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enter image description here

(Excuse the drawing quality)

What will happen if a shockwave has an angle less than that of the Mach cone?

In this picture, the black line represents the Mach cone. The orange line represents the shockwave angle that would be if it wasn't for the Mach cone making it be the same angle as the Mach cone itself. The yellow line is what this question is about.

So the shockwave can never be in front of the black line (like the orange one is). That's already established, here is an answer that explains it very well.

On a normal occasion, the shockwave would never be at the angle of the yellow line. Best stated from the answer linked above

The shock wave is therefore more "vertical" than the Mach cone

and that is always the case.


So now say that there is a shockwave at the angle of the black line and somehow it got bent past the angle of the Mach cone, just like the yellow line.

enter image description here

Picture of shockwave bending like linked answer above talks about.

If it bent the other way, I would understand how it would stay at the same angle as the Mach cone. But in this case, I don't see anything forcing the shock to be at the same angle as the Mach cone. To summarize, it makes sense why the shock wouldn't be at the angle of the orange line, but I don't see anything stopping it from being at the angle of the yellow line.


By "Certain Circumstances" I meant when the low pressure behind the object will bend the shockwave.

If I can improve this question in any way, please leave a comment. On that note, if the downvote was because this answer to my other question is too similar, I would understand but I don’t think it answers this specific question.

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  • $\begingroup$ Downvoter, what here did I not follow correctly? $\endgroup$
    – Wyatt
    Commented May 22 at 0:40

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No.

Below is an oblique (2D, Wedge) shock chart. It shows everything a 2D shock can do.

The top half of the chart is for strong shocks, the bottom half is for weak shocks. Let's just deal with weak shocks for now.

Consider one value of $M_1$. Say the $M_1=2.0$ line. Notice where this line hits the Y-axis. 30 degrees. For a deflection angle (x axis) of zero degrees.

This is the value of the Mach angle for $M=2.0$. So, an infinitesimal disturbance that does not turn the flow at all will make a 30 degree shock with a $M=2$ upstream -- this is the definition of the mach cone.

Any flow turning (deflection angle non-zero) and the shock wave angle increases.

Oblique shock chart

Here is the same sort of chart for a conical shock.

The values in this chart are different, but the general shape is the same. Importantly, the place where a given Mach number intersects the axis is the same -- the Mach angle.

The differences in these charts are due to the 3D relieving effect of a cone.

Conical shock chart

You should look into NACA Report 1135 -- it is a great source for many compressible flow things.

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  • $\begingroup$ I see. I understand that the shockwave angle will decrease with lower deflection angles (as stated in the graph), but since the Mach cone is purely a geometric entity, nothing could really change the angle of it besides speed. If the Mach cone stays constant, why does the shockwave not 'misalign' with the angles of it? Actually here is something I just thought of : Are you basically saying because there has to be a non 0 flow turning angle for a shock, and that the shock angle will always have to be greater than the Mach cone angle because of that? $\endgroup$
    – Wyatt
    Commented May 22 at 18:40
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    $\begingroup$ Yes (to the last part). Or, you could think of it this way -- an infinitesimally weak shock is at the Mach angle. It does not turn the flow. Any finite strength shock will be at a greater angle and will turn the flow. $\endgroup$ Commented May 22 at 19:34

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