Mathematical Analysis of an Ideal ZED

[MarkE]

Marsing, LarryC published it in message 182.

[LarryC]

Here is LarryC's latest spreadsheet with decomposition and reduction.  Larry is free to defend the physical basis he thinks justifies his model.

MarkE,


Impressive. I'm not anal enough to review, unless someone brings up an issue. If anyone would like to check, it is easy to see all the formulas at once in excel by pressing Ctrl`. The accent is next to the 1 key.

[TinselKoala]

How about this for an issue: FLOW ASSIST.

That sure sounds to me like an injection of energy from outside the system of the two zeds. So I must be wrong, because we have been told by Travis that the system has no input, no exhaust, just constant clean energy output.

So can someone please tell me what _exactly_ is meant by "FLOW ASSIST" in employee LarryC's scheme?

[mondrasek]

So can someone please tell me what _exactly_ is meant by "FLOW ASSIST" ... ?

Some of the Pressure left in a first ZED at the end of its "power stroke" is allowed to "Free Flow" to a second ZED that is at the bottom of its stroke and therefore at a lower Pressure.  This equalizes the Pressure between the two ZEDs and requires no additional Energy.  But it does not bring the second ZED up to the full Pressure necessary to perform its "power stroke."  So additional Energy is required and is called the "Flow Assist."  The Energy for the "Flow Assist" is a recycled portion of the excess energy that is supposed to be harvested during each "power stroke" where the rising ZED is pumping fluid under pressure into a hydraulic accumulator.  The harvesting portion of the system is not being shown in the spreadsheets currently being analyzed, AFAIK.

[TinselKoala]

Some of the Pressure left in a first ZED at the end of its "power stroke" is allowed to "Free Flow" to a second ZED that is at the bottom of its stroke and therefore at a lower Pressure.  This equalizes the Pressure between the two ZEDs and requires no additional Energy.  But it does not bring the second ZED up to the full Pressure necessary to perform its "power stroke."  So additional Energy is required and is called the "Flow Assist."  The Energy for the "Flow Assist" is a recycled portion of the excess energy that is supposed to be harvested during each "power stroke" where the rising ZED is pumping fluid under pressure into a hydraulic accumulator.  The harvesting portion of the system is not being shown in the spreadsheets currently being analyzed, AFAIK.

Uh-huh. So what part of pumping fluid into a hydraulic accumulator and getting it back out is "overunity"? You get the same work out from an accumulator as you put into it, minus losses. Did you check out the PDF file I linked to up above (or maybe in the other thread, this is pretty silly having this discussion in two threads.) I think, based on that PDF, that hydraulic systems are pretty darn well understood, and in fact in that PDF you will find illustrations of every individual piece of any Zed system that I have ever seen diagrammed, with the proper equations to compute pressures, volumes, flow rates..... and, after page 33, ENERGIES. 

So to get more work out of the accumulator than you are putting into it, you will need to supply some outside source. It really sounds to me like you need to get more _volume_ out of the accumulator than you are putting into it, and that situation will not last very long.... and the longest _confirmed_ reported run of any of Travis's devices that I can find is only about four hours.

It is obvious that the need for "flow assist" means that the output of the single Zed is not OU. So where does the pressure for the "flow assist" come from ON THE INITIAL CYCLE? The only possible places I can identify are from the pre-charge, which will eventually run out, or from outside the system. And the "flow assist" only resets the second zed back to the start condition, right? So _where_ is the excess whatever coming from? Do the zeds create fluid volume out of nothing?

That's why I asked my question about the Heron's Fountain that was ignored. If a small Heron's Fountain with 100 mL reservoirs will pump a head to, say, 10 cm above the highest level in the reservoirs, and continue doing that for, say, about a minute, as mine do.... what would happen if you had reservoirs of 10,000 liters, elevated by 5 meters above your reference level? How long would you expect it to run, pumping a head of, say, 1 meter? Long enough to impress the money, er, "heck" out of the observers? I've had educated people accuse me of faking Heron's Fountain runs on the tabletop system, it is so unbelievable to them.