Can we hear a supersonic plane?

[fizzy]

On the other hand for a thin airfoil, the pressure variation seems to vanish outside the Mach cone:

Since sound cannot propagate in front of the cone, isn't this always going to be zero, no matter what the form of the supersonic object?
This is why we do not hear anything until the perimeter of the mach cone passes the point of observation.

 

[DrDu]

Since sound cannot propagate in front of the cone, isn't this always going to be zero, no matter what the form of the supersonic object?
This is why we do not hear anything until the perimeter of the mach cone passes the point of observation.

For the front this is obvious, but for the rear?

 

[fizzy]

What do you mean by "the rear".

 

[DrDu]

What do you mean by "the rear".

Have a look at the figure in https://www3.nd.edu/~atassi/Teaching/ame 60639/Notes/supersonic_airfoil.pdf

 

[fizzy]

What do YOU mean by 'rear'? Behind the cone? Behind the plane of the aircraft perpendicular to line of flight? Inside the cone?

I really can't understand what you are trying to suggest.

 

[DrDu]

Sorry. I mean the region with x>c and small y in that figure.

 

[mfb]

The ghettoblaster/cannon is not emitting continuous sound. Let our "music" be composed of the individual gunfires. Each one creates its own pseudo-shockwave (which has a roughly spherical pattern if the sound duration is short enough, because we don't care about the airplane motion during that short period) - but those shockwaves emitted later are ahead of those created earlier. For shockwaves produced early enough, an observer on the ground will hear the later shockwaves first.

 

[fizzy]

mfb, you are still ignoring the fact that you can not model sound produced by the cannon as a spherical wave as it would be if it were stationary. That is the whole point of the confusion here, and the false notion that there is a sound wave propagating forwards behind the aircraft.

 

[fizzy]

Sorry. I mean the region with x>c and small y in that figure.

"On the other hand for a thin airfoil, the pressure variation seems to vanish outside the Mach cone:"

Isn't what you describe INSIDE the mach cone?

 

[DrDu]

Isn't what you describe INSIDE the mach cone?

Maybe I should have written "outside the region between the front and rear Mach cone".

 

[fizzy]

Ok, I don't see this discussion getting anywhere at all. I'm off to do other things.

Interesting post though.

 

[A.T.]

you are still ignoring the fact that you can not model sound produced by the cannon as a spherical wave as it would be if it were stationary.

They don't have to perfectly spherical for the reverse order detection to occur.

 

[mfb]

mfb, you are still ignoring the fact that you can not model sound produced by the cannon as a spherical wave as it would be if it were stationary. That is the whole point of the confusion here, and the false notion that there is a sound wave propagating forwards behind the aircraft.

The extremely fast cannon perturbes the air at a single point of spacetime (let's say it adds more air there from the explosion). Neglecting scattering from the aircraft, what shape do you expect? What breaks the symmetry?
What breaks it so massively that we don't have anything that looks like forward/downward propagation (as seen from the ground) any more?

 

[fizzy]

What breaks it so massively? The fact that the forward propagation is at mach II , not mach I

I have not done the maths on this but I guess that the pressure and density of the air in front of the aircraft has to be that at which the propagation of sound is twice as fast as it is in still air.

So the sphere is elongated. Moreover it is continually stretching not just scaling up in a linear fashion as with the animation posted earlier.

The shock wave is like a continuous explosion. Unlike a thunder clap which propagates and attenuates, the crack of supersonic flight is only a crack for stationary observer. The sharp almost step change in air pressure at the front of the plane continues to emit it's explosive energy as long as the craft is supersonic.

 

[A.T.]

The fact that the forward propagation is at mach II , not mach I

We are not interested in propagation exactly forward, like the bow shock at the very front of the plane. We are interested in gun/explosion sounds generated at the side of the plane propagating sideways with some forward component.

 

[fizzy]

There is a similar shock wave off the leading edge of the wings and a tiny one on the end of the cannon if you want to look at that aspect. ( even if we want to avoid confusing the issue by discussing the explosion of the charge ).

You simply can not pretend that the cannon , moving forwards at mach 2. will but pumping out nice little spherical wavelets propagating omni-directionally at mach 1 .

I can't believe that this discussion is still going on without advancing one iota.

 

[mfb]

I think we get to the main point.

The fact that the forward propagation is at mach II , not mach I

There is not even a forward propagation involved in the single, instant, explosion in the gun.

You imagine a continuous sound source, but we don't have that. Our sound source (gunfire, ghetto blaster, whatever) is not active continuously. It is not an obstacle in the wind (or at least that is not the part we care about).

 

[jack action]

@mfb , @A.T. :

I can hear the music backward now!

Slow, but got there.

 

[A.T.]

You simply can not pretend that the cannon , moving forwards at mach 2. will but pumping out nice little spherical wavelets propagating omni-directionally at mach 1 .

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.

I can't believe that this discussion is still going on without advancing one iota.

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.

 

[DrDu]

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

IMHO, the really interesting point is that you seem to need a strong perturbation for information to leave the conical shell between the front Mach cone and the rear Mach cone. In paragraph 3 they explicitly state that information can't leave the Mach wedge,
https://www3.nd.edu/~atassi/Teaching/ame 60639/Notes/supersonic_airfoil.pdf
although this analysis clearly refers to small perturbations, only.
An explosion, is certainly a strong perturbation while I doubt that a loud speaker in an aerodynamical plane will qualify as a strong perturbation.

 

[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.