Shock wave angle, deflection angle: Question bank

Before the traps, let's put every symbol on the page with a picture, so nothing is used before it's earned.

Look at the sketch: the incoming arrow, the tilted shock line, and the bent outgoing arrow are all we ever draw. Figure s01 shows how the single velocity arrow splits into an into-the-shock piece and an along-the-shock piece — that split is the whole trick.

Figure — Shock wave angle, deflection angle

Now the geometry step-by-step: only the normal piece gets squashed (like a normal shock), while the tangential piece glides through unchanged. Figure s02 shows this reduction and where $M_{n 1}$ , $M_{n 2}$ , and the turn $θ$ live.

Finally, the core relation the whole bank leans on — the parent $\theta$–$\beta$–$M$ relation:

This page also builds on its neighbours Mach Angle and Mach Waves, Rankine-Hugoniot Relations, Prandtl-Meyer Expansion, Detached Bow Shock, and Supersonic Wedge & Cone Flow.

True or false — justify

A normal shock is a special case of an oblique shock.

True — it is the oblique shock with

β = 9 0^{\circ}

, where the whole velocity is normal and none is tangential, so

θ \to 0

Increasing the wedge half-angle

θ

always increases the wave angle

β

on the weak branch.

True on the weak branch up to

θ_{m a x}

— larger turning needs a steeper shock; but past

θ_{m a x}

there's no attached

β

at all (the shock detaches).

For a fixed

M_{1}

, every deflection angle

θ < θ_{m a x}

corresponds to exactly one shock angle

β

False — there are two: a weak shock (smaller

β

, flow stays supersonic) and a strong shock (larger

β

, flow goes subsonic).

Tangential velocity is conserved across the oblique shock because the flow is slow along the shock.

False — it's conserved because the (inviscid) shock exerts pressure only perpendicular to its face, so there's zero tangential force regardless of how fast the tangential flow is.

Raising

M_{1}

at fixed

θ

lowers the weak-shock wave angle

β

True — a faster flow can turn the same amount through a more oblique (shallower) shock, so

β

drops toward the (also-shrinking) Mach angle.

The maximum deflection angle

θ_{m a x}

is the same for every Mach number.

False —

θ_{m a x}

grows with

M_{1}

; a faster stream can be turned through a larger attached-shock angle before detaching.

At the Mach angle limit (

β = μ

), the "shock" carries no compression.

True — there

M_{n 1} = M_{1} sin μ = 1

, so the normal Mach number is exactly 1: an infinitely weak wave (a Mach wave), no pressure jump,

θ = 0

Behind a weak oblique shock the flow can still be supersonic even though

M_{n 2} < 1

True — the normal component is subsonic (

M_{n 2} < 1

), but

M_{2} = M_{n 2} / sin (β - θ)

divides by a small sine, restoring a supersonic total speed.

Both the lower limit (

β = μ

) and the upper limit (

β = 9 0^{\circ}

) give zero deflection.

True — at

μ

the shock is vanishingly weak; at

9 0^{\circ}

it's a normal shock with purely head-on flow, so in both cases the streamline doesn't bend.

Spot the error

"Since the algebra matches a normal shock, plug

M_{1}

into the density-ratio formula directly."

Error: you must use the normal Mach number

M_{n 1} = M_{1} sin β

, not

M_{1}

, in every normal-shock relation — only the perpendicular component is shocked.

"The deflection angle equals the wave angle at the corner."

Error:

β

is measured from the incoming flow to the shock line,

θ

is the streamline bend; always

θ < β

, and at

β = 9 0^{\circ}

we have

θ = 0

— they are genuinely different.

"To recover the downstream Mach number, use

M_{2} = M_{n 2}

Error:

M_{n 2}

is only the component normal to the shock; the full speed is

M_{2} = M_{n 2} / sin (β - θ)

because the tangential component is still present downstream.

"The oblique-shock relation gives you

β

uniquely from

θ

and

M_{1}

Error: inverting the

θ

–

β

–

M

relation yields two

β

roots (weak and strong); a third "solution" at

β

near

μ

collapses to the Mach wave only when

θ = 0

"A larger wedge just makes a bigger, steeper, still-attached shock forever."

Error: once

θ > θ_{m a x}

no attached oblique shock exists — the shock detaches into a curved bow shock standing off ahead of the body (Detached Bow Shock).

"The shock slows the tangential velocity too, just less than the normal part."

Error: the tangential velocity is completely unchanged (

w_{2} = w_{1}

); zero tangential force means zero tangential deceleration — this is the cornerstone of the derivation.

"If

M_{1} sin β < 1

we still get a compression shock."

Error:

M_{n 1} = M_{1} sin β

must exceed 1 for a shock to exist; if it's below 1 no shock forms (the "

β

" you chose is shallower than the Mach angle, which is physically impossible for a shock).

Why questions

Why do we split the velocity into normal and tangential components at all?

Because only the normal component is compressed like a normal shock while the tangential glides through unchanged — the split (Figure s01) reduces a 2-D oblique problem to a 1-D normal-shock problem plus geometry.

Why is

M_{n 1} = M_{1} sin β

(sine, not cosine)?

β

is measured from the flow to the shock, so the component into the shock face is

V_{1} sin β

; dividing by the sound speed gives

M_{1} sin β

Why does the

θ

-vs-

β

curve return to zero at both ends?

β = μ

the shock is infinitely weak (no turning), and at

β = 9 0^{\circ}

the flow hits head-on with no sideways deflection — a curve rising from 0 and returning to 0 must peak in between, giving

θ_{m a x}

(Figure s03).

Why does exceeding

θ_{m a x}

force a detached shock rather than a very steep attached one?

No attached geometry can turn the flow more than

θ_{m a x}

; the flow "gives up" on turning abruptly at the wedge and instead compresses through a curved bow shock that stands off, where local turning stays within limits everywhere.

Why is the weak solution the one usually seen in external aerodynamics?

The weak shock leaves the flow supersonic and matches the low-back-pressure environment of free flight; the strong (subsonic) branch typically requires a downstream pressure constraint that external flow doesn't impose.

Why does mass conservation across the shock face use only the normal velocity?

Mass crosses the shock only by moving through its face, i.e. via the normal component;

ρ_{1} u_{1} = ρ_{2} u_{2}

, which is why

ρ_{2} / ρ_{1} = u_{1} / u_{2}

Why is the Mach angle

μ = sin^{- 1} (1/ M_{1})

the smallest possible

β

A shock needs

M_{n 1} = M_{1} sin β \geq 1

; the smallest

β

satisfying this has

sin β = 1/ M_{1}

, i.e.

β = μ

— any shallower and the normal Mach would drop below 1.

Edge cases

What happens to

θ

and the shock as

β \to 9 0^{\circ}

The shock becomes a normal shock, tangential component

\to 0

, and the deflection

θ \to 0

— strong compression but no turning (Normal Shock Waves).

What happens as

β \to μ

(the Mach angle)?

M_{n 1} \to 1

, the shock weakens to a Mach wave carrying no finite compression, and

θ \to 0

(Mach Angle and Mach Waves).

What if you plug in a

β

smaller than the Mach angle?

The formula gives

M_{1}^{2} sin^{2} β - 1 < 0

, so

tan θ

turns negative — physically meaningless; it signals that no shock exists below

μ

At exactly

θ = θ_{m a x}

, how many solutions are there?

Exactly one — the weak and strong branches merge at the peak of the

θ

–

β

curve, the unique attached shock right at the detachment threshold.

What is the weak-branch limiting behaviour as

M_{1} \to \infty

at fixed

θ

The weak wave angle

β

approaches a finite hypersonic value (not zero), so the shock lies close to the wedge surface even though the Mach angle

μ \to 0

What is the strong-branch limiting behaviour as

M_{1} \to \infty

The strong-shock

β

stays near

9 0^{\circ}

(approaching a normal shock) and

θ_{m a x}

approaches a finite ceiling (about

45. 6^{\circ}

for

γ = 1.4

) — a very fast stream can be turned more, but not without bound.

What happens at exactly sonic upstream,

M_{1} = 1

The Mach angle

μ = sin^{- 1} (1) = 9 0^{\circ}

collapses onto the upper limit, so the only allowed

β

9 0^{\circ}

and

θ_{m a x} \to 0

— no finite turning is possible; the wedge cannot be pierced by an oblique shock, marking the exact threshold below which shocks cease to exist.

What happens for a wedge in subsonic flow (

M_{1} < 1

No shock at all — the Mach angle isn't defined (

1/ M_{1} > 1

has no arcsine), signals travel upstream, and the flow adjusts smoothly instead of shocking (Supersonic Wedge & Cone Flow).

If the corner turns the flow away from itself (an expansion corner) instead of into a wedge, what forms?

Not a shock but a smooth Prandtl–Meyer expansion fan — the flow accelerates and cools through a continuous fan of Mach waves (Prandtl-Meyer Expansion).

Recall One-line self-test

Cover everything: name the two $β$ -solutions, their supersonic/subsonic status, and what replaces them past $θ_{m a x}$ . Answer ::: Weak (smaller $β$ , downstream supersonic) and strong (larger $β$ , downstream subsonic); beyond $θ_{m a x}$ a detached bow shock forms with no attached solution.