# Why does the angle of a shock change when the flow leaves the influence area of an object?

In this (great) answer, it says:

"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 θ is zero (i.e. the airflow goes straight) and the shock wave is 30°.

So the oblique shock is made because the flow needs to turn rapidly. It starts from the object turning the flow, then the shock made from that turns the flow and makes another bit of shockwave, then that process repeats. (See here for a more comprehensive explanation)

After the flow leaves the influence area of an object, why is this shock angle able to change? What makes it so the shock can achieve the same turning angle with a different shock angle?

Here is a photo to illustrate what happens after the flow leaves the influence area of an object:

On the right side, you can see the shock bends once it leaves the influence area. That shock still needs to achieve the same amount of flow turning, but it's at a different angle. What allows it to have a different shock angle but still have the same amount of flow turning?

Basically, the shock needs to turn the flow the same angle no matter what. What allows the shock to turn the flow the same amount even though the shock itself is at a different angle?

The best reason I have thought of is because the flow directions/patterns are different around the object, so the shock has to be at a different angle to achieve the same amount of flow turning (relative to the airflow).

I just re-read the answer I provided at the beginning of this question, and found a good thing to illustrate my question.

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.

"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"

Why is the turning smaller further from the nose?

• The 'influence area' of an object in supersonic flow extends downstream at the Mach angle -- forever. Likewise, the area that can influence a point in supersonic flow extends upstream at the Mach angle -- forever. I think your mental model of the 'influence area' is somehow limited to the vicinity of the body -- that is incorrect. Commented May 19 at 19:13

## 1 Answer

It seems we have 2 phenomena at work here, air being moved by the energy of the moving object and an energy pulse being transmitted by the air: the sound wave.

At the leading edge interface the (supersonic) object both moves the airmass and initiates the sound wave.

Behind the object (on the right), the low pressure area moves air inwards, resulting in a bending of the wave inwards.

Once the wave leaves the "influence area" it propagates away at the same angle.

The relationship of mass flow and energy waves can be seen analagously in tidal bores, which can make a wave moving into an estuary stationary because the outflow of the river matches the inward velocity of the (transmitted) wave.

• Downvoter please step forward, identify yourself, and read this reference. What do you think? Commented May 19 at 20:47
• It seems they downvoted my question also (I think). +1 from me though. Commented May 19 at 21:29