Why does the shock angle always seem to equal the "sound cone" angle?

Question

(I'm going to use the term "sound cone" for the area you can hear an aircraft when it's at Mach ≥ 1, illustrated by this picture. Not considering the shockwave.)

---

So as explained in this answer, the angle that an oblique shockwave will form at is dependent on 2 things.

1. The Mach number before the shock
2. How much the flow has to turn in the shock

---

So this means that the shock angle will change with speed, but you can also change the shock angle by changing the aircraft shape (which would change the flow turning angle, therefore changing the shock angle).

The sound cone angle depends only on the speed, if I'm correct. When going Mach ≥ 1, the sound waves can't travel forwards relative to the plane because the plane is going faster than the sound, but they can travel sideways and back, evidenced by the picture above.

My confusion is here : Why does the sound cone angle always equal the shock angle?

(You can't hear an aircraft until after the shockwave)

Is it just a coincidence that the shock angle (determined by the flow turning angle and Mach number) equals the sound cone angle which only depends on the speed?

For an example (at Mach 1 in example) -- If you were to only change the nose shape of an aircraft, that would change the flow turning angle, and consequently the shock angle. You didn't change your speed, so the sound cone angle (again pictured above) wouldn't change. Wouldn't this mean at some point the sound cone and shockwave would meet, because they are not parallel? If they did intersect, the shock would travel through the sound cone and in that case you might be able to hear the aircraft before the shockwave.

---

My theory up to this point is that sound (say sound from a jet engine) can't travel through a shockwave, so it wouldn't matter if the shock and sound cone were parallel, because the sound can't travel through it anyway.

Answer 1

Close to the body, the shock wave has an angle given by the oblique shock formula. Far from the body, the shock wave has an angle given by the mach cone formula. At the transition, the shock cone bends. See right photograph below for a very visible example of this bend.

Answer 2

My confusion is here : Why does the sound cone angle always equal the shock angle?

Close to the body they don't. But further away they do.

---

The angle of the Mach cone is a pure geometric entity: using the picture in your question as basis, if you have a circle (the sound source) whose radius grows with a speed $a$ and which is moving with a speed $V$ bigger than $a$, then it "trails" along a inclined path which has an angle $\mu = \arcsin(\frac{a}{V})$ (by definition of $sin$). If $a$ is the speed of sound then $\frac{V}{a}=M$ (again by definition).

---

On the other hand, the angle $\beta$ at which an oblique shock wave forms upon encountering a wedge of angle $\theta$ is calculated applying the usual physical principles of continuity (mass cannot be created nor disintegrated) and momentum's conservation (second and third Newton's laws). What you get is the following equation which is normally solved via its plot (both pictures from Wikipedia):

$\tan \theta = 2\cot\beta\ \frac{M_1^2\sin^2\!\beta-1}{M_1^2(\gamma+\cos2\beta)+2}$

This plot is important because it helps us understand why the inclination of the shock wave changes.

---

Let's do a practical example and see what happens with our Concorde flying at Mach 2. Let's say that its nose has a (semi-) angle $\theta$ of 15°. So:

1. From the equation of the Mach cone we get an angle $\mu$ of 30°.
2. In the plot of the $\theta-\beta-M$ relation we follow the line labelled with a "2" (the Mach number) till it crosses with the vertical line at $\theta=15°$ (the nose of the Concorde). They intersect at $\beta=45°$ and $\beta=80°$ (both circled in red). The first solution is the so-called "weak" one while the second one is the "strong" solution. The "strong" shock wave normally happens in confined spaces (inlets for example) and we are not interested in it. Therefore our shock wave leaves the nose of the Concorde at 45° (point a).

The shock wave is therefore more "vertical" than the Mach cone (45° vs 30°) and should reach the terrain first.

But it doesn't.

Why that? Because "far enough" from the Concorde's nose, the path followed by the airflow is not affected by its presence and it's just a straight line. Close to the nose we need a shock wave at 45° to make the airflow turn the 15° of the nose. Anyway, a bit further, the flow turns a bit less than 15° (let's say 10°) and therefore a shock wave of 40° is enough to achieve this inclination (point b). Still a bit further, the flow turns again a bit less (let's say 5°) and therefore a shock wave of 35° is this time enough (point c). And still a bit further, we reach the "far enough" distance where the airflow is just going straight: we reached the point c where $\theta$ is zero (i.e. the airflow goes straight) and the shock wave is 30°. Those 30° are exactly the same 30° of the Mach cone!

The end.

To recap: approaching the pointy nose of the Concorde the airflow must turn 15° and this is achieved via an oblique shock wave at 45°. The further we are from the nose, the smaller the turn of the airflow is and therefore the smaller the inclination of the shock wave is. Far from the nose we reach the limit where the flow just travels straight on and the inclination of the shock wave coalesces into the Mach cone at 30°: the sonic boom starts as an oblique shock wave but ends as a Mach cone.

Answer 3

Here's another diagram of the equations and terminology in sophit's answer.
(This isn't a proper answer as such.)