# Why will the flow turning angle be 0 (or close to) inside of the Mach cone?

(Disclaimer : This probably seems like my 56th question on the same topic, but this will be my last relating to Mach cone angle and shock angles, as I think I understand most of the topic)

So the reason a shock can never be at less of an angle than the Mach cone is because if it was the flow turning angle would be 0. If it's 0, then there is no shockwave. This makes sense, but this question is asking why the flow turning angle is 0 below the Mach cone angle. This answer provides some insight. Also, here's the graph showing shock angles at different deflection angles.

Also here's a picture I sketched up to illustrate my question:

(Obviously the yellow line would never be a shock with a turning angle of 0)

Why would that yellow line (if it was a shock) have a turning angle of 0 (or a turning angle so low that the shock didn't exist)? It would still have an angle relative to the plane, so it seems like it would still turn flow, allowing a shock to exist. For it to have a turning angle of 0, my intuition tells me the line would have to be parallel to the flow direction.

The answer linked above (here) gives some information about this subject, and it uses the graph above to do that. It explained that the shock angle will never be below the Mach cone angle, but I couldn't seem to find why the flow turning angle will be 0 below the Mach cone angle.

This situation would never occur normally, as the shock angle is always greater than the Mach cone angle, but assume the shock was somehow artificially bent to the angle pictured in the illustration.

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.

Why would the weakest possible shock be right above the Mach cone angle, and the point where the shock disappears be right at the Mach cone angle?

Graph source

Edit : In case there is some confusion, I know the flow always turns near the object because it has to, but this is talking about the flow turning further away from the object.

Why would that yellow line (if it was a shock) have a turning angle of 0 (or a turning angle so low that the shock didn't exist)?

There would be no shock.

There is no shock because there is no turning. The turning is determined by the obstacle. A zero degree obstacle has been shown to cause the Mach angle. More obstacle, more shock angle.

What obstacle is less than a zero obstacle?

Why would the weakest possible shock be right above the Mach cone angle, and the point where the shock disappears be right at the Mach cone angle?

Without a disturbance to cause the flow to turn, there is no shock.

Zero disturbance, zero shock.

The object, wall, thing in the flow is the disturbance. The flow must turn to go around the disturbance. The flow turning causes the shock.

No wall, no turning, no shock.

It is not the other way around...