[net.physics] Modification of membrane-cup PMM

js2j@mhuxt.UUCP (sonntag) (04/12/85)

     Having received mail redescribing the membrane-cup PMM, I sat down
to analyze it and discover why it wouldn't work.  My first step was to
modify it slightly to make it slightly easier to analyze.  I've replaced
the cup and the membrane with a cylinder, closed at one end, and a massive
piston:              A (inverted)                   B (rightside up)
		----------------		|		|
		|		|		|		|
		|		|		|		|
		|      air    	|		|		|
		|		|		|+-------------+|
		|+-------------+|		|| heavy piston||
		|| heavy piston||     water     |+-------------+|
		|+-------------+|		|	air	|
		|		|		|		|
		|		|		-----------------

      Well, I failed miserably at showing why it wouldn't work.  All I
was able to show was that the bouyancy is greater for the inverted
configuration than for the non-inverted configuration.  And if that's
true, I really can't figure out why one couldn't hook a lot of these to
a vertical loop of chain and have a PMM.  (I mean, nothing is moving
near the speed of light here, so I *should* be able to figure out why
it doesn't work!)  Maybe someone can point out my mistake?

	A:  Force pushing piston down:  F=p*A+M*g, where:
							p=pressure of air
1.)	    Force pushing piston up: F=g*d*A*D		A=area of piston
							M=mass of piston
	    Solving for p, p=g*(d*D - M/A)		g=gravitational acc.
							d=depth of piston surf.
							D=density of water
2.)	B:  Same steps give p=g*(d*D + M/A) for the non-inverted cells.

    We'll need to know the volume of that air:

3.)              V' = V  * p /p'          where: V= volume of air in piston
		       o    o                    (V  is initial volume, V' is
						   o         current volume.)
    Since we're concerned only with the *difference* in bouyancy between
inverted and non-inverted cells at the same depth, we can, in the following
calculations, ignore the mass of the system the the volume of everything
but the air, since these things remain constant.

4.)   Delta bouyancy is = D * (V[a] - V[b]).  Making the appropriate
      substitutions, we find that delta bouyancy, dB ((:-)), is:
		    D*V *p      
	    dB =       o  o        2*M
		    --------    ------------
		     g * A      (d*D)^2 - (M/A)^2

     I *know* it can't work, but I just can't figure out why.  Help!
-- 
Jeff Sonntag
ihnp4!mhuxt!js2j
    "Pulled a muscle in my ear!"-Penfold

matt@oddjob.UUCP (Matt Crawford) (04/14/85)

Would one of you debaters please take a cup covered with a flexible
membrane down to the bottom of a swimming pool and turn it over a
few times?  Report the results to the rest of the newsgroup.
_____________________________________________________
Matt		University	crawford@anl-mcs.arpa
Crawford	of Chicago	ihnp4!oddjob!matt