Shock wave angle, deflection angle: Visual walkthrough

Step 1 — What a supersonic flow "sees" when it meets a wedge

WHAT. Picture air streaming to the right at a steady speed, faster than sound. It runs into a solid wedge — a triangle pointing into the wind. The surface of the wedge tilts up by some angle. The air must get out of the way.

WHY. In everyday (slow) flow, air feels the wall coming and curves smoothly around it — pressure "messages" travel upstream and warn it. But these messages travel at the speed of sound. If the air moves faster than sound, no message can outrun it. So the air gets no warning: it slams into a razor-thin sheet and turns all at once. That sheet is the shock.

PICTURE. Two straight lines matter. One is the shock itself, leaning back from the wedge tip. The other is the wedge surface. Two angles are born here — and keeping them apart is the whole story.

Figure — Shock wave angle, deflection angle

Step 2 — Split the velocity into two honest arrows

WHAT. Take the incoming velocity — one arrow of length $V_{1}$ pointing straight into the wind. We replace it with two arrows that add up to it: one pointing straight into the shock face (perpendicular to the shock), one pointing along the shock face.

WHY. A shock is a wall of pressure. Pressure always pushes perpendicular to a surface (like water pressing flat against a dam). So the shock can only shove on the flow across itself — it has no grip along itself (we assume no friction, i.e. inviscid flow). That means:

the along-the-face part of the velocity sails through untouched;
only the into-the-face part gets slammed, squashed, and slowed — exactly like a plain normal shock.

This single insight is why the whole hard problem collapses into a normal-shock problem in disguise.

PICTURE. The right triangle formed by $V_{1}$ and its two pieces. Here $sin β$ = opposite/hypotenuse picks out the perpendicular piece, and $cos β$ = adjacent/hypotenuse picks out the along piece.

Step 3 — Downstream: the same two pieces, but turned

WHAT. After the shock, the flow has a new speed $V_{2}$ and a new direction — bent by $θ$ . We split this arrow the same way: a perpendicular piece $u_{2}$ and an along-piece $w_{2}$ .

WHY. We need to compare "before" and "after" on the same axes (perpendicular vs along the shock). Because the outgoing flow makes angle $(β - θ)$ with the shock face — not $β$ — the split uses $(β - θ)$ .

Where does $(β - θ)$ come from? The shock leans at $β$ from the incoming flow. The outgoing flow sits $θ$ closer to the shock (it bent toward the wall). So the angle left between the outgoing flow and the shock face is $β - θ$ . Read it straight off the figure.

PICTURE. The downstream right triangle, sitting at angle $(β - θ)$ to the shock, with its perpendicular and along legs labelled.

Step 4 — Conservation of mass gives a velocity ratio

WHAT. Mass can't pile up or vanish inside the thin shock. Whatever flows into the face per second flows out. In symbols, $ρ_{1} u_{1} = ρ_{2} u_{2}$ , where $ρ$ (Greek "rho") is density — how much mass sits in each cubic metre.

WHY only the normal piece? Mass crosses the shock by moving through it — that is the perpendicular motion. The along-the-face gliding doesn't carry anything across. So continuity uses $u$ , not $w$ .

PICTURE. A little box straddling the shock: same mass rate in the left face as out the right face.

Rearrange to get a ratio of the perpendicular speeds: $\frac{u _{2}}{u _{1}} = \frac{ρ _{1}}{ρ _{2}}$

Left side: how much the perpendicular speed dropped.
Right side: the inverse of how much the flow got compressed. Denser behind ⇒ slower perpendicular flow behind. That is just "traffic bunches up when it slows down."

Step 5 — Borrow the normal-shock density law

WHAT. For a normal shock, the Rankine–Hugoniot density jump depends only on the normal Mach number $M_{n 1}$ : $\frac{ρ _{2}}{ρ _{1}} = \frac{( γ + 1 ) M _{n 1}^{2}}{( γ - 1 ) M _{n 1}^{2} + 2}, M_{n 1} = M_{1} sin β$

WHY $M_{n 1} = M_{1} sin β$ ? The Mach number is speed ÷ sound-speed. Only the perpendicular piece of the speed feels the shock, and that piece is $V_{1} sin β$ . Divide by the same sound speed and you get $M_{1} sin β$ . We use $M_{n 1}$ — not $M_{1}$ — because a normal shock reacts to the flow hitting it head-on, and only the perpendicular part hits head-on.

WHY this tool? We needed a known relationship between density and the flow. The normal-shock law is that known thing — and we've earned the right to use it because Step 2 proved the perpendicular flow is a normal shock.

PICTURE. The disguise revealed: rotate your eyes so the shock is vertical, and the perpendicular flow is a textbook normal shock; the along-flow is just a constant sideways drift.

Combine with Step 4: $\frac{u _{2}}{u _{1}} = \frac{ρ _{1}}{ρ _{2}} = \frac{( γ - 1 ) M _{n 1}^{2} + 2}{( γ + 1 ) M _{n 1}^{2}}$ (We flipped the fraction because we want $ρ_{1} / ρ_{2}$ , the reciprocal.)

Step 6 — Turn speeds into angles with $tan$

WHAT. On each side, divide the perpendicular piece by the along piece: $tan β = \frac{u _{1}}{w _{1}}, tan (β - θ) = \frac{u _{2}}{w _{2}} = \frac{u _{2}}{w _{1}}$

WHY $tan$ , and why now? On a right triangle, $tan$ (angle) = opposite ÷ adjacent = perpendicular piece ÷ along piece. That is exactly the ratio of our two velocity components. So $tan$ is the tool that converts a velocity ratio into the angle we actually care about. We use it now because we finally have both components on both sides.

The key cancellation. On the downstream side $w_{2} = w_{1}$ (Step 2). So when we divide the two tangents, the along-piece $w_{1}$ cancels and only the perpendicular ratio survives: $\frac{t a n ( β - θ )}{t a n β} = \frac{u _{2} / w _{1}}{u _{1} / w _{1}} = \frac{u _{2}}{u _{1}}$

PICTURE. Both triangles share the same horizontal base $w_{1}$ ; only their heights ( $u_{1}$ vs $u_{2}$ ) differ. That shared base is why the ratio cleans up.

Now stitch Steps 4–6 together: $\frac{tan ( β - θ )}{tan β} = \frac{( γ - 1 ) M _{n 1}^{2} + 2}{( γ + 1 ) M _{n 1}^{2}}$ This is the θ–β–M relation — just not yet combed into its prettiest form.

Step 7 — Comb it into the standard compact form

WHAT. Substitute $M_{n 1} = M_{1} sin β$ , expand the tangents with $tan = sin / cos$ , and use the identity $cos 2 β = 1 - 2 sin^{2} β$ . After the algebra dust settles:

WHY this form? It isolates $tan θ$ alone on the left, so you can sweep $β$ from small to large and plot $θ$ — which is how we discover the weak/strong branches next.

PICTURE. The $θ$ -vs- $β$ curve for a fixed $M_{1}$ : rising from zero, cresting at a maximum, falling back to zero.

Step 8 — Every case, read off the curve

WHAT & WHY. The curve in Step 7 tells us how many shocks a given wedge produces. Walk across it:

PICTURE. The same curve, now with a horizontal line at your chosen $θ$ : it crosses the curve twice (weak + strong), grazes it once (at $θ_{m a x}$ ), or misses entirely (detached).

The one-picture summary

Everything on one canvas: incoming arrow split into perpendicular ( $u_{1}$ ) and along ( $w_{1}$ ); the shock as a leaning line at $β$ ; the outgoing arrow bent by $θ$ , its own split into $u_{2}$ (smaller) and $w_{2} = w_{1}$ (unchanged); and the reminder that the perpendicular story is a normal shock.

Recall Feynman retelling — the whole walkthrough in plain words

Air races into a wedge too fast to be warned, so it crashes onto a thin leaning line — the shock, tilted at angle $β$ . We split the air's speed into two arrows: one poking straight into that line, one sliding along it. The sliding one doesn't care about the shock and passes through unchanged — because the shock's push is all sideways, straight into its own face, with nothing gripping along it. Only the poking-in arrow gets crushed and slowed, exactly like a plain head-on (normal) shock, which we already know the rules for. "How much did it slow?" is the same as "how much did it get squished?" (traffic bunches when it slows) — and the squish law only needs the poking-in Mach number, $M_{1} sin β$ . Finally, dividing the poke by the slide on each side turns velocities into angles via $tan$ , the sliding arrow cancels because it's the same on both sides, and out drops the θ–β–M formula. Plot it and you see the whole life story: a gentle Mach wave at one end, a square normal shock at the other, two shocks (weak and strong) for any modest turn in between, and — if you demand too sharp a turn — no attached shock at all, just a bow shock floating out front.

#recall

Why does only the normal velocity component get changed by the shock?

Pressure acts perpendicular to the shock face and there's no friction, so no force acts along the face — the tangential (along) component is conserved, only the normal one is compressed.

Why is

M_{n 1} = M_{1} sin β

and not

M_{1} cos β

The perpendicular velocity is

V_{1} sin β

(the side opposite

β

on the velocity triangle); divide by the sound speed to get the perpendicular Mach number.

Why does the along-component

w_{1}

cancel in Step 6?

Because

w_{2} = w_{1}

, so when you divide

tan (β - θ) / tan β = (u_{2} / w_{1}) / (u_{1} / w_{1})

the common

w_{1}

cancels, leaving

u_{2} / u_{1}

β = 9 0^{\circ}

why is

θ = 0

cot 9 0^{\circ} = 0

, so

tan θ = 0

— a normal shock turns the flow by nothing.

What does

M_{1}^{2} sin^{2} β - 1

represent in the formula?

The shock strength

M_{n 1}^{2} - 1

; positive = real shock, zero = Mach wave.

For

θ > θ_{m a x}

, what forms?

A detached curved bow shock ahead of the body — no attached oblique shock exists.