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Overunity Machines Forum



Mathematical Analysis of an Ideal ZED

Started by mondrasek, February 13, 2014, 09:17:30 AM

Previous topic - Next topic

0 Members and 53 Guests are viewing this topic.

MarkE

Quote from: mrwayne on March 03, 2014, 03:07:28 PM
I am sorry MarkE,

You seemed to miss this post. I will post it again.
... Wall of pointless text ...

Nonsense video with no measurements:  http://www.youtube.com/watch?v=q-0TITC4Wrc



Wayne Travis
President
Zydro Energy, LLC
Mr.Wayne@ZydroEnergy.com
Every time you or TK, or the puppets come back.
If your last line is a promise to keep spamming your wall of text, I don't think Stefan will appreciate that.

MarkE

Quote from: mondrasek on March 03, 2014, 03:00:34 PM
MarkE, one might wonder why you are ignoring my two posts that point out a mistake in your presented values as well as a couple in the methods in your final Stage 3 presentation.  It is an invalid "Energy Balance" Analysis, as well as completely ignores that works was done in the form of F*ds.
You are absolutely wrong.  The spreadsheet correctly shows all of the math.  Each energy value in the spreadsheet explicitly calculates energy as the integral of F*ds.  All of the math reflects real physics.  The drawing had one transcription error, picking up a value one row off from the spreadsheet.

Here is are all the files again with the transcription error in the AR2 value that was pasted onto the State 3 drawing corrected.

You are free to show by competing analysis and / or demonstration that I got anything wrong in the spreadsheet.  If you do, then make certain that you apply actual physics and not contrived ideas such as you have so far.

The math shows exactly what I told you and that you refused to discuss:  That equalizing water column heights as occurs in the transition from State 2 to State 3 loses stored energy without doing any useful work.  The most efficient ZED is no ZED at all.

ETA: Also the spreadsheet is here.  No values or formulas were changed.  It adds more comments and corrected the State 3 graphic.  It also includes audit of the incompressible fluid volumes.  Got that Mondrasek:  The fluids are incompressible:  Their volumes are fixed.

MarkE

Quote from: mondrasek on March 03, 2014, 03:24:58 PM
In the condition you show in your State 3 drawing the pod is still submersed in 28.537mm of water.  It is displacing 89.652 cc of that water.  Are you saying that does not create a buoyant Force equal to ~89 grams?  What is keeping it from rising further?  Where is the equal and opposite force that is cancelling this buoyant Force that remains?

The three risers are likewise still positively buoyant due to the head difference of the water on their OD and ID, right?
I have shown all the work in the spreadsheet in convenient easy to read algebraic form.  If you aren't going to bother to read and understand the work, then there is little hope for you.  Either you accept your stipulation that the system was vented in State 1 so that there was no net force up or down on the risers, or you change your problem definition.  If you accept the former, then what you have is no different than objects floating on a volume of water at the end of each:  State 1 and State 3.  State 2 added volume to the water which because of the restraint on the Russian Dolls of Ignorance was forced into AR7.  Once the restraint was released at State 3, the added fluid effectively redistributed to an equal height across the entire cross-section as required by fluid behavior.  Ergo, we poured some water into Archimedes' bath and his rubby ducky, Spanish Armada playset all rose with added water.

MarkE

Quote from: mrwayne on March 03, 2014, 03:23:53 PM
MarkE,

I am still patient with you.

As far as your insinuations and slander, my motives have been clear and consistent from my first post.
QuoteYes you are selling lies to investors.  That has been clear for years now.

......

I know, our ZED system is tough to wrap your head around for some, you are very close - don't give up.
Despite the complications of detail that you try and use to obfuscate the fact that you have nothing of value, the fact is that at the end of the day, you lift and drop weights.  There is no energy gain to be had from doing such a thing.  You have never shown the slightest bit of evidence to the contrary.
Quote

If all you see is lifting and dropping weights - you missed the production which is removed before dropping, and the re use of each half cycles charge.
I see it for what it is, not the illusion that you try and misrepresent.
Quote

That is not dropping a rock twice.

Wayne
You should be more concerned about where your burned investors might want to drop rocks.

mondrasek

Quote from: MarkE on March 03, 2014, 03:58:07 PM
You are free to show by competing analysis and / or demonstration that I got anything wrong in the spreadsheet.  If you do, then make certain that you apply actual physics and not contrived ideas such as you have so far.

MarkE, your math is not in question.  But it results in a paradox, right?  If you solve to balance Energy, you end up with unbalanced Forces left over (this is what your analysis shows).  If you solve to balance the Forces, then the Energy balance is not correct.  Do you disagree?  Are there not unbalance buoyant Forces in State 3 of your analysis?

Quote from: MarkE on March 03, 2014, 03:58:07 PM
The math shows exactly what I told you and that you refused to discuss:  That equalizing water column heights as occurs in the transition from State 2 to State 3 loses stored energy without doing any useful work.  The most efficient ZED is no ZED at all.

There was useful work performed that you have ignored.  The ZED in this Analysis was not allowed to rise uninhibited by a load.  We agreed on this from post 249 of this thread:

I utilized the same analysis method for the output rise as was used for the input of the water charge:  F*ds as expressed for the case of a Volume of a Fluid that is being moved by a change in Pressure that either starts or ends at zero:  Paverage*V.  The riser initially will want to move with a Pressure that can be calculated from the buoyant force sum of the pod and risers.  That Pressure should drop linearly to zero as the ZED reaches equilibrium at the end of the rise.  The physical device that would restrain the initial Pressure and allow it to drop to zero while performing the rise is not important for the analysis I think.  Please let me know if you think otherwise.

You violated this first by NOT calculating the lift difference using the described method of an Energy balance.  You instead resorted to a Volume balance which is flat wrong for this open ZED.  Then you ignore that work output preformed upon the non-physical device agreed upon would provide the theoretical output F*ds as the Paverage*V, or simply Faverage*ds where ds is the lift distance.  So there is Work that can be performed by the change from State 2 to State 3 and you ignore that in your Analysis.