Re: If NOT a reactionless drive, will it work?
From: Jim Black (ghytrfvbnmju7654_at_mail.com)
Date: 07/18/04
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Date: 17 Jul 2004 17:18:32 -0700
vnemitz@pinn.net (vernonner3voltazim) wrote in message news:<42336979.0407170917.324bf497@posting.google.com>...
> ghytrfvbnmju7654@mail.com (Jim Black) wrote in message news:<cb623e6.0407161502.71b06250@posting.google.com>...
> > vnemitz@pinn.net (vernonner3voltazim) wrote in message news:<42336979.0407151933.18800f35@posting.google.com>...
> > > ghytrfvbnmju7654@mail.com (Jim Black) wrote:
> > > > What this is is a simplification of what would happen if
> > > > Superman gave it a whack on one end. In a more realistic
> > > > rod, the shock would eventually spread out over the entire
> > > > rod, as much of its energy gets converted into heat. The
> > > > other picture is the same thing, only this time, the
> > > > people all along the rod have exerted a force equal and
> > > > opposite to Superman's. In that case, there's a lot of
> > > > vibration, but the center of mass of the rod doesn't go
> > > > anywhere.
[snip]
> OK, I apologize for misinterpreting it AGAIN. And if I am
> not misinterpreting it a third time, then I am seeing some
> vibratory consequences of a single impact at one end. But
> don't you think those vibes are going to die down after a
> while, and the battering ram will have finished experiencing
> jerk and acceleration, and will then proceed smoothly forward?
Definitely. In fact, that's exactly what I said. The reason that
wasn't in the diagram is because it would be difficult to draw.
> \ \ \ \ \ \ \
> \ \ \ \ \ \ \
> \ \ \ \ \ \ \
> etc.
Not quite that fast:
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> Also, I'd still like to know what you think about
> what I wrote in the last portions of my prior
> message to you.
-> With respect to the center of mass, YES, if the battering
-> ram could respond all-at-once to Superman's applied force,
-> then ordinary oscillation would be the result (no overall
-> permanent displacement of the center of mass). The response
-> time that interferes with Superman's effort, however has the
-> effect of NOT allowing the ram's center of mass to move as
-> completely/easily as if the whole ram responded at once. That
-> is, Superman's quick-struck compression of the end of the ram
-> immediately constitutes SOME displacement of the center of
-> mass,
What happens is that the center of mass will start moving as the
compressed region moves through the ram:
\ | | | | | |
\ | | | | | |
| \ | | | | |
| | \ | | | |
| | | \ | | |
| | | | \ | |
| | | | | \ |
If p is the amount of momentum imparted to the ram by Superman, and m
is the total mass of the ram, this will cause the center of mass to
begin moving at a constant rate of p/m.
-> but not until that wave of compressed material reaches
-> the far end of the ram, and the ram decompresses, can we say
-> that the center of mass has any chance of moving as far as it
-> would-have-moved-at-once, if whole-ram-could-have-moved-at-once.
If people positioned all along the ram impart the same amount of
momentum, the whole ram will be moving at a rate of p/m. Whether the
ram is moving piece by piece or all at once, the velocity with which
the center of mass moves is the same.
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