Re: If NOT a reactionless drive, will it work?
From: vernonner3voltazim (vnemitz_at_pinn.net)
Date: 07/20/04
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Date: 19 Jul 2004 23:41:19 -0700
ghytrfvbnmju7654@mail.com (Jim Black) wrote:
> > vnemitz@pinn.net (vernonner3voltazim) wrote:
> ghytrfvbnmju7654@mail.com (Jim Black) wrote:
>
> 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 now be
moving at the final rate (assumed here to be 1 cm/sec) so it
therefore must now be displaced by 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. (I see that
this requires an assumption that the compression affects
the whole length of the ram, faster than the speed of sound.
But that does not really affect, I think, the problem pointed
out in the next paragraph -- same assumption applied.)
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 will
now attempt to analyze that in somewhat fine detail.
To begin, we imagine the whole ram balanced like a
see-saw at T=0.00. (Nitpicks about it maybe being 2km
long can be met with a pretense of this taking place
in a 0.0000001 G gravity field. :) Next, after one
end is compressed by Superman's impact, I'll start
with the OTHER data from T=0.05 above, where the ram
has compressed by 1mm. Since all of that compression
is on one side of the fulcrum, at the 1/20-length
mark, this compression is actually insufficient to
provide the required 1 cm/sec of motion of the center
of mass. Another way to think of this is, suppose I
HACKED OFF that first millimeter of the battering ram;
this immediately displaces the balance point by half
that. Well, the hacked material "belongs" at the 1/20
length mark, so setting it there (like a sliding
weight on a doctor's scale) cancels out some of that
displacement. So, I WILL accept a greater compression
(hack-off amount) to achieve your specified motion of
the center of mass (CM), for the first data-point here!
Now to figure what that is... Let me call the mass of
that 1mm slice as being M. In its original location,
its effect upon the balance point is the same as an
equal slice at the other end of the battering ram.
If this slice is moved 1/20 down the length of the ram,
well, that is 1/10 the way to the fulcrum, and so its
effect upon the balance point is 9/10 the effect of
that other still-attached slice at the far end. This
implies that 9/10 of that initial CM displacement is
cancelled out, and that the effective displacement is
only 0.005 cm. It now seems reasonable that all I need
to do is mulitply the slice by 10. That is, a total
initial compression of 1cm, otherwise treated as a 1cm
slice removed from one end, means a 0.5 displacement of
the CM. Attached, it is balanced by an identical slice
at the other end of the battering ram. Moved 1/10 of
the way to the fulcrum, it effectively cancels 9/10 of
that removal-caused 0.5cm displacement of the CM,
leaving the desired 0.05cm, which is what you want if
the CM is moving at 1 cm/sec after 1/20 sec. Whew!
OK, NOW what about that second time-point, where at T=0.10
the wave of compressed battering ram has moved 2/20 or
1/10 through the entire length? This is like sliding that
1cm slice 1/5 of the way toward the fulcrum, and you want
the CM to have been displaced 0.10cm. Can it? Well,
repeating that the removal of the 1cm slice means that the
CM is displaced by 0.5cm, sliding the slice 4/5 of the way
from the fulcrum to its original location means 4/5 of that
displacement is cancelled out, leaving a 0.1cm
displacement of the CM! Well, Well!!!!! Newtonian
Mechanics SEEMS safe for another day!
Unless I made an error in the preceding analysis! Let
me know! --AND....how likely is that scenario with real
ordinary materials? To whatever extent the battering
ram fails to be compressed that much (and not break),
then that is the extent to which the efforts of
Superman cannot perfectly cancel the efforts of the
2000 men. Even perfect Newtonian reasoning must bow
to the real world....
Thanks Again!
Vernon Nemitz
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