A Question About Shock Waves From an Airplane

[hutchphd]

Here is how it works.

I believe everyone understands this explanation.
But in 3D everything is spreading out and the overpressures which allow the shock to persist rapidly diminish with distance. The question on the table is "how rapidly does it diminish?". A stick of dynamite can produce a supersonic shock wave a metre away, but the boom I hear at 500 meters is not supersonic. It is a loud noise. Where is the transition and how is it defined?
Similarly for a supersonic projectile and the conical disturbances...at some point it is "just sound. I also wonder about the sound of nearby lightning strkes...there is often a prompt "sizzle-crack" which preceeds the "big boom" Is the "crack" supersonic?

 

[boneh3ad]

Sorry for what is about to be a very long-winded answer. We are hitting on some concepts that would really be easier to discuss with pen and paper and a week to give a crash course on shock theory. I will do my best.

The straight definition of a shock wave (>c) is fine and no one can argue with that. So, what would a probe, a few hundred metres away from a ss aircraft 'see'? Where would the supersonic flow be and in what direction? From the angle, the velocity of apparent propagation of the conical wave appears to be sonic.

This is now getting into the concept of moving shocks, which is usually the most difficult topic for students to grasp in an introductory gas dynamics course in my experience. It is easiest to start by considering a stationary normal shock with a flow approaching it normal to its wavefront. The incoming flow has Mach number ##M_1>1## and the outgoing flow has Mach number ##M_2<1## by definition. This is one of the most fundamental aspects of shock waves.

[Aside: Shocks do not alter velocities that occur parallel to their wavefronts, so if you superposed a parallel velocity onto the flow (e.g., in the case of oblique shocks), the jump conditions do not change so long as you only consider the Mach number of the velocity components normal to the shock and then add the parallel velocity to them later. This is generally how oblique shock theory is introduced.]

If you placed a probe sitting on the ground as a shock wave passed over it, you'd measure a whole bunch of things. Let's assume the probe measures temperature (##T##), pressure (##p##), and velocity (##\vec{v}##) and is located above the ground, say, 5 m/15 ft/whatever units you prefer.

It would measure "nothing" before the shock passes. Constant ##p## and ##T## and ##|\vec{v}| = 0##. After the shock passes, it would measure an increase in ##p## and ##T## and a nonzero velocity in the direction the shock was propagating (i.e. normal to the wavefront, in this case downward and in the direction the plane was flying). At first glance, this would appear to break the no penetration condition (i.e. you'd have flow going into the ground), but this is solved when the shock reflects off of the ground and turns the flow again the other way.

Whether or not ##|\vec{v}|## is supersonic depends on the shock Mach number. In most cases, the answer will be no. You need a fairly strong shock to accelerate the flow behind it beyond the local speed of sound. I emphasize local here because the increase in ##T## across the shock also increases the speed of sound, which you have to consider. These results probably feel a little bit unintuitive since thinking from the point of view of a stationary observer (rather than the moving vehicle) is not as straightforward.

If what you have told me is true then where does this come from? When / where is the transition from shock to sound? Everything I have written in this thread has been to do with the sound wave. Are there two waves hitting the ground.

Your post seems to deal with just the formation of the shock wave near the craft. I now understand that the shock wave will be oblique. That is interesting. Does it imply that the resulting wave will always have that tilt? Different planes will have a different Mach Angle? Or does it mean that the Mach angle will be established sooner?

Mach angle is inherently a local phenomenon. Any point in a supersonic flow has an associated Mach angle that is measured relative to the flow direction and is a function only of the Mach number. The way Mach cones and Mach waves are often introduce in a course is the use of an infinitesimally small "beeper" that emits a wave every ##\Delta t## with speed of sound ##a##. You can use that along with the velocity of the beeper itself to derive the definition of ##\mu##.

Ultimately, ##\mu## is primarily a measure of the region of influence emanating from a given point in a supersonic flow. Since information travels at the speed of sound but the flow is moving supersonically, it is likely no surprise that a disturbance has no upstream influence. If you are familiar with partial differential equations, then you'd also not be surprised that the equations governing inviscid supersonic flows are hyperbolic (as opposed to elliptic in subsonic flows). If you carry that to its logical conclusion, ##\pm\mu## defines the characteristic curves passing through a give point in a supersonic flow.

You can observe Mach waves propagating at the Mach angle in practice. Any disturbance small enough that it does not meaningfully change flow direction (##\theta##) in a manner that causes compression (turning the flow "into" itself) will produce what is essentially a Mach wave. This is why I made the point earlier that
[tex]\lim_{\theta\to 0}\beta = \mu.[/tex]
This might be a small ding in the surface of a wing or from a small hole like a pressure measurement port. Any disturbance that does cause a change in ##\theta## is going to instead produce an oblique shock. at some angle ##\beta > \theta##. As I mentioned before, ##\beta## depends on more parameters than ##\mu##, and that's a result of the fact that ##\beta## requires a change in flow angle. The shock angles propagating away from an aircraft are therefore geometry and Mach number dependent.

Typically, the leading shock angle will be determined by the nose cone shape. Where things get complicated is when you consider expansion waves. After the nose cone sets the shock angle, the flow will typically encounter a convex geometry as it passes over, for example, the canopy. That is an angle change, but one that produces an expansion rather than compression. This also produces Mach waves in a phenomenon called Prandtl-Meyer expansion. Those waves will propagate out at the local ##\mu## and eventually intersect the shock. The interaction will cause the shock to refract, generally back toward the vehicle.

There are generally various patterns of shocks and expansion waves that form over various portions of the surface of an aircraft that produce some pretty wild wave shapes. The two strongest, though, are the shock emanating from the nose and the shock emanating from the back of the aircraft when the flow all has to meet back up and turn parallel. Those two shocks will generally propagate the greatest distance and are what result in the characteristic "double thump" of most audible sonic booms.

This seems do deal with my question but what constitutes a great distance and what would the 'velocity', rather than 'speed' be?
Intuitively, I would think that ss air flow would dissipate energy in a short distance and leave you with a sonic speed. The power flux would follow an inverse law (square / linear?) as the cone widens.

Bottom line is can you answer my question about the probe at a distance. How will the air be moving? Will any of it be supersonic and in what direction? I think all this is so obvious to you that you can't see how a bear of vey little brain could have a problem with it. ;-)

I answered a bit of this in my response above, but to answer the question about what constitutes "a great distance," will will admit at this point that I don't have a good answer. It's not within my sub-area of expertise in the field of high-speed aerodynamics and I don't want to just make stuff up. I feel I am in danger of inadvertently doing that. suspect the guys at NASA and Lockheed working on sonic boom mitigation and the X-59 QueSST would be able to give a long, detailed answer, though.

If I get a chance before leaving town on Friday I might try to do some digging.

The air doesn't move, only its pressure, temperature and density are changing. Though, this "change" is moving.

Not really accurate. See above.

What is confusing is that "shock wave", "shockwave" and "shock" are synonyms (Wikipedia). A shock can be stationary (like in a nozzle). I guess "shock wave" sounds more appropriate when talking about a moving object where the shock moves with the object. But its apparent velocity depends solely on the velocity of the object.

It's all a matter of frame of reference. Shock "waves" are always propagating in some frame of reference, but are most easily defined and studied in the frame where they are stationary.

Now that is totally confusing me. A shock wave is, according to pretty much every source I can find, is generated when the air (/medium) travels faster than the local speed of sound.

The air doesn't necessarily have to travel faster than the speed of sound at any given point for a shock to form. It just has to be traveling faster than the speed of sound relative to an immersed object. This is why a supersonic plane moving through stationary air generates shocks despite the fact that the air never actually moves at a supersonic speed relative to the Earth. The interaction with any object is also an important part. Something has to compress the supersonic flow in order to generate the shock.

 

[russ_watters]

The sound from a passing sub sonic aircraft also follows a cone which also 'travels' with the aircraft. The wave that hits you is traveling at the speed of sound and not the speed of the aircraft so your argument is hardly conclusive.

At this point it's probably best to break out the animation or a sketch. You're not differentiating between speed parallel and perpendicular to the direction of motion. That's the key. It travels parallel at the speed of the plane and perpendicular at the speed of sound.

 

[hutchphd]

It's all a matter of frame of reference.

The natural frame of reference here is that of the medium. Which also happens to be the frame of the observor. Why would you choose otherwise ? And so the plane is supersonic and generates a shock front from any protuberance into the line of motion...this is not difficult. The question is the persistence of that structure any appreciable distance away. If you wish to call it "sound generated by the shock front" I have no quibble.

It travels parallel at the speed of the plane and perpendicular at the speed of sound.

Sorry I am not following this argument at all. What is "it" and Perpendicular to what?

The nomenclature (and perhaps the physics) seems very unclear to me.

 

[jack action]

A stick of dynamite can produce a supersonic shock wave a metre away, but the boom I hear at 500 meters is not supersonic. It is a loud noise. Where is the transition and how is it defined?
Similarly for a supersonic projectile and the conical disturbances...at some point it is "just sound. I also wonder about the sound of nearby lightning strkes...there is often a prompt "sizzle-crack" which preceeds the "big boom" Is the "crack" supersonic?

I'm not sure I understand what the interrogation is. It seems you think a shock wave will transform into a sound at some point? That the speed of a wave influences the fact that we hear it or not?

Without being an expert n the subject, a sound is not a question of pressure or velocity, it is a question of vibration, frequency. A shock is a disturbance, which causes a short vibration, thus the sound. I don't think the shock itself transforms into a sound. Like a hammer striking a metal plate doesn't transform those objects into a sound. The hammer and the plate vibrate, which causes the surrounding air to vibrate, and those vibrations activate your auditory sensors.

 

[sophiecentaur]

The air doesn't necessarily have to travel faster than the speed of sound at any given point for a shock to form. It just has to be traveling faster than the speed of sound relative to an immersed object. This is why a supersonic plane moving through stationary air generates shocks despite the fact that the air never actually moves at a supersonic speed relative to the Earth.

That was very useful for me. The earlier part of the post threw me into a panic because we had the air moving and the plane stationary and I had to shift my frame of intuition radically. But I now get the idea of stationary shocks (films of rocket launches) and that helps.
But you say that the air never moves faster than sound. That's difficult because I have to ask what about the air immediately in front of the plane? It has to get out of the way pretty damn quick. Can't that require a bulk displacement at greater than the speed of sound? Or will you introduce a temperature change to take care of that?

but to answer the question about what constitutes "a great distance," will will admit at this point that I don't have a good answer.

That's a shame as it's basically the first thing I wanted to know. The detailed stuff about shock waves and how they form came out in the process of the thread. But it only needs a ball park idea of distance because I really wanted a firm idea about the nature of the Boom and whether there is any 'disturbance' (or energy flow) that is actually faster than sound. There can be a 'virtual speed' as a cone intersects with the ground but that's not real, is it?

It seems you think a shock wave will transform into a sound at some point? That the speed of a wave influences the fact that we hear it or not?

That reads like a bit of a false dichotomy. Why should one not hear a shock wave? Perhaps the details of the construction of the sentence you were reading were not perfect.

Without being an expert n the subject, a sound is not a question of pressure or velocity, it is a question of vibration, frequency

A sound wave consists of variations in pressure and displacement. That's just basic Physics. The frequency (and amplitude) are quantities that are perceived as sound. The word "wave" is sometimes omitted when the meaning is assumed.

 

[russ_watters]

Sorry I am not following this argument at all. What is "it" and Perpendicular to what?

The nomenclature (and perhaps the physics) seems very unclear to me.

Like I said, a diagram is needed:

The sound pulses/waves travels radially out from the plane and are heard by the observer on the ground along a path perpendicular to the ground and direction of travel of the plane. So the sound from the plane traveled perpendicular to the direction of travel of the plane, from the plane at A to reach the observer at A1; At 300m/s (the speed of sound, rounded to the nearest hundred). In both cases.

In the first case, the plane is moving at 150 m/s and the point of arrival of the pulses also moves at 150 m/s. In the second case, 600 m/s.

I'll say it again:
1. The sound waves travel away from the plane at 300 m/s (speed of sound).
2. The point where the sound waves arrive at the ground moves at the speed of the plane.

 

[hutchphd]

We're not connecting here. If the first pulse is emitted 1s prior to the second then its circle will have a radius 300m bigger than the second at any subsequent time.

 

[russ_watters]

We're not connecting here. If the first pulse is emitted 1s prior to the second then its circle will have a radius 300m bigger than the second at any subsequent time.

The circles are drawn 1s after each pulse is emitted; t=1 for the first pulse, t=2 for the second pulse. I didn't draw a second circle for the first pulse at t=2.

 

[russ_watters]

I may be jumping backwards in the discussion with my diagram, but I'm not sure people have such basics straight. Ultimately though I think what this probably ends with is the issue in @boneh3ad 's post #26. It's in the difference between the mach wave/cone and shock wave/angle. I'll admit that I've had them muddy in my mind and looking at sources on the internet they seem to be mushed together. It may be useful then to work and actual example comparing the two for the same scenario.

 

[hutchphd]

The whole process repeat itself at time t+Δt. Pressure waves and reflected pressure waves add up together as they cross each other. This is more true inside pipes since a plane is in the atmosphere with not much to reflect the pressure waves from the source.

Scattering in 3 dimensions is very much different from 1D in a pipe. I simply cannot see how this would persist for any appreciable distance for a small source in a big world. What you are describing is a resonance effect, with no justification for the resonance,

 

[boneh3ad]

That's a shame as it's basically the first thing I wanted to know. The detailed stuff about shock waves and how they form came out in the process of the thread. But it only needs a ball park idea of distance because I really wanted a firm idea about the nature of the Boom and whether there is any 'disturbance' (or energy flow) that is actually faster than sound. There can be a 'virtual speed' as a cone intersects with the ground but that's not real, is it?

Don't worry. I'm going to give this a bit more thought and see if I can come up with a better response or source. It just might require me to whip up some more graphics to avoid another wall of text.

 

[boneh3ad]

The natural frame of reference here is that of the medium. Which also happens to be the frame of the observor. Why would you choose otherwise ? And so the plane is supersonic and generates a shock front from any protuberance into the line of motion...this is not difficult. The question is the persistence of that structure any appreciable distance away. If you wish to call it "sound generated by the shock front" I have no quibble.

That is the most natural reference to encounter a shock, but it is not the most natural to analyze them. The jump conditions are dictated by incoming Mach number and most readily produce the outflow conditions in the frame of the stationary shock. All other frames are frequently considered first by "freezing" the shock and then converting back to the original frame of interest later.

Part of the reason for this is for ease of analysis. By freezing the shock, you've converted the unsteady problem of a propagating shock to a steady problem of a stationary one. Once all ##\partial/\partial t## terms are thus eliminates, the conservation laws are much more tractable (e.g., the steady Euler equation instead of the unsteady one).

That shock persists a considerable distance. As I mentioned previously, I'll need to do a little more digging to define "considerable" more precisely. My background is in supersonic boundary layers and the shock fields closer to bodies.

I'm starting to wonder if this warrants me cooking up an Insight article since I've seen so much bad information online about all of this while having this discussion.

 

[A.T.]

It just might require me to whip up some more graphics to avoid another wall of text.

What would be helpful, is a non-local version of the graphic below, that shows what happens with the oblique shock with increasing distance. It should be based on a 3D object like a cone (the graphic being just a 2D cross section).
- Does the conical oblique shock maintain the same opening angle ##\beta## with increasing distance?
- Do the streamlines maintain the same turning angle ##\theta## with increasing distance?

 

[sophiecentaur]

since I've seen so much bad information online about all of this while having this discussion.

I was beginning to feel bad about keeping this going but, in the light of all the conflicting views that have emerged, I think it's been valuable. @boneh3ad I think you could usefully dig a lot of us out of the slough of misinformation with a concise, definitive few paragraphs.

 

[hutchphd]

I'm starting to wonder if this warrants me cooking up an Insight article since I've seen so much bad information online about all of this while having this discussion.

Yes please. I am woefully ignorant about this stuff and freely profess massive confusion! I do know it is not simple.

 

[boneh3ad]

What would be helpful, is a non-local version of the graphic below, that shows what happens with the oblique shock with increasing distance. It should be based on a 3D object like a cone (the graphic being just a 2D cross section).
- Does the conical oblique shock maintain the same opening angle ##\beta## with increasing distance?
- Do the streamlines maintain the same turning angle ##\theta## with increasing distance?

View attachment 291751

In the near field, yes, the conical version of an oblique shock maintains the same opening angle ##\beta## with increasing distance from the tip. Assuming a cone of infinite length, it will continue to do so. For a finite length cone, there will be other waves generated that eventually intersect that shock and cause it to bend (typically a reduction in ##\beta##). I should also note that ##\beta_{\mathrm{wedge}}\neq\beta_{\mathrm{cone}}##. It's the far field I am less sure of. I am intuitively able to convince myself either way depending on the argument, so I need to do some digging on that one.

The streamlines along a cone are trickier than a 2D wedge. Behind the shock for a 2D wedge, the flow is entirely 1D. That's not true for a cone, where the streamlines continue to converge a bit after passing through the shock. However, for a given ##\beta## and ##M_1##, the flow angle ##\theta## will always be the same immediately downstream of the shock.

 

[A.T.]

It's the far field I am less sure of. I am intuitively able to convince myself either way depending on the argument, so I need to do some digging on that one.

My intuitive assumption was that in 3D, for a finite size cone, the flow turn angle decreases with distance (the streamlines straighten out) and so the oblique shock cone angle tends towards the Mach cone angle.

Kind of like what is explained here for a blunt body at point 5 (time 11:56):

 

[boneh3ad]

My intuitive assumption was that in 3D, for a finite size cone, the flow turn angle decreases with distance (the streamlines straighten out) and so the oblique shock cone angle tends towards the Mach cone angle.

If (when) the shock weakens and dissipates, this would likely be the mechanism. The unclear answer to me is where this actually occurs. You can't really model it as a spherical wave deteriorating based on energy considerations proportionally to ##1/r## because it isn't spherical. But then again, it's not a plane wave (clearly has a 3D relieving effect behind it), nor is it truly cylindrical. Now, if the Mach number is high, you actually can approximate it as cylindrical using what is called blast wave theory (as part of the hypersonic equivalence principle), but that wouldn't really be appropriate at the Mach numbers involved here.

 

[hutchphd]

Allow me to stir up the fire a little and thereby shed light on my ignorance

In my mind the tip of the arrow is roughly the size of the airplane. We assume the velocity is parallel to the ground.

My intuitive assumption was that in 3D, for a finite size cone, the flow turn angle decreases with distance (the streamlines straighten out) and so the oblique shock cone angle tends towards the Mach cone angle.

I would be (and perhaps will be) very surprised if the "near field" distance is very much bigger than the object. And just to be definite, in the far field the Mach cone defines a leading edge that propogates at the local speed of sound. It intersects the surface of the Earth at the supersonic speed of the plane.

.

 

[sophiecentaur]

If (when) the shock weakens and dissipates, this would likely be the mechanism. The unclear answer to me is where this actually occurs. You can't really model it as a spherical wave deteriorating based on energy considerations proportionally to 1/r because it isn't spherical.

This rocket seems to be generating shocks and they only extend for just a few times its length. Could this be an indication as to the extent of actual shocks from a supersonic aircraft?

 

[boneh3ad]

This rocket seems to be generating shocks and they only extend for just a few times its length. Could this be an indication as to the extent of actual shocks from a supersonic aircraft?
View attachment 291791

That's a different physical mechanism, though, and one that would make a fun Insights article in its own right.

 

[russ_watters]

What would be helpful, is a non-local version of the graphic below, that shows what happens with the oblique shock with increasing distance. It should be based on a 3D object like a cone (the graphic being just a 2D cross section).
- Does the conical oblique shock maintain the same opening angle ##\beta## with increasing distance?
- Do the streamlines maintain the same turning angle ##\theta## with increasing distance?

View attachment 291751

...including a relationship with the mach cone/angle.

 

[sophiecentaur]

a different physical mechanism, though,

Standing wave?

 

[boneh3ad]

Standing wave?

Yes you can set up a standing wave that way. It happens frequently when doing static test stand fires of jet and rocket engines. You have a pressure leaving the nozzle that is different from atmospheric and requires a series of expansion and shock waves to allow it to come to equilibrium with the surroundings.

 

[sophiecentaur]

Yes you can set up a standing wave that way. It happens frequently when doing static test stand fires of jet and rocket engines. You have a pressure leaving the nozzle that is different from atmospheric and requires a series of expansion and shock waves to allow it to come to equilibrium with the surroundings.

But, for a supersonic aeroplane, is this substantially different for a blunt nose?

 

[boneh3ad]

But, for a supersonic aeroplane, is this substantially different for a blunt nose?

Yes and no. Fundamentally, the plane moving through air is creating a shock because it needs to turn the air out of the way and the air cannot react fast enough without a shock. For the engine, it's about an expansion that causes the high pressure gas in the combusted to exit at a pressure different from ambient.

 

[tmbouman]

Hi everyone, new user who stumbled on this awesome thread. I have been struggling and now believe that struggle is connected to the difference between mach angle & shock angle. Let me propose a problem:

A plane is traveling Mach 2 one mile above the ground. The plane has a cone shaped nose with alpha equal to 20 degrees. How far past the person on the ground (in the horizontal/parallel direction) will the plane be when the person hears the shockwave assuming it has not deenergized enough to be a soundwave? (Lets assume we live on a flat Earth )

 

[hutchphd]

How far past the person on the ground (in the horizontal/parallel direction) will the plane be when the person hears the shockwave assuming it has not deenergized enough to be a soundwave?

What does the phrase in red even mean??

Let me restate it as "when the person first hears the airplane". The answer I would give is the airplane will be two miles downrange (to within a few %). If that is not correct please explain it to me ! Please assume the ambient air to be uniform in T and P

oops: Please see next post by @jbriggs444

 

[jbriggs444]

What does the phrase in red even mean??

Let me restate it as "when the person first hears the airplane". The answer I would give is the airplane will be two miles downrange (to within a few %). If that is not correct please explain it to me ! Please assume the ambient air to be uniform in T and P

Let us treat it as a mathematical problem and assume that the sound of the passing plane travels to the listener at the speed of sound. Meanwhile, the plane is moving at twice the speed of sound.

It is tempting to assume (as you seem to have done) that the sound that the person first hears will have been generated when the plane passed directly overhead. From that assumption, the calculation is simple -- one mile from the sound, the plane must be two miles downrange.

However, the listener will hear the plane earlier than that.

The angle of the sound wave front will ideally be arcsin(1/2) from the horizontal. Thirty degrees. At the time of emission, the plane will have been ##\tan 30## miles before passing directly overhead. The sound wave will then travel ##\frac{1}{\cos 30}## miles to the listener. During this time the plane will have moved ##\frac{2}{\cos 30}## miles onward for a net of ##\frac{2}{\cos 30} - \tan 30## or approximately 1.73 miles downrange. (This is also equal to ##\frac{1}{\tan 30}## miles).

Edit, think I messed up lead calculation. Fixing.

 

[hutchphd]

It is tempting to assume (as you seem to have done) that the sound that the person first hears will have been generated when the plane passed directly overhead.

Absolutely that was my quick and dirty (and obviously incorrect) analysis. Thank you for gently pointing out the correct version.

 

[DrClaude]

@jbriggs444 Nice post, but that figure is unreadable.

 

[jbriggs444]

@jbriggs444 Nice post, but that figure is unreadable.

It is a poor workman who blames his tools. I blame mspaint. But I've zoomed in the figure now.

 

[DrClaude]

It is a poor workman who blames his tools. I blame mspaint.

One rarely errs when blaming MS*

 

[tmbouman]

The angle of the sound wave front will ideally be arcsin(1/2) from the horizontal. Thirty degrees. At the time of emission, the plane will have been tan⁡30 miles before passing directly overhead. The sound wave will then travel 1cos⁡30 miles to the listener. During this time the plane will have moved 2cos⁡30 miles onward for a net of 2cos⁡30−tan⁡30 or approximately 1.73 miles downrange. (This is also equal to 1tan⁡30 miles).

Thank you for taking the time to work this up. From what you wrote, it seems that the propagation is only a function of mach angle. I am confused on this point. Why do you not use the cone angle in your math and compute Shock angle as boneh3ad describes in post #26 either via the θ-β-M equation he listed or Taylor-Maccoll equations:

The Mach angle is defined very simply and is based on the speed of sound waves propagating relative to a supersonic source. It is
μ=arcsin⁡1M1
where M is the inflow Mach number. In contrast, the shock angle is defined very differently. For a simple 2D wedge, it is common to use the θ-β-M equation, which is quadratic in M12 and depends on θ (the flow turning angle or wedge angle) and β (the shock angle).
tan⁡θ=2cot⁡βM12sin2⁡β−1M12(γ+cos⁡2β)+2.
Clearly, β≠μ. Additionally, β>μ and μ+θ>β. Finally,
limθ→0β=μ.
The schematic below from Wikimedia commons (and the oblique shock Wikipedia page) lays out the variables. Note that γ=cp/cv is the ratio of specific heats of the gas.

I will note that the θ-β-M does not work for conical flows, so you have to use the more complicated Taylor-Maccoll equations to solve for β in that case, though the answers are similar to those for a wedge.