Hydro Differential pressure exchange over unity system.

[mondrasek]

Corrections to my previous diagrams and calcs when removing 5 in^3 of water from the full ZED model.  Everything is still buoyant, so I may try and see what happens if a few more in^3 are removed.

The air expansion due to lessened pressure is still not being accounted for accurately.

M.

[LarryC]

@LarryC, after thinking about it some more I think what I had done originally is fine, since all my Bouyany Force calculation were based off of the head pressure differentials, so the internal air pressures are already accounted for.  So now I'm having trouble understanding the spread sheet you have.  Could you draw up a diagram to explain where the dimensions and especially the "Next layer" are refering?

This is faster, but let me know if you need more.

The Next layer at 10A is the top outside surface of Riser 3. The Next layer at 13A is the top outside surface of Riser 4.

The Height of all water channel are generalized at 100 inches, same as the pod. The miner clearance between risers and pod are not considered as they would not make a significant difference in the force calculation.

Riser 4 is 2 inches in diameter larger than the pod. Riser 3 is .9 of 2, and so on.

Regards, Larry

[neptune]

Hi M,

Nice work. But think Wayne, he's the genius, I'm just hanging on for the ride and loving it.

The pod is not effected as it is Archimedes's, so the volume or pressure differentials can be used for the calculation.

For those that don't have a spreadsheet and want to understand the calculations in the spreedsheet file, OpenOffice can be downloaded for free and it has most of the Microsoft Office applications. 

The total force is without the system losses.  But think of it this way to help understand the great potential, allow that force to lift the external load 1 MM, then the water inlet refills that 1 MM and it is only sees the water head.

Regards, Larry

That last paragraph says it all. I am not 100% sure what system losses are involved. In a working machine there would be lots of losses, but if we are just talking about a ZED on its own those losses are going to be small. We need a pump to inject water into a ZED, which might consist of say a large syringe pushed down by a weight. That would involve some friction in the syringe. Or we could use a vertical pipe full of water with a valve at the bottom and a suitable head of water . Less losses . The losses within the Zed itself would be very small.I think what I am trying to say is that this device, on paper at least looks very promising indeed .
     I make no claims to be a mathematician, but I understand basic maths. I am just Gobsmacked by the amount of info that you guys have derived from next to nothing. One can not help but be impressed by the large forces in a model so small. Well done lads .


It seems to me that there are two main problems in building an efficient model of a ZED.
  1. Knowing what the water levels are in the various parts of the ZED , especially during set up.
  2. The smaller the model, the closer tolerances it will be necessary to work to. This will be mitigated by the fact that initially we are not trying to build the most efficient ZED ever. One would probably need to compromise between tight clearances and ease of building . Since we are not talking of a tiny amount of OU, there would seem to be some room for compromise .

[telecom]

That last paragraph says it all. I am not 100% sure what system losses are involved. In a working machine there would be lots of losses, but if we are just talking about a ZED on its own those losses are going to be small. We need a pump to inject water into a ZED, which might consist of say a large syringe pushed down by a weight. That would involve some friction in the syringe. Or we could use a vertical pipe full of water with a valve at the bottom and a suitable head of water . Less losses . The losses within the Zed itself would be very small.I think what I am trying to say is that this device, on paper at least looks very promising indeed .
     I make no claims to be a mathematician, but I understand basic maths. I am just Gobsmacked by the amount of info that you guys have derived from next to nothing. One can not help but be impressed by the large forces in a model so small. Well done lads .


It seems to me that there are two main problems in building an efficient model of a ZED.
  1. Knowing what the water levels are in the various parts of the ZED , especially during set up.
  2. The smaller the model, the closer tolerances it will be necessary to work to. This will be mitigated by the fact that initially we are not trying to build the most efficient ZED ever. One would probably need to compromise between tight clearances and ease of building . Since we are not talking of a tiny amount of OU, there would seem to be some room for compromise .

Hi Neptune,
can you or somebody else describe the operation of ZED, the cycle itself. I can easily now understand it for a variable geometry piston machine, but not for  this particular machine.
Alex

[neptune]

@telecom. I have yet to study the variable geometry piston. There are still parts of the actual cycle that I am struggling to grasp. I suggest you read reply 525 on page 35, as this is one of the best aids to understanding , as it takes us step by step from the basic Travis Effect to the working of a ZED. Having understood this, then the original sequence of diagrams by mrwayne will start to make more sense.
      The most telling thing about this device is that the resistance to the water injected is only the head of water it has to resist. And when that water is sucked out again as the load falls, it still has half of its energy recoverable. Mathematically, it seems to stack up perfectly. No one can say specifically where the excess energy is coming from . At this stage , that seems to me unimportant.
       To sum up, I have spent ages studying this, and for me there is no Eureka moment , just continuing small steps in understanding .