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Will the 3D effect change shockwave angle/strength?

Similar to how the 3D relieving effect will change the shockwave angle (see above), will this same effect change the shock angle and strength further away from the body? As a shock goes through different temperature air and density, the shockwave strength will be affected. Will the 3D effect also change the angle/strength?

Imagine you have a cone going supersonic like the one above. As the shock gets further from the body, it will continue in a straight line. However, shocks aren’t just 2d, they continue out in a cone shape.

Similar to how if a sphere was emitting sound in all directions, the sound would be forced to spread out as it got farther from the sphere. I’m asking if this same effect happens to shockwaves, considering shocks are in a 3d cone shape.

(If the question is hard to understand, comment and I’ll do my best to improve it; I found this hard to explain.)

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$\begingroup$ Hmm, okay. I’ll copy/paste what I said to Mr. Kämpf here, because I didn’t explain my question properly. Imagine you have a 2d circle. It is emitting sound waves in all directions. In this case for the sake of explanation, when the sound waves travel off, they go in a straight line and don’t interact with each other at all. Because each of those waves were emitted at a slightly different angle (because it was a circle) at some distance wouldn’t those waves start to separate from each other? I’m asking if that same thing applies to the (oblique) shocks that extend from a body. $\endgroup$
– Wyatt
Commented Jul 7 at 16:14
$\begingroup$ (Let me know if that didn’t make sense) $\endgroup$
– Wyatt
Commented Jul 8 at 1:52
$\begingroup$ The circular pressure wave that comes from a disturbance does not come at an angle, it goes in all directions equally. The reason they can coalesce into a Mach wave is because the disturbance is moving. It is the combination of the forward speed with the radial wave speed that causes the waves to build up in front of the disturbance and spread out behind. Again, this is for an infinitesimal disturbance -- not for a cylinder or sphere. $\endgroup$
– Rob McDonald
Commented Jul 8 at 4:56

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