Quote from: fletcher on August 20, 2012, 01:39:45 AM
You were reading my mind ;7) - I had a chance to have a closer read of your pdf this afternoon - I have some clarifications required if you don't mind.
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My feeling is that if an internal air pocket is used then it will transfer pressure - at the same time it will compress its volume somewhat & store PE to be returned later as KE [the spring analogy] once the riser is released & as the head increases - this would be a partial pneumatic element - but we know the energy input to compress a gas & increase its density & pressure is equal at best to its output potential...

It seems apparent that the current internal air packet can transfer pressure from the H2 height to the piston height & that the air packet pressure will be practically the same as the H2 head [this is the startup position of the cycle].

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Further observations & comments needing clarification :

1. the work done to raise the piston is by providing a force N.B. pressure is force / area therefore force = pressure times area - so the input force could be provide by a hydraulic piston, for example, which is well known.

2. the H2 to piston head depth determines the pressure head.

3. as the piston is forced upwards the hydrostatic pressures internally change - not only are there upward pressures acting on the all the riser parts [some of these can be combined], but also as the H2 levels rises in height proportional to the volumes transferred, & the hydrostatic pressure acting downwards on the piston increases [this is the hydrostatic paradox] - this increase in hydrostatic pressure acting downwards must be overcome & included in the work done input [I didn't recognize it in your net force formula's & diagrams].

Fletcher,

Thanks for taking a look.  I think your observations are along the same lines that I have been zeroing in on.  I think it will be easier to follow in my revision PDF that I should have updated later today.

I see the Piston transferring PE through H0 to the compressed air and through H1 to H2.  I think the net effect is that H0 and H1 heads are effectively stacked on top of each other as has been stated by others.  I think my latest formulas will represent the various internal heads and pressures in the system.

However, I have struggled with exactly which elements to include as net lift for the output.

I can now see the PSI of the water at the bottom of the Pod area as one.  The PSI of the water at the bottom of the Riser Wall area as another.  However, that gives a lift force that is less than a straight Archimedes un-convoluted riser.  The Riser air pocket area is not included, but somehow I think it needs to be.  If I include the raw airPSI in that area as part of the Riser lift, the total lift number becomes unbelievably large.

So my dilemma has been which potential areas of lift are to be included.  My current sim has every potential lift calculated and included on separarte lines so I can comment out each one to see the overall effect.  I have not found any combination that feels right yet.  Perhaps it is all in the stacking of the heads, and I will have to get into multiple internal layers before the ZED catches up and passes simple Archimedes.

You mentioned counteracting forces to the lift that you did not see me taking into consideration.  That was the basis of my original force calculations where I did not include forces that were relative to the Piston, but did include forces that were relative to the Tank.  This is the area that I think needs more conceptual work.

http://physics.kenyon.edu/EarlyApparatus/Fluids/Hydrostatic_Paradox/Hydrostatic_Paradox.html

It has nothing to do with overunity nor with surface instead of volume;
Buoyancy is still concerned by volume: one should take the right system into account:
1/ without insert
2/ with insert

in 1/ the glass with small volume of air will naturally have a weak buoyancy.
but in 2/ the glass and the insert are partially forming a global system: so you should consider the global volume of moved water: by air + partially by insert, since there is a direct contact inbetween, without water anymore.
Then it is said that global application point of buyancy is the centre of gravity of moved water. It's correct as a resultant force, but if you consider each differential bit of external surface, then you see that a big part of the surface submitted to the buoyancy force is ...the glass. So, the glass has partially "stolen" the buoyancy force of the insert.
There is no magics nor overunity !

Quote from: dgouttman on August 20, 2012, 11:13:57 AM
It has nothing to do with overunity nor with surface instead of volume;
Buoyancy is still concerned by volume: one should take the right system into account:
1/ without insert
2/ with insert

in 1/ the glass with small volume of air will naturally have a weak buoyancy.
but in 2/ the glass and the insert are partially forming a global system: so you should consider the global volume of moved water: by air + partially by insert, since there is a direct contact inbetween, without water anymore.
Then it is said that global application point of buyancy is the centre of gravity of moved water. It's correct as a resultant force, but if you consider each differential bit of external surface, then you see that a big part of the surface submitted to the buoyancy force is ...the glass. So, the glass has partially "stolen" the buoyancy force of the insert.
There is no magics nor overunity !

Hello DG,
No "magic" is right!
Thanks Wayne

First Picture: 3 U with water levels set to account for the air compression.

Second Picture: Calculated values for 3 U, with air compression calculated by SI of air channel.

Regards, Larry

Quote from: webby1 on August 20, 2012, 12:38:12 AM

I have intentions to build a better unit, more precise and all that BUT I am hoping that other builders out there, and you know there are more,  will be willing to come forward with results in line with what I have.

Hello Webby,
My team is currently planning the build for our desk top self running water pump.
I would love to see your teams creativity with the ZED technology!
How about a new challenge for you and your team..a $5K ZED  challenge.
Specifically designed by your team to ease measurements of input and output.
Two single five or six layer systems will do, or a combined system.
It does not need to be transparent.
Weight of the riser is not an issue works as a counterbalance - steel, aluminum may be used.
The risers may all be attached to one lid.
The system needs to be able to be measured and have access to each layer (tubing both air and water).

Part two: $10K ZED challenge ----5 layer, dual Z.E.D. water pump.
If your design team connects the system together for a self runner - I will double the 5K Challenge to 10K.
and each member of your team will receive a HER ownership certificate - valued at the current valuation and equal to the prize.

Five members max - pick them wisely.

Suggestions:
You do not have to beat our team - but time is important.
plan on 1/4 or 1/2 inch gaps between risers and ring walls - will work fine.
If you start with a pumping system and then size accordingly - a lot of time is saved.
The previous expectations apply - after a reasonable time of testing and reporting - you must send me the units, you are welcome to make an authorized extra set to keep for yourself.
Placard with your team members name and the name of the device Zydro Energy Device on all systems made.

What do you think? Up to it?

Wayne Travis