Gravity Mill - any comments to this idea?

[tbird]

hi Andi,

i have a few questions that i hope you have asked yourself and answered.

is only using half the cycle a benefit?  will this air supply down under be equal to or less energy intent than anything else?  will the system now depend on some other system to work?  if so, is it as or more reliable than the orignal design?

i'm sure there will be more, but before they can be asked, let's get past these.

tbird

[ooandioo]

Hi tbird.

I think, using only one part of the cycle is a benefit as the shuttle doesn't have to be that heavy in order to went down and push the water up. It only has to be a little heavier than the water itself displaces. On the way up theres also a benefit because of the weight of the shuttle.

The idea with the air supply is because of the practical use. Perhaps its easier to implement a baloon that can be filled with air under water. The needed energy will be the same i think.

Andi

[tbird]

hi again Andi,

let me see if i follow.

if you have a piston that consist of a disk (shape of the pipe, be it round or square) with a balloon attached that would deflate at the surface and allow the disk to turn vertical so they both can reach the bottom again in the fastest time (no water delivery), then once there, inflate the balloon to the size we talked about before (1 cubic ft.) and turn the disk horizontal again to form the seal.  since there is hardly any weight to be lifted, the balloon assy. would effectively increase the pressure 11-12 times the other system.  this would allow you to push the water higher or at least faster to the surface.  to get the same amout of water out, you would have to do this twice in the same time as the other system would take for 1 cycle. this might happen because of the speed coming down and if the water wasn't lifted any higher, faster going up.  in order to have the air pressure at the bottom to inflate the ballon, you would have a compressor running topside constantly to provide enough volume at the pressure needed.  this work would be done with the same (or less) 1 cubic ft of water per 2 cycles we set aside for 1 cycle of the other system.

does that sound about right?

tbird

[FreeEnergy]

i dont think air sealed shuttle is a good idea but i can be wrong. i think a sponge/foam shuttle is easier to push back down under water.

[hartiberlin]

here again comes back to my previous post.  leverage is the key.  the water being moved by the shuttle piston sideways has, basicly, no effect on the amount of pressure being applied.  the load comes from being higher than water level.  the pressure needed to change the volume of the shuttle piston depends on how deep and how much force you want the shuttle piston to go and supply, respectively.  in the 12" square example above, the figures i used were to keep it simple, but not accurate.  fresh water does have a specific weight.  1 cubic foot weighs 64.7 pounds (real close).  so 1/12th (1inch x 12inches x 12inches) of that is 5.39 pounds.  in the netural state (doesn't rise or sink) the shuttle piston would displace the same amount of water as it weights.  if we give our shuttle piston a 12" depth we would now have a 1 cubic ft piston weighing 64.7 pounds.  if we compress it 1 inch, it will have a negative buoyacy of 5.39 pounds which = (5.39lbs/12cubic.in.) .449psi.  not much pressure, but it will move water above water level.  now if we allow our piston to fall to a depth of 11.33 ft (1/3 of an atmosphere (14.7lbs psi)), we will have an increase of (14.7/3) 4.9psi.  if we allow our piston to return to its orginal size, it won't do it.  the pressure surrounding it will be greater than the .449psi we compressed it to in the beginning.  ok...what do we do?  we increase the pressure in the piston to 5psi to start with.  now we need to compress it to 5.449psi so it will sink with the afore mentioned pressure of .449psi.  now when we release the piston at 11.33 ft, it will expand to not only the orignal 1 cubic ft, but to a plus .1psi.  this will give it a positive buoyancy, so it will float back up.  the higher it travels, the more it will expand (if not contained), creating more pressure above.  this would be ok if we didn't expect continuous work to be done on this half of the cycle.  BUT, we are greedy.  so, let's go back to the start and increase the pressure inside the piston by the amount to produce the work we want.  anybody figure it out already?   5.349 would be right.  if we want the work to be exactly = in both directions, we wll have to limit our piston volume to 13 inches high.  once back to the top of the run, it will be time to recompress the shuttle piston.  now we know how much pressure is inside (at compressed state), we need to set up the leverage to recompress the shuttle piston.  now, how much weigh do we need to exert 5.349psi to 12 sq.in.?  12 x 5.349=64.188lbs.  that amounts to just under 1 cubic ft of water.  since we allowed our shuttle piston to go down 11.33ft and return that same 11.33ft, we should have 22.66 cubic ft of water available.  now the question becomes, how much do we take away from doing our main work?  if we allow for a little loss, we could use 1 cubic ft at 1 to 1 leverage and still leave 21.66 cubic ft to do work.  we must remember we didn't set this up to produce much pressure (.449psi).  as a result, the exit pipe would have to be fairly small to get to any favorable height.  the smaller the pipe, the slower the flow for a given pressure.

armed with this example, you can now think more clearly about the real project to build.

if i made any math mistakes, please rap me on the knuckles and let me know.

ooandioo & 2tiger, more to come.

Hi TBird,
many thanks.
What kind of formulars did you use to calculate this ?
I am pretty busy with other things right now and don?t have time
to look it up myself.
Maybe you can post as a summary all the basic formulas with
also the units in MSI system ( Meter,  Kg , Bar etc..)
so we can try to calculate specific systems ?

Maybe we can just use as the swimmer lifter unit an
inverted "U" shaped case, where we press via a hose at the bottom
of the water container air into it and if it has risen to the top,
we open a valve at the top of this inverted "U" case and all the air goes
out at the top and this device can dive again to the ground.

Maybe someone can calculate  how much energy you need to press e.g. 1 Liter of air
via an about 1 cm (0.39 inch) diameter hose e.g. 10 Meters deep unter water ?

Thanks.