Gravity powered water generator

[bw100007]

lets try to simplify it.  If you have a teeter totter with a bucket at each side and 1 side is pre filled.  When that is allowed to drop it will create X amount of energy.  if you lift the water back to the top that will take = to or more energy than is gained from it dropping due to losses. 


The advantage of the screw drive to move the water uses the fact that water always wants to seek its own level.  Lets say we modify the screw drive i added in the video above a little so that is more in the shape of a spring. 1 full wrap at the left, a helical wrap to shift the water from left to right and a full circle wrap on the right.  Think of the wraps on the right and left as your buckets or reservoirs. as the screw is rotated the water exists the left bucket and enters the transfer helical wrap that moves the water from left to right and another turn moves it into the right bucket.  If you mount the whole pump on the teeter totter ( lever and fulcrum) Then the water moving from left to right could move the lever.  The energy input is not in lifting the water but in rotating the screw the water will seek its own level all the way through the screw and into the other end.  This does change with the greater the angle of the screw and the higher the lift.   The other advantage is the fact that the farther you move the water out on the lever there is a gain due to leverage. 1 bucket does the work of 3 at 3 ft when taken at the fulcrum and not the dropped end of the teeter totter.  Reverse the direction of the screw and the water moves back to the left and the full cycle is completed. No water is ever added or removed once it is pre-filled. 

[antijon]

Hey guys, I've got a question if someone can help me out.

Reading a part of the argument here, I was wondering about the lifting energy of something compared to the impact energy of it falling. A simple way to find the impact energy is height x weight.

So a gallon of water at roughly 8 lbs. dropped from 10 ft. should have an impact of 80 lbs.

Going from that I found a formula that stated it takes 10J to lift 1 kg 1 m.

So a fat guy at 100 kg climbing a 3 meter ladder in 5 seconds used 600w of energy.
10j x 100kg x 3m / 5 s = 600w (notice that total energy is 3000 joules).
I then found a splat calculator here https://www.angio.net/personal/climb/speed.html that shows the fat guy falling from 3 meters will have an impact of 2940 joules. Obviously lower than the energy needed to climb the ladder, but whatever. My problem is converting that energy into watts. Certainly, the total impact time is less than one second. So if impact time is .5 secs total watts at impact would be 6kw. Does that sound right?
Physically, I know if I put rubber, or a spring, on the ground under the fat guy, I'm increasing the impact time and lowering the impact energy. So do I really get more watts out than in?

[bw100007]

This is really crude but shows the left wrap the transfer wrap and the right wrap. Sorry about the file size i am a newbie.  Any way i can correct this?

[Brutus]

webby1;
  in a closed system the air pressure change will be the same change value as the change in the water levels. 

Check out scan 0023.pdf  and let me know if you think the way I drew it will work.   This is my question.  Will the water once released from the top of tank B through the outlet and then falling into tank A which is connected with the same water source, ( It is all one tank actually but with a riser on the right side), Go back into tank B from the air pulling it as it tries to maintain its air pressure.

My point being,  If the water is caused to pull itself back into tank B through air pressure stabilization within tank B, then I can alleviate the need to pump water against gravity through mechanical/electrical means to bring it back to the water pump again.

Or maybe some other medium besides air which, until now, I had not considered might be used to keep a pressure point applied to do the same thing?   Like an oil?  No Gap, so the drawing power would be constant with no variance.  Or, better yet, no water or air gap at all just filled to the top with water. 

[bw100007]

This can work, but to rotate the tube will require a torque to be applied and this is where the manometric pump part comes into play.

IF you prime the left wrap almost full and when looking from the right side turn the cylinder clockwise 1 turn the fluid will move into the transfer helix.  keep rotating it clockwise an additional 1 turn and the fluid will move into the right wrap/bucket. The cylinder is rotated by a secondary source.  If it is 9 ft in dia then moving a ton of weight 6 ft from left to right with the fulcrum centered would give a ton of weight at 3 ft of leverage.  We should not have to lift it very high.


How would the manometric pump rotate the cylinder?