A shock that leans

A supersonic stream meets a wedge. It cannot turn gradually, because nothing has told it the wedge is there, so it turns abruptly — through a shock that sits at an angle to the flow rather than across it.

Everything about that shock follows from one observation, and the observation is worth stating before any algebra: the component of velocity along the shock is unchanged.

Why nothing happens along the shock #

The jump conditions of the previous ladder came from a control volume with faces parallel to the shock. Consider the momentum balance tangential to it.

Pressure acts normally to a surface, so it contributes nothing tangentially. Viscosity is absent from the control-volume argument. So the only tangential momentum flux is the one carried by the mass crossing the shock — and since mass conservation says the same mass crosses both faces, the tangential velocity must be the same on both sides.

That is the whole of it. The tangential component passes through as though the shock were not there.

The consequence is immediate and powerful: an oblique shock is a normal shock, applied to the normal component of the velocity. Resolve the incoming Mach number into a component across the shock, $M_1 \sin\beta$, and one along it. Put the first through the normal-shock relations. Leave the second alone. Reassemble.

Every pressure, density and temperature ratio across an oblique shock is therefore the ordinary normal-shock ratio evaluated at $M_{n1} = M_1 \sin\beta$, and there is nothing new to derive.

The θ–β–M relation, and its two roots #

What is new is geometry. The wedge specifies the deflection $\theta$, and the shock angle $\beta$ has to be worked out from it — because the deflection is what happens when the normal component is reduced and the tangential one is not.

$$\tan\theta = 2\cot\beta \, \frac{M_1^2 \sin^2\beta - 1}{M_1^2(\gamma + \cos 2\beta) + 2}$$

This is a relation between three quantities and it does not invert. Fixing $M_1$ and $\theta$ leaves an equation in $\beta$ with two roots.

Both roots are solved by bisection here rather than by any closed form, with the bracket for each placed entirely on one side of the maximum. Asking for a shock without saying which branch throws: the relation genuinely has two answers, and a solver that silently picked one would be making a physical decision in a numerical routine.

Weak and strong, and what actually appears #

The two solutions are not equally common, and the reason has nothing to do with the wedge.

The weak solution has the smaller shock angle, produces the smaller pressure rise, costs the least total pressure, and usually leaves the flow supersonic. It is what appears on a wedge in an unbounded free stream — every experiment, every photograph.

The strong solution stands nearly normal to the flow, produces a large pressure rise, is expensive, and always leaves the flow subsonic. It appears where the downstream pressure demands it: in a duct with a high back pressure, or near the axis of a detached bow shock, where the shock is locally almost normal.

At Mach 2 with a 10° wedge, the two are: $\beta = 39.31°$, $M_2 = 1.641$, $p_2/p_1 = 1.707$, total pressure ratio 0.9846; and $\beta = 83.70°$, $M_2 = 0.604$, $p_2/p_1 = 4.315$, total pressure ratio 0.7395. The strong shock costs seventeen times as much total pressure for the same turn.

Nothing in the θ–β–M relation says which occurs. That is a boundary condition’s job, and it is the same structure as the two branches of the area–Mach relation — a local relation with two solutions, and a distant condition selecting.

The tangential component is not a bookkeeping trick #

It is worth resisting the reading that resolving into components is merely a convenient change of variables. It is a statement about what the shock is.

A shock is a wave, and a wave propagates in one direction: normal to itself. Fluid crossing an oblique shock is being processed by a wave that is doing nothing whatever in the direction along its own front, so the fluid’s motion in that direction is simply carried through. Change frames — slide along the shock at the tangential speed — and the oblique shock becomes a normal shock exactly, because in that frame there is no tangential motion left.

That frame change is legitimate because the governing equations are Galilean invariant, and it is the cleanest way to see the result: there is only one kind of shock, and obliqueness is a property of the observer’s frame rather than of the wave.

Two consequences follow that are otherwise mysterious. The flow behind an oblique shock can be supersonic even though the flow behind a normal shock never is — because the normal component behind is subsonic in both cases, and in the oblique case the surviving tangential component is added back on. And the entropy rise depends only on $M_1\sin\beta$, which is why the same total turn is so much cheaper when the shock leans.

The limit that turns out to be exact #

The relations must reduce to the normal-shock case at $\beta = 90°$, and that is worth checking rather than assuming, because it exercises a lot of algebra at once.

assertObliqueReducesToNormal does two things. It evaluates the θ–β–M relation at exactly 90° and requires the deflection to be zero — a normal shock turns the flow through nothing. Then it takes the strong branch at a vanishingly small deflection and compares its pressure, density and total pressure ratios with the normal-shock values at the same Mach number, requiring agreement to $10^{-9}$ in each.

At Mach 2.5 the pressure ratio comes out as 7.125000 by both routes. That agreement is not guaranteed by construction: the oblique route computes $M_1 \sin\beta$ from a bisected $\beta$ and feeds it through, and the normal route uses $M_1$ directly.

The other limit, which is the Mach cone #

Run the deflection to zero on the weak branch instead and $\beta \to \arcsin(1/M)$: the Mach angle.

That is the connection this ladder shares with the wavefront construction. A Mach line is an oblique shock of zero strength, and the chart shows it — every curve leaves the axis at exactly the Mach angle of its Mach number, 41.8° at Mach 1.5 and 11.5° at Mach 5.

So the whole family is continuous. Turn the flow by nothing and get a Mach line carrying no pressure jump. Turn it a little and get a weak shock leaning slightly further forward. Turn it more and the shock leans further still and costs more, until the wedge asks for more than a shock can deliver.

Reading the chart, which repays the effort #

The θ–β–M chart looks like an engineering nomogram and is better than that: almost every qualitative fact about supersonic flow over a body can be read off its shape.

The curves start at the Mach angle and end at 90°. Both endpoints are zero-deflection shocks — one of zero strength, one of full normal strength turning the flow through nothing.

The maxima move right as the Mach number rises. At Mach 1.5 the largest turn a shock can manage is 12.1°; at Mach 2 it is 22.97°; at Mach 3, 34.07°; at Mach 5, 41.1°. Faster flow can be turned further, which is the opposite of what intuition suggests about a flow that is harder to influence.

The maxima converge. As $M \to \infty$ the limiting deflection approaches 45.6° and stops, so there is a turn no attached shock can ever accomplish at any speed whatever.

The curves crowd together near the top. A given deflection on the strong branch produces nearly the same shock angle at every Mach number, which is why the near-normal part of a bow shock looks much the same at Mach 2 as at Mach 6.

The cheapness of leaning #

The practical importance of oblique shocks is that they are much cheaper than normal ones for the same job, and the mechanism is visible in the geometry.

A shock at $\beta = 39.31°$ in a Mach 2 stream processes a normal component of $2 \sin 39.31° = 1.267$. The entropy it produces is that of a Mach 1.267 normal shock, not a Mach 2 one — and since entropy goes as the cube of shock strength, that is a very large saving.

Two oblique shocks and a final normal one at Mach 2 recover around 92 per cent of the total pressure, against 72 per cent for the single normal shock. That difference is the reason for the ramps, cones and wedges on the front of every supersonic intake ever built.

What a wedge does to the pressure, and hence to the force #

An oblique shock is also how a supersonic body carries load, and the pressure coefficient behind one is worth a number.

At Mach 2 the 10° wedge gives $p_2/p_1 = 1.707$, so

$$c_p = \frac{p_2/p_1 - 1}{\tfrac{1}{2}\gamma M_1^2} = \frac{0.707}{2.8} = 0.2523$$

A pressure coefficient of a quarter, produced by a ten-degree ramp, with no circulation, no Kutta condition and no lifting-line theory involved. Everything the subsonic half of this site says about where lift comes from is simply not the mechanism here — the force on a supersonic surface is the local pressure the local turn produced, and it can be read off face by face.

That fact is what makes supersonic aerofoil theory exact and elementary, and it is the deepest practical difference between the two speed ranges.

Where the two branches actually meet a body #

A single body frequently carries both branches at once, which is the best argument that neither is a special case.

Every face is a separate application of the same relation with a different deflection, and the resulting pressures differ by an order of magnitude across the section. That is how a supersonic body carries load, and it is the whole of shock–expansion theory — which is exact, because supersonic flow carries no information upstream and each face can be solved knowing only what happened before it.

The maximum, applied twice #

The maximum deflection is introduced above as the thing that ends the chart, and it does more work than that. Apply it to a second shock and it decides the whole structure of how shocks reflect.

Send an oblique shock from a wedge onto a flat wall. The flow behind it is running at angle $\theta$ to the wall, and the wall requires it to be parallel, so a reflected shock must turn it back by exactly $\theta$ again. Everything about that reflection follows from applying this essay’s relation a second time — at the post-shock Mach number $M_2$ rather than at $M_1$.

And $M_2$ is smaller, so its maximum deflection is smaller. The reflection is therefore possible only while

$$\theta \le \theta_{\max}(M_2),$$

and there is no reason a given incident shock should satisfy it. When it does, the picture is the familiar one: two straight shocks meeting at a point on the wall, a regular reflection.

When it does not, the flow cannot be turned parallel by any single reflected shock at any angle, and it does something else entirely. The two shocks meet above the wall at a triple point, and a third shock runs from there down to the surface — a nearly normal Mach stem, which turns the flow through whatever is required by being strong rather than by being angled. A slip line trails downstream from the triple point, separating gas that crossed one strong shock from gas that crossed two weaker ones: same pressure, same direction, different entropy, different speed, exactly the contact surface the shock tube produces by the same mechanism.

That is Mach reflection, and Mach found it in 1878 in spark photographs of two intersecting spherical waves — three decades before there was a theory of oblique shocks to explain it. It is what a blast wave does when its reflection off the ground can no longer be regular, and the stem’s near-normal strength is why the pressure under it is so much higher than the incident wave’s.

The criterion above is not the only candidate, and where the two candidates disagree the equations admit both structures at once — which is an essay of its own, and one of the few places in this collection where what decides the answer is the history of the experiment.

What the solver computes, and how it is checked #

obliqueShock takes a Mach number, a deflection and a branch. It first computes the maximum deflection available at that Mach number, by ternary search on the smooth unimodal θ(β) curve, and refuses immediately if the requested deflection exceeds it. Then it brackets the requested branch — between the Mach angle and the maximum for weak, between the maximum and 90° for strong — and bisects.

The returned object carries the normal components on both sides as well as the assembled Mach number, so a caller can see the disguise rather than take it on trust: at Mach 2 and 10°, $M_{n1} = 1.267$ and $M_{n2} = 0.8032$, and $M_2 = M_{n2}/\sin(\beta - \theta) = 1.641$.

The rejection tests cover the two ways of asking wrongly. Requesting a deflection past detachment throws, with a message naming the detachment angle. Requesting a branch that is not “weak” or “strong” throws rather than defaulting, because defaulting would be a physical assumption made silently.

Where the model stops #

The wedge is two-dimensional and infinite. A cone at the same half-angle produces a weaker shock at a smaller angle, because the flow can relieve itself in three dimensions rather than two — the Taylor–Maccoll problem, which is a different solve and is not implemented here.

The wedge has a shoulder somewhere. These relations describe the shock at the nose. What happens where the wedge ends is an expansion, and a real body is a sequence of turns rather than one.

No viscosity, so no shock–boundary-layer interaction. A real oblique shock striking a surface with a boundary layer on it produces a separation bubble, a lambda foot and a reflected shock system that this description knows nothing about.

Perfect gas, $\gamma = 1.4$. As on every rung of this field.

What the picture cannot show #

The wedge figure draws both solutions on one wedge, which is a deliberate falsehood in the service of a true point.

Only one of them occurs at a time. Drawing both is a way of showing that the relation has two roots, and it should not be read as a picture of a flow — no experiment ever produced that image.

The figure also draws the shock as attached exactly at the nose with a mathematically sharp apex, and real wedges have a nose radius. At a small enough scale every wedge is blunt, and there is always a tiny detached region at the tip with a curved shock over it. It is invisible at any scale a drawing can show, and it is the reason a real wedge’s shock never quite matches the theoretical angle near the apex.

And, as everywhere in this field, the drawn thickness of the shock line means nothing at all.

Who found it, and when #

The oblique shock relations follow from Rankine and Hugoniot’s work by the resolution argument, and that step belongs to Prandtl’s Göttingen group in the 1900s — Meyer’s 1908 dissertation, supervised by Prandtl, contains both the oblique shock and the expansion fan that bears his name with Prandtl’s.

Busemann drew the shock polar in 1929, which is the same information in a different coordinate system and is better than the θ–β chart for understanding shock interactions. He also proposed the supersonic biplane, an arrangement of two sections whose shock systems cancel each other’s wave drag at one design Mach number — a beautiful idea which works exactly as predicted at that Mach number and not at all elsewhere.

Where the ladder goes next #

Every curve in the θ–β–M chart has a maximum, and the maximum means something physical: past a certain deflection, no shock at any angle can turn the flow that far.

The flow does not stop being possible. It stops being attached — the shock detaches and stands off, curved, with a subsonic pocket behind it, and the elegant algebra of this rung is replaced by a problem that needed computers.