Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10.2 9/18/84; site cci-bdc.UUCP
Path: utzoo!watmath!clyde!burl!ulysses!allegra!bellcore!decvax!genrad!mit-eddie!cybvax0!cci-bdc!jlup
From: jlup@cci-bdc.UUCP (jrl devlt)
Newsgroups: net.physics
Subject: Re: A new improved perpetual motion machine
Message-ID: <176@cci-bdc.UUCP>
Date: Wed, 10-Apr-85 20:09:12 EST
Article-I.D.: cci-bdc.176
Posted: Wed Apr 10 20:09:12 1985
Date-Received: Thu, 18-Apr-85 04:31:23 EST
References: <489@vax2.fluke.UUCP>
Distribution: net
Organization: Computer Consoles, Inc., Cambridge, MA
Lines: 108

> Ok perpetual motion machine fans, see if you can explain this one.  This was
> presented to me several years ago by a friend, and even though we both "knew"
> it couldn't work, no one was ever able to tell us why.
> 
> Start with a vessel, like a cup, with a weight resting in the bottom and a
> flexible diaphragm stretched across the top.  The bottom and sides are rigid.
> It might look something like this-
> 
>             H-----------H
>             H     _     H
>             H    / \    H
>             H   |   |   H
>             \\   \_/   //
>               =========
> 
> Now if we invert the vessel and allow the diaphragm to stretch, the volume
> increases, while the internal air pressure decreases.
> 
>               =========
>             //         \\
>             H           H
>             H           H
>             H           H
>             H\    _    /H
>               \  / \  / 
>                \|   |/
>                 \\_//
>                   -
> Since buoyancy is related to volume displaced, the inverted vessel should be
> more buoyant than the upright vessel.  By making use of that property, we
> now construct our perpetual motion machine by linking many of these vessels
> together in a chain and submerging them under water (in theory any fluid
> should work).
> 
>               <-
> 
>               o
>             /   \
>            u     O
>           |       |
>           u       O
>           |       |
>       |   u       O   ^
>       v   |       |   |
>           u       O    
>           |       |
>           u       O
>           |       |
>            u     O
>             \ o /
> 
>               ->
> 
> There you have it, free energy!  The inverted vessels on the right provide
> greater buoyance than those on the left so the net result is an upward force
> on the right greater than that on the left.
> 
> Who can tell me what is wrong with this picture?
> 
> 	Craig Johnson			uw-beaver!	\
> 	John Fluke Mfg. Co., Inc.	decvax!microsof!  \
> 	Everett, Washington		ucbvax!lbl-csam!    >  fluke!vince
> 					allegra!	  /
> 					ssc-vax!	/

  The reason this will not produce any new energy (as might be required
in order to overcome frictional losses, etc.), is that the energy to
be gained by taking advantage of the added buoyancy is the same energy
used to increase the volume of the vessel.

At the top, there is a transition from the hi volume state to the low 
volume state. The energy balance for a box drawn around the top "pulley"
is, assuming vacuum inside and very slow speed (no kinetic effects - 
not a needed condition as kinetic effects cancel, but I don't want to 
bother showing this, someone else can):

1) the potential energy  of the upward bound vessel:

		i) energy stored in the deformed diaphram.
		ii) gravitational potential of the vessel.
		iii) gravitational potential of the weight.
		iv) bouyant potential of the vessel.

2) the potential energy of the downward bound vessel:

		i) energy *not* stored in the *undeformed* diaphram.
		ii) gravitational potential of the vessel.
		iii) gravitational potential of the wieght.
		iv) bouyant potential of the vessel.

Assuming that one upward bound vessel enters the top box at the same
time as a downward bound vessel leaves, energy conservation shows that
the difference between these two energy states is reflected in tensions
in the string.

   The point is that the operations at the bottom, with respect to 
energy balance, are the exact converse of this situation. The work used
in distending the diaphram is *exactly* equal to the weight of water 
displaced times the hieght of displacement, both at the top and the
bottom, and is thus exactly exactly equal the work obtained from the 
change in buoyancy (the weight of water displaced times the distance 
of translation). 

   This scheme works as well as any other "classical" system of 
perpetual motion: only without friction.


		-John