A Question About Shock Waves From an Airplane

[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.

 

[jbriggs444]

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:

In the very first sentence of my post, I wrote:

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.

Here I was trying to dodge any argument about mach angle versus shock angle and address only the geometry of a series of spherically propagating signals traveling at one speed from a moving source traveling at twice that speed.

The post to which I was responding had adopted that simple model and used it to arrive an an erroneous result. The intent was to focus narrowly on a corrected analysis under the simplistic model.

@boneh3ad has doubtless forgotten more about aerodynamics and shocks than I can ever hope to learn.

 

[tmbouman]

The post to which I was responding had adopted that simple model and used it to arrive an an erroneous result. The intent was to focus narrowly on a corrected analysis under the simplistic model.

Understood @jbriggs444 . No problem. Thanks again for your effort. Is it your understanding ideally we would calculate shock angle with the Taylor-Maccoll equations and use that instead of mach angle?

Is anyone willing and technically capable to help me understand the difference between mach and shock angle?

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:

When solving the previously stated example problem, what is the correct angle to use: Mach or shock?

 

[jbriggs444]

You (@tmbouman) have quoted @russ_watters as suggesting a speed of propagation "perpendicular to the direction of motion" and "at the speed of sound".

Certainly, a momentary signal (say the sound of a rifle shot) will propagate perpendicular to the plane's motion at the speed of sound. But what about the position of the wave front (think the series of sounds of a machine gun) as the rat-a-tat-tat approaches a stationary observer on the ground below.

Question: Is the vertical speed of this wave front greater than, equal to or less than the speed of sound?

Clearly, the propagation speed perpendicular to the wave front is at the speed of sound. This is the rate at which a wave progresses along a line perpendicular to the wave front. If one watches the wave propagate along a line at a diagonal, the wave front will take the same time to cover a longer line. So it will have a faster rate of propagation. (Think of an ocean wave almost parallel to the shore. The wave front will traverse the shore faster than the wave moves).

So the answer is that the wave front corresponding to the series of machine gun sounds coming from the passing plane will cover the perpendicular/vertical separation between plane and ground at a rate greater than the speed of sound.

It is mildly interesting to contemplate a series of machine gun pops: 1, 2, 3, 4, 5 from a supersonic craft which would then (hypothetically) be heard on the ground as 3, (2,4), (1,5).

 

[tmbouman]

@jbriggs444 You said the wave front travels at the speed of sound

Clearly, the propagation speed perpendicular to the wave front is at the speed of sound.

@boneh3ad in post 26# said it travels at the speed of the aircraft

This shock will propagate for a great distance (for as long as there needs to be a corresponding change in the direction of the flow) and will travel with the speed of the aircraft since it is attached to and continuously generated by it.

Which is correct?

My current understanding is that while the shockwave has sufficient enough energy to keep the shock occurring it will travel at a speed less than or equal to the speed of the plane and greater than the speed of sound, but as it losses energy it will slow down and eventually just be a sound wave traveling at the speed of sound. Is this the correct understanding?

 

[jbriggs444]

@jbriggs444 You said the wave front travels at the speed of sound

It does. It propagates along a line perpendicular to the wave front at the speed of sound.

@boneh3ad in post 26# said it travels at the speed of the aircraft

It does. It propagates along the line of the aircraft's flight at the speed of the aircraft.

The line of the aircraft's flight is not perpendicular to the wave front. [Unless the aircraft is moving exactly at the speed of sound and has flat plane wave front extending in a disc perpendicular to the flight path -- not sure how physically realistic that is]

The ocean waves on the shore analogy is applicable.

 

[tmbouman]

Shock waves travel faster than the speed of sound normal to their length. That's about as fundamental to shocks as it gets.

@jbriggs444 here is from the start of post #26. When he says "normal to their length" i read that as perpendicular to the wave front.

 

[jbriggs444]

@jbriggs444 here is from the start of post #26. When he says "normal to their length" i read that as perpendicular to the wave front.

This is in addition to the geometric arguments that I am making.

I make no assertion about shocks potentially propagating faster than the speed of sound due to the increased temperature, physical flow velocity or any other non-linear characteristics of the medium.

That said, your understanding matches my own that a shock (think about the portion of the shock directly in front of the plane's nose) must travel faster than the speed of sound. My understanding is that this will result in dispersion so that over time, the shock will dissipate into a more normal sound wave.

The resulting wave front then behaves according to the geometric arguments that I have making. In particular, a line through the wave front will progress along the flight path at the speed of the aircraft while progressing along a perpendicular to the wave front at the speed of sound and along a perpendicular to the flight path at a speed greater than sound.

 

[tmbouman]

Perfect. Thank you for your time! I just want to point out that @boneh3ad has mentioned the shock can propagate for many miles at times. So I think the analysis challenge is understanding when it changes, because it must also correspond to an angle change at that time too. Putting it another way, the solution to the problem stated above depends on the assumption of if it is still a shockwave traveling > c or if it is a soundwave traveling at c.

That said, your understanding matches my own that a shock

If anyone else reading this thinks we are missing something, please let us know.

 

[jbriggs444]

Perfect. Thank you for your time! I just want to point out that @boneh3ad has mentioned the shock can propagate for many miles at times. So I think the analysis challenge is understanding when it changes, because it must also correspond to an angle change at that time too. Putting it another way, the solution to the problem stated above depends on the assumption of if it is still a shockwave traveling > c or if it is a soundwave traveling at c.If anyone else reading this thinks we are missing something, please let us know.

The wikipedia article is unhelpful as to the time/distance scale over which the wave front is softened.

Over longer distances, a shock wave can change from a nonlinear wave into a linear wave, degenerating into a conventional sound wave as it heats the air and loses energy. The sound wave is heard as the familiar "thud" or "thump" of a sonic boom, commonly created by the supersonic flight of aircraft.

[emphasis mine]

 

[hutchphd]

I think thas been implied but not actually sfated: that the "Boom" represents the piling up of sounds emitted from the airplane/source at all previous times. That temporal pile up can occur only if the source is supersonic and explains the sharp wavefront (I still don't know if this is called a shockwave but here I fear we sail the terrible semantic sea ).

 

[sophiecentaur]

What does the phrase in red even mean??

I don't understand why there is a problem with the idea of the shock wave losing energy and eventually becoming indistinguishable from the wave from a large loudspeaker. I haven't found anything quantitative about this except that the term 'a few wavelengths' seems to be used. (i.e. in the order of the length of the craft)
The pressure effect that is sometimes experienced at ground level can be magnified impressively because there is a fast virtual wavefront where energy is built up as it progresses over the ground. A single event as from a 'low explosive' will only knock you over if you are quite close - much closer than the height of a ss jet.

The great RF analogue of this is the the Beverage Antenna which consists of a long wire (hundreds of metres, sometimes), at or just above ground level. The forward tilt of a traveling EM wave means that the received fields add up along the wire and produce significant gain over a few wavelengths (until the signals go out of phase).

 

[sophiecentaur]

I think thas been implied but not actually sfated: that the "Boom" represents the piling up of sounds emitted from the airplane/source at all previous times. That temporal pile up can occur only if the source is supersonic and explains the sharp wavefront (I still don't know if this is called a shockwave but here I fear we sail the terrible semantic sea ).

Semantics - yes. I think it's questionable whether the speed of this virtual wave as it hits the ground at an angle actually makes it a shock wave. Is the air actually moving faster than c anywhere? The only source to make it do that would be the air near the ground - unlike how a shock wave is produced by a fast moving solid object.

The ss boom is not from a single event but the energy is constantly supplied by the plane displacing the air and launching its shock wave continuously.

An observer in a balloon would only get one contribution of sound from this wave, with no build up due to propagation along the ground.

There will be dispersion to 'soften' the initial crack / impulse.

 

[sophiecentaur]

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

Draw software is better than Paint software. For a cheap drawing package, it's handy to use what Powerpoint (and the like) will give you. You can edit every object you put in, at any time. With Paint, you can't move things or rub out bits of them.
Sorry if I'm trying to teach you how to suck eggs.