Question bank — Oblique shock waves — θ-β-M relation

True — the normal component is compressed and slowed like a normal shock, and since the tangential part $w$ is unchanged, the total speed drops; $M_2<M_1$ always, even if $M_2$ stays supersonic.

True — only the normal component must go subsonic ($M_{n2}<1$); for a weak shock the surviving tangential slide $w$ can keep the resultant $M_2>1$.

True — in the relation $\cot \beta =0$, so $\tan \theta =0$; the whole velocity is normal to the shock, nothing slides, so there is no direction to bend into and the flow crosses straight through.

False — the θ-β curve is a hump, so each $\theta <{\theta }_{max}$ has two valid $\beta$: a weak (small $\beta$) and a strong (large $\beta$) solution.

False for the weak branch beyond a point, and outright false past ${\theta }_{max}$ — once the demanded turn exceeds ${\theta }_{max}$ no attached solution exists at all and the shock detaches.

True in spirit — there is no pressure gradient along the shock face, so no tangential force acts on $w$, hence tangential momentum (and speed) is unchanged.

True — as $\theta \to 0$ the numerator $M_1^2{\sin }^2\beta -1\to 0$, forcing $\sin \beta =1/M_1$, i.e. $\beta =\mu$ the Mach angle; the "shock" carries no jump. See Mach Angle and Mach Waves.

False — every compression shock raises entropy; weak shocks raise it very little (third-order small), but never decrease it. That is why expansion happens smoothly instead, via Prandtl-Meyer Expansion.

Wrong Mach number — only the normal component crosses like a normal shock, so you must use $M_{n1}=M_1\sin \beta$, which is smaller and gives the correct (weaker) compression.

No — the shock stands ahead of the surface at angle $\beta >\theta$; the flow between shock and wall runs parallel to the wall, but the shock itself is more steeply inclined than the wall.

Missing the geometry — $M_{n2}$ is only the normal piece; recover $M_2=M_{n2}/\sin (\beta -\theta )$, because downstream the normal direction makes angle $(\beta -\theta )$ with the flow.

Backwards for external flow — for a wedge in free flight the weak shock is realised because the downstream pressure is just the free-stream ambient value the weak branch produces; the strong branch requires a higher imposed back-pressure to hold it, so it survives only in confined/internal flows where a wall or throat supplies that pressure. "Stability" here is not an inherent property — it is whichever branch matches the actual downstream pressure.

Not quite — the tangential speed $w$ is conserved, but the temperature (and thus sound speed $a$) jumps up across the shock, so the tangential Mach number $w/a$ actually decreases.

No shock at all — $\sin \beta <1/M_1$ means $M_{n1}<1$, a subsonic normal component, which cannot support a shock; only $\sin \beta \ge 1/M_1$ is physical.

Not speed but turning — the bow shock appears when the required deflection $\theta$ exceeds ${\theta }_{max}$ for that $M_1$; a blunt body demands a turn no attached shock can supply.

Because the shock physics (the Rankine–Hugoniot jump) act purely across the shock face; splitting lets us reuse the entire normal-shock machinery on just the normal component $u_{n1}$.

The pressure jumps only across the shock, so the force acts only in the normal direction; there is nothing pushing sideways, so the sideways slide $w$ continues untouched.

A shock is a supersonic-compression phenomenon in its own crossing direction; if the normal component were subsonic there is no supersonic flow to shock, so the wave degenerates to a Mach wave or nothing.

Trace the numerator $M_1^2{\sin }^2\beta -1$ against the prefactor $\cot \beta$: at small $\beta$ the numerator is near zero so $\theta \approx 0$; at $\beta =90^{\circ }$ the $\cot \beta$ prefactor is zero so $\theta =0$ again. Since $\theta$ starts at zero, ends at zero, and is positive in between, it must rise to a peak ${\theta }_{max}$ and fall — that peak is the hump.

The two branches are both mathematically valid; which one appears is set by downstream boundary conditions — open flight relieves pressure and selects weak, while a required high back-pressure selects strong.

The relation's $\theta$ has two zeros — one where the numerator $M_1^2{\sin }^2\beta -1$ vanishes ($\beta =\mu$, an infinitely weak Mach wave) and one where the $\cot \beta$ prefactor vanishes ($\beta =90^{\circ }$, a full normal shock). Same zero deflection, opposite strengths.

Three-dimensional relief — flow around a cone can spread azimuthally, easing the effective turn, so cone shocks stay attached at angles that would detach a 2-D wedge.

$\beta$ approaches a finite minimum, not zero. In the relation, as $M_1\to \infty$ the $M_1^2$ terms dominate top and bottom and cancel, leaving $\tan \theta \to 2\cot \beta \frac{{\sin }^2\beta }{\gamma +\cos 2\beta }$ — an equation with no $M_1$ left. A fixed $\theta$ therefore fixes a finite $\beta$; the shock hugs closer to the wedge but never flattens onto it, because a real deflection still needs a real normal component $u_{n1}$.

The Mach angle $\mu =\arcsin (1/M_1)\to 90^{\circ }$, so even the weakest wave stands nearly normal, and ${\theta }_{max}$ shrinks toward zero — almost any turn detaches the shock.

Zero — this is the Mach-wave limit where $M_1^2{\sin }^2\beta -1=0$, the flow slides through with no compression and no turn.

No oblique shock forms — a compression shock can only turn the flow into itself; a convex turn is handled smoothly by a Prandtl–Meyer expansion fan instead.

Neither cleanly — the two roots merge into one at the hump's peak; a hair more turning and there is no attached solution, so it is the last attached shock before detachment.

A zero-strength Mach wave — $M_{n1}=1$ means no compression jump, so it marks the boundary between "genuine shock" ($M_{n1}>1$) and "no shock."

It is still slightly supersonic there. Follow $M_2$ along the curve: as $\beta$ grows the shock gets stronger and $M_2$ falls, reaching $M_2=1$ (sonic) only past the peak, a little way up the strong branch. Because that sonic point sits at a $\beta$ slightly larger than the one giving ${\theta }_{max}$, the flow at ${\theta }_{max}$ itself is not yet subsonic — the "$M_2=1$" and "$\theta ={\theta }_{max}$" points are close but not identical.