As the boyuancy force must be greater than the volume
of the water above the balloon swimmer unit, also if you go deeper,
say 10 Meters below seawater level you also have to lift 10 Meters of
water with the balloon, so you need at least a balloon filled with air that is also
10 Meters high in the pipe ! Otherwise you can not lift 10 Meters of water volume
above it !
If the balloon would only be 9 Meters high inside the pipe, the
water above it would be too heavy to get lifted  and all stays in place !

Is this correct ?

hi FreeEnergy,

it's not opening for air passege but for water. when the one-way-valve-shuttle goes down the pipe it opens for water to come in, then it closes when it goes up to push the water to the upper half of the pipe.

no air being pumped in my idea

in order to sink, the air either has to go out or be compressed.  if you are not pumping air in....  tell me about the air.  how do you handle it?

andi,

Thanks for the good calculation - very good and exact job.
There is one thing that makes the whole thing running: You calculated the energy gain and loss while the swimmer is 2 meters under the water, so it has to go 2 meters upward and pump 1m water in a reservoir. Then your calculations are right then energy gained=energy lost when compressing. I think, never mind how deep the swimmer is and how much water he has to push up, the needed energy is always the same, its always the energy we need for pumping 1m water in the reservoir, the 1m above sealevel. As long as sealevel is not reached by the shuttle, it has the same buoyancy which needs to be greater than the force of 1m water above the sealevel. Correct me if I'm wrong or simply tell your criticism!

I think, the problem is - the deeper the shuttle goes, the more pressure we need to reinflate it...

did you miss my last post to stefan?  let me address some things here.

to begin with, you can not get water out at the total lift height.  if lift = 1 meter, the exit would have to be < 1 meter and the storage area (if used) lower than that.  space available for the storage device is the only limitation for the size.  the shallower you can make it, the higher it can be mounted (water exit height from system max).  if the shuttle is at 10meters and has expanded to displace a positive 1 meter, the 10 meters above will be displaced to the system just under 1 meter above sealevel.  this does make a difference.  you now have almost 10 time the weight to do work.  that's a lot!  also the designs you are talking about don't reduce the exit pipe size, you don't take advantage of leverage for recompression, and stefan wouldn't let it go all the way to the top (actually could go above surface).

this is why i didn't want to get the cart before the horse.  there are a lot of ways to make this work, there are a lot of ways to make it fail, but before we draw up plans we should see how all fits together.  clearly type 1 and type 2 use different parts, so combining the 2 would just produce a type 3 that probably doesn't work.  maybe we should stick to the orignal design until everyone understands how it works.  then if we want to modify, for the better, everybody will know why it will or won't work.

tbird

There is a difference if a cylindrical balloon is under water inside a pipe
or inside the water free floating at the same water deepth !

If you have it in a pipe  it can not lift up, if the water volume above it
will be more than its own balloon volume.

In the free floating case you can have it several hundred Meters under water
and it will go up to the top, so there is the difference, that there the
water can move to the side ontop if it, so the boyuancy force F is always there.
For instance, if you have a balloon 100 Meters under water and hold it via a thread
and let it go to the top you just can get the Energy: boyuancy force F  x distance 100 Meters.

In the case of the gravity mill the violation would be, that it would be able to shift
10 times the weight force of all the water volume above it, which is
not the case, otherwise the energy would be e.g. 10 x boyuancy force F x distance,
which is not the case.

stefan,

you have to stop.  you are telling people things that are not true.

let's start over with the basics.

if you have a fixed object that displaces 1 pound more of water than it weights, what will it do at 10 meters?  if you said rise, you were right.  how far will it rise?  if you said, all the way to the surface and expose 1 pound worth of its mass, you were right.  how deep can you take this object before it won't rise?  if you said, there is no such depth, you would be right. it has a fixed shape so it would still displace 1 pound more than it weights. any questions?

now, if you place this object in a tube, how high can the tube be above water before the object would not rise?  if you said, the distance that would contain 1 pound of water, then you would be right.  if we doubled the mass, but not weight, how far could the tube be above water level before the object would stop rising and pushing water out?  to answer this question, we have to have more info.  lets say our object is 1 cubic ft. (12x12x12 inches) and the tube is 10 meters long and fits the object snug.  for our work here, we'll say no water gets by in either direction.  if it is 1 pound less than it is displacing (fresh water), that makes it weigh 63.7 pounds.  now back to the question.  if you said, at least 12 inches plus enough for 1 pound, you would be right.  i doubt if many here really got this right.  if they did, it was probably for the wrong reason.  if the top of the tube were lowered so only 11 inches was above water, the object would rise.  not because it can push over 9 meters, 12 inch square of water weighting over 600 pounds, because it can not.  it can only push 65.7 pounds (1 cubic ft (64.7lbs) we increased the mass plus our orignal positive 1 pound) up.  so how can it rise?  the water coming in behind the object has the balance of the force due to the pressure exserted by an external colume of water, 12 inches square (the tube prevents any more than that to be applied), from the surface.  this is why if the object were removed, the water level in the pipe would be the same as outside the pipe.

it's getting late here now.  study the above and if you have any questions about what we just discussed, feel free to post here.  if all are clear, i will continue tomorrow with what i believe to be problem areas.

tbird

Holy shit, this thing really works !

I just tried it in my bathtube !
The plastic-air-cylinder I had in a bit larger diameter tube (pipe) could lift at least 3 times
its own volume of water above it.. !
My tube(pipe) was too short, so I could not lift more water !
But it really works !

I have to go out and get some longer tubes ! This is really amazing !
If you go very deep you can really "pump" so much water with it upwards !
Now I also saw my fault.
As the hydrostatic preessure force is always bigger at the lower surface of the air-cylinder,
there is always a positive pressure and thus force to lift up the whole stack-volumeof water
above the aircylinder ! Thus if you have it 100 Meters deep, it will lift the whole 100 Meter column
of water above the aircylinder !
This has really a very big overunity factor !
Stay tuned.

Regards, Stefan.

Okay, here is now a quick video I just recorded
of my experiment via my PDA with just one hand , so it is a bit shaky !

The attached AVI movie  has Microsoft MPEG4-V2 video codec and GSM6.10
audio codec at 320x240 res.
You can only download it, if you are logged into the forum.

Let me know, if you can replicate it simularly !
Many thanks to TBIRD to getting me onto the right track !

Regards, Stefan.

P.S: To overcome 100 Meters of water pressure you need to pump
the air with about 10 bar in the hose to get down to 100 Meters deepth.

Now the question is, how much energy do we need to push a volume of 10 Meter x 1 Meter diameter
volume 10 bar air down there 100 Meters ?

hi stefan,

it sure feels good having you back onboard.  extra well done for really trying to get the definative and not just say "it doesn't work"!!  with this attiude, we can do anything!

Thus if you have it 100 Meters deep, it will lift the whole 100 Meter column
of water above the aircylinder !

we should restudy John Herrings designs.  using his method, you can produce the same weight (lift the same amount of water) in only 50 meters. after all this topic is called "Gravity Mill" (his work).

one easy design criterium is, that you can lift the water up over the
level of the main bassin as high as the swimmer body has as height !

this may be true, but it is not limited to this.  the size of the exit pipe determines how high you can push it.  remember the shuttle piston will displace above water level the positive buoyancy.  if your exit pipe were twice as large, it would only raise it half as high.  you still get the same weight.  if the exit pipe is half the volume, you can push it twice as high.  as a result though, you change the time it takes to get all the water out.  bigger pipe, more volume.  smaller pipe, less volume.  being able to lift the water to the height you want is a nice advantage.

Now the question is, how much energy do we need to push a volume of 10 Meter x 1 Meter diameter
volume 10 bar air down there 100 Meters ?

this can probably be done. that's why i didn't boohoo andi's design before, but why go to 100meters?  if you stay with John Herrings designs, you can have a continues (pretty much) flow from a more reasonable depth. remember, the deeper you go, the more pressure you have to create.  plus if you drop something, who's going to pick it up at 100 meters?

andi

hartiberlin - thats the question I'm asking all the time. Will the amount of stored water be enough to alow recompression or reinflating the shuttle?

if you use the John Herring design,  this is quite easy.  he uses leveage (could be big time) to do this work.  from the examples before if you took 1meter (not all 10) of water 10 feet away and attached it to a lever, the other end would now feel 10 times the weight.  the only question is how much of your water supply do you want (or have to because of space resriction) to use to do this job.  having said that, you should be able to use this same leverage to work your compressor.  it may not end up looking like a conventional compressor, but there should be a way.

i sure there are still unanswered questions and i probably just created more, so let's get to it.

tbird