Quote from: neptune on August 17, 2012, 05:48:35 AM
Referring to post 1376 on page 92 by LarryC. People have expressed doubts as to whether we can get the same head pressure from a number of layers in the diagram on the right as we would from a single tall column of water, and there has been talk of using submersible pressure gauges etc. Here is a very simple experiment to prove the point. Imagine we have the apparatus set up as in the right hand diagram. Now with the water levels as shown we extract the piston, and throw it away. The tube in which the piston fitted is now extended upwards to a hight of several feet. Now instead of depressing a piston with a weight, we pour water from a jug into the extended piston tube. We keep pouring until the water levels in the various U tubes move to the same position as happened when the piston was pushed down by the weight. At this stage, we find that the hight of the water column in the piston tube, is equal to the sum of the heights of the vertical columns in the U tubes. To me, this proves that the pressure of the column in the piston tube is equal to the pressure of the columns in the U tubes.

And if it is demonstrated to be true, which I don't doubt, in this manner, what _then_ does it prove? Buoyancy does not depend on the pressure of the liquid, only on the mass displaced by the displacing object. In the system with a locked riser, increased "head pressure" causes the water rise and to wet more of the riser, effectively displacing more liquid. The pressure isn't affecting the buoyancy directly, and if the riser is free to float, it doesn't affect its buoyancy at all, as long as the riser's density (or equivalently the wetted volume) doesn't change.

It's an elegant demonstration, though, nothing wrong with that. If I had another tube I'd put it together in a minute. But I can't find any more of those polycarbonate light bulb protectors, they have been replaced by much weaker and flimsy tubes. Maybe one of those will work for this particular purpose but I wouldn't trust them for external pressure loads.

see3d said,

QuoteOn a ZED in the down initialized state, add a 1 pound output load weight and a 1 pound input weight.  After the output load weight is just raised to the top stop, remove it and the input load weight.  The ZED will return to the initial down state.  If the input weight has dropped less than the output load weight has risen, then the cycle can repeat forever.

Really? If the input weight (that you have removed) has dropped less than the output load (that you have removed) has risen ... then the system is NOT reset to the beginning of a cycle with all levels and pressures the same as when started. Do you think that by bleeding off some pressure, doing work elsewhere, it will then be reset to the original state?

Quote from: neptune on August 17, 2012, 09:01:28 AM

Yes, you have too much hydraulic fluid. So the excess fluid is then used to drive a hydraulic motor, which , in turn drives an alternator, feeding a load. That is where the "free energy" comes from.

Well, I've been told in no more clearer terms that my conception of operation like this is wrong. But... just in case it's not, you still aren't dealing with what becomes of the fluid... sorry, hydraulic oil.... after it goes through the motor. How and where does it get back into the Zeds, at what pressure, how is this pressure obtained etc etc.
And I believe that big thing with fins on it, being driven by the hydraulic pump/motor, is actually a generator, not an alternator, built for wind power systems. I don't know its exact specs... but I sure would like to.

ETA: Where, in the diagrams shown, does the extraction of 30 volume units happen, and the injection of 15 back in while diverting 15 thru the motor? After it goes through the motor it winds up in the reservoir. How does it then get back into the system?

Quote from: LarryC on August 17, 2012, 10:30:15 AM
Attached is the third picture from my previous post with additional Psi pointer information.

This will be much harder to understand where the force and Psi are coming from than the 3 U water levels. If anybody has questions and they don't want to post here, please PM me.

Regards, Larry

Why are you showing this confusing diagram, when there exists a three layer system that is clearly overunity by itself?

Quote[cite]MrWayne[/cite]I will repeat what I have said - our current single unit is barely better than a hydraulic cylinder - you can measure the input and output  - we do not use weight - but volume and pressure. Yes - our three layer system is clearly overunity by itself.

http://www.overunity.com/10596/hydro-differential-pressure-exchange-over-unity-system/msg332026/#msg332026

Why not diagram _that_ system, that is clearly overunity by itself, and let us know how the clear overunity was determined?

I've asked for this information once or twice before, in several ways. Is it an unreasonable request? I don't find it so at all.
Certainly, if I had made such a statement, I'd be prepared to back it up with real data, and I think everybody that knows me would agree that that is what I'd do.  If you can't diagram it for some reason, then just tell me please how the clear overunity performance was determined for that three-layer (single?) system.

And while you are at it, I am also curious about the single system that is better than an hydraulic cylinder (barely).

Quote from: TinselKoala on August 17, 2012, 11:04:26 AM
Well, I've been told in no more clearer terms that my conception of operation like this is wrong. But... just in case it's not, you still aren't dealing with what becomes of the fluid... sorry, hydraulic oil.... after it goes through the motor. How and where does it get back into the Zeds, at what pressure, how is this pressure obtained etc etc.
And I believe that big thing with fins on it, being driven by the hydraulic pump/motor, is actually a generator, not an alternator, built for wind power systems. I don't know its exact specs... but I sure would like to.

ETA: Where, in the diagrams shown, does the extraction of 30 volume units happen, and the injection of 15 back in while diverting 15 thru the motor? After it goes through the motor it winds up in the reservoir. How does it then get back into the system?

TK,  in the picture below all the red hydraulic lines are filled with high pressure hydraulic oil.  All the blue hydraulic lines are filled with unpressurized hydraulic fluid that is also stored in a reservoir that is not represented pictorially.  You can see where each blue hydraulic line from the cylinder on top of each ZED and after the motor ends is the labeled "Res" for "reservoir.

To each of the three sections of the multichamber hydraulic cylinders above the ZEDs are two lines, one the high pressure red and the other the low pressure blue.  On each of these routes are also one way check valves.  So when a ZED is going up it is pressurizing the Hydraulic oil inside these cylinders.  That high pressure hydraulic oil can only flow through the red lines to the accumulator(s)(s).  The one way check valves make this the only open passage.  Then when the ZED is going down hydraulic fluid would be pulled back into those same hydraulic cylinder chambers above the ZED from the reservoir.  Again, the one way check valves make this the only open passage.

M.