stefan,

i'll try another angle.  if the pipe in your example were 2 meters tall (above water level) and still 1 meter in diameter, at depth, your ballon would stop going up when the displacement weight, 785,4kg, was above water level.  we figured this to be 1 meter, so with the pipe 1 meter taller, all the water would be still in the pipe but 1 meter above water level, right?  now in this state, one way we can make the balloon rise is to give it more volume, which displaces more weight (without increasing the balloon weight), thus giving the balloon the ablity to lift more WEIGHT, right?  or we could do the oppisite.  we could reduce the weight above.  either way, your balloon rising depends on how much WEIGHT is above water level.  as long as we keep this weight less than the lift of your balloon, we can take it as high as we want.  as long as the balloon has more LIFT than the column of water above water level has WEIGHT, the balloon will rise.  it doesn't matter what shape this weight is as long as it stays in column.  785.4kg is 785.4kg.  we can say the rest of the water, in tube and outside, equalize each other.

i hope you can see now.

tbird

How can I resolve( rewrite) the formula:
W=P1*V1*ln(V2/V1)

to:
V1= ....

Its Long time I have used mathematics..

Okay, I have now the solution.
The missing link was the gas law:

p1 x V1 = p2 x V2

So now I can calculate V1 !

V1= 2 bar x 785,4 Liter / 1 bar= 1570,8 Liters

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 ?

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.

Well, I had still another look at it how to do it,
if we start with equilibrium at
9 Meters under sealevel= 10 Meters under the whole water column
at 2 Bar and at 785,4 Liters,
where we have equilibrium. In this position the swimmer body will stay there and
don?t move and we have a water column of 1 Meter above the seawaterlevel,
which will not yet run out of the pipe.

Then we go  from 785,4 Liters to about 800 Liters
of volume in the swimmer body  by pumping a bit of additional air into the swimmer body,
then we can rise all the fluid above it and only
need an energy of 0,81 Watthours for the pumping this 14,6 Liters air into the swimmer body at 2 bar !

In contrast we will then get 21 Watthours of potential water energy
from all the water being pumped up
to 1 Meter up over the seawaterlevel !

So you see, how big the overunity factor is !

It just works !
Enjoy !
Regards, Stefan.

hi stefan,

it's amazing what a good nights rest will do for you.

here is how you prove to yourself a smaller diameter will make the water go higher.  first let's agree on a given.  if we have positive buoyance in our shuttle (balloon, piston, whatever the device is that moves the water up), we have created pressure on the water above, right?  if the water can't go around our shuttle, it has to go up, right?  now if we don't push the pipe too far out of the water, we will see flow above the waters surface (we proved this with test).  we all agree on this now, right?  here comes the proof.  take your garden hose with the water running and point it straight up.  this looks kinda like what our pipe would sticking out of the water, right?  now cover the end slowly starting from the side going to the other side (use your thumb).  did you beat me to the punch line?  without changing the water pressure, what does the water coming out do?  it's been my experience that the water goess farther up.  remove your thumb, the water height goes down.

now what have we proved?  the smaller the opening (exit diameter), the higher the water will travel.  do i need to say more?

tbird 

hi all,

to get the most (efficiency) from our machine, we should use both the up and down strokes.  after all there is energy in the fall of the shuttle too.  the reason we may consider using only the up stroke, would be to make building and assemble easier (simpler).  there is only so much water in the pipe we can move per energy used to compress the air.  if we use the down stroke too, the pipe will fill twice.  that means twice the water per cycle, per compression of the air.  if the sigle stroke operation can be done in at least half the time and consume half the recompression energy, then we should continue to explore that design.  this maybe possible, but before we go there, we need to get the basics of the machine (elsa) well understood.  let's not throw away the 30 years of work and research John Herring has already done.

the depth we use for elsa may or may not result in more or less efficiency.  if the formula stefan used last night is correct, work will be proportioned to the pressure needed. so if you only use 1/3 the depth, you only need 1/3 the pressure, but you only have 1/3 the water to do the work (if i am applying the formula right).  if everything is relavent, a smaller depth would allow easier building of a self contained unit.  this would eliminate the need for an ocean, lake, pond, or the like.  just so many square feet (meters) in the backyard (garden).

seeing how important the needed energy to compress the shuttle is, maybe we should concentrate on this part of the machine.  John Herring devotes several drawings to this job.  here again, i think we should take advantage of his experience.

farther down the road we will get to the actual use of the energy we have brought up to and above ground level.  how this is done is only limited to our imagination.

i'll stop here for now and let you do some studying.

if i have made a mistake in the early going, then the rest is probably wrong too.  please let me know.

tbird