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To: general_re

Oops, I'm so smart I posted this to myself ;)

Now for you. In your model, how does the front of the wave "know" what the final equalized pressure will be?

In fact, as air enters the tunnel it creates a bit of a vacuum at the entry compared to the surrounding area. The air molecules at the front of the wave bounce around, vice proceed on a direct path, thus, some of that leading wave is always turning back. If the pressure behind them isn't as great, as it was at the beginning, a forgone conclusion, then not as many bounce back in the initial direction. As time progresses this leads to a diminishing pressure. This effect is constant trough out the air mass, and thus you have the effect of slowly (relatively speaking), or perhaps better put, lineally increasing pressure in the path of propagation.
Despite the subject, this isn't really rocket science, but rather freshman (in HS) chemistry.


100 posted on 09/09/2005 2:23:03 PM PDT by SampleMan
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To: SampleMan
In your model, how does the front of the wave "know" what the final equalized pressure will be?

It's not going to be zero, and that's just about all that matters, innit?

If the pressure behind them isn't as great, as it was at the beginning, a forgone conclusion...

Do you imagine that this is enough to stop the flow to the zero pressure area farther down the tube? After all, whatever the pressure is behind the leading edge, it's not zero, right?

Basically, what you're telling me is that turbulence and M. Bernoulli - and presumably special relativity - are going to conspire to keep the molecules of air from accelerating to c. Well, that's fine, but they're still going to the zero pressure parts of the tube, and they're still going to meet up with your railgun sled somewhere in that tube as they go one way that the sled goes the other.

103 posted on 09/09/2005 3:41:24 PM PDT by general_re ("Frantic orthodoxy is never rooted in faith, but in doubt." - Reinhold Niebuhr)
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