Can we hear a supersonic plane?

[Nidum]

 

[fizzy]

Again, we don't need to assume perfect spheres to get the reverse effect.

You on the other hand, seem to claim that the entire forward hemisphere of the explosion shock will be deformed exactly such that it catches up, and perfectly merges with the shock cone of the plane. That seems the far more unlikely option, and I haven't seen a good reason or evidence for it, just mere assertions.Because you keep repeating the same non-argument: Stating that the explosion shock propagation won't be perfectly spherical, doesn't imply that its entire forward hemisphere will be part of the Mach cone.

No, I have said that the source of the sound ( eg the cannon ) is also moving forwards at mach II and must have its own shock wave. There is no catching up to do.

I think a lot of the misconception here, as illustrated by the anim which you posted, is that you are attempting to construct a totally unrealistic model where the plane has some kind of umbrella shaped shock wave and everything behind that can be modeled as ghetto blaster sitting the middle of a football field with sound propagating at mach 1. The idea of "catching up" comes from there as do the wavelets which lead to the image of the sound arriving backwards.

No one has addressed the issue that sound source itself is moving at mach 2 which I pointed out.

You accuse me of making "non arguments" and assertions but do not address what I have written. I have pointed out that there is a fatal flaw in the way the argument for reversed sound is being present. It is not incumbent on me to provide fully worked wave equation in three dimensions in order to contradict the spherical model. It does not work in the direction of travel and that is sufficient to disprove the applicability of the model being presented.

It is then incumbent on those suggesting reversed sound to provide a model which is not contradicted at least in the line of flight and then, if they wish to develop this model for off axis forward propagation and see what it shows.

 

[jack action]

No one has addressed the issue that sound source itself is moving at mach 2 which I pointed out.

The source emitting the sound moves at Mach 2, but the location where the sound was initially emitted does not move.

Say there is you running toward an observer standing still at the finish line. You throw a ball while running. Which one arrives at the finish line first, you or the ball?

If you run faster than the ball (say you've thrown it backward such that the speed of the ball is your running speed minus the speed you've thrown the ball, but thrown slow enough that the direction of motion is unchanged), the observer will see you arrive at the finish line first, then the ball.

Say you throw a second ball after the first one, and it goes at the same speed as the first one and you still run faster than both balls. The observer will see you arrive first, then the 2nd ball thrown, then the 1st ball thrown. Because both balls are at the same speed, the first one will never catch up the second one. Because you run faster than both balls, they'll never catch you up. The locations where the balls were initially thrown (which are fixed with respect to the observer and remain unchanged through time) play a big role in the order the balls will cross the finish line.

As soon as the sound emitted «gets out» of the plane, it is no longer part of the plane and it is completely on its own.

Imagine this other scenario: You are in a plane at Mach 2. You jump out of the plane with a jet pack on your back such that you can go at Mach 1 as soon as you get out of the plane. Before you even think of reaching the front shock wave of the airplane, the airplane will be long gone in front of you. So you will just travel you own way. If there is a second jumper in the plane that jumps after you and goes at Mach 1 too, you'll never catch him up and he will arrive at any location in front of you before you do. That's because he had a «free ride» at Mach 2 longer than you.

Does that make any sense? Because this is what I understood.

 

[fizzy]

Your ball analogy will work for someone with a radio on a bicycle, but in no way accounts for approaching speed of sound or mach 2. That is the problem with banal analogies which do not match what you wish of model mentally.

Now you want to examine sound "getting out" , what does that mean. Is the source of the sound waves now the cylindrical cabin of the plane? That will not project sound forwards. Now you need to account for how the sound will propagate in a medium traveling at mach 2 perpendicular to the emitting surface. Just starting to draw sub-sonic spherical waves again is not suitable.

The whole problem so far is still that all these silly little analogies are attempts to get around the need to look at how sound will propagate in these rather special circumstances. If someone wants to suggest some weird sound moving backwards in time oddity they are going to need to do some serious maths.
Kindergarten, bouncing balls and spherical waves are simply not going to be informative.

So far we have seen where the odd idea of sound playing backwards comes from and that they rely on simplistic inappropriate models.

Until someone comes up with a more rigorous demonstration which as a very minimum works for line of flight, I think we have to remain with the null hypothesis that sound is heard in the conventional sense at a all points of observation, albeit with some Doppler distortions.

 

[jack action]

If someone wants to suggest some weird sound moving backwards in time oddity they are going to need to do some serious maths.

Where is your math?

How will a sound emitted at location $(0,0,0)$ and time $t$, going at speed $v$, get in front of another sound emitted at location $(2v\Delta t,0,0)$ and time $t + \Delta t$, going also at speed $v$?

My math says that at time $t+\Delta t$, the first sound will have reached location $(v\Delta t,0,0)$ only. Afterward, there will always be a distance $v\Delta t$ between the two sounds (the second sound being ahead), since they both go at the same speed. From an observer directly ahead, the sounds arrive in reverse order. I can't see any reason why this shouldn't be true whether the sounds come from a moving source or from two fixed and grounded sources at two different locations.

That makes me think of another thought experiment:

Install a set of 3 speakers on the ground, all aligned and separated by 993 m (i.e. 3X the speed of sound times 1 second). Play a set of 3 notes on the speakers, but play the first note in speaker #1, the 2nd one in speaker #2 and the 3rd one in speaker #3. Each note played is separated by a one second interval. If there are 2 observers, one at one end of the speaker line-up and the other at the other end of the speaker line-up, what will they hear?

Let's look at the events:

t = 0: speaker #1 emits note #1 and reach observer #1
t = 1: speaker #2 emits note #2
t = 2: speaker #3 emits note #3 and reach observer #2
t = 4: note #2 reach observer #1 and observer #2
t = 6: note #1 reach observer #2
t = 8: note #3 reach observer #1

So observer #1 hears:

• t=0 -> #1
• t=4 -> #2
• t=8 -> #3

and observer #2 hears:

• t=2 -> #3
• t=4 -> #2
• t=6 -> #1

even though the actual notes played are:

• t=0 -> #1
• t=1 -> #2
• t=2 -> #3

So it is possible to hear music backward, even without a moving source.

With a moving source, there will probably be some compressibility phenomena that will affect the actual speed of propagation and the amplitude of the sound waves, but I cannot imagine it to be to the point of having the first one catching up the second one (especially considering the fact that I can set the location of the second event as far as I want). You are the one needing to show math to support such claim.

A sound source might go at twice the speed of sound, but the medium around it does not.

 

[fizzy]

Where is your math?

The speed of the shock wave in front of the plane is mach II. Therefore that is the speed of sound under those conditions. This shows that the idea of a spherical or even distorted but linearly expanding wavefront can not be applied. The animation and anything suggested by it is dead. You are still talking like "the speed of sound" is some universal constant.

All the ideas so far suggesting reversed sound have been based on the flawed notion that the sound we are supposed to be hearing is propagating at 332 m/s, ignoring that the supposed source of the sound is already moving through the medium at twice that speed.

If you ignore details like that you will anomalous results, no anomalous physical effects.

Anyone who has not realized that this reversed sound thing is a result of erroneous assumptions is beyond my help. Good luck with the physics.

 

[boneh3ad]

The speed of the shock wave in front of the plane is mach II. Therefore that is the speed of sound under those conditions.

No, no, no, no, no. The speed of sound, as it has always been, is $a = \sqrt{\gamma R T}$. The speed of sound is not Mach 2. That statement does not even make sense. For a Mach 2 wavefront, the shock is moving at twice the speed of sound relative to the air into which it is propagating. However, relative to the air behind the shock, which is the air it is propagating through, it is still moving subsonically. Sound doesn't move faster than the speed of sound in the medium through which it is propagating.

Consider, for example, shock with upstream Mach number, $M_1=2$. This means the downstream Mach number, $M_2 \approx 0.577$. If you imagine then the shock moving through a stationary medium at $M_1$, it is clear that it is dragging air behind it along with it. Let's assume for a moment that the still air is at $T_1 = 300\mathrm{ K}$, then the shock is moving at about 694 m/s and is dragging the air behind it along at about 434 m/s. The shock is propagating into region 1 at twice the local speed of sound, but it is propagating through region 2 at about 260 m/s relative to the air in region 2, which is only a little more than half of the speed of sound in that region.

 

[256bits]

No, no, no, no, no. The speed of sound, as it has always been, is $a = \sqrt{\gamma R T}$. The speed of sound is not Mach 2. That statement does not even make sense. For a Mach 2 wavefront, the shock is moving at twice the speed of sound relative to the air into which it is propagating. However, relative to the air behind the shock, which is the air it is propagating through, it is still moving subsonically. Sound doesn't move faster than the speed of sound in the medium through which it is propagating.

Consider, for example, shock with upstream Mach number, $M_1=2$. This means the downstream Mach number, $M_2 \approx 0.577$. If you imagine then the shock moving through a stationary medium at $M_1$, it is clear that it is dragging air behind it along with it. Let's assume for a moment that the still air is at $T_1 = 300\mathrm{ K}$, then the shock is moving at about 694 m/s and is dragging the air behind it along at about 434 m/s. The shock is propagating into region 1 at twice the local speed of sound, but it is propagating through region 2 at about 260 m/s relative to the air in region 2, which is only a little more than half of the speed of sound in that region.

The velocities are for a normal shock wave where the shock is perpendicular to the flow, in which case the after the shock is always subsonic.
Oblique shock waves have a flow normal to the shock as well as a tangential flow.
The Mach number for oblique flow can be supersonic, or subsonic.

For the oblique flow, since sound moves at M1, and normal to the shock is subsonic, is does stand to reason that any sound produced by the object would eventually encounter the shock, merge and become part of it. How much time that takes should be calculable.

 

[jack action]

any sound produced by the object would eventually encounter the shock, merge and become part of it.

A sound wave will most likely be altered going through a shock wave, but it would still go through it, not become part of it. This point was already discuss in post #50.

 

[256bits]

A sound wave will most likely be altered going through a shock wave, but it would still go through it, not become part of it. This point was already discuss in post #50.

What?
Sound from the plane eventually becomes part of the shock wave
Consider by traveling through, it would come out the other side at Mach 2 into the oncoming stream of fluid.
Surely that cannot be a possibility.
Leaving the sonic wedge is impossible.

 

[A.T.]

If someone wants to suggest some weird sound moving backwards in time oddity they are going to need to do some serious maths.

This was linked here already I think:

https://www.researchgate.net/publication/230702229_Reproduction_of_Virtual_Sound_Sources_Moving_at_Supersonic_Speeds_in_Wave_Field_Synthesis

"...the wave field contains a component carrying a time-reversed version of the source’s input signal...”

 

[A.T.]

Sound from the plane eventually becomes part of the shock wave.

All the sound from the plane? You cannot hear the engine of a super sonic jet, even inside the mach cone? What about a tail gun firing blanks?

 

[nsaspook]

I can't add much to this thread but I thought it would be informative to see how what I believe to be a normal shockwave builds as air speed increases on a airfoil until it becomes an Oblique shock.
https://www.grc.nasa.gov/www/k-12/airplane/oblique.html

 

[256bits]

All the sound from the plane? You cannot hear the engine of a super sonic jet, even inside the mach cone? What about a tail gun firing blanks?

Yeah, I guess that was too strong of a statement.
I clarify it to mean that since sound moves at Mach1, and the shock is moving away at a speed of subsonic velocity M2 ( less than Mach 1) , then depending upon the angle α which is a function of velocity v of the object, the arc angle ( for the picture ) of the sound that will eventually encounter the shock will vary with the speed of the object. Sound at an angle great then that will be heard anytime by anyone within the cone, baring 1/r^2.
Make more sense now I hope.I just realized that the picture might be labeled incorrectly, Do they mean Ma is the velocity of the shock? Or something else?
Ma is always < 1. That is the normal velocity of the shock should be subsonic within the cone.

With respect to Boneh3ad, showing what he mentioned previously.
Here is normal shock table:
https://en.wikipedia.org/wiki/Normal_shock_tables
or, same thing
http://www.cchem.berkeley.edu/cbe150a/normal_shock.pdf

 

[256bits]

I can't add much to this thread but I thought it would be informative to see how what I believe to be a normal shockwave builds as air speed increases on a airfoil until it becomes an Oblique shock.

Ok
That was neat.
No wonder, Chuck Yeager, thought his plane was breaking apart.
Notice that the second segment has not a shock on the bottom, producing difference in forces top and bottom, and thus lift.

I can't add much to this thread.

Neither can I compared to the others, but as a way of understanding, either I get shot down, or not, so I posted.

 

[jack action]

What?
Sound from the plane eventually becomes part of the shock wave
Consider by traveling through, it would come out the other side at Mach 2 into the oncoming stream of fluid.
Surely that cannot be a possibility.
Leaving the sonic wedge is impossible.

I was actually referring to the general case of a sound wave crossing a shock wave. In fact, it is the general consensus in this thread (except maybe for @fizzy ) that no sound waves emitted by the airplane will ever reach the front shock wave.

Yeah, I guess that was too strong of a statement.
I clarify it to mean that since sound moves at Mach1, and the shock is moving away at a speed of subsonic velocity M2 ( less than Mach 1) , then depending upon the angle α which is a function of velocity v of the object, the arc angle ( for the picture ) of the sound that will eventually encounter the shock will vary with the speed of the object. Sound at an angle great then that will be heard anytime by anyone within the cone, baring 1/r^2.
Make more sense now I hope.I just realized that the picture might be labeled incorrectly, Do they mean Ma is the velocity of the shock? Or something else?
Ma is always < 1. That is the normal velocity of the shock should be subsonic within the cone.

Ma is the Mach number of the moving object, which is the object velocity divided by the local speed of sound.

This is why I wrote post #27 where I made the distinction between pressure waves and sound waves. One is due to aerodynamics (the airplane pushing the air) and the other is from a sound source. The shock wave is caused by the pressure waves. It is possible to design an object that will NOT produce a shock wave (see post).

The Mach cone in your post is true for both pressure waves and sound waves. But it will correspond to a shock wave only in the case of pressure waves. In reality, it often look more of a bow shape than a cone shape (see bow shock), it depends on the shape of the object:

With sound waves only, there shouldn't be any shock waves.

 

[A.T.]

I clarify it to mean that since sound moves at Mach1, and the shock is moving away at a speed of subsonic velocity M2 ( less than Mach 1) , then depending upon the angle α which is a function of velocity v of the object, the arc angle ( for the picture ) of the sound that will eventually encounter the shock will vary with the speed of the object. Sound at an angle great then that will be heard anytime by anyone within the cone, baring 1/r^2.

Do you mean emission at "greater than α" or "greater than 90deg-α" for the sound that will be heard inside the cone?

Or can you clarify this: When a supersonic plane shuts down it's engine, just when a near detector is at 90deg-α of the flight path, will the detector record any engine sounds after entering the Mach cone? Or just the crack(s) of the aerodynamic shock waves, and then silence again?

 

[A.T.]

It is possible to design an object that will NOT produce a shock wave (see post).

Or at least make them very weak compared to the distinct explosion sounds we create with that object. That's why I also don't see why those aerodynamic shock waves would completely erase the explosion sounds, when passing them.

 

[boneh3ad]

The velocities are for a normal shock wave where the shock is perpendicular to the flow, in which case the after the shock is always subsonic.
Oblique shock waves have a flow normal to the shock as well as a tangential flow.
The Mach number for oblique flow can be supersonic, or subsonic.

For the oblique flow, since sound moves at M1, and normal to the shock is subsonic, is does stand to reason that any sound produced by the object would eventually encounter the shock, merge and become part of it. How much time that takes should be calculable.

Here's the thing, though. This is sort of a matter of perspective. An oblique shock is still essentially propagating normal to itself, not at an angle. Any flow moving tangentially to the shock is not really factored into the shock propagation speed. For example, for an object traveling at several times the speed of sound, you might view that as a continuous acoustic source that is constantly pumping energy into the shock wave, which then propagates away from the plane at the same rate as that tangential flow. In essence, then, the shock is essentially stationary relative to the flow tangential to it, and is propagating as I described above in the direction normal to the shock front.

It is absolutely true that sound approaching the shock from behind will eventually catch up to it and merge with it, and it is a fairly common homework problem to calculate something similar to that in most introductory compressible flow courses. It's also a very interesting subject from the applied mathematics/wave theory perspective. For example, https://www.amazon.com/dp/0471359424/?tag=pfamazon01-20 is a pretty fascinating text (at least by my nerdy standards) that covers a lot of the more mathematical side of the subject.

 

[woody stanford]

Ok, hopefully close enough to topic that I don't get scolded but I do rocketry (and you can also see this on youtube videos as well) but why don't rockets when they go transonic make a sonic boom. I am serious, they really dont. And can someone explain this claim even if it perceptual or something?

 

[A.T.]

why don't rockets when they go transonic make a sonic boom.

You have to be in the right position to be passed by the Mach cone and hear the boom. You also don't hear the sonic booms from the supersonic bullets you fire.

 

[A.T.]

When a sound wave interacts with a shock, it is very unlikely that it will come out the other side unchanged. Most likely the wave will undergo some degree of additional attenuation because shocks are highly dissipative. They wouldn't necessarily be completely destroyed, though.

To add to the above, see the below pictures of shock waves passing through each other, without canceling or dissipating. I see no reason why explosion shocks generated by a gun at the side wouldn't survive passing through the aerodynamic tail shock waves.

Source: http://stl-www.aero.kyushu-u.ac.jp/research/index_r_e.htm

 

[fizzy]

You have to be in the right position to be passed by the Mach cone and hear the boom. You also don't hear the sonic booms from the supersonic bullets you fire.

Ah, thanks AT. That is what the sound waves in the anim should look like. The circle is typical speed of sound in the static medium: the bang of the gun.
The conic part will get progressively longer in proportion to the spherical wave since it propagating faster. ie this shape will extend, not just scale up in a linear fashion. I made this point a couple of pages back.

 

[A.T.]

Ah, thanks AT. That is what the sound waves in the anim should look like.

Not really, it's a different situation with a single explosion, not several ones produced by the super sonic object along the path.

Also, you claimed that the explosions wouldn't produce spherical waves propagating forward inside the Mach cone, while the rifle picture clearly show them.

 

[fizzy]

A sound wave will most likely be altered going through a shock wave, but it would still go through it, not become part of it. This point was already discuss in post #50.

Many things have been 'discussed', that does not make them established fact. Actually what was said there does not relate to the building shock wave in the forward direction which I was referring to.

A sound can pass through a subsonic shockwave. it can not pass through the mach cone since this would involve it traveling faster than the ( elevated ) speed of sound in that medium. The observation that the incredible noise from the engines cannot be heard in front of the mach cone should be sufficient to make that obvious without the need for special expertise or deep reflection.

 

[fizzy]

Not really, it's a different situation with a single explosion, not several ones produced by the super sonic object along the path.

Also, you claimed that the explosions wouldn't produce spherical waves propagating forward inside the Mach cone, while the rifle picture clearly show them.

Thanks. The anim you posted would be like a series of such snapshots. The gun blast being like the aircraft's cannon (blanks) producing the animation. The waves here are spherical because the shotgun is stationary. My earlier point was for the cannon which itself if moving at mach 2.

 

[A.T.]

A sound can pass through a subsonic shockwave. it can not pass through the mach cone since this would involve it traveling faster than the ( elevated ) speed of sound in that medium.

The tail shockwave cone can pass the spherical shock from the gun explosion without destroying it, as shown in post #122. So the spherical shocks from the gun can continue to propagate forward, and reach a detector in reverse order of their generation.

 

[fizzy]

No, no, no, no, no. The speed of sound, as it has always been, is $a = \sqrt{\gamma R T}$. The speed of sound is not Mach 2. That statement does not even make sense. For a Mach 2 wavefront, the shock is moving at twice the speed of sound relative to the air into which it is propagating. However, relative to the air behind the shock, which is the air it is propagating through, it is still moving subsonically. Sound doesn't move faster than the speed of sound in the medium through which it is propagating.

What is R and T , not much point is posting undefined terms.

I'm guessing that T is temperature. What do you imagine the temperature of the air 1 cm in front of the nose cone is? What does this tell us about the pressure there. What does that imply about the speed of sound at that point?What is making the air move at mach 2 just in front of the nose cone ? It is not in contact with the plane. It is the air pressure behind it which makes it move at mack 2. Sound is a compression wave. This speed is, by definition, the speed of sound at that point in the medium.

 

[fizzy]

So the spherical shocks from the gun can continue to propagate forward, and reach a detector in reverse order of their generation.

Oh great, so we now go back about 3 days and start again.

Your shotgun photo is a static bang. The cannon ( or other on board sound source ) is also moving at mach2 and has its own bow wave and mach cone. The froward propagation is like that of the craft itself. This whole sound backwards thing comes from assuming these little spherical wave-fronts and this is not appropriate. The speed of sound depends upon the air pressure and density of the medium. This in NOT isotropic for sound emitted by a supersonic sound source.

This, rather old looking paper talks about stagnation temperatures of up to 11,000 F in missiles. Presumably up from an ambient of about 220 kelvin. That puts pressure up around 10 bar , it seems.
ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19930090162.pdf

Speed of sound more than doubles at 1500 F even at 1 bar.
http://www.engineeringtoolbox.com/air-speed-sound-d_603.html

The air closer to the nose cone gets progressively heated and pressurised and becomes more dense. This means that the speed of sound nearer to the cone is slightly higher than just a little further in front. Hence effect of sound energy 'piling up'.

Like I said several days ago, the reason that the sound in front of the craft is propagating at mach 2 is because that is THE SPEED OF SOUND at that point in the medium.

 

[mfb]

it can not pass through the mach cone since this would involve it traveling faster than the ( elevated ) speed of sound in that medium

No one claimed (or even suggested) that sound would pass through the mach cone from the front of the aircraft. I don't know why you keep arguing about that point.

It does not matter if a source of sound is moving if the sound emission happens within a negligible time interval. There is no mach cone from a single explosion, it doesn't matter if the explosives are attached to the aircraft or explode directly next to the aircraft at any other speed.