The face Newton left in shadow

The theory that forbade flight established that Newton’s sine-squared pressure law is hopeless at the speeds he argued about and nearly exact behind a strong shock. The only theory simple enough to optimise found the property that keeps it in use — every point’s pressure depends on nothing but its own angle — and then separated out the one part of the rule that is not physics at all. A surface that cannot be reached in a straight line from upstream is shaded, and Newtonian theory gives it a pressure coefficient of exactly zero. That essay called the rule geometry bolted onto a physical argument, said the error it makes is systematic and in a known direction, and ended by naming the calculation that would say how large.

This essay makes that calculation on the one body where the shaded face has an exact answer: a flat plate in inviscid supersonic flow, whose windward face sits behind an attached oblique shock and whose leeward face sits after a centred Prandtl–Meyer expansion. The result is not that the zero is a bad approximation. It is that its quality is decided by a single parameter, which is not the Mach number, and that a plate flying at its best lift-to-drag ratio is always on the wrong side of it.

A pressure coefficient has a floor #

The pressure coefficient is $C_p = (p - p_\infty)/\tfrac12\rho_\infty V^2 = (2/\gamma M^2)(p/p_\infty - 1)$, and an absolute pressure cannot go below zero. So no surface in any flow can carry a coefficient below

$$C_{p,\text{vac}} = -\frac{2}{\gamma M^2},$$

which is −0.159 at Mach 3, −0.057 at Mach 5, −0.0143 at Mach 10 and −0.0036 at Mach 20. The Newtonian zero is a coefficient of zero — the shaded face at free-stream pressure — so the error it can make on a leeward face lies between nothing and that floor. That is the same point Nothing sucks made about a wing’s upper surface: the “suction” is a pressure below the free stream, and it is bounded because a fluid cannot pull.

The conventions are these. The gas is perfect with γ = 1.4; the flow is inviscid and two-dimensional; the plate is at incidence α to a stream at Mach M; and the windward Newtonian coefficient is the modified one, $C_{p,\max}\sin^2\alpha$ with $C_{p,\max}$ the exact stagnation value behind a normal shock — 1.809 at Mach 5, 1.832 at Mach 10 and 1.837 at Mach 20. The quantity that organises everything below is the hypersonic similarity parameter $K = M\sin\alpha$.

Both faces of a plate, exactly #

The windward face turns the stream into itself by α, through an oblique shock. The leeward face turns it away by α, through a fan of Mach waves that costs no total pressure at all — the asymmetry Turning the other way is free measured, and the reason the leeward pressure can be computed exactly from the Prandtl–Meyer function.

At Mach 5 and 10° the windward face carries 0.117 and the leeward face −0.043, already three quarters of the way to the floor at −0.057. The leeward face supplies 27 per cent of the plate’s normal force. At 20° it is at −0.055, 96 per cent of vacuum, and supplies 14 per cent. The figure also shows the other thing wrong at this Mach number: the Newtonian windward coefficient at 10° is 0.055 against the exact 0.117, so Newtonian theory’s whole normal force is 34 per cent of the exact one. At Mach 5 and ten degrees the plate is not yet in the regime where Newton is right about anything.

Scaled by the incidence, both faces are functions of K #

That regime has a name, and the pressures show it directly.

Divide each face’s coefficient by $\sin^2\alpha$ and plot it against $K$, and the curves for Mach 3, 5, 10 and 20 fall on top of one another — the Mach 3 and 5 windward curves peeling away only where their shocks approach detachment. They collapse onto the hypersonic small-disturbance forms, which hold as the Mach number goes to infinity at fixed $K$:

$$\frac{C_{p,w}}{\alpha^2} = \frac{\gamma+1}{2} + \sqrt{\Big(\frac{\gamma+1}{2}\Big)^2 + \frac{4}{K^2}}, \qquad \frac{C_{p,l}}{\alpha^2} = \frac{2}{\gamma K^2}\Big[\Big(1 - \frac{\gamma-1}{2}K\Big)^{2\gamma/(\gamma-1)} - 1\Big].$$

At $K = 2$ they give 2.762 and −0.347. The leeward form reaches vacuum at $K = 2/(\gamma - 1) = 5$, and it is nearly there long before: at $K = 3$ it is −0.1585 against the vacuum value −0.1587. As $K$ grows the windward form tends to γ + 1 = 2.4, which is the strong-shock value Newton’s 2 is the fraction 2/(γ + 1) of — the constant factor the optimisation essay relied on — and it is the same limit A shock that lies on the body found setting a blunt nose’s pressure.

The fan runs out of room at high Mach number #

Why the leeward face reaches its floor so quickly at high speed is worth a paragraph, because it is the fan rather than the shock that sets the pace. A supersonic stream can turn away from itself only until its pressure reaches zero, and the total turn available from Mach one to vacuum is the Prandtl–Meyer function’s limit, 130.45° for γ = 1.4. A stream already at a high Mach number has used most of that on the way up. At Mach 3 there are 80.7° of turning left before vacuum; at Mach 5, 53.5°; at Mach 10, 28.1°; and at Mach 20 only 14.3°. A plate at Mach 20 inclined at more than 14.3° has a lee face at vacuum exactly, and the fan leaves a region the stream simply does not reach.

Expressed in the similarity parameter those limiting turns are K = 2.96, 4.02, 4.72 and 4.93 — closing on the value 2/(γ − 1) = 5 at which the small-disturbance leeward form reaches vacuum. The fan is a family of Mach waves, each carrying a small turn at its own Mach angle, which is the structure the method of characteristics propagates information along; at high Mach number those angles are shallow, the waves crowd against the surface, and a small geometric turn exhausts the whole expansion. That is the leeward half of hypersonic similarity, and it is why the same K that makes the windward shock strong makes the leeward fan run dry.

The shaded face’s share, and why only K shrinks it #

The share of the normal force carried by the leeward face is the quantity Newtonian theory sets to zero, and it is the cleanest way to say how wrong the zero is.

It collapses on $K$ just as the pressures do. In the hypersonic limit it is 35.7 per cent at $K = 0.5$, 24.2 per cent at 1, 11.2 at 2, 5.8 at 3 and 3.5 at 4. The zero is right to within five per cent of the force only above $K \approx 3$. And $K$ is the product of Mach number and incidence, so raising the Mach number helps only if the incidence does not fall to compensate. A plate at Mach 20 and 2° has $K$ = 0.70 and a leeward share of 31 per cent — more than the same plate at Mach 5 and 10°.

That is the structural point. Newtonian theory’s windward pressure becomes accurate as $K$ grows, because a large $K$ means a strong shock lying close to the surface. Its leeward zero becomes accurate as $K$ grows, because a large $K$ means an expansion that has run all the way to vacuum and a vacuum coefficient that is small beside the windward one. Both halves of the theory are governed by the same parameter, and a flight condition with a small $K$ is wrong on both faces at once — with errors of opposite sign, since the windward face is under-predicted and the leeward face’s contribution omitted, which adds rather than cancels.

Friction gives the plate a best incidence, and the shaded face moves it #

The optimisation essay found that with a skin-friction coefficient of 0.001 on both faces, Newton’s plate has its best lift-to-drag ratio of 5.22 at 7.2°. The figure it drew had no leeward pressure in it, and the number is exactly where a leeward pressure would matter most.

The best ratio is found where the pressure force, growing with incidence, stops outrunning a friction force that does not. The leeward pressure adds normal force at no cost in friction, and it adds proportionally most at small incidence, so the exact plate reaches its best ratio earlier and higher. At Mach 5 it peaks at 10.13 at 2.84°, against the modified Newtonian 5.04 at 7.46°. At Mach 20 it peaks at 6.03 at 5.76°, against 5.07 at 7.42°. Above about 15° the exact and Newtonian curves run together, because there the leeward face has reached its floor and the floor is small.

Across Mach number the exact best ratio falls from 13.29 at 2.15° at Mach 3 through 10.13 at Mach 5, 8.07 at Mach 8, 7.34 at 4.15° at Mach 10, 6.41 at 5.14° at Mach 15 and 6.03 at Mach 20, to 5.86 at 6.11° at Mach 25. At each of those optima the leeward face carries 47, 43, 35, 30, 18, 11 and 7 per cent of the normal force: the plate chooses a small $K$ — 0.11 at Mach 3, 0.72 at Mach 10, 2.0 at Mach 20 — because a small $K$ is where the lift-to-drag ratio is best. The modified Newtonian plate reads 4.99 to 5.07 at about 7.4° throughout, and Newton’s own coefficient gives 5.22 at 7.21° at every Mach number, because it does not know one. The exact curve approaches the Newtonian one only slowly, and at Mach 25 is still 15 per cent above it.

That qualifies the optimisation essay’s strongest claim. It argued that a theory with a constant bias orders shapes correctly even when it gets every number wrong, and that the ordering is safe for shapes the theory can see all of. A plate at incidence cannot be seen all of — one face is shaded — and the ordering of incidences is exactly what fails: Newtonian theory puts the best incidence at 7.5° at Mach 5 and the exact answer is 2.8°, a factor of 2.6. The bias is not constant across incidence, because the omitted term scales as $1/K^2$ and the incidence is half of $K$.

Why the controls go on the windward side #

The floor has a design consequence that follows directly from the numbers above. A control flap works by changing the pressure on the face it sits on. Deflected into the stream on the windward side, it raises that pressure without limit as its angle grows — the windward coefficient of a surface at 10° is 0.117 at Mach 5, 0.087 at Mach 10 and 0.077 at Mach 20. Deflected away from the stream on the leeward side, it can lower the pressure only as far as vacuum, and the whole of that floor is 0.057, 0.0143 and 0.0036 at the same three Mach numbers.

So at 10° a windward flap can command 2.0 times the largest force change a leeward flap could ever produce at Mach 5, 6.1 times at Mach 10 and 21 times at Mach 20. The leeward flap does not merely lose effectiveness at hypersonic speed; it is bounded by a number that falls as the square of the Mach number, while the windward flap’s authority does not fall at all. That is why re-entry vehicles put their pitch control on the windward surface — a body flap under the aft fuselage rather than elevons working in the shadow — and it is a place where the Newtonian zero gives the right design answer for a reason the theory does not state: the shaded face really is nearly powerless once K is large, and the calculation here says how large.

A wedge’s base can cost more than its whole forebody #

A plate at incidence has one shaded face. A wedge at zero incidence has a shaded base, and Newtonian theory gives it the same zero — the choice that Drag in the theory that forbids it made for a free-streamline wake, where it produced a drag coefficient of 0.8798 from nothing else. The exact base pressure of a wedge depends on the separated flow behind it, which inviscid theory cannot supply, but its worst case can: a base at vacuum.

For a 5° half-angle that ratio is 2.20 at Mach 3, 1.24 at Mach 5, 0.49 at Mach 10 and 0.16 at Mach 20. For 10° it is 0.95, 0.49, 0.16 and 0.047; for 20°, 0.36, 0.17, 0.048 and 0.0125. Once the forebody’s shock is strong its pressure tends to $(\gamma+1)\theta^2$, so the ratio approaches $2/((\gamma+1)\gamma K^2)$ — five-sixths of the dashed $1/(\gamma K^2)$, with $K = M\sin\theta$ — which is the same parameter again: at 20° and Mach 20 it is 0.0125 against 0.0153. Where the bound exceeds one, a slender wedge’s base could carry more drag than its entire nose, and a real base, which sits somewhere between vacuum and the free-stream pressure, carries a substantial fraction of that. For a slender body at moderate supersonic speed the zero on the base is not an approximation to the answer; it omits a term the size of the answer.

The exact faces against the limits they must reach #

The shock and the fan come from this collection’s own oblique-shock and Prandtl–Meyer solvers, and the plate built from them is checked against three limits that share none of their algebra.

At 0.2° of incidence both faces agree with Busemann’s second-order theory, $C_p = \pm C_1\alpha + C_2\alpha^2$ with $C_1 = 2/\sqrt{M^2 - 1}$ and $C_2 = ((\gamma+1)M^4 - 4(M^2-1))/2(M^2-1)^2$, to 5.8 × 10⁻⁵ at Mach 2, 3 and 5. The second term is not optional: at Mach 5 the linear term alone is out by 1.04 per cent even at 0.2°, which is $C_2\alpha/C_1$ with $C_1$ = 0.408 and $C_2$ = 1.219. At Mach 60 the windward face matches the small-disturbance form to 6.4 × 10⁻⁴ and the leeward face to 8.2 × 10⁻⁴ for $K$ from 0.5 to 3, and the leeward share at Mach 20 matches the share at Mach 40 at $K$ of 0.7, 1.5 and 3 to 6.9 × 10⁻⁵. No leeward face computed at Mach 3, 6, 12 or 24 and incidences from 2° to 20° falls below the vacuum floor. The Newtonian plate is the collection’s own, with its frictionless ratio exactly cot α. And the calculation refuses a subsonic stream, a windward shock past detachment, a negative incidence and a best ratio without friction, where none exists.

What the inviscid plate cannot show #

The boundary layer on the lee side. At hypersonic speed a leeward boundary layer is thick, hot and prone to separate, and it changes the effective shape the expansion turns round. The inviscid leeward pressure is a bound on what the fan can do, not a prediction of what a real lee surface reads.

A real base. The base of a wedge or a cone is a region of recirculating, low-speed flow whose pressure is set by mixing along its edge, and nothing inviscid supplies it. The vacuum bound says how large the omission could be, not how large it is.

Three dimensions. A cone or a lifting body at incidence has crossflow round its sides, and its leeward pressure is set by that crossflow separating rather than by a two-dimensional fan.

A gas that is not perfect. At the temperatures behind a strong shock γ falls, as When gamma stops being a number computed, and both the vacuum floor and the similarity forms move with it.

Still open: two parts of a vehicle that can see each other #

The optimisation essay named two ways locality fails, and this essay has computed one. The other is interference: a shock from one part of a vehicle striking another. Where an oblique shock from a wing or a ramp crosses the bow shock of a leading edge, the flow between them can form a supersonic jet that runs into the surface, and the peak pressure and heating there — a flux driven from the recovery temperature of a stream that has lost less total pressure than one normal shock would cost it — can be many times the undisturbed stagnation values. Computing the total pressure a stream keeps through an oblique shock and a terminating normal shock, against a single normal shock, gives an upper bound on that amplification from the same shock solver used here; placing the jet needs the pressure–deflection polars of both shocks. Either would put a number on the price of a theory that cannot see one part of a body from another.