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Overunity Machines Forum



Gravity Mill - any comments to this idea?

Started by ooandioo, November 03, 2005, 06:13:20 AM

Previous topic - Next topic

0 Members and 2 Guests are viewing this topic.

tbird

stefan,

that's great!  what does it mean in plain english?
It's better to be thought a fool than to open your mouth and prove it!

hartiberlin

Okay, now with this information:

P1= 1 bar
V1= 1570,8 Liter
P2 = 2 bar
V2= 785,4 Liter we can now calculate the work-energy to compress
the 1570,8 Liter from 1 bar at the top to 2 bar at 10 meters deep, which will
then have a volume of 785,4 liter :
With
W=P1*V1*ln(V2/V1)= P2*V2*ln(V2/V1)
we get

W= -30,24 Watthours.

So we need more energy to compress the air, then we got from
the reservoir 1 meter above seawaterlevel...

But now is the question, if we really need 785,4 Liter of air inside
the cylinder in 10 meters deepth as TBird also suggests ?
Stefan Hartmann, Moderator of the overunity.com forum

tbird

stefan,

i think i follow somewhat.  maybe i'm just tired.  better go to bed.  one last thought.  how much would you need if it had a 2 foot lever?  5 foot lever?  10 foot lever? ???

good night all.

tbird
It's better to be thought a fool than to open your mouth and prove it!

hartiberlin

Hmm, in my setup there the air-cylinder under water must be at least 1 Meter high
at 9 Meters deepth, otherwise it will not be able to lift the 1 Meter water column
ontop of the water surface, which weight 785,4 Kg or Liter.

Also if you make this diameter of this water column much smaller and have thus
less water in it, you would still need the same air volume at 9 meters deepth,
cause we have the hydrostatic paradoxon, so that 1/2 of the water diameter column,
so 50 cm in diameter , will
weight the same as if we use the full 1 Meter diameter water column ontop the seawater level.

I just tried to glue a smaller pipe onto the bigger pipe I used yeasterday and it seems to
confirm this, that the water column ontop the seawaterlevel is only this high
as the airvolume height inside the swimmer body will be.
Stefan Hartmann, Moderator of the overunity.com forum

hartiberlin

Well, I guess I found a solution to the problem.
I assumed, that I must pump ALL the air into the cylindrical plastic-balloon
container down at 10 Meters.
This must not be !

If we define a different START condition, we need only to add some
air at 9 Meters deep to the swimmer body there being in equilibrium with the water weight column
ontop the water surface,
so the swimmer can rise and move the water again
and when have reached seawaterlevel remove some air, so the swimmer can sink again.

Then we need way less pump energy !
I will calculate this also soon, but now I have to go to bed.
Stay tuned, this is going to be exciting !
Regards, Stefan.
Stefan Hartmann, Moderator of the overunity.com forum