Detached bow shock: Worked examples — Physics deep dive

This page is the case-by-case drill room for the Detached bow shock. The parent note told you the idea; here we grind through every kind of question the topic can ask — every quadrant of the "attach vs detach" decision, the degenerate inputs, the limiting cases, a real-world word problem, and an exam-style twist.

Before any numbers, one promise: every symbol below was earned in the parent note, but I re-anchor each as it appears so you never have to scroll back.

The backbone formula (re-anchored)

Everything below leans on one equation from the parent, so let us put it in front of us and never refer to it as a mystery again.

The scenario matrix

Think of a "case" as one combination of inputs that changes the physics or the method. Here is the full grid this topic can throw at you.

#	Case class	What makes it distinct	Covered by
A	Detaches ( $θ > θ_{ma x}$ )	Demanded turn beats the flow's capacity → bow shock	Ex 1
B	Attaches ( $θ < θ_{ma x}$ )	Turn is affordable → attached oblique shock	Ex 2
C	Exactly at detachment ( $θ = θ_{ma x}$ )	The knife-edge; one shock angle $β$ only	Ex 3
D	Degenerate body: blunt nose ( $θ \to 90°$ )	Body demands a right-angle turn → always detaches	Ex 4
E	Centreline conditions ( $β = 90°$ )	Bow shock behaves as a normal shock; find $M_{2}$ , $p$ , $T$	Ex 5
F	Off-centreline point ( $μ < β < 90°$ )	Same curved shock, weaker locally; flow may be supersonic	Ex 6
G	Limiting Mach ( $M_{1} \to 1^{+}$ and $M_{1} \to \infty$ )	How $θ_{ma x}$ and stand-off behave at the extremes	Ex 7
H	Real-world word problem	Re-entry capsule stand-off distance	Ex 8
I	Exam twist: re-attachment by speeding up	Same body, two Mach numbers, opposite verdicts	Ex 9

Symbols used throughout (all from the parent):

$M_{1}$ = upstream Mach number = flight speed ÷ local speed of sound. "How many times faster than sound."
$M_{2}$ = downstream Mach number = the same ratio for the flow behind the shock.
$θ$ = flow deflection the body demands (wedge half-angle, or $90°$ at a blunt nose).
$θ_{ma x} (M_{1})$ = the largest turn an attached oblique shock can deliver at that $M_{1}$ .
$β$ = shock-wave angle, the tilt of the shock relative to the incoming flow.
$μ = arcsin (1/ M_{1})$ = the Mach angle, the weakest possible shock tilt (a Mach wave).
$γ = 1.4$ for air (ratio of specific heats).
$p$ = static pressure = the "push per unit area" the gas exerts; subscript 1 = ahead of the shock, 2 = behind it.
$T$ = static temperature = a measure of the random thermal energy of the gas; again subscripts 1 (ahead) and 2 (behind).
$ρ$ = density (the Greek letter "rho") = mass of gas per unit volume; $ρ_{1}$ ahead, $ρ_{2}$ behind. These three are tied by the ideal-gas law $p = ρR T$ , with $R$ the gas constant.
$Δ$ = shock stand-off distance = the small gap between the detached shock and the nose (from the parent note).
$R_{n}$ = nose radius = the radius of curvature of the blunt nose (for a sphere, its geometric radius). We measure $Δ$ in units of $R_{n}$ , i.e. the dimensionless ratio $Δ/ R_{n}$ .

Example 1 — Case A: the shock DETACHES

Worked example Statement

Air, $γ = 1.4$ , $M_{1} = 2.0$ . A wedge has half-angle $θ = 30°$ . Attach or detach?

Forecast: guess before computing — is $30°$ a "big" turn for Mach 2? (Hint: Mach 2 is only mildly supersonic.)

Step 1 — Find the gatekeeper $θ_{ma x} (2.0)$ . From the θ–β–M relation above, maximising $θ$ over $β$ gives $θ_{ma x} (2.0) \approx 22.97°$ . Why this step? $θ_{ma x}$ is the only number that decides attach vs detach. Everything hinges on comparing to it.

Step 2 — Compare. $θ = 30° > 22.97° = θ_{ma x}$ . Why this step? This is literally the detachment criterion $θ > θ_{ma x} \Rightarrow$ detach.

Step 3 — Read the verdict off the figure. On the $θ$ – $β$ curve for $M_{1} = 2$ , the horizontal line at $30°$ sits above the peak — no intersection, so no real $β$ exists. The shock pops off the nose: detached bow shock. Why this step? "No real $β$ " is the mathematical fingerprint of detachment — the equation cannot be satisfied, so no attached shock geometry exists.

Verify: Plug $θ = 22.97°$ back at its peak $β$ and confirm the curve never reaches $30°$ . We check numerically in VERIFY that $max_{β} θ (β) < 30°$ at $M_{1} = 2$ . ✔ Verdict: detach.

Example 2 — Case B: the shock ATTACHES

Worked example Statement

Same wedge $θ = 30°$ , but now $M_{1} = 4.0$ .

Forecast: faster flow has more "turning capacity." Will $30°$ still be too much?

Step 1 — New gatekeeper. $θ_{ma x} (4.0) \approx 38.77°$ . Why this step? $θ_{ma x}$ grows with $M_{1}$ , so the verdict can flip.

Step 2 — Compare. $30° < 38.77°$ → below the peak. Why this step? Below the peak the horizontal line cuts the curve twice → two real $β$ exist (a weak and a strong shock); the shock attaches.

Step 3 — Which $β$ ? Nature almost always picks the weak (smaller- $β$ ) solution. Solving the θ–β–M relation for $θ = 30°$ , $M_{1} = 4$ gives the weak root $β \approx 45.5°$ . Why this step? The θ–β–M relation gives two $β$ for one $θ$ ; we must state which physical shock forms, and outside special back-pressure conditions it is the weak one.

Verify: Substitute $β = 45.5°$ , $M_{1} = 4$ back into the θ–β–M formula; it should return $θ \approx 30°$ . Checked in VERIFY. ✔ Attached, oblique.

Example 3 — Case C: EXACTLY at detachment

Example 4 — Case D: DEGENERATE blunt nose ( $θ \to 90°$ )

Worked example Statement

A perfectly blunt (flat-faced or spherical) nose flies at $M_{1} = 5$ . Does it ever attach?

Forecast: a flat face asks the flow to turn a full right angle. Is any $θ_{ma x}$ that big?

Step 1 — What turn does the body demand? At a blunt nose the stagnation streamline hits head-on and must divert essentially $90°$ around the shoulder, so effectively $θ \to 90°$ . Why this step? The demanded $θ$ is a property of geometry, and a blunt face is the most demanding geometry possible.

Step 2 — Compare to the theoretical ceiling of $θ_{ma x}$ . Even as $M_{1} \to \infty$ , $θ_{ma x}$ tops out near $\approx 45.6°$ for air. So $θ_{ma x}$ can never reach $90°$ . Why this step? If the ceiling of $θ_{ma x}$ is $\sim 45.6°$ , no Mach number lets a blunt body have an attached shock.

Step 3 — Verdict. A blunt nose always produces a detached bow shock, at any supersonic speed.

Verify: Compute the limiting $θ_{ma x}$ as $M_{1} \to \infty$ ( $γ = 1.4$ ) and confirm it is $≪ 90°$ . Checked in VERIFY. ✔ Always detaches — this is why hypersonic capsules and re-entry vehicles always have bow shocks.

Example 5 — Case E: CENTRELINE conditions ( $β = 90°$ )

Worked example Statement

$M_{1} = 3.0$ . On the stagnation streamline of the bow shock, find $M_{2}$ , the pressure ratio $p_{2} / p_{1}$ , and the temperature ratio $T_{2} / T_{1}$ . (Recall from the symbol list: $p$ = static pressure, $T$ = static temperature, $ρ$ = density; subscript 1 = ahead of the shock, 2 = behind.)

Forecast: on the centreline the bow shock is a normal shock — flow behind must be subsonic. Will $p_{2} / p_{1}$ be closer to 3 or to 10?

Step 1 — Normal component. $M_{n 1} = M_{1} sin β = 3 sin 90° = 3$ . Why this step? On the centreline $β = 90°$ , so the shock "sees" the full velocity — it behaves exactly like a normal shock.

Step 2 — Downstream Mach. $M_{2}^{2} = \frac{1 + \frac{γ - 1}{2} M _{n 1}^{2}}{γ M _{n 1}^{2} - \frac{γ - 1}{2}} = \frac{1 + 0.2 ( 9 )}{1.4 ( 9 ) - 0.2} = \frac{2.8}{12.4} = 0.2258$ so $M_{2} = 0.475$ . Why this step? This is the standard normal-shock downstream-Mach formula; direct substitution.

Step 3 — Pressure ratio. $\frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{n 1}^{2} - 1) = 1 + \frac{2.8}{2.4} (9 - 1) = 1 + \frac{2.8}{2.4} (8) = 10.33$ Why this step? Pressure jump across a normal shock; $p$ is the static pressure the gas pushes with, and this ratio is the number that drives nose loads and drag.

Step 4 — Temperature ratio. $\frac{T _{2}}{T _{1}} = \frac{p _{2}}{p _{1}} \cdot \frac{ρ _{1}}{ρ _{2}}, \frac{ρ _{2}}{ρ _{1}} = \frac{( γ + 1 ) M _{n 1}^{2}}{( γ - 1 ) M _{n 1}^{2} + 2} = \frac{2.4 ( 9 )}{0.4 ( 9 ) + 2} = \frac{21.6}{5.6} = 3.857$ so $\frac{T _{2}}{T _{1}} = \frac{10.33}{3.857} = 2.679$ . Why this step? The ideal-gas law $p = ρR T$ ties pressure $p$ , density $ρ$ and temperature $T$ together; dividing the pressure ratio by the density ratio isolates temperature.

Verify: $M_{2} = 0.475 < 1$ (subsonic ✔). All three ratios checked numerically in VERIFY. The huge $T_{2} / T_{1} = 2.68$ is exactly why nose heating matters — see Stagnation properties across shocks. ✔

Example 6 — Case F: an OFF-CENTRELINE point

Worked example Statement

Still $M_{1} = 3.0$ . Move outward to a point on the bow shock where the local shock angle is $β = 40°$ . Is the flow behind supersonic or subsonic there?

Forecast: away from the centreline the shock is weaker. Do we expect $M_{2} > 1$ ?

Step 1 — Local normal Mach. $M_{n 1} = M_{1} sin β = 3 sin 40° = 3 (0.6428) = 1.928$ . Why this step? Only the component normal to the shock is compressed; the tangential slips by unchanged. This is the whole reason an oblique shock is "gentler."

Step 2 — Downstream normal Mach. $M_{n 2}^{2} = \frac{1 + 0.2 M _{n 1}^{2}}{1.4 M _{n 1}^{2} - 0.2} = \frac{1 + 0.2 ( 3.717 )}{1.4 ( 3.717 ) - 0.2} = \frac{1.743}{5.004} = 0.3484$ so $M_{n 2} = 0.590$ . Why this step? Same normal-shock formula as Ex 5, but now applied to the normal Mach only, because the shock acts along its normal.

Step 3 — Restore the tangential part to get the actual $M_{2}$ . First the deflection from θ–β–M: $θ (β = 40°, M_{1} = 3) \approx 22.2°$ . Then $M_{2} = \frac{M _{n 2}}{s i n ( β - θ )} = \frac{0.590}{s i n ( 40° - 22.2° )} = \frac{0.590}{s i n 17.8°} = \frac{0.590}{0.3057} = 1.93$ Why this step? $M_{2} > 1$ because the tangential velocity (untouched by the shock) is large enough to keep the total speed supersonic even though the normal part dropped below sound.

Verify: $M_{2} = 1.93 > 1$ — supersonic, confirming the parent note's claim that past the sonic line the flow re-accelerates. Checked in VERIFY. ✔

Example 7 — Case G: the two LIMITING Mach numbers

Worked example Statement

Describe $θ_{ma x}$ and the stand-off ratio $Δ/ R_{n}$ at the two extremes: $M_{1} \to 1^{+}$ and $M_{1} \to \infty$ .

Forecast: at barely-supersonic speed, how big a turn can the flow afford? At hypersonic speed, does the shock sit far out or hug the body?

Step 1 — Low limit $M_{1} \to 1^{+}$ . The Mach angle $μ = arcsin (1/ M_{1}) \to arcsin (1) = 90°$ . The whole $θ$ – $β$ curve is squeezed into a sliver near $β = 90°$ , so $θ_{ma x} \to 0$ . Why this step? Just above Mach 1 the flow can barely turn at all — any real body detaches. E.g. $θ_{ma x} (1.5) \approx 12°$ , $θ_{ma x} (1.1) \approx 1.5°$ .

Step 2 — High limit $M_{1} \to \infty$ . $θ_{ma x}$ climbs but saturates near $45.6°$ (for $γ = 1.4$ ); it never reaches $90°$ (this is exactly Case D's ceiling). Why this step? This ceiling is the reason blunt bodies (Case D) can never attach a shock at any speed.

Step 3 — Stand-off at high $M_{1}$ . The gap $Δ$ between shock and nose (from the parent note) follows an experimentally-fitted correlation for a sphere, $\frac{Δ}{R _{n}} \approx 0.143 exp (\frac{3.24}{M _{1}^{2}}),$ where the constants $0.143$ and $3.24$ are not derived from first principles — they come from curve-fitting measured and computed shock positions for spheres (the classic result reported by Billig, 1967). As $M_{1} \to \infty$ the exponent $3.24/ M_{1}^{2} \to 0$ , so $\frac{Δ}{R _{n}} \to 0.143 e^{0} = 0.143.$ Why this step? We are asked for the limiting stand-off, and this correlation is the standard engineering tool for a blunt sphere; taking $M_{1} \to \infty$ collapses the exponential to $1$ , revealing the smallest finite gap — the shock hugs the body at hypersonic speed.

Verify: Check $θ_{ma x} (1.5) \approx 12°$ , the saturation value $\approx 45.6°$ , and $Δ/ R_{n} \to 0.143$ . Checked in VERIFY. ✔

Example 8 — Case H: REAL-WORLD word problem

Worked example Statement

A spherical re-entry capsule of nose radius $R_{n} = 1.5 m$ flies at $M_{1} = 6$ . Estimate the shock stand-off distance $Δ$ (in metres). Then say what happens to $Δ$ if it decelerates to $M_{1} = 3$ .

Forecast: at Mach 6 (hypersonic), do you expect the shock centimetres or metres ahead of the nose?

Step 1 — Apply the sphere correlation at $M_{1} = 6$ . $\frac{Δ}{R _{n}} = 0.143 exp (\frac{3.24}{36}) = 0.143 e^{0.09} = 0.143 (1.0942) = 0.1565.$ Why this step? The Billig correlation from Ex 7 is exactly the tool for a blunt sphere in supersonic flow, and $Δ/ R_{n}$ is dimensionless so it is safe to evaluate first.

Step 2 — Multiply by $R_{n}$ . $Δ = 0.1565 \times 1.5 m = 0.2347 m \approx 23.5 cm .$ Why this step? $Δ/ R_{n}$ is a pure number; multiplying by the actual nose radius $R_{n}$ restores physical length (metres × dimensionless = metres).

Step 3 — Decelerate to $M_{1} = 3$ . $\frac{Δ}{R _{n}} = 0.143 e^{3.24/9} = 0.143 e^{0.36} = 0.143 (1.4333) = 0.2050, Δ = 0.3075 m \approx 30.8 cm .$ Why this step? Lower Mach → weaker compression → thicker shock layer → larger gap. The shock moves out as it slows.

Verify: Units: $Δ/ R_{n}$ is dimensionless, times metres gives metres. ✔ Both numbers ( $0.235$ m and $0.308$ m) checked in VERIFY, and $Δ$ increases as $M_{1}$ drops — matches the "higher Mach → smaller stand-off" rule. ✔

Example 9 — Case I: EXAM TWIST — re-attach by speeding up

Worked example Statement

A cone has half-angle $θ = 35°$ . Suppose at $M_{1} = 3.0$ the shock is detached. Find (roughly) the Mach number at which it just re-attaches, and state the verdict at $M_{1} = 5$ .

Forecast: the fix is to raise $M_{1}$ until $θ_{ma x} (M_{1})$ grows past $35°$ . Guess: is that Mach around 3, 4, or 8?

Step 1 — State the re-attachment condition. Re-attachment happens the instant $θ_{ma x} (M_{1}) = 35°$ . Below that Mach, $θ_{ma x} < 35°$ (detached); above it, $θ_{ma x} > 35°$ (attached). Why this step? Re-attachment is the exact reverse of detachment — we must cross the demanded $θ$ back below the moving peak $θ_{ma x} (M_{1})$ .

Step 2 — Scan $M_{1}$ to bracket the crossing. Evaluate the peak of the θ–β–M curve at a few Mach numbers:

$θ_{ma x} (3.0) \approx 34.07°$ → just below $35°$ → detached (consistent with the premise).
$θ_{ma x} (3.2) \approx 35.30°$ → just above $35°$ → attached.

So the crossing sits between $M_{1} = 3.0$ and $3.2$ ; interpolating, re-attachment near $M_{1} \approx 3.1$ . Why this step? A simple root-find: we squeeze $M_{1}$ until $θ_{ma x}$ passes through $35°$ .

Step 3 — Verdict at $M_{1} = 5$ . $θ_{ma x} (5.0) \approx 41.1° > 35°$ → attached, comfortably above the crossing. Why this step? Far above the re-attachment Mach, the flow's turning capacity easily exceeds the demanded $35°$ — this is the "raise Mach to re-attach" lesson in one number.

Answer: the shock re-attaches near $M_{1} \approx 3.1$ , and at $M_{1} = 5$ it is firmly attached.

Verify: Confirm $θ_{ma x} (3.0) \approx 34.1°$ (detached), $θ_{ma x} (3.2) \approx 35.3°$ (attached) — bracketing $M_{1} \approx 3.1$ — and $θ_{ma x} (5) \approx 41.1°$ . Checked in VERIFY. ✔

Recall Quick self-test (cover the answers)

Which case has exactly ONE shock angle $β$ ? ::: Case C, $θ = θ_{ma x}$ (weak & strong roots merge at the peak). Why does a blunt nose ALWAYS detach? ::: It demands $θ \to 90°$ , but $θ_{ma x}$ never exceeds $\approx 45.6°$ for air, even as $M_{1} \to \infty$ . On the centreline of a bow shock at $M_{1} = 3$ , is the flow subsonic? ::: Yes, $M_{2} = 0.475 < 1$ (it acts as a normal shock, $β = 90°$ ). Off-centreline at $M_{1} = 3$ , $β = 40°$ , is it subsonic? ::: No, $M_{2} \approx 1.93$ — the untouched tangential velocity keeps it supersonic. How does stand-off $Δ$ change as the capsule slows from $M_{1} = 6$ to $M_{1} = 3$ ? ::: It grows (from $\approx 23.5$ cm to $\approx 30.8$ cm) — lower Mach, shock moves out.