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It's been (well) established that a shockwave can only be at the angle of the Mach cone. (I now understand this part). However, in this picture:

enter image description here

you can see that the shock is at 2 angles. I also understand why this happens. Mainly because of the 3D relieving effect, and also because the shock can never be at a greater angle than the Mach cone.

The question:

Why is the shock allowed to essentially have 2 angles in this case? Towards the rear and further away from the object, the shock is at a different angle than it is near the object. I understand why this happens, but I don't understand why a single continuous shock is allowed to have 2 different angles. Essentially, because the shock can only ever be at the same angle as the Mach cone, why is it seen at 2 different angles in this picture, considering the Mach cone is only at one angle in this picture?

Another example:

Here is a picture of a T-38C:

enter image description here

If you look closely, near the plane some of the shockwaves are at different angles than the others. When they get further out, they all tend to the same angle (the Mach cone). Why are the shockwaves allowed to have different angles close to the body? Why aren't they forced to the Mach cone angle, like the shocks are further out?

Chart for shock angles: enter image description here

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You know the answer to this.

The shock angle is determined by how much it turns the flow. Recall the oblique shock chart I've posted multiple times -- we call it the theta-beta-Mach chart. Theta is the turn angle, beta is the shock angle, Mach is the upstream Mach number.

Different places around the body need different amounts of turning.

The 3D relieving effect means you need less turning away from the body than near the body.

Edit: Include oblique shock chart.

enter image description here

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  • $\begingroup$ I see. I'll ask this elsewise: Imagine the Mach cone is say 30 degrees. The shock further away from the body will also be 30 degrees. Now closer to the body, the shock is at 35 degrees. (See first picture for example) In the picture, the shock is clearly allowed to exist (the one closer to the body), why is that considering it's at a higher angle than the Mach cone? Asked differently, why doesn't the Mach cone force the shock to be 30 degrees? Is it because of the different flow around the object, compared to freestream? $\endgroup$
    – Wyatt
    Commented Jun 8 at 3:22
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    $\begingroup$ If the Mach angle is 30 degrees, consulting the theta-beta-Mach chart, we see that the freestream is Mach 2.0. For a Mach 2.0 freestream and a 35 degree shock, we see that the turning angle is approximately six degrees. So, near the body, where the shock is 35 degrees, the flow is turning six degrees. There is no inconsistency. $\endgroup$
    – Rob McDonald
    Commented Jun 8 at 5:51
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    $\begingroup$ The flow turning angle is whatever it needs to be. This might be caused by the immediately local geometry, other nearby flow features (fans, shocks),etc. But together, the turning angle and upstream Mach determine the shock angle. Period. Full stop. When you look at the chart, the zero-degree deflection point for each Mach number curve is the Mach angle. All the finite deflection points are higher angles, so yes, the 'rules' are never broken. $\endgroup$
    – Rob McDonald
    Commented Jun 8 at 23:29
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    $\begingroup$ @Jpe61 Sorry for leaving it out. It is just the Oblique shock chart. $\endgroup$
    – Rob McDonald
    Commented Jun 8 at 23:30
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    $\begingroup$ That is correct. It is forced to a normal shock -- and the flow downstream of a normal shock is subsonic (the flow downstream of an oblique shock is still supersonic, just slower). That subsonic pocket of flow can then deal with whatever the obstacle is. $\endgroup$
    – Rob McDonald
    Commented Jun 9 at 19:43

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