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Author Topic: Mathematical Analysis of an Ideal ZED  (Read 749522 times)

MileHigh

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1020 on: March 15, 2014, 03:03:59 AM »

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1021 on: March 15, 2014, 12:34:38 PM »
It looks like this thread is winding down.  I'll leave this drawing for interested parties to chew on.  It shows the complete linear equation set that describes the restoring force of the compression spring behavior exhibited by the "ideal ZED".  The r4 spreadsheet previously posted and that at least webby1 has been playing with calculates effectively using these equations.  The equations do not handle annular ring overflow or underflow. 

As has been pointed out many times already:  A ZED is a device that just lifts and lowers weights.  Gravity is a conservative force:  The gravitational energy potential of an object at some height is the same no matter what path the object takes to reach that height.  One can lift and lower a weight all day, all month, or for a millenium and one will not gain any potential energy traversing some path from a given starting height back to the same ending height.  This holds true no matter what the Bible thumping fraud Wayne Travis or his lackeys try to say about the matter.

minnie

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1022 on: March 15, 2014, 12:53:45 PM »



    Good question MarkE,
                                  the sun's set but will it rise again?
   Will this thing keep on with its wax and Wayne cycle?
   Will it wEBBy and flow again?
   Mike we see some more input?
   Will Larry carry on with the flow assist?
   It all remains to be seen!
                   John.
[size=78%]    [/size]

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1023 on: March 15, 2014, 01:07:53 PM »
I have little doubt that Wayne Travis will carry on the fraud as long as he possibly can.  Investors can and should sue his professional engineer Kevan Riley for promoting HER/Zydro's obviously false claims.  Unless he has written records disclaiming his support of Wayne Travis / HER / Zydro's blatantly fraudulent claims, I think that he is completely exposed to investor lawsuits.  Unless I am mistaken, he continued to support this fraud.  Depending on how much E&O insurance Riley carries the investors may get a good percentage of their money back via that path.  Other avenues of recovery are less certain.  The money that HER/Zydro has already spent is gone.  It is unlikely that a lawsuit against them would yield much.  That would leave recovery down to the personal assets of our Bible thumping free energy promoting fraudster: Wayne Travis.  Pictured here, he resembles a funeral director in a cemetery garden.  It's fitting as he's seen to it that investor funds have gone off to their final resting place.

powercat

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1024 on: March 15, 2014, 01:15:22 PM »
This thread will join the many others that have been unable to prove their claim, and have been shown to be fraudulent, to the more experienced members it would seem so obvious when the inventor fails to show a continuous running device, but these fraudulent inventors always have excuses and diversionary tactics.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1025 on: March 15, 2014, 01:30:36 PM »
One can always hope that people pointing out a fraud in action creates some resistance to that fraud's continuation.  Fraud is often profitable because it is seldom prosecuted.  As John Rohner has found out:  There are exceptions.  I think he has a court date next week for a hearing on charges of contempt in the civil case.  The corporate defense attorneys revealed in public filings that there is a criminal case under seal.  Some of these fraudsters are so full of themselves that they never see the authorities coming.

mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1026 on: March 15, 2014, 02:03:33 PM »
MarkE,

Thank you so much for providing all your time and hard work!  Your Analysis is a masterpiece and I admit that I came nowhere close to your level of work and presentation.  I especially appreciate where you found the error in my belief that State 1 would have a net buoyant Force of zero.  You are correct that the system would have a net positive buoyancy due to the water being displaced by the riser walls that were assumed to have an SG=0.  So State 1x is correct.  Nice catch.

How much Energy needs to be supplied to the system to go from State 3 back to State 1x?  There is no reason to go from State 3 back to State 2 and then back to State 1x, AFAICS.  If we "pull the plug" from the bottom of the pod chamber I think the system should transition from State 3 back to State 1x.  Do you agree?  Would we need to supply Energy to do that?

M.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1027 on: March 15, 2014, 02:59:04 PM »
MarkE,

Thank you so much for providing all your time and hard work!  Your Analysis is a masterpiece and I admit that I came nowhere close to your level of work and presentation.  I especially appreciate where you found the error in my belief that State 1 would have a net buoyant Force of zero.  You are correct that the system would have a net positive buoyancy due to the water being displaced by the riser walls that were assumed to have an SG=0.  So State 1x is correct.  Nice catch.

How much Energy needs to be supplied to the system to go from State 3 back to State 1x?  There is no reason to go from State 3 back to State 2 and then back to State 1x, AFAICS.  If we "pull the plug" from the bottom of the pod chamber I think the system should transition from State 3 back to State 1x.  Do you agree?  Would we need to supply Energy to do that?

M.
Thanks.  Physics always prevails.  The soda bottle / water bottle experiment was proof of the State 1 fallacy.

To go from State 3 to State 1X, let the water out of the pod chamber.  The internal energy decreases from 3.6054mJ of State 3 to 3.4029mJ of State 1X. losing 0.202454mJ. 

mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1028 on: March 15, 2014, 03:14:29 PM »
To go from State 3 to State 1X, let the water out of the pod chamber.  The internal energy decreases from 3.6054mJ of State 3 to 3.4029mJ of State 1X. losing 0.202454mJ.

And from what you have taught us all, theoretically, exactly half of that Energy lost (0.5*0.202454mJ) could be transferred to a second Ideal ZED that starts at State 1 (while the former starts at State 3) if, instead of simply letting the water run out of the pod chamber to nowhere, it was coupled to the pod chamber of that second Ideal ZED?  In a similar fashion as how two ZEDs are coupled in LarryC's diagrams in his spreadsheets?
« Last Edit: March 15, 2014, 08:09:38 PM by mondrasek »

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1029 on: March 15, 2014, 03:48:30 PM »
And from what you have taught us all, theoretically, exactly half of that Energy lost (0.5*.202454mJ) could be transferred to a second Ideal ZED that starts at State 1 (while the former starts at State 3) if, instead of simply letting the water run out of the pod chamber to nowhere, it was coupled to the pod chamber of that second Ideal ZED?  In a similar fashion as how two ZEDs are coupled in LarryC's diagrams in his spreadsheets?
Why would you want to do that?  Adding more complications or instances of under unity devices just increases percentage loss. Because the risers are unrestrained, going from ST1X to ST3 or back is effectively just the same as pumping water into a column of some dimension or draining it from same, respectively.  The pair reduces to a glorified "U" tube with some water in it.  The system will naturally seek an equilibrium state between ST1X and ST3 on both sides that in total holds less internal energy than when one is at ST3 and the other is at ST1X.

mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1030 on: March 15, 2014, 08:05:45 PM »
Why would you want to do that?

Because that would mean the Ein needed to bring a second ZED from State 1x to State 2 would be reduced by 0.5*0.202454mJ I think.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1031 on: March 16, 2014, 12:30:31 AM »
MarkE,

Are you saying that I can ALWAYS vent the fluid out of the pod chamber and have the system return to state 1?
If you open the valve at the bottom of the pod chamber you change the thing into a glorified crazy straw.  In any state, the force that reflects back to the pod chamber is the sum of the relative areas times the heads as previously shown in LarryC's 0+3+3, and 1+2+4 model.  When there is water in the pod chamber, this is like his 1+2+4 and there is net force required to hold the water in the pod chamber.  Heads in odd AR#'s add to the reflected force, heads in the even AR#'s subtract and there are additional terms due to the nested risers.  The thing will drain so long as the reflected sums of the other annular ring forces and the weight in the pod chamber is positive.  Once the machine has been set-up to State 1 with the associated venting, the neutral condition is State 1X not State 1.   You can play with the spreadsheet and see if that's what you get.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1032 on: March 16, 2014, 12:50:41 AM »
Because that would mean the Ein needed to bring a second ZED from State 1x to State 2 would be reduced by 0.5*0.202454mJ I think.
You can change the conditions, and you can change the states, but what you can't do is find any combination that will give you a closed cycle with an energy gain over that cycle.  Adding complexity only worsens the situation.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1033 on: March 16, 2014, 01:48:51 AM »
I am a little confused.  If I can pull the plug on the pod chamber and it will fall apart and go back to state 1,, will it be able to do it with these settings.
Why do you ignore what I write? 

Quote
The thing will drain so long as the reflected sums of the other annular ring forces and the weight in the pod chamber is positive.  Once the machine has been set-up to State 1 with the associated venting, the neutral condition is State 1X not State 1.   You can play with the spreadsheet and see if that's what you get.

What is confusing you? You've entered drastically different dimensions, so now State 1X is a higher energy state than State 3.  There should be no big surprise that the system will not have a tendency to go from State 3 to State 1X for those conditions.
Quote

Click on the cell number and enter these values

B27 RiserWallThickness   257.5
Are you making bank vaults?  10" wide riser walls???
Quote
B28 RingWallThickness    0.25
B29 PodOD                400
B30 PodHeight           4000
B31 ST1_PrefillHeight   3000
B111 ST2_AR1Height      2000

This setup leaves lots of room for any further movement of the air\water columns and has no over or under issues with the columns and volumes.

I arranged these numbers to make it easy for me to see so you will not see them all together like this in the spreadsheet.

Input                   89.4066834062J
Would you be so kind as to either refer to cell #'s or use the actual names in the spreadsheet?  This value is ST2_EnergyAdded  the total energy in state 2 for your conditions:  ST2_EnergyAddedPlusST1E = 1070.4582073J
Quote

Output                 289.5823115248J
This is cell ST2_ST3_External_Work_Performed.
Quote

Input to Output        325.07%   
This is cell ST2_ST3_PCT_ENERGY_LOSS vs ST2_ENERGY_ADDED.  There should be no big surprise since you made made the risers gigantic there is now lots more energy stored in State 1, and State 1X than incrementally added going to State 2. 
Quote
      
Uplift distance          4.4593613399mm
Total uplift force  130348.603561497N

State 1 Energy         981.0515238844J
State 2 Energy        1070.4582072907J
State 3 Energy         779.8224455746J
Again, what is surprising to you here?  And why is it that you choose to pick cells an internal energy change and label that as OUTPUT? B225 is clearly named as: ST2_ST3_ENERGY_LOSS.  B226 is clearly named as: ST2_ST3_External_Work_Performed.  In your set-up, ST2_ST3_ENERGY_LOSS = 290.6357617J while ST2_ST3_External_Work_Performed = 289.5823115J.  The spreadsheet continues to demonstrate the inherently lossy behavior even with your bank vault riser walls.


MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1034 on: March 16, 2014, 02:11:09 AM »
This is my modified file.

I took MarkE's file and removed the graphics and then down at the bottom of state 3 I added some things so you can change the dimensions,, I pointed the actual cells up above to these cells, and I included the height levels so you can see those changes,, no red numbers are allowed,, that would be blowing something :)

I am sure MarkE would do a much better job of it,, I am not a spreadsheet person.
Since all the formula references are to named cells you could have just cut and pasted the lines you wanted to group together.  For some reason you renamed many cells with a _1 suffix.  It looks like you left the form of the calculations intact.  So you should still be getting the right answers.  What you label as Input at B230 is:  ST2_EnergyAdded_1 (_1 your relabel) and what you label as Output is:  ST2_ST3_External_Work_Performed_1 (_1 your relabel).  The real "input" is the internal energy lost in the cell immediately above:  ST2_ST3_ENERGY_LOSS_1 (_1 your relabel).  As you relabeled that cell, I think you should be aware of it.