Gravity Mill - any comments to this idea?

[tbird]

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

[hartiberlin]

think about it this way.  what stops the piston (ballon, whatever) from rising? either it weighs more than it is displacing or in our case in the tube, a larger weight above water level than the piston is displacing.  right?  so it doesn't matter if it is all placed in the first meter or divided in 2 and placed on top of each other.  both shapes weight the same, but 1 is only half as wide as the other.  so as long as the volume of water in the tube above water level weighs less than the piston is displacing, the piston will rise.  your test was a little awkward for you   but if you could actually put a smaller tube on top at water level, the water will come out higher.  like getting a flat tire on your car, the smaller the hole, the longer it takes to go flat, but it will go flat.

Well could be, but we have to watch the hydrostatic paradoxon
with it, as only the height plays a role what water pressure weight
is ontop the sewaterlevel.
Can you calculate your setup completely ?

[hartiberlin]

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

to:
V1= ....

Its Long time I have used mathematics..

[tbird]

hi stefan,

Well could be, but we have to watch the hydrostatic paradoxon
with it, as only the height plays a role what water pressure weight
is ontop the sewaterlevel.

this is not true.  hydrostatic paradoxon, i think this is your label for why the balloon doesn't have to raise the total weight of the column.  i can't remember hearing this term before, but did notice it in some of the reading material.  what i know is the water at and below water level does is not involved in lifting the water above water line.  the closest thing you can say is that it nutralizes the water above the balloon, but only to the top of the water level.  the only thing that pushes the water above water level in a column is the balloon displacing that amount of water.  even if the balloon didn't have positive buoyancy, it would displace it's volume above water level if contained in a column with bottom closed.

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

to:
V1= ....

i'm afraid i don't understand this formula.  is there a drawing it refers to at the link you sent?  if you can explain what each stands for, it might help.

[hartiberlin]

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