Why is there no sonic boom when shooting an air rifle at 1400 feet per second?

  • Thread starter curtmorehouse
  • Start date
In summary, the speed of sound in air at 20 degrees celsius is about 1126 feet per second. When shooting a rifle, there is a noticeable crack that is believed to be the 'sonic boom'. This is caused by the conical shockwave behind the bullet, which never passes the ears of the shooter. However, when shooting an Air Rifle (bb gun) that shoots bb's at an advertised rate of 1400 feet per second, there is no noticeable crack because the bullet is not traveling faster than the speed of sound. The shooter hears the explosion and associated pressure wave, not the shock and sonic boom associated with the bullet. When bullets are fired from a distance, they may be heard as a whistling
  • #1
curtmorehouse
6
0
I read that the speed of sound in air at 20 degrees celsius is about 1126 feet per second.
When I shoot my rifle, there is a noticeable crack which I believe to be the 'sonic boom'. I also have 'sub-sonic' ammo that does not cause the CRACK, because the bullet is traveling slower than the speed of sound.

My question is, if I shoot an Air Rifle (bb gun) that shoots bb's at an advertised rate of 1400 feet per second, why is there no CRACK (sonic boom) ?

Is there a MASS requirement for me to hear the boom?
 
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  • #2
curtmorehouse said:
When I shoot my rifle, there is a noticeable crack which I believe to be the 'sonic boom'.
I don't think that you hear the sonic boom of a bullet that you have fired yourself.

wQZOj2K1Gfm9oex22NHmpmkQo1_500.jpg


When the round shockwave reaches your ears, you hear the "bang" from the shot. The conical shockwave behind the bullet that causes the sonic boom never passes your ears. They are inside the geometrically extended cone from the start (or behind the cone if you will). So the expanding cone surface (sonic boom shockwave) never passes them.
 
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  • #3
Oh nice reply!
A.T. said:
because they are inside the cone.
Should be "because they are NOT inside the cone"?
 
  • #4
No, he meant inside the cone. The sonic boom is a result of the pent up sound pressure at the shockwave coming off of the object (in this case the bullet). You start in an area where that shockwave will never cross your ears since it starts in front of you, effectively placing you inside the cone.

Of course, there is the shockwave generated at the barrel by the explosion, and that is where the difference is between the aforementioned guns. You are really just hearing the explosion and associated pressure wave, not the shock and sonic boom associated with the bullet. The bullet itself wouldn't have much of a sonic boom simply because it isn't putting much noise into the flow.
 
  • #5
fluidistic said:
Should be "because they are NOT inside the cone"?
No I meant inside the cone in more abstract geometrical terms (infinite cone). I modified the post to make it more clear.
 
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  • #6
For those of you who got some military training (don't try it with your riffles!): if someone shoots in your direction from the distance, then, as the bullet passes you by (so the boom cone should cross my ears):
1. subsonic bullets (e.g. pistol ones) are almost quite silent - that is pretty understandable;
2. supersonic bullets (e.g. from sniper's carabine) may be heard not as a single 'boom', but rather as a whistling sound, lasting for 1/2s or so - I don't understand the mechanism behind this whistle.
 
  • #7
xts said:
2. supersonic bullets (e.g. from sniper's carabine) may be heard not as a single 'boom', but rather as a whistling sound, lasting for 1/2s or so - I don't understand the mechanism behind this whistle.
Well, they are not supersonic forever. So at large distance they might already gone subsonic. Also: The conic shock wave is affected by obstacles like vegetation and reflected by the ground. This might "blur" it.
 
  • #8
A.T. said:
Well, they are not supersonic forever. So at large distance they might already gone subsonic.
True, but you hear 'whistle' rather than 'boom' when bullets are definitely still supersonic, e.g. if you are crawling on no-man's land, and your colleagues continue carabine fire over your head from a distance of 100m or so.

The bullet of 800m/s muzzle velocity, having effective range of over 1km, cannot slow down to subsonic on first 100m.

The blur by reflections from the ground seems to be convincing explanation...
 
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  • #9
Bullets are tiny and fast-moving, meaning that the sound they generate will be high-pitched due to the fact that the perturbations they create in the air are small and high-frequency.
 
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  • #10
boneh3ad said:
Bullets are tiny and fast-moving, meaning that the soun they generate will be high-pitched due to the fact that the perturbations they create in the air are small and high-frequency.

Excellent answer. Sonic boom just means that the sound generated by an object has piled up into a shock wavefront. But the sound does not magically appear because an object is going fast; it still must create the sound. An object that is very quiet at sub-sonic speeds (e.g. a bullet slicing through the air) will have a quieter sonic boom. Similarly, an object that creates high-pitched sounds normally will create a high-pitched sonic boom. The classic sonic boom heard when a supersonic jet passes overhead is the low-pitch roaring sound of the engines piled up so that it is much louder and sudden.
 
  • #11
chrisbaird said:
Similarly, an object that creates high-pitched sounds normally will create a high-pitched sonic boom. The classic sonic boom heard when a supersonic jet passes overhead is the low-pitch roaring sound of the engines piled up so that it is much louder and sudden.

No, it seems to me, the whole thing about a sonic boom is that it is a single wavefront of pressure, not a sustained sound. Regardless of the length of an object (be it a face-on manhole cover or a jet plane) it creates a single pressure wave dependent on the area and on the velocity. i.e. sonic booms do not have a pitch, so small object do not make "high-pitched" sonic booms and large objects do not make "low-pitched" sonic booms.

(I grant that objects can create a multitude of shock waves, such as leading and trailing edges, but that does not change the principle that any given shich wave is a single pressure wave.)
 
  • #12
chrisbaird said:
Excellent answer. Sonic boom just means that the sound generated by an object has piled up into a shock wavefront. But the sound does not magically appear because an object is going fast; it still must create the sound.

Not true. The sonic boom appears exclusively because the object is traveling faster than the speed of sound. The object can be emitting no sound of its own, and it will still create shockwaves simply because it is traveling through the air faster than sound.
 
  • #13
cjl said:
Not true. The sonic boom appears exclusively because the object is traveling faster than the speed of sound. The object can be emitting no sound of its own, and it will still create shockwaves simply because it is traveling through the air faster than sound.

Perhaps we are getting into semantics, but I consider a shockwave a sound, therefore this statement is contradictory.
 
  • #14
DaveC426913 said:
sonic booms do not have a pitch, so small object do not make "high-pitched" sonic booms and large objects do not make "low-pitched" sonic booms.
I agree in respect to the initial wavefront. See the conical shock wave in the picture. This is not a "whistling sound, lasting for 1/2s or so". This is something you will hear as a crackle or loud click.

The whistling sound as described by xts is then heard in the cone and blurs with the initial crackle. But without actual recording we are just guessing around, about human perception.
 
  • #15
chrisbaird said:
Perhaps we are getting into semantics, but I consider a shockwave a sound, therefore this statement is contradictory.

A shockwave can be considered a sound of sorts, but your prior statement implied that the sonic boom was the result of other sounds emitted by the object traveling supersonically. For example, you claimed that jet aircraft creating a sonic boom did so because the sound of the engine "piled up". This is false - the reason jet aircraft make loud, rumbling sonic booms has more to do with their size and the resultant intensity of the shock wave produced, which is why (for example) the sonic boom created by the space shuttle during reenty sounds very similar to the one generated by the concorde, even though one of them has jet engines and the other is gliding when the boom is created.
 
  • #16
DaveC426913 said:
No, it seems to me, the whole thing about a sonic boom is that it is a single wavefront of pressure, not a sustained sound. Regardless of the length of an object (be it a face-on manhole cover or a jet plane) it creates a single pressure wave dependent on the area and on the velocity. i.e. sonic booms do not have a pitch, so small object do not make "high-pitched" sonic booms and large objects do not make "low-pitched" sonic booms.

Any object moving supersonically will create shockwaves, but the shape and intensity most certainly is dependent on more than just the area. In fact, the area has very little to do with it and much more on the overall shape. For example, a sharper object creates a sharper Mach cone. A blunt object creates a bow shock. There are many things that play into it. However, the physical reason for a shockwave is what you aren't grasping.

An object moving through the air perturbs the air around it, which manifests itself as pressure fluctuations. These pressure fluctuations are sound. The wavelength (hence frequency) of these fluctuations depend on a great many things, but in particular an objects size, speed, shape and any noise-generating devices on the object. When the objects move faster than the speed of sound, these sound perturbations can't keep up and the ones moving forward start falling behind. As more sound waves are created, they coalesce with the previous waves and eventually form a shockwave. This shockwave most definitely will have a characteristic frequency to it based on what frequency the soundwaves generated by the object were before they coalesced. With a plane, you hear and feel an low boom. With a supersonic bullet, you hear and feel a much smaller and higher-pitched crack. Any sustained whistling afterwards is as a result of the sound created normally through the passing of the bullet through the air.

cjl said:
Not true. The sonic boom appears exclusively because the object is traveling faster than the speed of sound. The object can be emitting no sound of its own, and it will still create shockwaves simply because it is traveling through the air faster than sound.

True, and the frequency of that sound will be dependent on what frequency the perturbations generated by the object are. Smaller objects will generally have higher frequency sound (from things such as shed vortices). With a plane, the sound of the engines and other things just add to the effect (though it certainly won't do anything like double the effect or anything).

chrisbaird said:
Perhaps we are getting into semantics, but I consider a shockwave a sound, therefore this statement is contradictory.

A shockwave is sound. However, you can generate sound purely aerodynamically as opposed to with engines.
 
  • #17
boneh3ad said:
Any object moving supersonically will create shockwaves, but the shape and intensity most certainly is dependent on more than just the area. In fact, the area has very little to do with it and much more on the overall shape. For example, a sharper object creates a sharper Mach cone. A blunt object creates a bow shock. There are many things that play into it. However, the physical reason for a shockwave is what you aren't grasping.
I am grasping it quite fine.

boneh3ad said:
An object moving through the air perturbs the air around it, which manifests itself as pressure fluctuations. These pressure fluctuations are sound. The wavelength (hence frequency) of these fluctuations depend on a great many things, but in particular an objects size, speed, shape and any noise-generating devices on the object. When the objects move faster than the speed of sound, these sound perturbations can't keep up and the ones moving forward start falling behind. As more sound waves are created, they coalesce with the previous waves and eventually form a shockwave. This shockwave most definitely will have a characteristic frequency to it based on what frequency the soundwaves generated by the object were before they coalesced. With a plane, you hear and feel an low boom. With a supersonic bullet, you hear and feel a much smaller and higher-pitched crack.

An object that is otherwise making no noise will still create a shock wave and thus a sonic boom.
An object that has a simple geometry, and a correspondingly simple shock envelope will still create a shock wave and thus a sonic boom.

As I pointed out, there are overtones, because real-wrold objects are not simple and silent, however, in principle the shockwave is a single pressure wave, and has no frequency.

When chrisbaird said this:
Sonic boom just means that the sound generated by an object has piled up into a shock wavefront.
he was wrong.
 
  • #18
DaveC426913 said:
I am grasping it quite fine.

Clearly not based on the rest of your post.

DaveC426913 said:
An object that is otherwise making no noise will still create a shock wave and thus a sonic boom.

Objects moving through the air never make no noise. Just because you can't hear the noise doesn't mean it isn't being made. Even the tiniest of pressure fluctuations, which are present as a result of quite literally any object moving through a viscous fluid, are going to result in some degree of noise. This sound is what coalesces into the shock.

DaveC426913 said:
An object that has a simple geometry, and a correspondingly simple shock envelope will still create a shock wave and thus a sonic boom.

I don't believe anyone ever said it wouldn't. However, your originally said the shock was dependent on area (only sin a sense) and Mach number (it is), which was incomplete and partially incorrect. I was correcting this.

DaveC426913 said:
As I pointed out, there are overtones, because real-wrold objects are not simple and silent, however, in principle the shockwave is a single pressure wave, and has no frequency.

Not true. A shock is a pressure wave, as you put it, but it is a result of the coalescence of many sound waves, each one of which contains a frequency (it is sound after all). The very fact that it can be described in terms of a wave means it has a frequency. If it didn't have a frequency, you couldn't hear it. I could go on and on.

DaveC426913 said:
When chrisbaird said this:
chrisbaird said:
Sonic boom just means that the sound generated by an object has piled up into a shock wavefront.
he was wrong.

No he wasn't. The key point is that any object moving through a viscous medium produces sound. It doesn't just come from the engines or other similar sources.
 
  • #19
Two more points about 'whistling sound':

- I am not sure if it is 0.5s or 0.1s, but definitely it lasts long enough to be noticed as lasting, rather than single 'click'

- maybe the 'whistle' is somehow related to bullet rotation? Carabine bullets usually make one rotation every 25-35cm, which, at 800m/s gives 2-3 kHz. That might be a pitch of bullet 'whistle'. But, on the other hand the bullet is pretty symmetrical...
 
  • #20
boneh3ad said:
Clearly not based on the rest of your post.
This is rude and uncalled for. Knock it off.

boneh3ad said:
Objects moving through the air never make no noise. Just because you can't hear the noise doesn't mean it isn't being made. Even the tiniest of pressure fluctuations, which are present as a result of quite literally any object moving through a viscous fluid, are going to result in some degree of noise. This sound is what coalesces into the shock.
The point is that the noise is incidental. A single pressure wave - one compression - is enough to be a shock wave. That is the essence of a shock wave, regardless of the details you've mentioned.
 
  • #21
DaveC426913 said:
This is rude and uncalled for. Knock it off.

Then stop claiming you know things based on what appear to be (educated) guesses.
DaveC426913 said:
The point is that the noise is incidental. A single pressure wave - one compression - is enough to be a shock wave. That is the essence of a shock wave, regardless of the details you've mentioned.

Noise is not incidental. Noise is the source of the shock. One compression is not necessarily enough to be a shock. It has to be of such a strong magnitude that certain properties are effectively discontinuous across the shock (in reality, a shock is of finite thickness, but so tiny that it can be treated as infinitesimal). This happens when a whole bunch of sound waves (which are incidentally smaller compression waves but not shocks) coalesce. In the case of an object moving through the air, this is because the sound emitted by the object cannot keep up with the object and the sound waves coalesce (see image).

The shock is sound, therefore it has a frequency (or pitch) associated with it based on the sound that generated it. If this wasn't true, the sound made by a bullet would not differ from that made by a plane save in magnitude
 
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  • #22
boneh3ad said:
The shock is sound, therefore it has a frequency (or pitch) associated with it based on the sound that generated it.

Here is the caption of the animation you posted, taken from:
http://en.wikipedia.org/wiki/Doppler_effect
The sound source has now broken through the sound speed barrier, and is traveling at 1.4 times the speed of sound, c (Mach 1.4). Since the source is moving faster than the sound waves it creates, it actually leads the advancing wavefront. The sound source will pass by a stationary observer before the observer actually hears the sound it creates. As a result, an observer in front of the source will detect
2100ebfcb4a7a26a2a817952cefb4f99.png
Try different values for f0 and see if it affects f.
 
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  • #23
That equation isn't correct though. [itex]c-v_{s} \not= 0[/itex], so [itex]f \not= \infty[/itex].

Of course, that equation isn't even valid when you reach Mach 1, so you really can't take anything on Wikipedia as automatically true.

Of course, I may be slightly off in my explanation. Perhaps it is better to say the shock doesn't have a frequency, per se. The higher frequencies generated by a smaller object will still be heard just behind the shockwave and would explain the whistling and the smaller sound of the shock is probably better explained by the smaller pressure wave generated by a small object on account of less air being displaced. My mistake.

Still, any sound source on the object will add to the shock since the radiated sound will be of greater magnitude; it just won't add that much.
 
  • #24
boneh3ad said:
Objects moving through the air never make no noise. Just because you can't hear the noise doesn't mean it isn't being made. Even the tiniest of pressure fluctuations, which are present as a result of quite literally any object moving through a viscous fluid, are going to result in some degree of noise. This sound is what coalesces into the shock...

Not true. A shock is a pressure wave, as you put it, but it is a result of the coalescence of many sound waves, each one of which contains a frequency (it is sound after all). The very fact that it can be described in terms of a wave means it has a frequency. If it didn't have a frequency, you couldn't hear it. I could go on and on.

The physical picture you are promoting, that (a) a shock wave is the result of the coalescence of small perturbations, and that therefore (b) the shock wave retains some kind of information from the source of the perturbations, is incorrect.

Consider that the typical sonic boom is the result of the passing of the oblique shocks attached to a supersonic object. In the frame of the object, the atmosphere (quiescent in the Earth frame) is approaching uniformly at the flight speed of the object. The oblique shock arises from the need to turn the flow in a way that is compatible with the physical requirement that pressure waves can only travel at the speed of sound. Shocks are nature's way of matching upstream and downstream boundary conditions when the upstream flow is supersonic. There is also typically an oblique shock at the tail of the object, as the flow has to straighten out again.

Shock layer thickness is equal to only a few mean free paths. In practical terms this means that the shock is a step function. The Fourier transform of a step function goes like -i/omega, so it has contributions from all frequencies. There is no substructure that incorporates tiny oscillations emanating from the supersonic body. In fact shock are highly dissipative; they destroy information and generate entropy.

Sonic booms incorporate the pressure jumps from both the tip and tail of the flying object. Consider the shock structure for the paradigmatic diamond-wing airfoil. If the flight Mach number is sufficiently high, there is an attached oblique shock at the tip. There will be a Prandtl-Meyer expansion fan at the mid-chord corner, which allows the flow to accelerate around the corner. And then there is a compression shock at the tail as the flow has to straighten out again. On the ground, the timing of the shocks will depend on the length of the object, its altitude, and its flight speed. The typical sonic boom is the result of the passing of this "N wave". Larger objects create a bigger lag between tip and tail and may be experienced as a lower-pitched boom, although the shocks themselves are step functions.
 
  • #25
boneh3ad said:
Noise is the source of the shock.

Dopplereffectsourcemovingrightatmach1.4.gif

Taken from Wikimedia commons.

The animation you linked from Wikipedia is mislabeled. I have informed the author and I hope he corrects this mistake.

The pretty picture labeled "sonic boom" is not a boom and is not really even a shock wave. The oblique wave you see in the picture is properly called a "Mach wave" and is equivalent to an oblique shock in the limit of zero strength (i.e. no turning of the flow and no rise in pressure). You can create a Mach wave in a wind tunnel by roughening the surface of a test specimen and the perturbations in flow will propagate and coalesce, very much like your description of shock waves. However, you would hardly notice the passage of a Mach wave, because there is no static pressure rise such as the rise you get across a true shock.

However, shock waves do not require the coalescence of perturbations to form. An oblique shock will form in a supersonic flowfield pretty much whenever the flow has to turn compressively (i.e. into itself). A normal shock will form when a supersonic flow needs to change speed abruptly without turning. The F-16, famously, has a normal-shock inlet which slows the flow abruptly from supersonic to subsonic to enable it to be further slowed down in the subsonic diffuser. These "true" shocks -- as opposed to Mach waves -- have nothing to do with coalescence of tiny perturbations.
 
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  • #26
bbbeard said:
The pretty picture labeled "sonic boom" is not a boom and is not really even a shock wave.
The picture is used in the main article as well:
http://en.wikipedia.org/wiki/Sonic_boom

bbbeard said:
These "true" shocks -- as opposed to Mach waves -- have nothing to do with coalescence of tiny perturbations.
Maybe you don't have to interpret the circles in the animation as perturbations, bet merely as distance markers that show how fast a perturbation would advance. Would this be a correct description of a shock wave, in contrast of a Machian wave? However, the mention of Doppler effect in the labeling would have to go, because it refers to actual perturbations.
 
  • #27
A.T. said:
Maybe you don't have to interpret the circles in the animation as perturbations, bet merely as distance markers that show how fast a perturbation would advance. Would this be a correct description of a shock wave, in contrast of a Machian wave? However, the mention of Doppler effect in the labeling would have to go, because it refers to actual perturbations.

I should say first that it is very common to find a diagram similar to the Wikipedia animation in pedagogical developments of shock theory. It has the virtue of making clear the derivation of the Mach angle (arcsin(1/M)). It is probably worthwhile to note that the Mach angle is independent of the frequency of the "beeper" that is producing the waves. Since the paradigmatic Mach wave is a continuous wave at a fixed angle, it's slightly misleading to portray the development of the Mach wave this way -- taken literally, the Mach cone would have a bumpy contour that is the union of all the sonic spheres emitted by the beeper. That is, the Mach wave forms a geometric cone only in the high-frequency limit.

As I pointed out, the Mach wave is the zero-strength limit of an oblique shock. As such, I feel obligated to amend my strenuous objection to the "coalescence" language applied to the development of shock waves. After all, there is probably a way to write down the shock relations in terms of something like a Huygens superposition integral of all the perturbations (not "noise") from the body, essentially a complexified version of the beeper animation. However, I've never run across that treatment. It is more efficient to derive the shock equations from the conservation of mass, momentum, and energy, deploying geometric compatibility relations as needed.

I stand by my assertion that the shock structure is such that any noise emitted by the body is dissipated. There is no "memory" in the shock that would store oscillations emanating from the body. And I still maintain that any perceived "pitch" of a sonic boom is related to the wavelength of the "N wave" that is typical of sonic booms. Small objects like bullets produce compact N waves and large objects like supersonic aircraft produce extended N waves. But the shock waves that make up the N waves are very sharp discontinuities.
 
  • #28
bbbeard said:
The physical picture you are promoting, that (a) a shock wave is the result of the coalescence of small perturbations, and that therefore (b) the shock wave retains some kind of information from the source of the perturbations, is incorrect.

Consider that the typical sonic boom is the result of the passing of the oblique shocks attached to a supersonic object. In the frame of the object, the atmosphere (quiescent in the Earth frame) is approaching uniformly at the flight speed of the object. The oblique shock arises from the need to turn the flow in a way that is compatible with the physical requirement that pressure waves can only travel at the speed of sound. Shocks are nature's way of matching upstream and downstream boundary conditions when the upstream flow is supersonic. There is also typically an oblique shock at the tail of the object, as the flow has to straighten out again.

Shock layer thickness is equal to only a few mean free paths. In practical terms this means that the shock is a step function. The Fourier transform of a step function goes like -i/omega, so it has contributions from all frequencies. There is no substructure that incorporates tiny oscillations emanating from the supersonic body. In fact shock are highly dissipative; they destroy information and generate entropy.

Sonic booms incorporate the pressure jumps from both the tip and tail of the flying object. Consider the shock structure for the paradigmatic diamond-wing airfoil. If the flight Mach number is sufficiently high, there is an attached oblique shock at the tip. There will be a Prandtl-Meyer expansion fan at the mid-chord corner, which allows the flow to accelerate around the corner. And then there is a compression shock at the tail as the flow has to straighten out again. On the ground, the timing of the shocks will depend on the length of the object, its altitude, and its flight speed. The typical sonic boom is the result of the passing of this "N wave". Larger objects create a bigger lag between tip and tail and may be experienced as a lower-pitched boom, although the shocks themselves are step functions.

bbbeard said:
The animation you linked from Wikipedia is mislabeled. I have informed the author and I hope he corrects this mistake.

The pretty picture labeled "sonic boom" is not a boom and is not really even a shock wave. The oblique wave you see in the picture is properly called a "http://en.wikipedia.org/wiki/Mach_wave" [Broken]" and is equivalent to an oblique shock in the limit of zero strength (i.e. no turning of the flow and no rise in pressure). You can create a Mach wave in a wind tunnel by roughening the surface of a test specimen and the perturbations in flow will propagate and coalesce, very much like your description of shock waves. However, you would hardly notice the passage of a Mach wave, because there is no static pressure rise such as the rise you get across a true shock.

However, shock waves do not require the coalescence of perturbations to form. An oblique shock will form in a supersonic flowfield pretty much whenever the flow has to turn compressively (i.e. into itself). A normal shock will form when a supersonic flow needs to change speed abruptly without turning. The F-16, famously, has a normal-shock inlet which slows the flow abruptly from supersonic to subsonic to enable it to be further slowed down in the subsonic diffuser. These "true" shocks -- as opposed to Mach waves -- have nothing to do with coalescence of tiny perturbations.

I already admitted in a previous post that some of both what I said and the way I said it was inaccurate. I take back what I said about a shock retaining frequency information from the sound wave generated by the object. That was misguided.

However, your definition of a shock has some holes. If your definition was true, then a normal shock would not be possible. For example, there is no turning of the flow in a shock tube. Similarly, in the case of a bow shock, the flow physically can't turn, but the shock forms anyway detached from the surface and then the flow turns once it is slowed by the bow shock. In other words, shocks can form without the need for turning.

Now I will admit that before, I misspoke in equating Mach lines and the Mach angle with the shock angle. You are correct in saying that I was describing Mach lines (waves). However, I still maintain that the shock itself (of any sort) is a coalescence of compression waves. The difference between what I said earlier and what is correct, is that I was attributing the shock to sound waves rather than compression waves. If you look at a flow in terms of a series of discrete compressions (as illustrated by Seitzman at Georgia Tech http://soliton.ae.gatech.edu/people/jseitzma/classes/ae3450/shocks.pdf), you can see that strong compression waves form and eventually catch up to one another, forming a shock. With a continuous compression as in the case of an actual shock tube or a plane or anything else in practice, you would get a continuous compression wave that eventually bunches up, or coalesces, into a nearly discontinuous step.

In other words, yes, you are right that I misspoke in saying it was sound waves. However, it is the coalescence of waves, and while the necessary turning of the flow determines the angle for an oblique shock, it is not the physical basis of the shock. The need to turn into itself would be one reason for the formation of the compression wave that forms a shock, but it is still that wave that coalesces into the shock.

I also feel it necessary to point out that it is actually the shock itself that is on the order of several mean free paths in thickness (about 6 if I remember correctly). The shock layer is a different concept where, at very high mach numbers in the hypersonic regime, the shock comes so close to the surface of the object that you end up with strong entropy gradients close to the wall and in the boundary layer.

Also, it seems you are correct on the Wikipedia thing. Serves me right for hastily pulling something off of Wikipedia without fact checking; a cardinal sin to be sure. Even in the Doppler effect article you linked it had incorrect captions since the equation used is invalid for [itex]M \geq 1[/itex]. Good catch.

---
Just as a side note, I have to say I am happy to have this discussion. It is, in fact, why I browse these forums: to educate and be educated. This has certainly helped sort some thoughts in my own head out. I have to say, thanks and good day, sir.

A.T. said:
Maybe you don't have to interpret the circles in the animation as perturbations, bet merely as distance markers that show how fast a perturbation would advance. Would this be a correct description of a shock wave, in contrast of a Machian wave? However, the mention of Doppler effect in the labeling would have to go, because it refers to actual perturbations.

It would be the correct description of Mach lines (or waves, same thing), but not a shock. bbbeard is correct in saying that the sound waves do not coalesce into shocks, and the compression waves that do do not propagate in spheres from the objects (roughly) as do sound waves.
 
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  • #29
boneh3ad said:
However, your definition of a shock has some holes. If your definition was true, then a normal shock would not be possible. For example, there is no turning of the flow in a shock tube. Similarly, in the case of a bow shock, the flow physically can't turn, but the shock forms anyway detached from the surface and then the flow turns once it is slowed by the bow shock. In other words, shocks can form without the need for turning.

Well, happily enough, a normal shock is a "strong" oblique shock in the limit of zero deflection angle. So I would say we are both right. If you look at the oblique shock relation plot,

[PLAIN]http://upload.wikimedia.org/wikipedia/en/thumb/c/c7/ObliqueShockAngleRelation.png/793px-ObliqueShockAngleRelation.png [Broken]

...Mach waves are the "weak" oblique shocks in the limit of zero deflection (the vertical axis for beta<90), and normal shocks degenerately fall on the point theta=0, beta=90. So, yes, normal shocks don't need turning -- to form, they just need the appropriate upstream and downstream boundary conditions. But they can be analyzed exactly with the oblique shock relations -- after all, an oblique shock is just a normal shock moving sideways. In terms of analyzing sonic booms, though, I think we only deal with oblique shocks with finite deflection angles.

boneh3ad said:
... I still maintain that the shock itself (of any sort) is a coalescence of compression waves. The difference between what I said earlier and what is correct, is that I was attributing the shock to sound waves rather than compression waves. If you look at a flow in terms of a series of discrete compressions (as illustrated by Seitzman at Georgia Tech http://soliton.ae.gatech.edu/people/jseitzma/classes/ae3450/shocks.pdf), you can see that strong compression waves form and eventually catch up to one another, forming a shock. With a continuous compression as in the case of an actual shock tube or a plane or anything else in practice, you would get a continuous compression wave that eventually bunches up, or coalesces, into a nearly discontinuous step.

Yes, I think you can discuss shock formation in terms of the coalescence of perturbations due to the body in a supersonic flowfield.

boneh3ad said:
In other words, yes, you are right that I misspoke in saying it was sound waves. However, it is the coalescence of waves, and while the necessary turning of the flow determines the angle for an oblique shock, it is not the physical basis of the shock. The need to turn into itself would be one reason for the formation of the compression wave that forms a shock, but it is still that wave that coalesces into the shock.

Fair enough. I think of the physical basis of a shock being that (in a streamtube, say) there are two solutions for compressible flow that have the same mass flow, momentum, and energy (provided the subsonic solution's Mach number is above a certain threshold, or there will only be a subsonic solution). So supersonic flow can suddenly and discontinuously change to subsonic flow without violating any conservation laws. Because the entropy rises across a shock, there are no "shocks" that go from subsonic to supersonic. (Of course, a convergent-divergent nozzle will take you smoothly from subsonic to supersonic, albeit without the discontinuity.)

I would submit that flow turning is the "mechanical" basis for an oblique shock. The turning is not as fundamental a part of the picture as the conservation laws. But the reason this is useful is that we have occasion to analyze, say, the curved shocks that form in front of a blunt or rounded body. The streamlines that go through the curved shock turn, and that turning angle is what we use to compute the variation of, say, the static pressure behind the shock, using the oblique shock relations.

boneh3ad said:
I also feel it necessary to point out that it is actually the shock itself that is on the order of several mean free paths in thickness (about 6 if I remember correctly). The shock layer is a different concept where, at very high mach numbers in the hypersonic regime, the shock comes so close to the surface of the object that you end up with strong entropy gradients close to the wall and in the boundary layer.

True.

Thanks!
BBB
 
Last edited by a moderator:
  • #30
Hoping the wiki article might help here:

... the energy of a shock wave dissipates relatively quickly with distance. Also, the accompanying expansion wave approaches and eventually merges with the shock wave, partially cancelling it out. Thus the sonic boom associated with the passage of a supersonic aircraft is the sound wave resulting from the degradation and merging of the shock wave and the expansion wave produced by the aircraft. ...

http://en.wikipedia.org/wiki/Shock_wave

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.

http://en.wikipedia.org/wiki/Shock_wave#In_supersonic_flows

Actual shock wave "crack" (followed by engine sounds) heard from a close flyby of a supersonic aircraft in the second half of this video:

http://rcgldr.net/real/f14flyby.wmv
 
  • #31
boneh3ad said:
I already admitted in a previous post that some of both what I said and the way I said it was inaccurate. I take back what I said about a shock retaining frequency information from the sound wave generated by the object. That was misguided.
So you might call it an "educated guess"...
 

1. Why is there no sonic boom when shooting an air rifle at 1400 feet per second?

The sonic boom is created when an object travels faster than the speed of sound, which is approximately 767 miles per hour at sea level. An air rifle shooting at 1400 feet per second is only traveling at about 954 miles per hour, which is below the speed of sound. Therefore, there is no sonic boom.

2. Can an air rifle ever produce a sonic boom?

Yes, an air rifle can produce a sonic boom if it is shooting at a speed greater than the speed of sound. This typically requires specialized ammunition and can be dangerous to use.

3. Why do some guns produce a sonic boom while others do not?

The speed of sound is affected by various factors such as air temperature, elevation, and humidity. Some guns, such as high-powered rifles, are capable of shooting at speeds that exceed the speed of sound, resulting in a sonic boom. Air rifles, on the other hand, typically do not have enough power to reach these speeds.

4. Is it possible to modify an air rifle to produce a sonic boom?

Technically, it is possible to modify an air rifle to shoot at speeds that exceed the speed of sound and produce a sonic boom. However, this is not recommended as it can be dangerous and illegal in some areas.

5. Are there any other factors that can affect the production of a sonic boom from an air rifle?

Air resistance and the design of the air rifle can also play a role in the production of a sonic boom. Rifles with a larger caliber and a more streamlined design are more likely to produce a sonic boom due to their ability to travel at higher speeds.

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