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(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.)

enter image description here


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.

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    $\begingroup$ Because they are the same thing. You are just using different words to describe the same thing. $\endgroup$ May 11 at 2:02
  • $\begingroup$ @MichaelHall Hmm okay. So the sound cone angle is determined from the speed of the aircraft. The shock angle is determined by a different set of factors. Is that incorrect? (Sorry if this seems obvious) $\endgroup$
    – Wyatt
    May 11 at 2:51
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    $\begingroup$ I will defer to the experts to explain how they are calculated if it's relevant, but what is the shockwave other than sound? Hasn't that been firmly established in the answer to one of your previous questions here? Why wouldn't they be the same? $\endgroup$ May 11 at 3:07
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    $\begingroup$ @MichaelHall also do you know why this question is getting downvoted? I would improve it if I knew what was wrong. $\endgroup$
    – Wyatt
    May 11 at 3:51
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    $\begingroup$ I downvoted it because I thought you just weren't getting it and keep asking very similar questions on shock waves. I thought the answer was obvious, but I am doubting myself now because Sophit is very smart on these matters. However, I need a better explanation. Anyway, I went to remove my downvote just now but it is locked in unless the question is edited... $\endgroup$ May 11 at 16:30

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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. enter image description here

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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):

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

enter image description here

enter image description here

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.

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    $\begingroup$ Nor do I understand why it might occur in the manner I described in my first comment to Sophit. $\endgroup$ May 11 at 16:27
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    $\begingroup$ @MichaelHall: I totally agree that my answer is thin, I'm expanding it with an example but I need some (free) time 😉 $\endgroup$
    – sophit
    May 11 at 16:48
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    $\begingroup$ @Wyatt: yes always. The lines in the plot are always above the values where they meet the y-axis and those values correspond to the angle of the Mach cone. If you follow the line at Mach 2 of my example, you'll see that it meets the y-axis at 30° $\endgroup$
    – sophit
    May 11 at 20:21
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    $\begingroup$ @ROIMaison: that's written at the end of my answer 🙃 $\endgroup$
    – sophit
    May 14 at 14:10
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    $\begingroup$ @ROIMaison: Ah, now I get it. I've analysed only what happen close to the body while you mean what happens far from it. Yep right, I'll update the answer 👍 $\endgroup$
    – sophit
    May 14 at 18:43
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Here's another diagram of the equations and terminology in sophit's answer.
(This isn't a proper answer as such.)
enter image description here

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  • $\begingroup$ Haha love that diagram 😂 It does give some good info. $\endgroup$
    – Wyatt
    May 14 at 0:31
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    $\begingroup$ The angle of the oblique shock wave is not gives so that a comparison cannot be done 😉 $\endgroup$
    – sophit
    May 14 at 5:53

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