Oblique shock waves — θ-β-M relation: Exercises

Intuition How to use this page

Work each problem before opening its solution. The problems climb a ladder: first you just recognise the pieces of the θ-β-M relation, then you apply it, then you analyse curves and limits, then you synthesise across a wedge, and finally you reach mastery with shock detachment and multi-shock reasoning (L5.1–L5.3).

Every symbol used here is defined in the parent note. Quick reminder of the cast:

$M_{1}$ = upstream Mach number (flow speed ÷ speed of sound, before the shock).
$β$ = shock-wave angle (tilt of the shock line vs the incoming flow).
$θ$ = flow deflection angle (how far the streamline actually bends).
$μ$ = Mach angle $= arcsin (1/ M_{1})$ — the tilt of the weakest possible wave (a sound wave riding on the flow); it is the smallest $β$ can ever be. Think " $μ$ = the angle of a whisper, $β$ = the angle of a shove."
$M_{n 1} = M_{1} sin β$ = the normal Mach component that does normal-shock physics.
$M_{n 2}$ = the normal Mach component behind the shock; $M_{2}$ = the full downstream Mach.
$V_{1}, V_{2}$ = the actual flow speeds upstream/downstream; $u_{n 1}, u_{n 2}$ = their pieces perpendicular to the shock; $w$ = the piece along the shock (unchanged).
$p$ = pressure, $ρ$ = density, $a$ = local speed of sound (so any Mach $= V / a$ ).
$γ = 1.4$ for air (ratio of specific heats).

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

Read the master diagram now, line by line — every problem leans on it. Each element is named by both colour and shape/label, so the description works even in grayscale:

The arrows on the far left, labelled " $M_{1}$ " (blue) are the incoming flow, all pointing horizontally.
The steeply rising solid line, labelled "shock" (red) is the shock; the angle between it and the incoming flow is $β$ .
The shallow solid line, labelled "wedge wall" (gray) is the wall; the angle between it and the incoming flow is $θ$ — the deflected arrow labelled " $M_{2}$ " (green) rides along this wall.
The short arrow labelled " $M_{n 1}$ " (orange) is the piece of the incoming velocity aimed straight into the shock; its length is $M_{1} sin β = M_{n 1}$ . That little right triangle (horizontal side, normal side) is the whole reason $M_{n 1} = M_{1} sin β$ : the orange side is the "opposite" side of the angle $β$ in that triangle.

Level 1 — Recognition

L1.1 — Name the three angles

Problem. A supersonic stream hits a wedge. On the master diagram you can see three angles: one between the incoming flow and the shock, one between the incoming flow and the deflected flow, and one between the deflected flow and the shock. Name each in θ-β-M language, and write the third one in terms of the other two.

Recall Solution

Look at the master figure above as you read this.

Incoming flow (arrows labelled " $M_{1}$ ") ↔ shock (line labelled "shock"): this is $β$ (the shock angle).
Incoming flow ↔ deflected flow (arrow labelled " $M_{2}$ "): this is $θ$ (the deflection angle).
Deflected flow ↔ shock line: this is $(β - θ)$ .

Why $(β - θ)$ ? The shock sits at $β$ from the incoming flow. The flow then bends toward the shock by $θ$ , so it closes the gap and now makes the smaller angle $β - θ$ with the shock. Nothing new is needed here — it is just $β$ minus $θ$ read straight off the picture.

L1.2 — Which Mach number crosses the shock?

Problem. For $M_{1} = 2.5$ and $β = 4 0^{\circ}$ , compute the Mach number that actually governs the normal-shock physics.

Recall Solution

Only the normal component (the short arrow labelled " $M_{n 1}$ " in the master figure) crosses the shock like a normal shock: $M_{n 1} = M_{1} sin β = 2.5 sin 4 0^{\circ} = 2.5 (0.6428) = 1.607.$ Since $M_{n 1} > 1$ , a real shock forms.

L1.3 — Shock or Mach wave?

Problem. For $M_{1} = 2$ , is $β = 2 5^{\circ}$ a valid shock angle? What about $β = 3 5^{\circ}$ ?

Recall Solution

A real shock needs $M_{n 1} = M_{1} sin β > 1$ , i.e. $sin β > 1/ M_{1} = 0.5$ , i.e. $β > 3 0^{\circ}$ .

$β = 2 5^{\circ}$ : $sin 2 5^{\circ} = 0.423 < 0.5$ → $M_{n 1} = 0.845 < 1$ . Not a shock — angle too shallow, this is inside the Mach cone.
$β = 3 5^{\circ}$ : $sin 3 5^{\circ} = 0.574 > 0.5$ → $M_{n 1} = 1.147 > 1$ . Valid shock. The boundary $β = 3 0^{\circ}$ is the Mach angle $μ = arcsin (1/ M_{1})$ — recall from the symbol list above that $μ$ is the tilt of the weakest possible wave (a plain sound wave), the smallest tilt a "shock" can have. See Mach Angle and Mach Waves.

Level 2 — Application

L2.1 — Deflection from a known shock angle

Problem. Air, $M_{1} = 3$ , shock angle $β = 4 0^{\circ}$ . Find the deflection $θ$ produced.

Intuition Where each factor of the θ-β-M relation comes from (before you plug in)

The relation $tan θ = 2 cot β \frac{M _{1}^{2} s i n ^{2} β - 1}{M _{1}^{2} ( γ + c o s 2 β ) + 2}$ is not a magic formula — every factor is a piece of geometry or physics you already met in the parent note. Read it as three chunks (and glance at the master figure):

$M_{1}^{2} sin^{2} β - 1 = M_{n 1}^{2} - 1$ in the numerator. This is the "shock strength" — how far the normal Mach exceeds $1$ . If $M_{n 1} = 1$ the flow only just reaches the sound barrier normal to the shock, so it turns by nothing: the numerator vanishes and $θ = 0$ . This is why the Mach-wave limit lives in the numerator.
$cot β$ prefactor. The turning happens because the normal speed drops while the tangential slide stays fixed. On the velocity triangle, the amount of visible "bend" for a given drop in the normal component scales with how the two components trade off — that ratio is a cotangent of $β$ . A steep shock ( $β \to 9 0^{\circ}$ , $cot β \to 0$ ) has almost no tangential slide to bend against, so it barely turns the flow — hence $cot β$ multiplies the whole thing and kills the deflection at $β = 9 0^{\circ}$ .
$M_{1}^{2} (γ + cos 2 β) + 2$ denominator. This is just the tidy algebraic residue of substituting the normal-shock density ratio and expanding $tan (β - θ)$ (Steps 4–6 of the parent derivation). It never changes sign for a real shock, so it only sets the magnitude of $θ$ , never whether the flow turns.

So the sign and existence of a turn are decided by the numerator (shock strength) and the $cot β$ prefactor (geometry); the denominator just scales.

Recall Solution

Plug straight into the θ-β-M relation (no inversion needed here — $β$ is given): $tan θ = 2 cot β \frac{M _{1}^{2} s i n ^{2} β - 1}{M _{1}^{2} ( γ + c o s 2 β ) + 2} .$ Pieces:

$sin 4 0^{\circ} = 0.6428$ , so $M_{1}^{2} sin^{2} β = 9 (0.4132) = 3.719$ ; numerator $= 3.719 - 1 = 2.719$ .
$cos 2 β = cos 8 0^{\circ} = 0.1736$ ; denominator $= 9 (1.4 + 0.1736) + 2 = 9 (1.5736) + 2 = 16.16$ .
$cot 4 0^{\circ} = 1/ tan 4 0^{\circ} = 1.1918$ . $tan θ = 2 (1.1918) \frac{2.719}{16.16} = 2.3836 (0.1683) = 0.4011 \Rightarrow θ = 21. 9^{\circ} .$

L2.2 — Downstream normal Mach and pressure ratio

Problem. For L2.1 ( $M_{1} = 3, β = 4 0^{\circ}$ ), find $M_{n 1}$ , then $M_{n 2}$ and $p_{2} / p_{1}$ using the normal-shock relations.

Intuition Why the pressure-ratio formula looks the way it does

Look at the little "shock box" in the figure below. The three Rankine–Hugoniot jump laws each say a flux is the same just before and just after the thin shock:

Mass: $ρ_{1} u_{n 1} = ρ_{2} u_{n 2}$ (same amount of gas per second crosses).
Momentum (normal): $p_{1} + ρ_{1} u_{n 1}^{2} = p_{2} + ρ_{2} u_{n 2}^{2}$ (pressure + "ram" is conserved). Rearranged, the pressure jump equals the loss of ram momentum, $p_{2} - p_{1} = ρ_{1} u_{n 1}^{2} - ρ_{2} u_{n 2}^{2}$ — this is literally where the factor $(M_{n 1}^{2} - 1)$ is born.
Energy: total enthalpy is conserved, which pins the temperature side. Combining these three and writing everything with $M_{n 1} = u_{n 1} / a_{1}$ collapses to the compact result $\frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{n 1}^{2} - 1) .$ Read it physically: at $M_{n 1} = 1$ the bracket is $0$ so $p_{2} / p_{1} = 1$ (no shock); the stronger the normal Mach, the bigger the pressure jump. The coefficient $2 γ / (γ + 1)$ is just how efficiently a gas of stiffness $γ$ converts that excess ram into pressure.

Recall Solution

Only the normal component crosses, so use $M_{n 1}$ everywhere: $M_{n 1} = M_{1} sin β = 3 (0.6428) = 1.928.$ Normal-shock downstream Mach:

=\frac{0.4(3.718)+2}{2.8(3.718)-0.4}=\frac{3.487}{10.01}=0.3484,$$ so $M_{n2}=0.590$. Pressure ratio (from the momentum jump above): $$\frac{p_2}{p_1}=1+\frac{2\gamma}{\gamma+1}(M_{n1}^2-1)=1+\frac{2.8}{2.4}(3.718-1)=1+1.1667(2.718)=4.171.$$

L2.3 — Recover the true downstream Mach

Problem. Continue L2.2: find $M_{2}$ (the full downstream Mach). Is the flow still supersonic?

Recall Solution

Behind the shock the normal component obeys $u_{n 2} = V_{2} sin (β - θ)$ , so dividing by the local sound speed $a_{2}$ gives $M_{n 2} = M_{2} sin (β - θ)$ . We undo that geometry: $M_{2} = \frac{M _{n 2}}{s i n ( β - θ )} = \frac{0.590}{s i n ( 4 0 ^{\circ} - 21. 9 ^{\circ} )} = \frac{0.590}{s i n 18. 1 ^{\circ}} = \frac{0.590}{0.3107} = 1.899.$ $M_{2} = 1.90 > 1$ : still supersonic — the hallmark of a weak oblique shock. Even though $M_{n 2} < 1$ , the tangential slide carries enough speed to keep the total supersonic.

Level 3 — Analysis

L3.1 — Which root is which?

Problem. For $M_{1} = 3$ , $θ = 2 0^{\circ}$ the θ-β-M relation has two roots, $β \approx 37. 8^{\circ}$ and $β \approx 8 2^{\circ}$ . Without computing $M_{2}$ fully, argue which is the weak shock and which is the strong shock, and predict the sign of $(M_{2} - 1)$ for each.

Recall Solution

In the figure above, the three humps are labelled directly on each curve with their Mach number (" $M_{1} = 2$ ", " $M_{1} = 3$ ", " $M_{1} = 5$ "); use the label, not the colour. Follow the curve tagged " $M_{1} = 3$ ". The horizontal dashed line at $θ = 2 0^{\circ}$ cuts it in two places — the two arrows mark exactly the roots $β = 37. 8^{\circ}$ (left of the peak) and $β = 8 2^{\circ}$ (right of the peak).

Smaller $β = 37. 8^{\circ}$ = left of the peak = weak shock. Smaller $β$ ⇒ smaller $M_{n 1}$ ⇒ milder compression ⇒ downstream usually supersonic ( $M_{2} > 1$ ).
Larger $β = 8 2^{\circ}$ = right of the peak = strong shock. Larger $β$ ⇒ $M_{n 1}$ close to $M_{1}$ ⇒ near-normal-shock ⇒ downstream subsonic ( $M_{2} < 1$ ). Nature normally selects the weak root in free flight (downstream pressure is not artificially raised).

L3.2 — The two limiting shocks

Problem. For any $M_{1}$ , the hump touches $θ = 0$ at two values of $β$ . Find both and say what each physically is.

Recall Solution

On the same figure, notice each curve starts on the horizontal axis at the left (its Mach angle) and returns to the axis at $β = 9 0^{\circ}$ on the right — those two axis touchdowns are exactly the $θ = 0$ solutions. Algebraically, set $tan θ = 0$ ; the fraction is zero (or the prefactor is) when:

Numerator zero: $M_{1}^{2} sin^{2} β - 1 = 0 \Rightarrow sin β = 1/ M_{1} \Rightarrow β = arcsin (1/ M_{1}) = μ$ . This is the Mach wave — the weakest possible disturbance, no compression (the left touchdown).
$cot β = 0$ : $β = 9 0^{\circ}$ . This is a normal shock — maximum compression, flow not turned because the shock is perpendicular (the right touchdown). So the hump is bookended by Mach wave on the left and normal shock on the right, both giving zero deflection for opposite reasons.

L3.3 — Mach angle at three speeds

Problem. Compute the Mach angle $μ = arcsin (1/ M_{1})$ for $M_{1} = 1.5, 2, 5$ . What trend do you see, and what does it mean physically for a fast vehicle?

Recall Solution

$M_{1} = 1.5$ : $μ = arcsin (0.667) = 41. 8^{\circ}$ .
$M_{1} = 2$ : $μ = arcsin (0.5) = 30. 0^{\circ}$ .
$M_{1} = 5$ : $μ = arcsin (0.2) = 11. 5^{\circ}$ . Trend: as $M_{1}$ grows, $μ$ shrinks — the Mach cone squeezes tight around the flight direction. You can see this on the θ-β figure too: the faster curve (labelled " $M_{1} = 5$ ") starts its left touchdown further left (smaller $β$ ). Physically, the faster you go, the more the pressure disturbances pile up into a thin sliver ahead; the weakest wave lies almost along the flow. See Mach Angle and Mach Waves.

Level 4 — Synthesis

L4.1 — Full wedge solve

Problem. Air at $M_{1} = 2.5$ flows over a symmetric wedge of half-angle $θ = 1 5^{\circ}$ . Find the weak shock angle $β$ , then $M_{n 1}$ , $p_{2} / p_{1}$ , and the downstream Mach $M_{2}$ .

Intuition How to actually invert θ-β-M for

$β$ (root-finding, done by hand) The relation gives $θ$ explicitly from $β$ , but here we know $θ$ and want $β$ — an implicit equation. There is no clean closed form, so we hunt the root. Define the "residual" $f (β) = θ predicted by this β 2 cot β \frac{M _{1}^{2} sin ^{2} β - 1}{M _{1}^{2} ( γ + cos 2 β ) + 2} - tan θ_{wanted} .$ We want $f (β) = 0$ .

Bracket first. Any real shock has $μ \leq β \leq 9 0^{\circ}$ . Here $μ = arcsin (1/2.5) = 23. 6^{\circ}$ . At $β = μ$ , $f = - tan 1 5^{\circ} < 0$ (predicted $θ = 0$ ). At $β = 9 0^{\circ}$ , again predicted $θ = 0$ so $f < 0$ . In between $f$ rises above $0$ (the hump exceeds $1 5^{\circ}$ ) and comes back down — so $f = 0$ has two roots, one on each side of the peak $θ_{ma x}$ .
Pick the weak root: it is the one with smaller $β$ , i.e. the crossing before the peak. Bracket it between $μ = 23. 6^{\circ}$ and the peak location (for $M_{1} = 2.5$ the peak is near $β \approx 6 5^{\circ}$ , so bracket $[2 4^{\circ}, 6 5^{\circ}]$ ).
Bisection. Evaluate $f$ at the midpoint; keep the half where $f$ changes sign; repeat. A few steps: $f (3 0^{\circ}) < 0$ , $f (4 0^{\circ}) > 0$ ⇒ root in $[3 0^{\circ}, 4 0^{\circ}]$ ; $f (3 7^{\circ}) < 0$ , $f (37. 5^{\circ}) > 0$ ⇒ root near $36. 9^{\circ}$ . That converges to the weak root $β \approx 36. 9^{\circ}$ . (Taking the bracket $[6 5^{\circ}, 9 0^{\circ}]$ instead would land on the strong root near $8 2^{\circ}$ .)

Recall Solution

Step A — invert θ-β-M for $β$ using the bracket-and-bisection recipe above, weak (smaller- $β$ ) root: $β \approx 36. 9^{\circ}$ . Step B — normal component. $M_{n 1} = 2.5 sin 36. 9^{\circ} = 2.5 (0.6004) = 1.501$ . Step C — pressure ratio. $\frac{p _{2}}{p _{1}} = 1 + \frac{2.8}{2.4} (1.50 1^{2} - 1) = 1 + 1.1667 (1.253) = 2.462.$ Step D — downstream Mach. First $M_{n 2}$ from the normal-shock relation: $M_{n 2}^{2} = \frac{( γ - 1 ) M _{n 1}^{2} + 2}{2 γ M _{n 1}^{2} - ( γ - 1 )} = \frac{0.4 ( 2.253 ) + 2}{2.8 ( 2.253 ) - 0.4} = \frac{2.901}{5.908} = 0.4911,$ so $M_{n 2} = 0.4911 = 0.7008$ . Then undo the geometry to get the full downstream Mach: $M_{2} = \frac{M _{n 2}}{s i n ( β - θ )} = \frac{0.7008}{s i n ( 36. 9 ^{\circ} - 1 5 ^{\circ} )} = \frac{0.7008}{s i n 21. 9 ^{\circ}} = \frac{0.7008}{0.3730} = 1.879.$ Result: $β \approx 36. 9^{\circ}$ , $M_{n 1} \approx 1.50$ , $p_{2} / p_{1} \approx 2.46$ , $M_{n 2} \approx 0.701$ , $M_{2} \approx 1.88$ (still supersonic). This is the whole wedge-flow recipe end to end.

L4.2 — Compression vs expansion sign check

Problem. In L4.1 the flow turned into itself and pressure rose. If instead the wall turned away (a convex corner) by $1 5^{\circ}$ , would a shock form? What physics governs that case?

Recall Solution

A convex (expansion) corner turns the flow away from itself. Information can escape, so no shock forms — the flow accelerates smoothly through a fan of Mach waves (drawn in the figure above: a spread of rays fanning out from the corner, each a weak Mach line, gradually bending and speeding the flow). Pressure and temperature drop, Mach number rises. That is Prandtl-Meyer Expansion, the mirror-image process to an oblique shock: instead of one abrupt line, the turn is shared smoothly across the whole fan. The θ-β-M relation only applies to concave/compression turns where the flow is forced into itself.

Level 5 — Mastery

L5.1 — Detachment threshold

Problem. For $M_{1} = 2$ , the maximum deflection is $θ_{ma x} \approx 22.9 7^{\circ}$ . A wedge of half-angle $2 5^{\circ}$ is placed in this flow. What happens, and why does the θ-β-M relation "fail" to give an answer?

Recall Solution

Since $2 5^{\circ} > θ_{ma x} = 22.9 7^{\circ}$ , the wedge demands more turning than any attached oblique shock can supply. The θ-β-M equation has no real root for $β$ — algebraically the required RHS can't reach $tan 2 5^{\circ}$ for any $β$ (the whole hump peaks below $2 5^{\circ}$ ). Physically the shock detaches and stands off the body as a curved bow shock: near the centreline it is essentially a normal shock (subsonic behind), curving to a Mach wave far out. The "failure" of the formula is the mathematics correctly telling you the assumed attached-shock geometry cannot exist.

L5.2 — $θ_{ma x}$ is speed-dependent

Problem. Explain qualitatively why $θ_{ma x}$ grows as $M_{1}$ increases (e.g. $θ_{ma x} \approx 22.9 7^{\circ}$ at $M_{1} = 2$ but $\approx 34.0 7^{\circ}$ at $M_{1} = 3$ ), and what this means for wedge design at higher speeds.

Recall Solution

Faster flow carries more momentum normal to the shock, so it can be forced through a sharper turn before the geometry becomes impossible. Hence the hump in the θ-β curve gets taller as $M_{1}$ rises: $θ_{ma x}$ increases (approaching about $45. 6^{\circ}$ as $M_{1} \to \infty$ ). You can read this straight off the θ-β figure — the curve labelled " $M_{1} = 3$ " peaks higher than the one labelled " $M_{1} = 2$ ". Design meaning: a wedge that would detach the shock at low supersonic speed may run with a clean attached shock at higher Mach — so the same geometry behaves differently across the flight envelope.

L5.3 — Double-wedge / multi-shock reasoning

Problem. Flow at $M_{1} = 3$ passes a $2 0^{\circ}$ wedge (from the parent's Worked Example 1, giving $β = 37. 8^{\circ}$ , $M_{2} \approx 1.99$ ). Downstream the flow meets a second $1 0^{\circ}$ compression turn. Set up (do not fully invert) how you would find the second shock angle $β_{2}$ , and note one subtlety.

Recall Solution

Setup. The upstream Mach for the second shock is the local Mach after the first shock, $M_{2} \approx 1.99$ , not $M_{1} = 3$ . Apply θ-β-M again with $M_{2}$ and $θ_{2} = 1 0^{\circ}$ : $tan 1 0^{\circ} = 2 cot β_{2} \frac{1.9 9 ^{2} s i n ^{2} β _{2} - 1}{1.9 9 ^{2} ( 1.4 + c o s 2 β _{2} ) + 2},$ and take the weak (smaller- $β_{2}$ ) root, bracketing between $μ_{2} = arcsin (1/1.99) = 30. 2^{\circ}$ and the second hump's peak. Numerically $β_{2} \approx 34. 3^{\circ}$ . Subtlety. $β_{2}$ is measured relative to the local flow direction after the first turn, which itself is already deflected by $2 0^{\circ}$ from the original freestream. To draw the shock in the lab frame you must add that $2 0^{\circ}$ back. Also check $M_{2} > 1$ still holds and $θ_{2} < θ_{ma x} (M_{2})$ so the second shock stays attached (here $θ_{ma x} (1.99) \approx 22. 9^{\circ} > 1 0^{\circ}$ , so it does).

Recall Self-test summary (reveal after finishing)

What bounds $β$ ? ::: $arcsin (1/ M_{1}) \leq β \leq 9 0^{\circ}$ . Which Mach enters normal-shock laws? ::: $M_{n 1} = M_{1} sin β$ . How do you recover $M_{2}$ from $M_{n 2}$ ? ::: $M_{2} = M_{n 2} / sin (β - θ)$ . What if $θ > θ_{ma x}$ ? ::: No attached shock — a detached bow shock forms. Weak vs strong root? ::: Smaller $β$ = weak (often supersonic behind); larger $β$ = strong (subsonic behind).