure #

The pressure is what this calculation computes, and it is not what destroys structures. Heating does.

Stagnation-point heat transfer scales roughly as the square root of the stagnation pressure divided by the effective radius of the surface the stream stagnates on. The jet’s stagnation pressure is up to the Mach number times the undisturbed value, and the jet is narrow, so the effective radius it stagnates against is much smaller than the leading edge’s own. Both effects raise the heating, and together they make the heating amplification larger than the pressure amplification. Edney’s measurements, and those that followed, report heat-transfer peaks well over ten times the undisturbed stagnation value in type IV interactions — a borrowed figure here, since a heating calculation needs the jet’s width, which the polars do not give.

The best-known consequence is the X-15’s flight of October 1967, at Mach 6.7 with a dummy ramjet mounted on a pylon below the fuselage. The shock from the ramjet’s spike struck the pylon, the heating burned through it, and the ramjet was lost in flight. The structure had been designed with the heating of the undisturbed flow in mind, which is to say locally.

A ledger of the states #

The checks are the ones that could have caught a sign or a branch error in the polars. A zero incident turn must return an amplification of exactly one, since the stream then crosses only the bow shock; it returns 1.000000014, the precision of an oblique shock found by bisection at a turn of 10⁻⁷ degrees. The one-turning-shock path must lie below the lossless ceiling at every Mach number, and does. And the jet must beat the one-turning-shock path, since it contains that path and adds a stage; it beats it by a factor of 2.0 at Mach 6, 2.4 at Mach 8 and 2.7 at Mach 10.

What the polars cannot show #

The jet itself. The polars fix the states at the shock crossing and nothing downstream of it. The real jet is bounded by shear layers, curves towards the surface, and passes through a train of expansion and compression waves before its terminal shock, which is itself curved. Taking the terminal shock as normal at the jet’s initial Mach number is the simplest closure and probably overestimates the jet’s losses slightly, since a curved terminal shock is oblique over part of its span; the peak amplification from full numerical solutions is of the same size.

Where the jet lands. The polars say nothing about position. The jet’s impingement point moves by a fraction of a leading-edge radius as the incident shock moves, which is why peak heating in a type IV interaction is so sensitive to small changes in incidence or Mach number.

Unsteadiness. Type IV jets are often unsteady, flapping at high frequency, which spreads the heating over a larger area and lowers its peak.

Real gas. Above about Mach 8 in air the gas behind strong shocks is no longer calorically perfect, and the normal-shock recovery — the reference for every amplification here — changes. The ratios would move; that the jet beats the bow shock would not.

Viscosity. The shear layers, the boundary layer at the impingement point, and the heating itself are all viscous, and none of them is in a calculation of inviscid shock polars.

Who found it #

Barry Edney, at the Aeronautical Research Institute of Sweden, published the classification of shock–shock interactions in 1968, having measured the pressure and heat transfer on blunt bodies with impinging shocks and sorted the flow patterns into six types by where the incident shock crossed the bow shock. Type IV, with its supersonic jet, produced the largest peaks. The X-15 incident had happened the year before the report appeared. The analysis of shock–shock interactions with polars is older and belongs to the treatment of shock reflection and intersection going back to von Neumann’s work of the 1940s.

What connects them to Newton is a sentence in the theory that forbade flight: that a theory can be good at forces for a reason entirely unrelated to whether it is good at pressures. Newtonian theory’s force predictions survived three centuries of being wrong about the physics. Its pressure predictions fail at a spot whose area no force calculation notices.

Still open: how wide the jet is #

The peak heating needs the jet’s width, and the polars do not give it. The width is set by where the shear layers bounding the jet go after the crossing, which depends on the pressure field around the stagnation region of the blunt body the jet strikes — a subsonic region whose shape is not local either.

The next calculation couples the polar solution at the crossing to a model of the bow shock’s standoff and curvature near a blunt leading edge, and follows the two shear layers as streamlines of the subsonic region until they meet the surface. What it would give is the jet’s width at impingement, and with it an estimate of the heating amplification from the square-root scaling — the number the X-15’s pylon needed, computed rather than borrowed.