Oblique shock waves — θ-β-M relation: Worked examples

Intuition What this page is for

The parent note gave you the machinery. This page drills it against every case that can appear — every branch of the θ-β "hump", every degenerate limit, a word problem, and an exam twist. If you can do all eight examples below, no oblique-shock question can surprise you.

This page is self-contained. Everything we use is restated here as we need it:

$θ$ = deflection angle (how much the flow physically turns at the corner).
$β$ = shock-wave angle (the tilt of the shock line relative to the incoming flow).
$M_{1}$ = upstream Mach number (flow speed ÷ upstream speed of sound).
$M_{n 1} = M_{1} sin β$ = the normal component of the upstream Mach number — this is the only piece that crosses the shock and "does" normal-shock physics.
$γ = 1.4$ = ratio of specific heats for air (a fixed gas property).
The master rule connecting them, plotted below, is the θ-β-M relation: $tan θ = 2 cot β \frac{M _{1}^{2} s i n ^{2} β - 1}{M _{1}^{2} ( γ + c o s 2 β ) + 2} .$
Two normal-shock facts we will reuse (both use $M_{n 1}$ , never the full $M_{1}$ ): downstream normal Mach $M_{n 2}^{2} = \frac{( γ - 1 ) M _{n 1}^{2} + 2}{2 γ M _{n 1}^{2} - ( γ - 1 )}$ , pressure jump $\frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{n 1}^{2} - 1)$ .

The scenario matrix

Think of the θ-β-M relation as a machine with three dials ( $θ$ , $β$ , $M_{1}$ ) and one knob ( $γ$ ). Every question fixes some dials and asks for the rest. The hump curve (θ rising, peaking at $θ_{ma x}$ , then falling) is the map — every example below lands on a labelled spot of it.

Figure 1 — the θ-β hump and where every case lives.

Figure — Oblique shock waves — θ-β-M relation

How to read Figure 1. Two humps are drawn. The pale-yellow curve is for $M_{1} = 3$ ; the chalk-blue curve is for $M_{1} = 2$ . The horizontal axis is the shock angle $β$ (in degrees) and the vertical axis is the deflection $θ$ (in degrees) that the flow undergoes. Refer back to Figure 1 by name from any example:

The left foot (yellow curve leaves the axis at $β = 19.4 7^{\circ}$ , labelled "Cell D") is the Mach-wave limit: $θ = 0$ , the weakest possible shock.
The right foot (curve returns to the axis at $β = 9 0^{\circ}$ , labelled "Cell E") is the normal shock: again $θ = 0$ , but now head-on.
The filled dot at the top of each curve is $θ_{ma x}$ — the peak (labelled "peak = theta_max, Cell F if exceeded"). Ask for more turn than this and no point on the curve reaches you, so the shock detaches.
The rising (left) branch holds every weak root — "Cell A" is labelled at $β = 37. 8^{\circ}, θ = 2 0^{\circ}$ .
The falling (right) branch holds every strong root — "Cell B" is labelled at $β = 82. 2^{\circ}$ for the same $θ = 2 0^{\circ}$ , showing the curve is double-valued.
Cell C simply reads the height of a curve at a chosen $β$ ; Cells G and H reuse the same map at different Mach numbers. Keep Figure 1 beside you for every example.

Cell	Case class	What is fixed / what is degenerate	Example
A	Standard invert: find weak $β$	$M_{1}, θ$ given, $θ < θ_{ma x}$	Ex 1
B	Same $θ$ , the strong root	pick the larger $β$	Ex 2
C	Forward: $β$ given, find $θ$	no inversion needed	Ex 3
D	Degenerate low: $β = μ$ (Mach wave)	numerator $= 0$ , $θ = 0$	Ex 4
E	Degenerate high: $β = 9 0^{\circ}$	normal shock, $θ = 0$	Ex 5
F	Peak / over-limit: $θ > θ_{ma x}$	no real $β$ → detaches	Ex 6
G	Real-world word problem	full downstream chain $p_{2}, T_{2}, M_{2}$	Ex 7
H	Exam twist: reflection off a wall	two shocks, sign of the second turn	Ex 8

Cell A — standard inversion (weak root)

Worked example Ex 1 ·

$M_{1} = 3$ , wedge half-angle $θ = 2 0^{\circ}$ , find the weak $β$ . Forecast: guess — will $β$ be nearer the Mach angle ( $arcsin \frac{1}{3} \approx 19. 5^{\circ}$ ) or nearer $9 0^{\circ}$ ? Write your guess before reading on.

Step 1 — Set up the equation to solve for $β$ . $tan 2 0^{\circ} = 2 cot β \frac{9 s i n ^{2} β - 1}{9 ( 1.4 + c o s 2 β ) + 2} .$ Why this step? The wedge forces $θ$ ; the only unknown left is $β$ . This is a root-find, not an algebra solve, because $β$ sits inside three trig functions at once.

Step 2 — Scan from the left foot upward. The left foot is the Mach angle $μ = arcsin (1/3) = 19.4 7^{\circ}$ where $θ = 0$ . As $β$ climbs from $19.4 7^{\circ}$ , $θ$ rises to a peak then falls back to $0$ at $9 0^{\circ}$ . The first crossing of $θ = 2 0^{\circ}$ is the weak root. Why this step? Weak means the smaller $β$ , on the rising (left) branch of Figure 1. In free flight nature selects this weaker root.

Step 3 — Numerical root. Solving gives $β \approx 37.7 6^{\circ} .$

Verify: plug back: $9 sin^{2} 37.7 6^{\circ} = 9 (0.6124)^{2} = 3.375$ , so numerator $= 2.375$ ; denominator $= 9 (1.4 + cos 75. 5^{\circ}) + 2 = 9 (1.4 + 0.2504) + 2 = 16.85$ ; $2 cot 37.7 6^{\circ} = 2 (1.292) = 2.585$ . Product $= 2.585 \times 2.375/16.85 = 0.364 = tan 20. 0^{\circ}$ . ✓

Cell B — same deflection, the strong root

Worked example Ex 2 · Same

$M_{1} = 3, θ = 2 0^{\circ}$ — find the strong $β$ , then prove it is subsonic. Forecast: it must be bigger than $37. 8^{\circ}$ and smaller than $9 0^{\circ}$ . Guess where.

Step 1 — Keep scanning past the peak. After the hump's peak, $θ$ falls. It passes $2 0^{\circ}$ again on the way down — that second crossing is the strong root. Why this step? The θ-β curve is double-valued: one $θ$ , two $β$ . Same equation, second root, on the falling (right) branch of Figure 1.

Step 2 — Root. $β \approx 82.1 5^{\circ} .$

Step 3 — Normal component. $M_{n 1} = 3 sin 82.1 5^{\circ} = 2.972$ — nearly the full $M_{1}$ , so this shock is almost head-on and strongly compressive.

Step 4 — Compute the actual downstream Mach $M_{2}$ (don't just assert "subsonic"). First the downstream normal component from the normal-shock relation: $M_{n 2}^{2} = \frac{( γ - 1 ) M _{n 1}^{2} + 2}{2 γ M _{n 1}^{2} - ( γ - 1 )} = \frac{0.4 ( 8.833 ) + 2}{2.8 ( 8.833 ) - 0.4} = \frac{5.533}{24.33} = 0.2274,$ so $M_{n 2} = 0.4769$ . Then restore the geometry — the flow leaves the shock at angle $(β - θ)$ , so $M_{2} = \frac{M _{n 2}}{s i n ( β - θ )} = \frac{0.4769}{s i n ( 82.1 5 ^{\circ} - 2 0 ^{\circ} )} = \frac{0.4769}{s i n 62.1 5 ^{\circ}} = 0.5395.$ Why this step? "Big $β$ ⇒ subsonic" is only a heuristic; the honest check computes $M_{2}$ and finds $M_{2} = 0.54 < 1$ . Now the subsonic claim is substantiated.

Verify: numerator $9 sin^{2} 82.1 5^{\circ} - 1 = 7.833$ ; denominator $9 (1.4 + cos 164. 3^{\circ}) + 2 = 5.934$ ; $2 cot 82.1 5^{\circ} = 0.2757$ . Product $= 0.364 = tan 2 0^{\circ}$ ✓. And $M_{2} = 0.54 < 1$ ✓ — genuinely subsonic.

Cell C — forward evaluation ( $β$ given)

Worked example Ex 3 ·

$M_{1} = 2.5$ , shock observed at $β = 4 0^{\circ}$ . Find $θ$ . Forecast: no inversion — just plug. Will $θ$ be small (near a Mach wave) or large?

Step 1 — First check a shock even exists. Need $M_{n 1} = M_{1} sin β > 1$ : $2.5 sin 4 0^{\circ} = 1.607 > 1$ . ✓ Why this step? If $M_{n 1} \leq 1$ the "shock" is really a Mach wave or nothing — the formula would hand back $θ \leq 0$ , which is unphysical to call a shock.

Step 2 — Direct substitution. $tan θ = 2 cot 4 0^{\circ} \frac{6.25 s i n ^{2} 4 0 ^{\circ} - 1}{6.25 ( 1.4 + c o s 8 0 ^{\circ} ) + 2} .$ Numerator $= 6.25 (0.4132) - 1 = 1.582$ ; denominator $= 6.25 (1.4 + 0.1736) + 2 = 11.83$ ; $2 cot 4 0^{\circ} = 2.384$ . So $tan θ = 2.384 \times 1.582/11.83 = 0.3189$ . Why this step? With $β$ and $M_{1}$ both known, every symbol on the right of the θ-β-M relation is now a number — no equation to solve, we just evaluate the machine at one point on the $M_{1} = 2.5$ hump. This is Cell C: read the curve's height in Figure 1.

Step 3 — Result. $θ = arctan 0.3189 = 17.6 9^{\circ}$ . Why arctan here? We know the ratio $tan θ$ ; arctan answers "which angle has this tangent?" — it undoes tan, and since $0 < θ < 9 0^{\circ}$ there is no quadrant ambiguity.

Verify: $tan 17.6 9^{\circ} = 0.3189$ ✓ and $M_{n 1} = 1.607 > 1$ so this $β$ is a valid shock. ✓

Cell D — degenerate low limit (Mach wave)

Cell E — degenerate high limit (normal shock)

Worked example Ex 5 ·

$M_{1} = 3$ , set $β = 9 0^{\circ}$ . Show $θ = 0$ and recover the normal shock. Forecast: a "vertical" shock — does the flow turn at all?

Step 1 — Evaluate the numerator's cofactor. At $β = 9 0^{\circ}$ , $cot 9 0^{\circ} = 0$ , so the whole product is $0$ regardless of the fraction ⇒ $tan θ = 0$ ⇒ $θ = 0$ . Why this step? $cot β = cos β / sin β$ and $cos 9 0^{\circ} = 0$ . This is the right foot of the hump in Figure 1.

Step 2 — Identify the physics. With $θ = 0$ the flow is not deflected: normal and full velocity coincide, $M_{n 1} = M_{1} sin 9 0^{\circ} = M_{1}$ . The oblique shock is a normal shock. Why this step? This is the sanity anchor: the general relation must contain the special case, or it's wrong.

Step 3 — Downstream Mach (normal-shock relation). $M_{2}^{2} = \frac{( γ - 1 ) M _{1}^{2} + 2}{2 γ M _{1}^{2} - ( γ - 1 )} = \frac{0.4 ( 9 ) + 2}{2.8 ( 9 ) - 0.4} = \frac{5.6}{24.8} = 0.2258,$ so $M_{2} = 0.4752$ — subsonic, as a normal shock demands.

Verify: $tan θ = 2 cot 9 0^{\circ} (\dots) = 0$ ✓ and $M_{2} = 0.475$ matches the standard Rankine–Hugoniot value for $M_{1} = 3$ . ✓

Cell F — over the limit (detachment) + how to find $θ_{ma x}$

Worked example Ex 6 ·

$M_{1} = 2$ , wedge half-angle $θ = 2 5^{\circ}$ . Find $β$ — and first, derive $θ_{ma x}$ . Forecast: for $M_{1} = 2$ the peak $θ_{ma x} \approx 22.9 7^{\circ}$ . We are asking for more turn than the flow can attach to. What happens?

Step 1 — WHY a derivative? Locate the peak with $d θ / d β = 0$ . The hump's top is where $θ$ stops rising and starts falling — i.e. where the slope of the curve is flat. The tool that measures "slope of a curve" is the derivative. So the peak $θ_{ma x}$ is the $β$ where $\frac{d θ}{d β} = 0.$ Why this tool and not another? A maximum of a smooth function is exactly a point of zero slope; nothing but the derivative pins that down. Setting it to zero converts "where is the top?" into a solvable equation.

Step 2 — The closed-form maximum condition. Differentiating the θ-β-M relation and setting $d θ / d β = 0$ gives the standard maximum-deflection condition (a quadratic in $sin^{2} β$ ): $sin^{2} β_{ma x} = \frac{1}{γ M _{1}^{2}} [\frac{γ + 1}{4} M_{1}^{2} - 1 + (γ + 1) (\frac{γ + 1}{16} M_{1}^{4} + \frac{γ - 1}{2} M_{1}^{2} + 1)] .$ Why this step? This is the algebra that solving $d θ / d β = 0$ produces — a formula you can evaluate, no root-scanning needed.

Step 3 — Plug in $M_{1} = 2, γ = 1.4$ . The bracket gives $sin^{2} β_{ma x} = 0.8175$ , so $β_{ma x} = 64.6 7^{\circ}$ . Feeding this back into the θ-β-M relation gives the peak height $θ_{ma x} = 22.9 7^{\circ} .$ Why this step? Now we have the peak without guessing — and $2 5^{\circ} > 22.9 7^{\circ}$ .

Step 4 — Conclude: no real solution ⇒ detachment. Since the demanded $2 5^{\circ}$ exceeds $θ_{ma x} = 22.9 7^{\circ}$ , the θ-β-M relation has no real $β$ . The attached-shock assumption fails and the shock detaches: a curved bow shock stands off the wedge nose — see Figure 2.

Figure 2 — attached shock vs detached bow shock.

How to read Figure 2. Left panel: $θ < θ_{ma x}$ , a straight oblique shock (yellow line) hugs the wedge nose at a definite angle $β$ . Right panel: $θ > θ_{ma x}$ , the shock cannot attach — a curved pink bow shock stands off ahead of the nose, near-normal (subsonic pocket, shaded) on the axis and weakening to a Mach wave far out on the flanks.

Verify: $θ_{ma x} (M_{1} = 2) = 22.9 7^{\circ}$ from the closed form and $2 5^{\circ} > 22.9 7^{\circ}$ ⇒ demanded > max attainable ⇒ detached. ✓

Cell G — real-world word problem (full downstream chain)

Worked example Ex 7 · A supersonic intake ramp turns air

$θ = 1 5^{\circ}$ at $M_{1} = 3$ , $p_{1} = 20 kPa$ , $T_{1} = 220 K$ . Find $β$ , $p_{2}$ , $T_{2}$ , and $M_{2}$ (weak shock). Forecast: compression ⇒ $p_{2} > p_{1}$ , $T_{2} > T_{1}$ , and $M_{2}$ drops but should stay supersonic (weak). Guess $M_{2}$ : above or below 2?

Definition The three "downstream speed" symbols (defined before use)

$V_{2}$ = the magnitude of the downstream velocity — literally how many metres per second the air moves after the shock, as one arrow.
$a_{2}$ = the downstream speed of sound — how fast a tiny pressure ripple travels in the post-shock gas.
$M_{2} = V_{2} / a_{2}$ = downstream Mach number — the ratio of the two.

Where they enter the geometry (see Figure 3): the downstream velocity arrow $V_{2}$ leaves the shock at angle $(β - θ)$ . Its component perpendicular to the shock is $u_{n 2} = V_{2} sin (β - θ)$ . Dividing through by $a_{2}$ turns velocities into Mach numbers: $M_{n 2} = M_{2} sin (β - θ)$ , i.e. $M_{2} = M_{n 2} / sin (β - θ)$ . We never need the raw $V_{2}$ or $a_{2}$ — they cancel — but naming them makes the geometry honest.

Figure 3 — the downstream velocity triangle.

How to read Figure 3. The blue arrow is $V_{2}$ , leaving the shock (yellow line) at angle $(β - θ)$ . Drop a perpendicular to the shock: the pink arrow is the normal part $u_{n 2} = V_{2} sin (β - θ)$ , the white arrow along the shock is the tangential part (unchanged across the shock). Divide every arrow's length by $a_{2}$ and the same picture reads in Mach numbers.

Step 1 — Invert θ-β-M for the weak $β$ . Solving $tan 1 5^{\circ} = 2 cot β \frac{9 sin ^{2} β - 1}{9 ( 1.4 + cos 2 β ) + 2}$ gives $β \approx 32.2 4^{\circ}$ . Why this step? Downstream properties all key off the normal component, which needs $β$ .

Step 2 — Normal Mach. $M_{n 1} = 3 sin 32.2 4^{\circ} = 1.600$ . Why this step? Only $M_{n 1}$ crosses the shock like a normal shock: the full $M_{1}$ would overstate the compression.

Step 3 — Pressure ratio (Rankine–Hugoniot). $\frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{n 1}^{2} - 1) = 1 + \frac{2.8}{2.4} (2.56 - 1) = 2.820,$ so $p_{2} = 2.820 \times 20 = 56.4 kPa$ . Why this step? Pressure is set by how hard the normal momentum is stopped, so the Rankine–Hugoniot pressure jump uses $M_{n 1}$ , not $M_{1}$ . The harder the normal component hits, the bigger the pressure spike.

Step 4 — Temperature ratio. $\frac{T _{2}}{T _{1}} = \frac{[ 2 γ M _{n 1}^{2} - ( γ - 1 )] [( γ - 1 ) M _{n 1}^{2} + 2 ]}{( γ + 1 ) ^{2} M _{n 1}^{2}} = \frac{( 6.768 ) ( 3.024 )}{( 5.76 ) ( 2.56 )} = 1.388,$ so $T_{2} = 1.388 \times 220 = 305.4 K$ . Why this step? The stopped normal kinetic energy is dumped into heat, raising $T$ — the squished-up air behind the shock is both denser and hotter. Again driven only by $M_{n 1}$ .

Step 5 — Downstream Mach. First the normal component after the shock: $M_{n 2}^{2} = \frac{( γ - 1 ) M _{n 1}^{2} + 2}{2 γ M _{n 1}^{2} - ( γ - 1 )} = \frac{3.024}{6.768} = 0.4468 \Rightarrow M_{n 2} = 0.6684.$ Then restore geometry using the triangle of Figure 3: $M_{2} = \frac{M _{n 2}}{s i n ( β - θ )} = \frac{0.6684}{s i n ( 32.2 4 ^{\circ} - 1 5 ^{\circ} )} = \frac{0.6684}{s i n 17.2 4 ^{\circ}} = 2.256.$ Why divide by $sin (β - θ)$ ? Because $u_{n 2} = V_{2} sin (β - θ)$ ; dividing by $a_{2}$ gives $M_{n 2} = M_{2} sin (β - θ)$ , and inverting isolates $M_{2}$ .

Verify: $M_{2} = 2.256 > 1$ (weak stays supersonic ✓), $p_{2} > p_{1}$ ✓, $T_{2} > T_{1}$ ✓. All ratios use $M_{n 1} = 1.600$ , self-consistent. Units: kPa×(dimensionless ratio)=kPa, K×(ratio)=K ✓.

Cell H — exam twist: shock reflection off a wall

Worked example Ex 8 · A weak oblique shock (from a

$θ = 1 0^{\circ}$ ramp, $M_{1} = 3$ ) hits a flat wall parallel to the original flow. Find the deflection the reflected shock must produce, and its incident angle. Forecast: the wall must send the flow back parallel to itself. So the reflected shock has to undo the turn. By how much?

Step 1 — Incident shock. For $M_{1} = 3, θ = 1 0^{\circ}$ , the weak root is $β_{1} \approx 27.3 8^{\circ}$ . Behind it $M_{n 1} = 3 sin 27.3 8^{\circ} = 1.380$ , giving $M_{n 2} = 0.7472$ and $M_{2} = M_{n 2} / sin (β_{1} - 1 0^{\circ}) = 2.505$ . Why this step? We need the new upstream Mach $M_{2}$ feeding the reflection.

Step 2 — Reflected turn. The wall is a solid boundary: the flow must leave it parallel. The first shock turned the flow $1 0^{\circ}$ toward the wall, so the reflected shock must turn it $1 0^{\circ}$ back — same magnitude, opposite sign of turn. Why this step? This is the whole trick: the deflection for the reflection is again $θ = 1 0^{\circ}$ , but relative to the new flow at $M_{2} = 2.505$ , not $M_{1}$ .

Step 3 — Reflected shock angle. Invert θ-β-M with $M = 2.505, θ = 1 0^{\circ}$ : the weak root is $β_{2} \approx 31.8 3^{\circ}$ (measured from the local post-first-shock flow direction). Why this step? Lower Mach ⇒ steeper shock for the same turn — the reflected shock is more inclined than the incident one ( $31.8 3^{\circ} > 27.3 8^{\circ}$ ), a classic exam catch.

Verify: incident $β_{1} = 27.3 8^{\circ}$ gives $θ = 1 0^{\circ}$ at $M_{1} = 3$ ✓; reflected $β_{2} = 31.8 3^{\circ}$ gives $θ = 1 0^{\circ}$ at $M_{2} = 2.505$ ✓; $31.8 3^{\circ} > 27.3 8^{\circ}$ confirms "lower $M$ ⇒ larger $β$ ." ✓

Recall Which cell was which? (self-test)

Given only "find the strong shock", which foot/branch of the hump? ::: Right (falling) branch, larger $β$ (Ex 2 / Cell B). How do you locate $θ_{ma x}$ in general? ::: Set $d θ / d β = 0$ ; the closed form gives $sin^{2} β_{ma x}$ , then evaluate θ-β-M there (Ex 6 / Cell F). $θ$ demanded exceeds $θ_{ma x}$ — outcome and note? ::: Detached bow shock, Detached Bow Shocks (Ex 6 / Cell F). Which single Mach number enters every downstream ratio? ::: $M_{n 1} = M_{1} sin β$ (Ex 7 / Cell G). Reflection off a wall — what sets the second deflection? ::: The wall forces the flow parallel again, so the reflected turn equals the incident $θ$ but at the new $M_{2}$ (Ex 8 / Cell H).

For expansion turns (the opposite sign of $θ$ , a convex corner) see Prandtl-Meyer Expansion; for the 3-D nose analogue see Supersonic Wedge and Cone Flow.