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Re: toroidal perpetual motion machine



Okay, I see your point to my explanation, and I see that I failed to add something
in my analysis. I will add it here.

There is a greater force on the outside portion of the tube pulling up than there
is on the inside portion of the tube pulling up (against the water to thus form the
meniscus). This greater force would tend to make the tube rotate up at the outer
perimeter and down on the inside. This is the opposite of the direction you stated
in your problem. I don't see how there is a net force? In your tug-o-war example,
you seem to pit the inside meniscus against the outside meniscus. Fine, there are
unbalanced forces, but now add the inside rubber against the outside rubber like
so:
OR___ ______OM___
OR___X--- IM--------XX______OM __ X-----IR

The X is a joining of ropes. The outer rubber and inner meniscus pull in one
direction and the inner rubber and outer meniscus pull in the opposite direction.
Look at the "left" side cross section and consider clockwise rotation to be
positive. For simplicity, consider the forces along the outer perimeter to be
twice the forces on the inner perimeter. Sum forces like this:

Inside rubber (IR) pulling up = -1
Outside rubber (OR) pulling up = +2
Inside meniscus (IM) pulling down = +1
Outside meniscus (OM) pulling down = -2
mass = 1
Now solve F=ma

IR + OR + IM + OM = a
-1 + 2 + 1 - 2 = a
a=0

Where exactly am I going wrong???


John Denker wrote:

At 04:22 PM 6/29/00 -0400, Hans G. Ammitzboll wrote:

Statement 3, claims an imbalance of force due to a longer meniscus on the
outside pulling down on the rubber (statement 2),

Yes, there is a longer meniscus and therefore a larger contribution to the
force budget.

however, the first half of
statement 1 clearly points out that the rubber must pull UP on the water to
form the meniscus.

Yes, at each point on the meniscus the rubber pulls up and the water pulls
down.

Therefore, there is a longer section of rubber on the
outside pulling up on the water then there is on the inside. Yet one more
unbalance set of forces. However, it should be clear by now that when the
upward forces of the rubber on the water are compared to the downward
forces of
the water on the rubber, all forces will balance out.

All the aforementioned forces balance _locally_ at each point.

The result would be
that, regardless of viscosity, material of the toroid, or any other factor not
involving some external energy source, this setup will NEVER produce any sort
vortex rotation, much less perpetual motion (that is after initial microscopic
oscillations induced during the placement of the toroid die out).

No, that general result does not follow. That analysis does not capture
the correct physics. That analysis contains logical fallacies. Here is an
analogy that may help elucidate the fallacies:

Consider a tug-of-war game, with mismatched teams and a Y-shaped rope.

___________B__B
=========*---------------
A A C B B

Team A has two people pulling leftward; team B has two people on each
sub-rope (four people total) pulling rightward. Each of the people pulls
equally hard. At point C I have a clamp that keeps things from
moving. According to the HGA argument, one would conclude there is no net
force because the force exerted by each team member's hands is locally
balanced by an equal and opposite force from the rope. But this is
nonsense; in fact there _is_ an overall rightward force on the clamp, and
if I release the clamp people will go flying.

The same reasoning applies to the inner-tube: there is a local force
balance at each point, but there are more points on the outer
radius. Result: net torque.

Is this your mis-direction? Kind of like the hidden division by zero in the
"proof" that 1+1=3 ?

Why is everybody so sure there's misdirection involved? If you don't
believe my analysis, just ignore it! Do your own analysis from
scratch. It's just physics.

Huge hint: Essentially all I've done is omit part of the story. The parts
of the story that I have told are basically true and not even particularly
misleading. You'll never find the solution by micro-analyzing and
cross-examining my notes. The solution is to find the physics -- plain old
physics -- that hasn't yet been brought to bear on the problem.

--
Hans G. Ammitzboll hammitzboll@graphnet.com
Sales Executive hansa@graphnet.com
Graphnet, Inc. physics@mindless.com
60 Hudson St.
Suite 1014
New York NY 10013
212-584-1000
ext. 229