Worked examples — Shock wave angle, deflection angle

This page is the "put it to work" companion to the parent topic. We will not re-derive the $\theta$–$\beta$–$M$ relation here — we use it, on every kind of input it can be handed. Before we start, one reminder of the two tools we will lean on repeatedly:

Every one of these formulas comes straight from Normal Shock Waves applied to the perpendicular component, plus the Rankine-Hugoniot Relations that connect the two sides.

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The scenario matrix

Every problem this topic can throw at you falls into one of these cells. The examples below each announce which cell they cover so you can see the whole space is filled.

#	Case class	What is special about it	Example
A	Forward problem ($\beta$ given → find $\theta$)	Plug straight in, no root-finding	Ex 1
B	Inverse, weak root ($\theta$ given → smaller $\beta$)	The usual external-aero case	Ex 2
C	Inverse, strong root ($\theta$ given → larger $\beta$)	Same $\theta$, subsonic exit	Ex 3
D	Degenerate: $\theta =0$	Two limits — Mach wave & normal shock	Ex 4
E	Limiting: $\beta =90^{\circ }$	Reduces exactly to a normal shock	Ex 5
F	Over-turning: $\theta >{\theta }_{\max }$	No attached solution → detached bow shock	Ex 6
G	Real-world word problem	Extract angles from a physical picture	Ex 7
H	Exam twist: find ${\theta }_{\max }$ itself	Where weak and strong roots merge	Ex 8

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Example 1 — Cell A: forward problem ($\beta$ given)

Forecast: $\beta =40^{\circ }$ is a fairly shallow shock at high Mach — expect a healthy turn, somewhere in the $20$–$25^{\circ }$ range. Guess before reading on.

1. Compute the numerator piece $M_1^2{\sin }^2\beta -1$. $M_1^2{\sin }^2\beta =9{\sin }^{2}40^{\circ }=9(0.6428{)}^2=3.719$, so the numerator is $3.719-1=2.719$. Why this step? This factor is the "shock strength" — it is $M_{n1}^2-1$, positive only when the normal Mach exceeds $1$. A positive value confirms a genuine shock, not a Mach wave.

2. Compute the denominator $M_1^2(\gamma +\cos 2\beta )+2$. $\cos 2\beta =\cos 80^{\circ }=0.1736$, so $9(1.4+0.1736)+2=9(1.5736)+2=16.16$. Why this step? This is just the algebra of the compact formula; nothing physical to interpret, but it must be built before we divide.

3. Assemble $\tan \theta$. $\cot 40^{\circ }=1/\tan 40^{\circ }=1.1918$. $\tan \theta =2(1.1918)\frac{2.719}{16.16}=0.4011\Rightarrow \theta =\arctan (0.4011)=21.9^{\circ }.$ Why this step? The relation is written to give us $\theta$ directly once $M_1,\beta$ are fixed — that is the entire point of the "forward" direction.

Verify: $\theta =21.9^{\circ }<\beta =40^{\circ }$ ✓ (the turn is always the smaller angle). And $M_{n1}=3\sin 40^{\circ }=1.93>1$ ✓ — a real compression shock.

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Example 2 — Cell B: inverse problem, weak root

Forecast: $\beta$ must lie between the Mach angle $\mu =\arcsin (1/2)=30^{\circ }$ and $90^{\circ }$. The weak root sits near the low end. Guess ~$45^{\circ }$.

1. Bracket $\beta$. The Mach angle $\mu =30^{\circ }$ is the floor; the weak root is the first $\beta$ (as $\beta$ climbs from $30^{\circ }$) that satisfies $\tan 15^{\circ }=0.2679$. Why this step? The $\theta$–$\beta$ curve rises then falls, so two $\beta$ give the same $\theta$. The smaller one is the weak branch — pick the root below ${\theta }_{\max }$'s location.

2. Solve $0.2679=2\cot \beta \frac{4{\sin }^2\beta -1}{4(1.4+\cos 2\beta )+2}$. Numerically this gives $\beta =45.3^{\circ }$ (weak). Why this step? No closed form for $\beta$ exists, so we root-find. This is the branch that occurs on real wedges in supersonic flight.

3. Check the exit Mach number. $M_{n1}=2\sin 45.3^{\circ }=1.421$. Feed into the normal-shock formula: $M_{n2}^2=\frac{1.421^2+5}{7(1.421^2)-1}=\frac{7.019}{13.14}=0.534\Rightarrow M_{n2}=\mathrm{0.731.}$ Then $M_2=\frac{0.731}{\sin (45.3^{\circ }-15^{\circ })}=\frac{0.731}{\sin 30.3^{\circ }}=\mathrm{1.45.}$ Why this step? $M_{n2}<1$ always (any shock makes the normal flow subsonic), but the total $M_2=1.45>1$ — that is the signature of the weak branch.

Verify: $M_{n1}=1.421>1$ ✓, $M_{n2}=0.731<1$ ✓, $M_2=1.45>1$ ✓ — all three consistent with a weak oblique shock. Matches parent note.

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Example 3 — Cell C: inverse problem, strong root

Forecast: The strong root lies near $90^{\circ }$ (close to a normal shock). Guess ~$80^{\circ }$.

1. Take the second root of the same equation $\tan 15^{\circ }=0.2679$. Root-finding above the curve's peak gives $\beta =79.8^{\circ }$ (strong). Why this step? Same $\theta$, same $M_1$, but the shock is now nearly perpendicular — much stronger compression.

2. Normal Mach upstream. $M_{n1}=2\sin 79.8^{\circ }=1.969$. Why this step? A steeper shock means a larger normal component, hence a stronger shock.

3. Exit Mach. $M_{n2}^2=\frac{1.969^2+5}{7(1.969^2)-1}=\frac{8.877}{26.14}=0.3396\Rightarrow M_{n2}=\mathrm{0.583.}$ $M_2=\frac{0.583}{\sin (79.8^{\circ }-15^{\circ })}=\frac{0.583}{\sin 64.8^{\circ }}=\mathrm{0.644.}$ Why this step? Here the total exit Mach drops below $1$ — the strong branch turns the flow subsonic.

Verify: $M_2=0.644<1$ ✓ subsonic exit — the defining feature of the strong branch. Same $\theta =15^{\circ }$ as Ex 2 but a completely different flow state, exactly as the two-root picture predicts.

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Example 4 — Cell D: degenerate input $\theta =0$

Forecast: Zero turning should collapse the two roots onto the two extreme shock types. Guess: the Mach angle and $90^{\circ }$.

1. Set the formula's numerator to zero. $\tan \theta =0$ forces either $\cot \beta =0$ or $M_1^2{\sin }^2\beta -1=0$. Why this step? A product equals zero when a factor is zero — that separates the two limits cleanly.

2. Factor 1: $M_1^2{\sin }^2\beta =1$. $\sin \beta =1/M_1=1/2.5=0.4\Rightarrow \beta =\mu =23.6^{\circ }$ — the Mach angle. Why this step? This is the vanishingly weak "shock" — a Mach wave that turns nothing.

3. Factor 2: $\cot \beta =0\Rightarrow \beta =90^{\circ }$. A perpendicular shock — a normal shock — which also produces zero turning. Why this step? At $\beta =90^{\circ }$ the entire velocity is normal; there is no tangential component to redirect, so $\theta =0$.

Verify: Both limits give $\theta =0$: $\mu =\arcsin (0.4)=23.6^{\circ }$ ✓ and $\beta =90^{\circ }$ ✓. These are exactly the lower and upper bounds of $\beta$ stated in the parent note.

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Example 5 — Cell E: limiting case $\beta =90^{\circ }$

Forecast: With the shock perpendicular, $M_{n1}$ should equal $M_1$ itself, and $M_2$ should be the standard normal-shock answer (~$0.475$ for $M=3$).

1. Deflection is zero. From Ex 4's Factor 2, $\beta =90^{\circ }\Rightarrow \theta =0$. Why this step? Confirms we're on the "normal shock" endpoint — no turning.

2. Normal Mach = full Mach. $M_{n1}=3\sin 90^{\circ }=3=M_1$. Why this step? When the shock faces the flow head-on, all the velocity is normal — the projection does nothing.

3. Exit Mach. $M_{n2}^2=\frac{9+5}{7(9)-1}=\frac{14}{62}=0.2258\Rightarrow M_{n2}=\mathrm{0.4752.}$ $M_2=\frac{M_{n2}}{\sin (90^{\circ }-0^{\circ })}=\frac{0.4752}{1}=\mathrm{0.4752.}$ Why this step? With $\beta -\theta =90^{\circ }$, the divisor $\sin (90^{\circ })=1$, so $M_2=M_{n2}$ — the oblique formula is the normal-shock formula here.

Verify: $M_2=0.475$ is the textbook normal-shock exit Mach for $M_1=3$ ✓. The oblique framework contains the normal shock as its $\beta =90^{\circ }$ special case.

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Example 6 — Cell F: over-turning ($\theta >{\theta }_{\max }$)

Forecast: From Ex 8 we'll find ${\theta }_{\max }\approx 23^{\circ }$ for $M_1=2$. Since $25^{\circ }>23^{\circ }$, expect no real $\beta$.

1. Compare to the maximum. For $M_1=2$, ${\theta }_{\max }=22.97^{\circ }$ (computed in Ex 8). The demand $25^{\circ }$ exceeds it. Why this step? The $\theta$–$\beta$ curve never reaches $25^{\circ }$ at this Mach, so the equation $\tan 25^{\circ }=(\dots )$ has no real root in $\beta$.

2. Interpret physically. The wedge is too blunt for the flow to negotiate with a single straight attached shock. Why this step? The flow cannot turn that sharply in one step at this speed, so it must find another arrangement.

3. State the outcome. A curved detached bow shock forms, standing off ahead of the wedge tip. Near the centreline it is essentially a normal shock (subsonic pocket behind it), curving to weaker oblique shocks further out. Why this step? This is the physical resolution the maths signals when no attached solution exists — see Supersonic Wedge & Cone Flow.

Verify: $25^{\circ }>{\theta }_{\max }(M=2)=22.97^{\circ }$ ✓ → detachment. Consistent with the parent note's "bigger wedge ⇒ shock detaches" warning.

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Example 7 — Cell G: real-world word problem

Forecast: $\beta =35^{\circ }$ is only a little above the Mach angle for $M=2.5$ ($\mu =23.6^{\circ }$), so expect a modest wedge, maybe $10$–$15^{\circ }$.

1. Forward-solve for $\theta$. $M_1^2{\sin }^2\beta -1=6.25{\sin }^{2}35^{\circ }-1=6.25(0.3290)-1=\mathrm{1.056.}$ Denominator: $6.25(1.4+\cos 70^{\circ })+2=6.25(1.4+0.3420)+2=\mathrm{12.89.}$ $\cot 35^{\circ }=1.428$. $\tan \theta =2(1.428)\frac{1.056}{12.89}=0.234\Rightarrow \theta =13.2^{\circ }.$ Why this step? We measure $\beta$ from the photo and read $M_1$ from the tunnel — the geometry then delivers the wedge angle, which we'd compare to the machined part.

2. Sanity via the Mach-wave floor. The Mach angle is $\mu =\arcsin (1/2.5)=23.6^{\circ }$. Our $\beta =35^{\circ }>23.6^{\circ }$ ✓, so a genuine shock (not a Mach wave) is present. Why this step? If the measured $\beta$ had been $\le \mu$, the "shock" would be an infinitesimal Mach line and the photo mislabelled.

Verify: $\theta =13.2^{\circ }$, wedge full-angle $\approx 26.4^{\circ }$ — a physically sensible test wedge. $M_{n1}=2.5\sin 35^{\circ }=1.434>1$ ✓, so compression is real.

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Example 8 — Cell H: exam twist, find ${\theta }_{\max }$

Forecast: The peak of the $\theta$–$\beta$ curve for $M_1=2$ sits around $\beta \approx 64^{\circ }$ with ${\theta }_{\max }\approx 23^{\circ }$.

1. Scan $\theta (\beta )$ over the valid range. Evaluate the relation for $\beta$ from the Mach angle $30^{\circ }$ up to $90^{\circ }$ and locate the largest $\theta$. Why this step? ${\theta }_{\max }$ is by definition the top of the hump; it separates "two attached solutions" from "detached shock."

2. Locate the peak. The maximum occurs at $\beta =64.7^{\circ }$, giving $\tan \theta =0.4239$. Why this step? At this $\beta$ the weak and strong roots merge into one — the frontier of attachment.

3. Read off the maximum turn. ${\theta }_{\max }=\arctan (0.4239)=22.97^{\circ }$. Why this step? This is the number Example 6 relied on to declare $25^{\circ }$ impossible.

Verify: ${\theta }_{\max }=22.97^{\circ }$ at $\beta =64.7^{\circ }$ for $M_1=2$ ✓ (matches standard oblique-shock charts, which give ~$23^{\circ }$). Confirms the detachment threshold used in Ex 6.

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Recall Which cell was which?

A forward ::: Ex 1 ($M=3,\beta =40^{\circ }\Rightarrow \theta =21.9^{\circ }$) B weak root ::: Ex 2 ($\beta =45.3^{\circ }$, $M_2=1.45$) C strong root ::: Ex 3 ($\beta =79.8^{\circ }$, $M_2=0.644$) D degenerate $\theta =0$ ::: Ex 4 (Mach angle $23.6^{\circ }$ and $90^{\circ }$) E $\beta =90^{\circ }$ ::: Ex 5 (reduces to normal shock, $M_2=0.475$) F over-turn ::: Ex 6 ($25^{\circ }>{\theta }_{\max }$ → detached bow shock) G word problem ::: Ex 7 (measured $\beta =35^{\circ }\Rightarrow \theta =13.2^{\circ }$) H find ${\theta }_{\max }$ ::: Ex 8 ($22.97^{\circ }$ at $\beta =64.7^{\circ }$)

See also: Prandtl-Meyer Expansion for the opposite problem — turning flow away from itself, where it accelerates instead of shocking.