Hydro Differential pressure exchange over unity system.

[TinselKoala]

@LarryC:
In the left-hand drawing in your image..... is the tube on the right open at the top? If it's not.... how is it possible to achieve the final state with all the water in the right tube and the weight fully down? Especially if you are using mercury and water, instead of water and air?

If it is open at the top... then what about the other two systems? Are they open at the top, of the last tube, as well?

You say each system makes a "7 foot water head". That, to me, means that the final pressure, measured at the bottom of the weight where it contacts the water it's pressing on, is the same in all three cases. Am I interpreting this correctly?

[mrwayne]

What are these items? Do they appear on any of the patent drawings or the other illustrations we have been given?

Simple - the Initialization system.
Not in the patent - for the Validation team only.
I will recap since this has been covered and gets tossed up again:
When we moved the system inside - from Mark's visit we spent $26,000 adding a set up system so that RIchard could program the initializations system.
We have - Separate from the Zed - except the program which controls the separate operation -
Water pump attached to every layer in the system
An Air compressor attached to every layer
and a Hydraulic pump attached to the accumulator - you will see that in the hydraulic drawing.
You see, when Mark was here I had to disassemble, drain and vent every component.
Then I used a standby generator, waterhose, and Log splitter pump to set the system up to the precharge condition - covered well in the patent.
Mark was a patient man, it took all day and into the night to perform this task.
p.s. So I had the team design an automation system so that the validation team could use it to measure the set up energy.
So now - we push a button and the system drains itself - vents itself - measures and displays every aspect within the system and then steps you thru a set up procedure to refill everything to the beginning run state.
Glad you asked - I meant to restate this again when a liar claimed that we pumped air and water and hydraulics in the system to fool everyone.
TK- you said you put me in the idiot file -
I respected you to answer ...
It was a lot of work and changes to install air and water pressure lines to every layer, chamber - there is a  pressure transducer on every line - we do use those during run to monitor the process.
I think the Engineers have the Initialization mapped - here is mine.

[LarryC]

@LarryC:
In the left-hand drawing in your image..... is the tube on the right open at the top? If it's not.... how is it possible to achieve the final state with all the water in the right tube and the weight fully down? Especially if you are using mercury and water, instead of water and air?

If it is open at the top... then what about the other two systems? Are they open at the top, of the last tube, as well?

You say each system makes a "7 foot water head". That, to me, means that the final pressure, measured at the bottom of the weight where it contacts the water it's pressing on, is the same in all three cases. Am I interpreting this correctly?

The last tube on the right of each system is open, and you're right a water mercury would require a different design.

You say each system makes a "7 foot water head". That, to me, means that the final pressure, measured at the bottom of the weight where it contacts the water it's pressing on, is the same in all three cases. Am I interpreting this correctly?
Yes.

Regards, Larry

[TinselKoala]

All right, since you did me the courtesy of answering about the air compressor and the pump --- I'll ask you politely and directly one more time.

How is the ratio of input work or energy, to the output work or energy, determined for the three layer system that is clearly overunity itself?

Please tell me what the total input energy is to initiate and sustain a cycle, and what the total output energy is for a complete cycle.

What _numbers_ do you compare, to determine the clearly overunity performance of this system itself?

If you again say you put in 15 units of fluid and get 30 units back... I say again that it sounds like you are creating fluid from nothing, or that the cycle is not complete at this point, or that pretty soon the system will be empty and all the fluid will be somewhere else.

[TinselKoala]

The last tube on the right of each system is open, and you're right a water mercury would require a different design.

You say each system makes a "7 foot water head". That, to me, means that the final pressure, measured at the bottom of the weight where it contacts the water it's pressing on, is the same in all three cases. Am I interpreting this correctly?
Yes.

Regards, Larry

OK, thanks. Now I have something I can test.

ETA: Before I do, though, let me ask another question.

If I put an outlet valve at the crossconnection between the first (piston) tube and the second tube, and pipe that into another, identical tube.... then open this valve when the piston is in the final fully depressed state, the water will run from the full tube through this valve into the new, empty tube. Right?
How high will the water rise in this new tube?

In the diagram on the left it would rise up until the two columns of water were at equal heights, I think.... but I can't quite picture what would happen in the third, righthand case.