Oblique shock waves — θ-β-M relation: Visual walkthrough

Step 1 — The picture before any symbols: a fast flow meets a corner

WHAT. A stream of air is moving fast (faster than sound) along a wall. The wall suddenly bends into the flow by a small angle. The air cannot flow around a sharp inward corner smoothly, so a thin, straight, tilted line forms in the air — a shock. The flow crosses that line, bends, and slows.

WHY start here. Everything below is bookkeeping on one drawing. If you hold this drawing in your head, no formula will surprise you. We are just naming three angles.

PICTURE. Look at the figure.

The cyan arrow coming in from the left is the incoming velocity. Call its length $V_{1}$ (speed 1 = "upstream").
The amber line is the shock. The angle it makes with the incoming arrow is $β$ (the "shock-wave angle"). We measure $β$ from the incoming flow up to the shock.
After crossing, the flow follows the new wall. The angle between the old direction and the new direction is $θ$ (the "deflection angle" — how much the flow got bent).

Figure — Oblique shock waves — θ-β-M relation

This is exactly the setup in Supersonic Wedge and Cone Flow: the wall bend is the wedge half-angle.

Step 2 — The one clever move: split each velocity into "into the shock" and "along the shock"

WHAT. Take the incoming arrow and break it into two pieces using the shock line as the reference:

a piece perpendicular to the shock (pointing straight into it), and
a piece parallel to the shock (sliding along it).

We do the same for the outgoing arrow.

WHY this tool, and not something else? A shock is a wall of sudden pressure change. Pressure pushes perpendicular to a surface. So the only direction where the shock can shove the flow is straight across itself. The sliding-along direction feels no push. Splitting into perpendicular + parallel is therefore the natural coordinate system — it separates "the direction that changes" from "the direction that doesn't." Any other axes would tangle the two together.

PICTURE. The right-angle boxes show the split.

$u_{n 1} = V_{1} sin β$ — the normal (perpendicular) part of the incoming arrow. Here $sin β$ appears because the perpendicular leg of a right triangle with hypotenuse $V_{1}$ and angle $β$ at the corner is (hypotenuse) $\times sin β$ .
$w_{1} = V_{1} cos β$ — the tangential (along-shock) part. The adjacent leg is (hypotenuse) $\times cos β$ .

The subscript convention: $n$ = normal, $t$ or $w$ = tangential, $1$ = upstream, $2$ = downstream.

Step 3 — The tangential slide is untouched (this is the master fact)

WHAT. The along-the-shock speed is the same before and after: $w_{1} = w_{2} ⟹ V_{1} cos β = V_{2} cos (β - θ) .$

WHY. No force acts along the shock (Step 2 reason: pressure only pushes across it). No sideways force ⇒ no change in sideways momentum ⇒ the tangential speed is conserved. This single sentence is the whole trick behind oblique shocks.

WHY $(β - θ)$ appears. The outgoing flow has been bent toward the wall by $θ$ . So the angle between the outgoing arrow and the shock line is no longer $β$ — it is $β$ minus the bend $θ$ . Look at the figure: the downstream triangle sits at angle $(β - θ)$ to the same amber shock line.

Step 4 — Turn the two velocity facts into one ratio of angles

WHAT. Divide the two normal-component equations, then kill $V_{2} / V_{1}$ using the tangential fact.

Start with the ratio of normal speeds: $\frac{u _{n 2}}{u _{n 1}} = \frac{V _{2} s i n ( β - θ )}{V _{1} s i n β} .$

From Step 3, $\frac{V _{2}}{V _{1}} = \frac{cos β}{cos ( β - θ )}$ . Substitute: $\frac{u _{n 2}}{u _{n 1}} = \frac{c o s β}{c o s ( β - θ )} \cdot \frac{s i n ( β - θ )}{s i n β} = \frac{t a n ( β - θ )}{t a n β} .$

WHY do this. We want an equation with no speeds left — only angles and Mach number. The speeds $V_{1}, V_{2}$ are unknown and annoying; the ratio trick cancels them. What survives is a pure-geometry statement: the fractional slow-down across the shock equals a ratio of tangents.

WHY tangent specifically? $tan = \frac{sin}{cos} = \frac{opposite}{adjacent} = \frac{across-shock part}{along-shock part}$ . Tangent is exactly the ratio of the two legs we split into. So $tan β$ and $tan (β - θ)$ literally measure "how steeply the flow dives into the shock" before and after.

PICTURE. The figure stacks the upstream triangle and downstream triangle sharing the same base $w$ (because $w$ is conserved). The heights are $u_{n 1}$ and $u_{n 2}$ . Same base, shorter height downstream ⇒ the flow got flattened against the shock ⇒ $tan (β - θ) < tan β$ .

Step 5 — Feed in the physics: the normal shock hiding inside

WHAT. So far it was all triangles. Now use one physics law. Along the perpendicular direction, the flow behaves exactly like a normal shock with Mach number $M_{n 1} = M_{1} sin β .$ The mass-conservation + Rankine–Hugoniot result for a normal shock gives the velocity drop: $\frac{u _{n 2}}{u _{n 1}} = \frac{ρ _{1}}{ρ _{2}} = \frac{( γ - 1 ) M _{n 1}^{2} + 2}{( γ + 1 ) M _{n 1}^{2}} .$

WHY we may borrow the normal-shock law. The whole point of splitting in Step 2 was this: the perpendicular slice is a self-contained normal shock, riding on an unchanged sideways slide. So every normal-shock formula applies if you feed it $M_{n 1}$ , not $M_{1}$ .

Reading the symbols.

$M_{n 1} = M_{1} sin β$ ::: the effective upstream Mach for the normal problem. Only this part "sees" the shock.
$γ$ ::: ratio of specific heats, $1.4$ for air — it sets how squishy the gas is.
The right-hand fraction ::: is always $< 1$ when $M_{n 1} > 1$ (flow slows), and $= 1$ when $M_{n 1} = 1$ (no shock).

PICTURE. The figure isolates the perpendicular strip and shows it as a mini normal-shock: incoming $u_{n 1}$ , a thin shock, slower $u_{n 2}$ , with the sideways slide $w$ drawn faded to remind us it's just going along for the ride.

Step 6 — Equate the two ratios and read the θ-β-M relation

WHAT. Step 4 (geometry) and Step 5 (physics) are two expressions for the same ratio $u_{n 2} / u_{n 1}$ . Set them equal: $\frac{t a n ( β - θ )}{t a n β} = \frac{( γ - 1 ) M _{1}^{2} s i n ^{2} β + 2}{( γ + 1 ) M _{1}^{2} s i n ^{2} β} .$

WHY. A ratio equals itself. Geometry made one side, physics made the other; forcing them to agree is exactly the condition that a real oblique shock exists for these angles.

Cleaning up. Expanding $tan (β - θ)$ and solving for $tan θ$ (routine trig algebra, verified below) collapses everything into the compact form:

The figure plots $θ$ against $β$ for a fixed $M_{1}$ : the famous hump. Trace it left to right and read off the two special crossings named in the next step.

Step 7 — Every edge case, read straight off the numerator and the hump

The single term $(M_{1}^{2} sin^{2} β - 1)$ controls all the degenerate cases. Watch its sign.

Case A — $sin β = 1/ M_{1}$ (the weakest shock, a Mach wave). Here $M_{1}^{2} sin^{2} β - 1 = 0$ , so $tan θ = 0$ ⇒ $θ = 0$ . No bending. The shock is infinitely weak: it's a Mach wave at the Mach angle $μ = arcsin (1/ M_{1})$ . This is the leftmost foot of the hump.

Case B — $β = 9 0^{\circ}$ (straight-on). $cot 9 0^{\circ} = 0$ , so $tan θ = 0$ ⇒ $θ = 0$ again. A shock perpendicular to the flow bends nothing — it's a pure normal shock. This is the rightmost foot of the hump.

Case C — the peak $θ_{m a x}$ . Between those two feet the curve rises to a maximum $θ_{m a x}$ and comes back down. For any $θ < θ_{m a x}$ the horizontal line cuts the hump twice:

the weak solution (smaller $β$ , left of peak) — usually what nature picks, downstream often still supersonic;
the strong solution (larger $β$ , right of peak) — downstream subsonic.

Case D — $θ > θ_{m a x}$ . The line misses the hump entirely: no real $β$ exists. The attached shock cannot form; it pops off the body as a curved bow shock.

Case E — $sin β < 1/ M_{1}$ (numerator negative). $tan θ$ would come out negative — an unphysical "expansion turn." That regime belongs to Prandtl-Meyer Expansion (a convex corner), not a shock. So we simply forbid $β$ below the Mach angle.

The one-picture summary

This final figure compresses the whole chain: one incoming arrow → split into normal + tangential → tangential conserved, normal does a normal shock → equate the two velocity ratios → the θ-β-M curve with its Mach-wave foot, its hump peak $θ_{m a x}$ , and its normal-shock foot. Trace the arrows in the order s02 → s07.

Recall Feynman retelling of the whole walkthrough

A fast river of air runs into a slanted line (the shock). I split the river's speed into two: the part driving straight into the line, and the part sliding along it. The sliding part doesn't care about the shock — nothing pushes sideways — so it stays exactly the same on both sides. The straight-in part gets slammed, and it obeys the same rules as a head-on (normal) shock, but only if I feed it its own smaller speed, $M_{1} sin β$ . Now I have two ways to write how much the straight-in part slowed: one from pure triangle geometry ( $tan (β - θ) / tan β$ ), one from shock physics (the density formula). They describe the same slow-down, so I set them equal. Rearranging gives one tidy equation linking the turn angle $θ$ , the shock tilt $β$ , and the speed $M_{1}$ . Reading its numerator $M_{1}^{2} sin^{2} β - 1$ tells me the whole story: zero at the gentle Mach wave and at the head-on shock (no bending either way), a hump in between whose peak is the biggest turn a clean shock can make — ask for more and the shock gives up and curves off the body.