# What happens in order to make a shock when the flow first turns?

Imagine you have a really zoomed in view of a 2d ramp, which is at 20 degrees. Zoom in on the exact point the ramp turns from horizontal to 20 degrees.

When the first supersonic air molecules reach where the ramp turns, they rapidly compress into a shock. My question is, what determines what angle this shock will be? What mechanism makes the shock be at a certain angle?

After this first section of shock is made, I understand why it will continue outwards, but at the exact point where the first section of shock is first made, I don’t understand how it will choose its own angle, considering different turning angles and speeds.

For example if you look at the oblique shock chart, there are different angles the shock will be with different speeds/turning angles. This question is asking how those shocks will ‘choose’ their angle based on flow turning and speed, or what happens to make them at said angle.

When the first air molecules get to where the wedge turns to 20 degrees, what exactly do the air molecules do to make a 54 degree shock (at Mach 2)?

(54 degrees according to the oblique shock chart)

(This (great) answer was explaining something different; it didn’t talk about what exactly happens at the point where the flow turns)

### Photos to help:

• first supersonic air molecules doesn't make sense. Air molecules faster then the fastest air molecules . Tip: avoid mixing the mental model about "atomic/molecular gases", with pressure/temperature (and sonic speed): statistics is not intuitive (also for statisticians) so risk to do things wrong are high. Either you do very methodological and mathematical, or use one or the other mental models. Commented Jun 14 at 7:14
• Doesn't the shock form at the point where the molecules go supersonic? ("Turning" can involve acceleration) Commented Jun 14 at 13:56

So, first a disclaimer. The oblique shock charts are based on inviscid flow. This works great away from the wall, but exactly at the corner (if you zoom in like you're saying), there is of course a boundary layer. That boundary layer messes with reality.

If we stick with inviscid flow, then the streamline at the wall is no different from the rest of the streamlines. The inviscid wall is what we call a slip wall (a viscous wall is a no-slip wall).

The wall is just like another streamline. It is a streamline that turns twenty degrees. The next one out must do the same thing, etc. If you accept that the other streamlines turn and form a shock, then the same thing applies to the wall.

In this case, the angle of the oblique shock is strictly determined by \theta and Mach number.

Post Script If that answer isn't satisfactory (which it probably isn't), then to really get into it, you have to go through the derivation and understand all the math.

It is only conservation of mass, energy, and momentum -- with a perfect gas equation of state. But connecting the dots from those simple principles to the end state is non trivial.

You need to read a textbook on gas dynamics or an aerodynamics textbook. Either one should do a good job with the oblique shock derivation. Don't just jump to that chapter, start at the beginning (of the compressible flow part if an aerodynamics book) and go through 1D flow, normal shocks, and then oblique shocks.

Edit: Trying to explain the shock angle a little more...

Here is something that blew my mind when I learned it.....

When you consider an oblique shock, you decompose the velocity vectors into two components -- the component normal to the shock and the component tangential (parallel) to the shock. (Don't worry about the fact that we don't know what the shock angle is yet).

Here the incoming velocity is V1. Like usual, the shock angle is $$\beta$$. The component of V1 normal to the shock is $$V_{n1}$$ and the component of V1 tangent to the shock is $$V_{t1}$$.

Likewise, after the shock, the velocity is V2, with components $$V_{n2}$$ and $$V_{t2}$$.

Here is the mind blowing part...

$$V_{t2}=V_{t1}$$

The tangential velocity after the shock is the same as the tangential velocity before the shock!

Here is the next mind blowing part...

In the direction normal to the shock -- the flow behaves exactly as if it was going through a normal shock!

The normal Mach number before the shock $$M_{n1}$$ is greater than 1.0 (supersonic) and the normal Mach number after the shock $$M_{n2}$$ is subsonic. The pressure, density, and temperature jumps all behave exactly like a normal shock as if the tangential velocity was not there.

Here is another view that may help...

If you were observing in a coordinate system attached to the shock, with the flow moving by you left to right, this is what you would see.

Next, imagine you were moving vertically at $$V_{t1}$$ such that $$V_{t1}$$ appeared to be zero to you. You would see exactly the normal shock situation. This is what the flow normal to the shock experiences.

So, if your guess was right (about the shock being parallel to the ramp), then $$V_{n2}$$ would have to be zero. The flow can't penetrate the ramp.

This is what I meant by 'where would the flow go'?

$$V_{n2}$$ must be non-zero and $$V_{t2}=V_{t1}$$. Together, these two components of velocity must form a triangle at the angle $$\theta$$.

Remember when I said "Don't worry about the fact that we don't know what the shock angle is yet". This is where the math gets a bit tricky.

To solve this problem (figure out all the angles, etc), we end up with a system of multiple equations and multiple unknowns.

We know $$M_1$$ and $$\theta$$, but we don't know $$\beta$$. Since we don't know $$\beta$$, we can't decompose $$V_1$$ into $$V_{n1}$$ and $$V_{t1}$$

We know that $$V_{n2}$$ and $$V_{t2}$$ must make the angle $$\theta$$, but we don't know their magnitudes.

We know that $$V_{n1}$$ and $$V_{t1}$$ must be the legs of a right triangle with magnitude $$V_1$$.

Then we know the normal shock relations for how $$V_{n2}$$ must relate to $$V_{n1}$$.

All this is non-trivial to solve and doesn't really lend itself to a simple physically intuitive solution.

In the days before computers, solving this was a real pain. So, really smart people at NACA (the predecessor to NASA) did the calculations for us and produced the famous theta-beta-Mach chart. They also produced numeric tabulated results. These were printed in volumes like NACA 1135. Paper copies of these were distributed widely -- to libraries at universities and research institutions as well as companies working on aerospace problems.

There are plenty of online copies available, but they're typically scans made from an old microfiche. They'll do, but they're grainy and low resolution. If you can lay your hands on an original paper copy (say on eBay), you'll find that the charts were quite beautiful.

• This answer is definitely satisfactory, but it would 100% make sense if the shock angle was equal to the flow turning angle, which it’s not. Essentially I understand why the shock will form, but it doesn’t make as much sense for the shock to not equal the flow turning angle. (It does make sense, but I don’t know how it forms at not the flow turning angle) Commented Jun 14 at 15:55
• @Wyatt it might make a lot more sense to consider that the object and the attached boundary layer are supersonic, and when they are moved into surrounding air, a shock wave will form. The amount of turning may be related, but not exactly matching, to the oblique shock angle. Commented Jun 14 at 17:40
• @Wyatt If the shock was the same as the turning angle, then where would the air go? The shock is not an impenetrable barrier. It is not a wall, it is a rapid change in properties. Commented Jun 14 at 18:12