vince@fluke.UUCP (Craig V. Johnson) (04/01/85)
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! /boris@mit-athena.UUCP (Boris N Goldowsky) (04/09/85)
I think that this is why the pm machine won't work.
The suggestion was to put jars on a conveyor belt in water such that
on the "up" side a weight in each one stretched a membrane thus
increasing the volume of the jar and increasing the buoyancy,
on the "down" side the weight would fall on the other (rigid)
side of the jar and the buoyancy would decrease.
The way to attack these problems is by considering energy. Nothing will
by itself go to a higher energy state.
Thus you can think of buoyancy as a force tending to lower the energy state
of something immersed in a fluid. The pressure at the bottom of the tank
of water is greater than the pressure at the top. The higher pressure
on the bottom of the jar is what pushes the jar upwards, toward a position
where it would have a lower potential energy.
We know that at least as much energy must be lost in one complete
circuit around the belt as is gained, but it is not obvious where. Energy
is gained moving the enlarged jars up and also letting them fall back down.
We might guess then that the same amount is lost in the *transition* between
the two states. It takes a good deal of work (energy) to make the jars
bigger... because you have to displace that volume of water up to the
surface against gravity. This work is actually done by gravity, when the
weights inside the jars fall and push down the membranes. But then
the machine has to do that same amount of work pulling the weights back
up.
Thus the amount of energy gained is equal to the amount of energy lost:
they are both the difference in energy between the extra volume of water
at the top vs. the bottom of the belt:
( E=Vgh=Volume*density of water*accel. of gravity*height)
and you lose a little to friction so the device won't work at all.
"This isn't speculation, this is predicted fact!"
--
Boris Goldowsky decvax!mit-athena!boris
boris@mit-athena.arpa
Goldowsky@mit-multics.{arpa,bitnet,mailnet}jlup@cci-bdc.UUCP (jrl devlt) (04/11/85)
> 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