Mathematical Analysis of an Ideal ZED

[MarkE]

   Hi,
      if you imagine a see-saw, equally weighted, try taking say one eighth of the weight from
one side. The now lighter side will stubbornly rise, that's what will happen if you try
this with your ZED.
     The pod can be discounted because it's just Archimedes, and the rest of the thing will
just behave like any ordinary telescopic ram, if you have massless, incompressible air.
    Anyone who seems to be getting more out than is put in has seriously got to hunt for
glitches in their work.
     Anyone proving that you really can get free energy out of this sort of device is going to
become a very famous person!
   I'm not saying that it can't be done, but it sure is one hell of a task
                            John.

I am saying that it cannot be done.  Gravity is a conservative field.  No amount of tinkering with:  weights, sloshing liquid volumes, compressed gas volumes, or other mechanical devices addresses that fundamental behavior of gravity.  Wayne Travis talks out of his hat when he claims as he does on the HER/Zydro web site that "Our Technology produces clean energy Mechanically, by altering the once believed conservative field of gravity - allowing s to supply endless and abundant clean Energy."  The statement is a bald-faced lie.  HER/Zydro have no evidence and never had any evidence that the claim was ever true.  They cannot demonstrate any means within their possession to alter the conservative nature of gravity.  "Slapstick" was yet another very entertaining work of fiction by Vonnegut.

[MarkE]

MarkE,

I was wondering why you have od squared of the outside of the risers as the area,, I think that is what is meant in the spreadsheet?

I calculate circular areas for many things in the spreadsheet, including the surface areas of the riser top surfaces identified as: RiserNODArea, where N is 1, 2, or 3, and each of the annular rings, ring walls, and the riser walls. You will see a note in column D that the areas are calculated as circular areas, IE the area that a square with the same width as the diameter would occupy.  This is merely a convenience for calculations where the ratio of the true area to the circular area:  pi/4 ends up common to a number of terms.  Consequently, that ratio is just rolled up in a constant when needed, or not used at all when it would appear in both the numerator and the denominator of some relationship.  Avoiding inclusion of irrational numbers like pi into equations when it cancels out reduces numerical error. 

[MarkE]

Would using the lightest workable liquid vs. air, cause us to use even more input due to the liighter liquid would have a lot more weight added to the system? More weight to move around, and more in before we get anything out. As compared to air. 

Was thinking also, being the air is compressible, initial input would be less loaded at the moment, rather than an abrupt requirement of 'full' input(surge) just to get things going. Just thinking. The total amount of possible excess input to begin a rise, as compared to an incompressible air substitute, might not be so much  in because of its compressibility. The incompressible air substitute definitely creates a much closer weight difference between it and the water, so more needs to be input than can be output because we are taking away the weight difference of the water, as compared to using air.

Fill a balloon with the oil and one with air and see which rises to the top of the pool from the bottom first.  So in the system being discussed, more oil would have to be moved than the amount of air mass, equals more loss to the system. Heck, might as well just use all water, if we dont care about the the object of changing the levels if weighted liquid in linked separate chambers as being discussed. In that case, then yes, buy a hand jack.

Also, once that air is up to pressure, that pressure contains the energy that put it there. So if we dont try to capture that after the riser is at bottom, then we just wasted it stupidly if it is just released foolishly without applying some use to it at release.  And beyond that, arrange the system to not release or use that compressed air that is at the pressure level of just before the riser is about to rise. Then we eliminated initial loss of compressing to the point of initial rise, again and again. We only have to go through 'that' loss 1 time. Pressure activated check valve?  Like a zener diode, wont let the current flow if below say 12v, for a 12v zener.


Mags

The best case for maximum force generation in a given volume is a very dense working fluid in place of water and a massless, incompressible fluid in place of the air.  The best case for efficiency is to replace both fluids with a massless, incompressible fluid. 

The real air being compressible creates pumping losses, making the already stupid contraption even worse than the "ideal ZED".  More work has to be exerted changing the volume because it goes into thermal losses compressing / decompressing the air.

See the discussion where Mondrasek came up with a clever way to only lose 31% of the energy each cycle.  His scheme identically loads the riser with just the mass that the riser can lift at any moment less a little tiny bit that we ignored in the analysis to ensure that it does rise.  It still loses 31%.

[Magluvin]

The best case for maximum force generation in a given volume is a very dense working fluid in place of water and a massless, incompressible fluid in place of the air.  The best case for efficiency is to replace both fluids with a massless, incompressible fluid. 

The real air being compressible creates pumping losses, making the already stupid contraption even worse than the "ideal ZED".  More work has to be exerted changing the volume because it goes into thermal losses compressing / decompressing the air.

See the discussion where Mondrasek came up with a clever way to only lose 31% of the energy each cycle.  His scheme identically loads the riser with just the mass that the riser can lift at any moment less a little tiny bit that we ignored in the analysis to ensure that it does rise.  It still loses 31%.

"The best case for efficiency is to replace both fluids with a massless, incompressible fluid.  "

So your saying it is no more than a hydrolic jack with a squishy piston.  And to improve on that, replace the piston with a rigid one and say aero gel for fluid.  Dont remember if it is compressible. lol, Im so beat, I should not be trying to think and post. late nights. You know.

Mags

[TkMhMefan]

Mags

Why didn't you jump in sooner   I'm not sure what your motive is on this topic.  Yep i'm a noob and you drew me out.

--R