Shock waves — Mach number, Mach cone — - CRITICAL for rockets -: Worked examples

This is the drill-yard for the Mach-cone topic. The parent built the one formula that runs this whole page:

Everything below is just this equation, pushed into every corner of its behaviour. Before we solve anything, let us list every kind of question it can produce.

The scenario matrix

Cell	Case class	What is unknown	Trap / feature	Example
A	Forward: given $M$ , find $θ$	$θ$	plain substitution	Ex 1
B	Inverse: given $θ$ , find $M$ then $v_{obj}$	$M$ , $v_{obj}$	must invert the sine	Ex 2
C	Degenerate $M = 1$	$θ$	cone flattens to $θ = 9 0^{\circ}$	Ex 3
D	Forbidden $M < 1$	(none)	$sin θ > 1$ → no cone	Ex 3
E	Limiting $M \to \infty$ (hypersonic)	$θ$ trend	cone $\to$ paper-thin	Ex 4
F	Real-world word problem: boom delay	time $t$	geometry $x = h / tan θ$	Ex 5
G	Changing medium ( $v_{s}$ varies with altitude)	new $M$ , new $θ$	same $v_{obj}$ , different $v_{s}$	Ex 6
H	Exam twist: two observers / find the height	height $h$	back out geometry	Ex 7
I	Sanity/limit check: is a "boom at Mach 0.9" possible?	yes/no	conceptual + numeric	Ex 8

The nine cells cover: the forward map, its inverse, both boundary values ( $M = 1$ and $M < 1$ ), the runaway limit, a full word problem, a variable-medium case, an inverted-geometry exam question, and a pure conceptual trap. Nothing the topic can ask lives outside this table.

The two geometries you will reuse

Before the examples, pin down the two pictures. Every example points back to one of these.

Picture 1 — the cone triangle (where $sin θ = 1/ M$ is born):

Figure — Shock waves — Mach number, Mach cone — - CRITICAL for rockets -

The object sits at apex $B$ . A ripple it emitted a time $t$ ago is a sphere of radius $v_{s} t$ (that is the opposite side of the right triangle). In that same time the object travelled $v_{obj} t$ (the hypotenuse $A B$ ). The angle at $B$ between the flight path and the cone wall is $θ$ , and "opposite over hypotenuse" is exactly $sin θ = v_{s} t / (v_{obj} t) = 1/ M$ .

Picture 2 — the ground geometry (where $x = h / tan θ$ is born):

The jet flies level at height $h$ . Its cone wall leans back at angle $θ$ from the flight line. Drop a vertical of length $h$ ; the wall meets the ground a horizontal distance $x$ behind the jet. In that skinny triangle, $tan θ = h / x$ (opposite $h$ over adjacent $x$ ), so $x = h / tan θ$ . That trailing distance $x$ is why you hear the boom after the jet is overhead.

The worked examples

Worked example Ex 3 — Cells C & D: the boundary

$M = 1$ and the forbidden $M < 1$ (a) What is $θ$ exactly at $M = 1$ ? (b) What does the formula say at $M = 0.8$ , and what does that mean physically?

Forecast: At $M = 1$ the object just keeps pace with its own ripples — guess the cone has "opened all the way up" to $9 0^{\circ}$ . Below that, guess "no cone at all."

(a) $sin θ = 1/1 = 1 \Rightarrow θ = sin^{- 1} (1) = 9 0^{\circ}$ . Why this step? $sin θ = 1$ is the largest value a sine can reach, so $θ$ is at its maximum — the cone wall stands perpendicular to the flight path. The "cone" is a flat plane: the sound barrier.
(b) $sin θ = 1/0.8 = 1.25$ . Why this step? Plug the sub-sonic value straight in and read the result.
A sine can never exceed $1$ , so $sin θ = 1.25$ has no real solution — there is no angle $θ$ . Why this step? The mathematics itself refuses to produce a cone; physically the source is still inside its own expanding ripples, so nothing piles up. No $M < 1$ shock cone exists.

Verify: $sin 9 0^{\circ} = 1 = 1/1$ ✓. For (b): $1/0.8 = 1.25 > 1$ , outside the range of sine ✓ — confirming the "no real cone" verdict.

Worked example Ex 5 — Cell F: real-world boom delay (uses Picture 2)

A jet at $M = 1.8$ flies level at height $h = 3 km$ in air with $v_{s} = 330 m/s$ . How many seconds after it passes directly overhead do you hear the boom?

Forecast: Faster and lower-cone than the parent's Mach-1.5 example, so the delay should be a few seconds — guess around 4–6 s.

Mach angle: $sin θ = 1/1.8 = 0.5556 \Rightarrow θ = 33.7 5^{\circ}$ . Why this step? The cone's lean-back angle is fixed by $M$ alone; we need it before touching the ground geometry.
Trailing distance on the ground (Picture 2): $x = \frac{h}{tan θ} = \frac{3000}{tan 33.7 5 ^{\circ}} = \frac{3000}{0.6682} = 4490 m$ . Why this step? The cone wall meets the ground this far behind the jet; $tan θ = h / x$ because $h$ and $x$ are the two legs of that ground triangle.
Jet ground speed: $v_{obj} = M v_{s} = 1.8 \times 330 = 594 m/s$ . Why this step? You hear the boom when the jet has flown far enough that its trailing cone reaches your spot — so we need how fast the jet (and its dragged cone) advances.
Delay: $t = \frac{x}{v _{obj}} = \frac{4490}{594} = 7.56 s$ . Why this step? The cone wall's ground intercept is rigidly $x$ behind the jet; the jet closes that gap over your position in time $x / v_{obj}$ .

Verify: $tan 33.7 5^{\circ} \approx 0.668$ ; $3000/0.668 \approx 4490 m$ ✓. $594 m/s$ and $4490/594 = 7.56 s$ ✓. Units: $m / (m/s) = s$ ✓. (A little slower than my forecast — the lower cone angle stretched the lag; the Doppler-bunching intuition of "wall trailing far behind" holds.)

Worked example Ex 6 — Cell G: same speed, different medium

A missile flies at a fixed $v_{obj} = 700 m/s$ . Compare its Mach cone (i) near the ground where $v_{s} = 340 m/s$ and (ii) high up where the cold thin air has $v_{s} = 295 m/s$ .

Forecast: Same speed, but sound is slower up high, so the missile beats its ripples by more ⇒ higher $M$ ⇒ narrower cone up high. Guess a smaller $θ$ at altitude.

Ground: $M_{1} = 700/340 = 2.059$ , so $sin θ_{1} = 1/2.059 = 0.4857 \Rightarrow θ_{1} = 29.0 6^{\circ}$ . Why this step? $M$ is speed relative to the local sound speed; we recompute it for each layer because $v_{s}$ changed, not $v_{obj}$ .
Altitude: $M_{2} = 700/295 = 2.373$ , so $sin θ_{2} = 1/2.373 = 0.4214 \Rightarrow θ_{2} = 24.9 3^{\circ}$ . Why this step? Same missile, colder air ⇒ smaller $v_{s}$ ⇒ larger $M$ ⇒ smaller angle, exactly as forecast.

Verify: $700/340 = 2.059$ and $sin 29.0 6^{\circ} \approx 0.4857 = 1/2.059$ ✓; $700/295 = 2.373$ and $sin 24.9 3^{\circ} \approx 0.4214 = 1/2.373$ ✓. And $θ_{2} < θ_{1}$ : cone narrower at altitude ✓. This is why the local sound speed must be quoted with any Mach figure.

Worked example Ex 7 — Cell H: exam twist, back out the height

Ground crew hears a supersonic drone's boom exactly $5.0 s$ after it passes overhead. The drone flies at $M = 2.0$ with $v_{s} = 340 m/s$ . How high is it flying?

Forecast: This is Ex 5 run backwards — we know the delay, want the height. At $M = 2$ the cone is $3 0^{\circ}$ ; a 5-second lag at ~680 m/s means a few km of trailing distance, so expect $h$ around 2 km.

Ground speed: $v_{obj} = M v_{s} = 2.0 \times 340 = 680 m/s$ . Why this step? The delay converts to a trailing distance only through the jet's ground speed.
Trailing distance from the delay: $x = v_{obj} t = 680 \times 5.0 = 3400 m$ . Why this step? The cone wall sits $x$ behind the drone; the drone crosses that gap over you in the measured $5.0 s$ , so $x = v_{obj} t$ (rearranged from Ex 5 step 4).
Mach angle: $sin θ = 1/2 \Rightarrow θ = 3 0^{\circ}$ . Why this step? We need the cone's lean to link horizontal $x$ to vertical $h$ .
Height from Picture 2: $h = x tan θ = 3400 \times tan 3 0^{\circ} = 3400 \times 0.5774 = 1963 m$ . Why this step? We invert $x = h / tan θ$ into $h = x tan θ$ because now $h$ is the unknown leg.

Verify: $h / tan θ = 1963/0.5774 = 3400 m = x$ ✓, and $x / v_{obj} = 3400/680 = 5.0 s$ ✓ — round-trips back to the given delay. $h \approx 1.96 km$ , matching the ~2 km forecast. Units: $m \times (dimensionless) = m$ ✓.

Worked example Ex 8 — Cell I: conceptual + numeric sanity trap

"A friend claims they measured a Mach cone of half-angle $θ = 4 8^{\circ}$ for a jet at $M = 0.9$ ." Is this possible? If a real cone exists at some $M$ giving $θ = 4 8^{\circ}$ , what is that $M$ ?

Forecast: $M = 0.9$ is subsonic, so from Ex 3(b) there should be no cone at all — the claim smells wrong. Then a genuine $4 8^{\circ}$ cone needs some supersonic $M$ ; guess just above 1.

Test the claim: at $M = 0.9$ , $sin θ = 1/0.9 = 1.111 > 1$ — impossible. No cone forms; the friend cannot have measured one. Why this step? We check whether the required $sin θ$ even lies in $[- 1, 1]$ ; subsonic $M$ pushes it above $1$ , killing any real angle.
Now the honest question — which $M$ gives $θ = 4 8^{\circ}$ ? $sin 4 8^{\circ} = 0.7431$ , so $M = 1/0.7431 = 1.346$ . Why this step? A real $4 8^{\circ}$ cone demands $sin θ = 1/ M$ with $sin θ < 1$ , forcing $M > 1$ ; we invert to find it.

Verify: $1/0.9 = 1.111 > 1$ confirms "no subsonic cone" ✓. $sin 4 8^{\circ} = 0.7431$ and $1/1.346 = 0.7431$ ✓, with $M = 1.346 > 1$ (supersonic, as required) ✓. This nails the parent's coherent pile-up rule: a shock wall only exists once the source outruns its own ripples.

Recall Which cell was which?

A forward substitution ::: Ex 1 Inverting the sine for speed ::: Ex 2 The $M = 1$ flat cone and $M < 1$ no-cone ::: Ex 3 The hypersonic $θ \to 0$ limit ::: Ex 4 Boom-delay word problem ::: Ex 5 Same speed, changed sound speed ::: Ex 6 Back-out-the-height exam twist ::: Ex 7 Spot-the-impossible-claim trap ::: Ex 8

Connections

Speed of sound in a medium — every $M$ above hides a local $v_{s}$ (Ex 6 makes this explicit).
Doppler effect — the trailing-cone lag in Ex 5/7 is the $M \geq 1$ face of wavefront bunching.
Superposition & constructive interference — why a cone forms at all (Ex 8).
Wave drag and aerodynamic heating — consequence of the hypersonic sliver in Ex 4.
De Laval nozzle — supersonic Mach numbers by design.
Compressible flow / Bernoulli limits — the $M \approx 1$ boundary of Ex 3.