Timeline for Why does the gas cloud of an underwater gunshot pulse?
Current License: CC BY-SA 3.0
8 events
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| Mar 31, 2014 at 17:14 | comment | added | user12029 | @Hoytman, so my point really is just that to understand the dynamics of the system, you should realize that the compression of the air is very important, while the compression of the water is not. | |
| Mar 31, 2014 at 17:12 | comment | added | user12029 | @Hoytman, So, the water stores potential energy (as it is pushed away, working against the pressure of the rest of the water) and kinetic energy (at times when it has velocity), but my point phrased another way is: If you took a container of water at the bottom of the sea, lifted it up to the surface, and poked a hole in it, the water would release a tiny bit of energy and that would be the end of it. If you took a container of air at sea level and lifted it up to space, then poked a hole in it, you would get a lot of energy from it. | |
| Mar 31, 2014 at 12:54 | comment | added | Hoytman | you said " for each pound of force applied, compressed water barely stores any energy" - although it is hard to get water to store energy via compression, it does appear to store it quite well in the video. Otherwise, the pulse would only happen once and the pulse resonance would die away very quickly. In the video, the pulse reoccurs (resonates?) and dissipates over the course of several pulses. This demonstrated that both the gas and the water are storing and emitting energy. Is this true or is there something else going on here (I have no training in this field so I may be way off.) | |
| Mar 30, 2014 at 1:51 | comment | added | user12029 | @leftaroundabout What can I say - I think that gross oversimplification is fine for non-technical questions, and I think that discussion of oscillatory phenomena in general adds to an understanding of this. Overshoot/undershoot from equilibrium, even though there may be no equilibrium, among other problems. | |
| Mar 30, 2014 at 1:26 | comment | added | leftaroundabout | Well, that wiki article is about "ordinary" gas bubbles, the ones that are permanently stable under their own inner pressure. Those behave in many ways quite different from cavitation vapour bubbles. | |
| Mar 30, 2014 at 1:01 | comment | added | user12029 | @leftaroundabout Okay, I'm not doing pulsing bubbles justice at all, but this page en.wikipedia.org/wiki/Liquid_bubble#Pulsation seems to imply that the amplitude is independent of initial displacement from equilibrium [in that specific case] (I say nothing about the bubble radius). I'm not familiar with much about nonlinear ODEs but this seems fine to me. (if I was familiar I'd try to do things more formally to see whether I'm right or wrong, but at the moment I'd probably spend way too long on it before I got anywhere) | |
| Mar 29, 2014 at 23:28 | comment | added | leftaroundabout | Where do you you see a small amplitude here? I think this is in fact quite a good example of an oscillation that's totally unrelated to the usual 2nd-order differential equation. I know little about the processes involved in cavity collapse, but pretty sure it's very much more complicated. For one thing, a prediction of a $\ddot x \propto - x$ model would be that the frequency is constant, but pretty clearly big bubbles oscillate slower than small ones. | |
| Mar 29, 2014 at 22:25 | history | answered | user12029 | CC BY-SA 3.0 |