Re: If NOT a reactionless drive, will it work?
From: vernonner3voltazim (vnemitz_at_pinn.net)
Date: 07/18/04
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Date: 18 Jul 2004 01:42:31 -0700
ghytrfvbnmju7654@mail.com (Jim Black) wrote:
> > > vnemitz@pinn.net (vernonner3voltazim) wrote:
> > > > 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
OK, thanks. (the preceding can now be snipped)
> wasn't in the diagram is because it would be difficult to draw.
>
> > \ \ \ \ \ \ \
> > \ \ \ \ \ \ \
> > \ \ \ \ \ \ \
> > etc.
>
> Not quite that fast:
>
> | | | | | | |
> | | | | | | |
> | | | | | | |
> | | | | | | |
> | | | | | | |
> | | | | | | |
> | | | | | | |
> | | | | | | |
Agreed. (more snipping to do)
> > 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:
>
> \ | | | | | |
> \ | | | | | |
> | \ | | | | |
> | | \ | | | |
> | | | \ | | |
> | | | | \ | |
> | | | | | \ |
I agree, EXCEPT that you have to be careful about
the portion of the ram for which the applied
force has yet to arrive. (more below)
> 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.
I think this is the crux. In one sense, we DO want
the total force applied by Superman to have the same
final effect upon the battering ram as the 2000 men,
but in another sense, what we really want is to find
out just exactly how that applied force REALLY causes
it to move. To see what I'm getting at, let's PRETEND
for a moment that the ram is infinitely rigid. This
allows Superman's force to instantly traverse its
length and affect all parts of it, better even than
the gang of men can do when it is only ordinarily rigid.
But what infinite rigidity lets us do is decide exactly
the value of Superman's applied force: That quantity
which leads to perfectly regular ordinary oscillation
of the battering ram, when alternately struck by the
2000 men and Superman.
WITH the value of the applied force known, we can now
drop the pretense of infinite rigidity, and study the
consequences of Superman's application of the force upon
the end of the battering ram. (more below)
> -> 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.
Here is where the debate can get hot and heavy. At first
glance, you seem to bemaking the unwarranted assumption that
the far end of the ram moves immediately, rather than waiting
for the applied force to arrive at the speed of sound. If so,
then of COURSE the ram's motion has its center of mass moving
at the appropriate momentum-related velocity. (Let's pick a
convenient velocity, say 1 cm/sec.)
On second glance, we have already been describing a wave of
compressed material in the battering ram, so that first
glance cannot be correct. BUT, does this second glance
REALLY allow the center of mass to move that way? That is,
if Supermon applies force at T=0.00, then AT that moment the
center of mass is exactly half-way along the battering ram.
In the next moment, say T=0.05 second, the wave of compression
is now moving inside the body of the ram, and is 1/20 down its
length (due to specification of 1 second for whole length).
Meanwhile, you are requiring the center of mass to have been
displaced 1/20 cm -- what amount of compression allows this?
After some BOTEC I get 2/20 cm, or 1 mm. If correct, I will
not object to that.
But now let's look at T=0.10 second, where the compression
wave has reached the 1/10 mark along the length of the
battering ram. The center of mass is now supposed to be
displaced 1/10 cm. What amount of compression allows THAT?
I get 2/10 cm or 2 mm. I DO OBJECT to that! Superman
applied a short sharp impulse, and we want to study how the
ram responds to it. Superman is NOT still compressing the
ram more and more, after that initial impulse! Yet exactly
such additional compression seems necessary to meet your
claim that the center of mass must immediately move at the
final rate! At the moment, unless you have some other
basis than the compression of the battering ram, to explain
your specified motion of the center of mass, it seems like
Newtonian Mechanics is faced with a dilemma.
Dealing with that dilemma is the whole purpose of this
Thread. I do know of one other basis to support your
argument, which concerns the increased density of the
battering ram in the region of the compression wave,
and its "place" with respect to Center of Mass and the
Law of the Lever. Returning to this sketch you drew
(I added two arrows):
\ | | | | | |
\ | | | | | |
| \ | | | | | <--
| | \ | | | |
| | | \ | | |
| | | | \ | |
| | | | | \ | <--
Notice that while the wave of force is moving down
the ram's length, the ram's overall position is static
for the duration of time that I marked with arrows.
In this other-basis argument, The motion of the
compressed-mass wave is supposed to be the reason
why the center of mass can move at 1 cm/sec. I'm not
sure I'm up to the task of analyzing this in fine
detail at 4:30am. But I would sure like to know the
result! (Will it be as unacceptable as the first
argument? Stay tuned, folks!)
--And, Thanks Again!
Vernon Nemitz
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- Previous message: Ballisticus: "Re: SR's velocity addition -- ANY Experimental Evidence?"
- In reply to: Jim Black: "Re: If NOT a reactionless drive, will it work?"
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