Mathematical Analysis of an Ideal ZED

[Pirate88179]

A very wise man said "you can not put new wine in old skins"

He was answering the Pharisees on their confusion and Jealousy toward Jesus, the Pharisees felt they should have been chosen (by God) to usher in the "Way" it was of course their educated duty..... they forgot thay were keepers of the truth - not directors of the truth.

So truth coming form someone else stepped on their toes...... (and puppets).

Whether it be God lifting a veil, or educational evolution - it has always been those with self worth and status that have felt that change was an attack on them......as it is here.

The Pharisees claimed they were protecting people also - from these disruptors with the "New truths"......

They were misguided as the trolls are here today.


Wayne Travis

Didn't the Bible also have this thing called the 10 Commandments?  Isn't one of them something like..."Thou shalt not lie"?  Maybe I am confused but I thought that was in there somewhere.

So, placing new wine in new skins is fine, as long as you do not lie about it.  Hmmmmm.

Bill

[mondrasek]

The spreadsheet already incorporates the uplift on the risers.  You are double dipping if you add that up lift again.  The spreadsheet includes auditing for the various volumes and forces.

If MarkE is correct, the F_start value is only 1.471061N.  And again the amount of Work performed by the system as it lifts from State 2 to State 3 (integral of F*ds) resolves to 0.5*1.471061N*.002590m = 1.905mJ.  That exactly equals the amount of Energy lost from the system as calculated by the difference of the Energy inside the system at State 3 and State 2.

If so, the Ideal ZED is, in fact, Ideal and not lossy at all?  (Edited to change that to a question and not a statement.  I am interested in what others think about this.)

[MarkE]

If MarkE is correct, the F_start value is only 1.471061N.  And again the amount of Work performed by the system as it lifts from State 2 to State 3 (integral of F*ds) resolves to 0.5*1.471061N*.002590m = 1.905mJ.  That exactly equals the amount of Energy lost from the system as calculated by the difference of the Energy inside the system at State 3 and State 2.

If so, the Ideal ZED is, in fact, Ideal and not lossy at all?  (Edited to change that to a question and not a statement.  I am interested in what others think about this.)

The "ideal ZED" like any other machine is fundamentally lossy.  In order to extract energy going from State 2 to State 3, the peak weight of the spillway load must be less than the total uplift force at the end of State 2 or else the risers will never lift the spillway load.  As a direct consequence, the effective spring rate and starting force of the spillway load are both smaller than those of the "ideal ZED" and less than 100% energy transfer results even with our pretense that the system is frictionless and the risers are massless etc.

[MarkE]

Yep,, it sure does for your lift calculations.

I was running a sanity check by volume of displaced water, not using your uplift values but comparing what you have for lift to what the volume of displaced water is.

Your lift numbers are higher than the displaced volume, so I went and added the riserwall lift force to the buoyant force value of the displaced water, and since within your calculations you are including those forces as an input and reset cost they are real enough to be added to the buoyant lift value, that and they are from the physical structure under the water.

You are talking so much nonsense Tom.  In the R4 spreadsheet, the lift distance is exactly accounted for in the volume.  If you take a correct model such as that represented by the R4 spreadsheet and alter it's behavior, then you break the model.

[MarkE]

Here is a screenshot a little higher showing the output work to input work using the calculated lift.


Edit to add
That lift is 3.0351985426mm

You've shown a bunch of numbers on a page.  That is less than useless.