Mathematical Analysis of an Ideal ZED

[MarkE]

MarkE,

All column energies are accounted.


The End Col 1 PSI in the spreadsheet is 1.30. The PSI for  Col's 1, 12" of water is .43. The 1.30 is calculated as Col 1 PSI - Col 2 PSI + Col 3 PSI = .43 - .87 + 1.73 = 1.30 after rounding. Col 2 is subtracted from Col 3 to give the water head pressure.


Pavg*V works for all cases from 1 to infinity number of columns or risers.

Larry look at your own problem above.  Unless I am reading your intent wrong, you tried to equate pumping 1ft of water into the serpentine structure as equivalent to pumping 3ft into a single column.  They do not end up with the same energy.  The 3ft column requires three times as much energy to fill as the 1ft working against the serpentine that starts with 3ft in each the second and third columns.  So while you can get the right answer for going from equilibrium to some height in the single column by figuring out that column's pressure, your analogy fails, as would averaging pressure across columns to find energy in the whole structure.  Pave_col(m)*Vcol(m) works for any single column m.  You have to go column by column.

[MarkE]

It takes me a while to learn new things,, like your spreadsheet.  After I have played with it for a while now, I now look at the cell formula and I see what it is saying,, where as before I saw a bunch of "stuff" and I needed to go find out what that "stuff" was.

I am sure that most people could figure out what you are doing very easy, and in a shorter time period than it took me, it is very well written and very self explanatory.  I like the side comments next to a few spots where the cell is in, like the circ units, but the side column gives the actual number that people like me would understand easier.

I do not understand what you mean by this: "but the side column gives the actual number that people like me would understand easier."  What would you like to see?  What would make it easier to understand?

[MarkE]

E 81 gives me the number that I can understand,, where as

3108mm^3   circ mm^2*mm  does not mean much to me

2441.0174918393  tells me right off of the bat that it is 2.44cc

So you have already made some things easy.

You can think of circular mm2 as a world where pi = 4, or in other words where circular sections are squares.  In calculations where it is the relative area that matters it is convenient for several reasons to use circular units instead of absolute units.  This leaves multiplying by pi()/4 to the end when we need absolute values such as force or energy.

E81 was a stray value that I forgot to delete. 

[LarryC]

Larry look at your own problem above.  Unless I am reading your intent wrong, you tried to equate pumping 1ft of water into the serpentine structure as equivalent to pumping 3ft into a single column.  They do not end up with the same energy.  The 3ft column requires three times as much energy to fill as the 1ft working against the serpentine that starts with 3ft in each the second and third columns.  So while you can get the right answer for going from equilibrium to some height in the single column by figuring out that column's pressure, your analogy fails, as would averaging pressure across columns to find energy in the whole structure.  Pave_col(m)*Vcol(m) works for any single column m.  You have to go column by column.

MarkE,


The intent was to show that the two methods ends up with the same Total PSI 1.3. No OU.


But, if you have a system that utilizes PSI to do work, would you want want to take 3 times as long and require 3 times the input to cycle it?


My Input foot lbs is equal to your Ein after correcting the .65 to .43.

The spreadsheets title start with 'Analysis of flow'. So Input and Output foot lbs or Ein and Eout is our only concern as we need to be able to teach people with a high school educations. So stored energy initial and final is not required, as it would be confusing to most. Our intent is not to confuse, unlike some people here.

Showing the starting and ending water levels in the drawing is sufficient for anyone with common sense to understand the stored energy is the same. But, if that's the only way you can understand a system, please continue and you can double check your Ein and Eout against my results.

[MarkE]

MarkE,


The intent was to show that the two methods ends up with the same Total PSI 1.3. No OU.

Pressure is not energy.  At the end of the day we are interested in determining the energy into and out of an alleged free energy machine.  Free pressure machines are called fixed weights.

But, if you have a system that utilizes PSI to do work, would you want want to take 3 times as long and require 3 times the input to cycle it?

Pressure is not work.  Pressure is not a capacity to do work.  Pressure is simply force per unit area.  How fast a process should go depends on the process. 

If it's speed your after, these buoyancy schemes are a very poor choice, because they combine large masses with low effective spring rates.  That ~$1. spring from Amazon that outperforms the "ideal ZED" also has a self resonant frequency that is close to 1kHz.   The "ideal ZED" has a self resonant frequency that is orders of magnitude lower.  Mark up another win for the ~$1. spring.

My Input foot lbs is equal to your Ein after correcting the .65 to .43.

The density of the fluid is really rather immaterial to the problem.  If you always integrate then you will never make a mistake due to an invalid simplifying assumption as to how to calculate the work.  Unless I am mistaken, the spreadsheet of yours that I decomposed averaged pressure across multiple columns.  That's not going to calculate energy correctly.  You can use P_{AVE_COL(m)} * V_COL(m) to obtain the energy column by column and then add those energies to get valid results.

The spreadsheets title start with 'Analysis of flow'. So Input and Output foot lbs or Ein and Eout is our only concern as we need to be able to teach people with a high school educations. So stored energy initial and final is not required, as it would be confusing to most. Our intent is not to confuse, unlike some people here.

Either you balance the books or you don't.  The history of free energy claims is riddled with bad accounting.  The audience does not determine the characteristics of a claim.  First and foremost you need to prove that you've got what you claim.  If the proof is not accessible to an audience of interest, you can later figure out how to explain your proof to a broad audience.  If you don't have a valid proof then who might find the material accessible is irrelevant.

Showing the starting and ending water levels in the drawing is sufficient for anyone with common sense to understand the stored energy is the same. But, if that's the only way you can understand a system, please continue and you can double check your Ein and Eout against my results.

Are you claiming to have something that shows an OU result?  If you are, then kindly point me to which post links your spreadsheet or other work that yields the claimed result.