Visual walkthrough — Mach number M = V - a — subsonic ( - 1), transonic (~1), supersonic ( - 1), hypersonic ( - 5)

Before we start, three words we will lean on, each in plain language:

Recall the two speeds from the parent topic:

• $V$ = how fast the source moves,
• $a$ = how fast each ripple spreads (the speed of sound),
• and $M=V/a$ is their ratio.

We will watch three characters — a slow dot, a dot moving at sound speed, and a fast dot — and let the pictures decide what happens.

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Step 1 — One ping, one growing circle

WHAT. Freeze a single ping the instant it is born, then let time $t$ pass. It becomes a circle whose radius grows steadily.

WHY. Everything complicated later is many of these simple circles overlapping. If you understand one circle honestly, the cone is free.

PICTURE. In the figure, the ping was born at the black dot. After time $t$ the circle's edge sits a distance $at$ away in every direction.

The circle is symmetric because still air carries sound equally fast every way. Hold onto that: the wavefront doesn't care where the source went next. Once a ping is launched, it forgets its parent.

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Step 2 — A still source: rings like a bullseye

WHAT. Let the source sit still and ping repeatedly at equal time intervals.

WHY. This is the $V=0$ baseline — the "nothing moves" case. Every other case is a distortion of this one, so we need to see it clean first.

PICTURE. Concentric circles, all sharing the same centre — a perfect bullseye. No crowding anywhere.

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Step 3 — A slow source ($M<1$): rings crowd ahead but never touch

WHAT. Now let the source creep forward, slower than sound. Ping at times $t_0,t_1,t_2,\dots$ from successive positions.

WHY. This is every airliner in cruise — the subsonic world. We want to see why the air still parts smoothly.

PICTURE. Each new circle starts a little further along the flight line. The centres march forward, so circles pack closer in front and spread behind — but crucially the source stays inside every circle it made. No two edges pile onto a single line.

Because the fastest news always stays ahead of the source, the air downstream is pre-warned and slides aside gently. The crowding in front is just the Doppler effect (higher pitch ahead) — annoying, not violent. Still no shock.

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Step 4 — The exact edge case ($M=1$): all circles kiss one flat wall

WHAT. Speed up until the source moves exactly at sound speed, $V=a$, so $M=1$.

WHY. This is the knife-edge between "news stays ahead" and "source outruns news." Edge cases reveal the switch; skipping them leaves a hole in your understanding.

PICTURE. Now $Vt=at$ exactly. The source sits right on the leading edge of every circle it ever made. All the front edges line up into a single vertical wall passing through the source — a flat plane of piled-up pressure.

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Step 5 — A fast source ($M>1$): the source escapes, and a cone appears

WHAT. Push past sound speed: $V>a$, so $M>1$. Ping repeatedly again.

WHY. This is the whole point of the parent note — supersonic flight. We want the cone to emerge on its own from the circles.

PICTURE. Now the source races ahead of every circle. Each old circle is left behind, smaller circles nested inside bigger ones but all trailing the dot. Their outer edges no longer surround the source — instead they all lean on a common straight line (in 3D, a cone) that springs from the source. That common tangent line is the Mach cone; along it, pressure piles up into a shock.

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Step 6 — Reading the angle straight off the triangle

WHAT. Draw the right triangle hiding inside Step 5's picture and read off the cone's half-angle $\mu$.

WHY. A picture that gives a formula for free is the best kind. We want $\mu$ in terms of $M$ — no guessing.

PICTURE. Take the oldest ping (born at time $0$). In time $t$:

• the source travelled a distance $Vt$ along the flight line — this is the triangle's long side (hypotenuse),
• that oldest circle grew to radius $at$ — this is the triangle's short side, drawn perpendicular to the cone's edge, reaching from the source's start point out to where the cone touches the circle.

The Mach cone edge is the hypotenuse's companion; the angle between the flight line and the cone edge is $\mu$.

Now check the picture against every case we drew:

That last bullet is the deepest check: the formula itself knows there is no cone below $M=1$. Beautiful.

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The one-picture summary

This final figure stacks all four worlds — still, slow, exactly sonic, fast — side by side, so you can see the wavefronts go from a calm bullseye to a piled-up cone as $M$ climbs past $1$.

Recall Feynman retelling — say it in your own words

Picture a dot that keeps dropping stones in a pond, one per second, while it walks. Each stone makes a ripple that spreads at the same speed no matter what the dot does — call that speed $a$. The dot walks at speed $V$.

If the dot walks slower than the ripples spread ($V<a$), it stays inside its own ripples: the water ahead always sees a ripple first, so it's warned and moves aside gently. No pile-up.

If the dot walks exactly as fast as the ripples ($V=a$), it always sits right on the front edge of its ripples — they stack into one flat wall in front of it.

If the dot walks faster than the ripples ($V>a$), it leaves its ripples behind. The old ripples, all growing at the same rate while their centres marched forward at the same rate, line up along a straight edge — a cone. The water ahead gets no warning: it's ambushed, and that sudden pile-up is the shock.

The angle of that cone is set by a right triangle: the ripple grew $at$ while the dot walked $Vt$, so the sine of the half-angle is $at/Vt=a/V=1/M$. That's the whole story — circles and a triangle.

Recall Predict-first checkpoints

Which regime has the source inside all its circles? ::: Subsonic, $M<1$ (Step 3). At $M=1$, what shape do the wavefronts make in front of the source? ::: A single flat wall, half-angle $\mu =90^{\circ }$ (Step 4). Why do the growing circles share a straight tangent when $M>1$? ::: Same growth speed $a$ and same march speed $V$ make nested same-ratio circles, which always share a common tangent line — the cone edge (Step 5). Which trig function gives $\sin \mu =1/M$ and why that one? ::: Sine, because the known circle radius $at$ is opposite the angle $\mu$ and the source travel $Vt$ is the hypotenuse (Step 6). What does $\sin \mu =1/M$ tell you when $M<1$? ::: $1/M>1$, impossible for a sine, so no real cone exists — the formula itself forbids a shock below Mach 1.

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Related deepening: the shock that forms along this cone is analysed in Normal Shock Waves (the $M=1$ flat wall) and Oblique Shocks & Mach Cone (the tilted cone edge). The $M\approx 0.3$–$0.8$ smooth-flow correction lives in Prandtl–Glauert Correction, and viscous effects (a different dimensionless ratio entirely) in Reynolds Number.