Hydro Differential pressure exchange over unity system.

[neptune]

@mrwayne. Thanks for the detailed explanations. I am a bit unsure about your explanation about where the energy comes from. As I understand it , you are saying that we have an excess of energy because we have saved energy at the input, by using exhaust from the descending ZED. Sort of saying that there are two ways to increase your bank balance, one is by earning more, and one is by spending less.
       However, I believe that you said it is possible to build an OU machine using a single ZED, and although it is less efficient than a two ZED machine, it is still OU. That being the case, there is something else happening here. I suppose that conventional physics would say that it must in some way extract energy from outside itself. Just what form that energy takes, is hard to say. It could be gravity, or something else. But this is the question that sceptics will ask. It is possible that at this stage, this is a question to which you do not know the answer. There is no shame in that. Ultimately, all that matters is that it can be shown to work.

[mrwayne]

Hello Neptune - it is time to test so i have to go,
But let me say - a single Zed system can demonstrate a gain over the operating cost - but it is not of much point -

Our Goal is to provide abundant excess energy - the two Zed system is better for that.
The "cannot beat" o/u comes from the simple travis effect as shown in Tom's Videos.

The whole operation is dependent on differential exchanges - namely caused by gravity.

The Physics of our system are simple - not magical as Seamus arrogantly alluded to in the naming of his drawing.

The cost of the entire operation is reduced by reusing the energy stored in the head - after the buoyancy has been captured.

The cost is less than the capture - before the resuse of the head (exhaust) because of the layering system.

You can check that with one of the calculatorslike Larry provided - compare to a common hydraulic cylinder - surface area x pressure = lift

we can lift more than a Hydraulic cylinder - which means if we were using a hydraulic cylinder to create the lift pressure and volume - we can capture more than used.

Remember - we use a 1"Si cylinder (input) to operate a 5"Si cylinder (Output) at the same pressure.

if you build a chart showing lift from 2 pounds to 9 pounds you will see a graph that climbs at a higher rate than a cylinder of matching surface pressure -

From an old note: 707 Si x 10psi would lift  7070 pounds - we lift 7700 with 9psi in our current three layer system.

BUT -on the low side 0-5 psi we are worse than a  hydraulic cylinder 5 psi only lifts 2500 pounds.

We use that to our advantage - it only takes 2500 pounds to create 5 psi hydro, exhaust

psi - the weight puts us in the operating range of being "above the hydraulic cylinder" at no loss.

Do the math - I have to go - crew standing by..

No magic - simple mechanical and operational uniqueness - mind boggling simple.....very good for this our energy needs.

Wayne

[see3d]

@Neptune.  Yes, if a single ZED shows OU, then, even though not as efficient as a dual, the loop can be closed to self run (by definition) -- as long as the loop closing mechanism does not lose all the excess in friction.  It has to be a relatively simple and pretty much friction free feedback system.  The dual setup has a nice straight forward symmetry for practical applications.  If an educational model is created as a single ZED, there is no particular reason that two could not be connected together for a dual ZED educational model.

[Cisco]

Hi Wayne,

You noted, "Free flow can be easily calculated (for those following) as the average between the top of stroke head and the bottom of stroke head (In my examples offered on line 6.7 psi hydro)" 

I presume the examples you refer to are shown in your post #373, "ZED through complete cycle-1"?
In that illustration, you show the pressure at the bottom of the head stroke as 4.8 psi, and at the top, 8.0 psi. And just as you said above, the post equalization free flow pressures are shown there as 6.7 psi.

For the sake of clarification, isn't the average between 4.8 and 8.0 not 6.7 but 6.4--or have I misunderstood/miscalculated?

[mrwayne]

Hello Cisco,
Thanks for the clarity - most of the time when I am answering questions - I answer directly related to the operating pressures of that moment in time - it does not explain in detail all of the lessons learned and changes to the system - just the pressure.
6.7 was the free flow pressure when Mark was here - it was increased from the 6.4 by adding weight to the system.
So the range if ideal usage changes - which effects the value of the free flow - in pressure volume and speed of delivery.
The formula still works to each specific range of usage.
If you lowering pressure is 4.0 and your overcoming resistance pressure is 8.0 ----- 8 4 / 2 = 6  predictable Free flow pressure
Thank you for the correction.

Wayne