Mathematical Analysis of an Ideal ZED

[MarkE]

Rotary is less conversion to electricity

Zeds do not have to run fast - they produce extremely large force -

The Rotary TAZ has a much larger foot print per KW

Plus and minus of both

ZEDs produce no net energy.

[MarkE]

When I set the system back to use MarkE's numbers and set the riserwall to 0mm there is a small discrepancy between the calculated pressure lift and the displaced volume lift force.

1.2641758762N displacement lift

1.2619003597N calculated pressure lift

At 0mm thickness I would assume that these two values should be the same.

I love how you obfuscate by relabeling things.  1.261900N is the total uplift force at the end of State 2 using zero thickness riser walls and all other dimensions as Mondrasek depicted them.  1.2641758762N does not occur in the correct R4 spreadsheet.

This also brings the output performed to just under the input loses.

The extractable work is always less than the input work.  That's what fundamentally lossy means.

2.9863713965mm lift distance

0.0018843J loss

0.0018842J performed

Using the buoyant lift value changes things

2.9917565583mm lift distance using displacement lift

0.0018842J loss

0.0018910J performed

If this system can then get back to state 1X by itself there would be a nominal gain.

Those must be numbers after your non-physical double dipping changes.  The R4 spreadsheet correctly calculates:

ST2_ST3_INTERNAL_ENERGY_LOSS   0.0018843   J
ST2_ST3_External_Work_Performed   0.0018797   J

Note that 1.8797mJ is less than 1.8843mJ.

[MarkE]

Lets pretend you did not read all the posts - and did not follow the whole story.... and you are not just a troll...



I am sorry you think I lied - I work on being as clear as possible - but I make errors in guestimates - that is why they are called "objectives..."

Wayne it is just awesome how: freely, effortlessly, and unashamedly that you lie.  I see that you are continuing your progrom of follower inoculation.  Yes, that's it:  Tell everyone that all those false claims that you've made to free energy machines weren't lies.  They were just "guestimates".  LOL.

Look up the word - you pretend that our efforts are something else.

Now I ask you:

What is your intent here? protecting someone from a presumed liar --- or just sticking to your guns no matter what the truth.

Who are your protecting? What has My company or Myself asked form you?

That's right - Due diligence - nothing more.

The fraud has gone full defensive.

[MarkE]

If I set both the ringwall thickness and the riserwall thickness to 0mm I then need to drop the ST2 fill height to 35.5mm to stop "blowing" ar2 but this still leaves the displaced lift force higher than the calculated pressure lift force and still has the work performed higher than the loses.

0.0020460J loss
0.0020533J performed

Again, this is with using the displace water volume force equivalency in N for the lift calculation and therefor distance.

With both set to 0mm I am assuming that then there is no physical device interaction.

This also leaves a positive pressure under r1 at end of state 3 and a zero pressure under r1 at the end of state 1X, so in this condition the system will collapse back down into state 1X without outside input.

Those must be numbers from your fantasy spreadsheet.  Making the changes you state to the correct R4 spreadsheet yields:

ST2_ST3_INTERNAL_ENERGY_LOSS   0.0017148   J
ST2_ST3_External_Work_Performed   0.0017099   J

[MarkE]

Alrighty MarkE,

Here is the spreadsheet WITHOUT the VerGAp change to the volume check in the state 3 audit.

Without this change you can not change the VerGAp setting.

I have set the riserwallthickness to 0 and changed the lidthickness to 0 to match,, just as if I had no changes to the height section.

It is using the lift force by displacement only and as you can see it is showing a force audit error of -0.0022755N

You will also see that there is no air volume errors.

You can point cell B164 to use cell E164, this is MarkE's calculated pressure lift, or you can point it to use cell F238, which is where it is pointing right now which is the displacement lift force,, or you can point it to cell F256 which is the riserwallthickness lift force plus the displacement lift force.

If you choose to change the riserthickness to 1 then also change the lidthicknes to one to keep MarkE's relationship in tact.  Those are cells B25 and B28.

Now the only thing that might not be happy with MarkE is that there are the other forces that have not been identified by his spreadsheet and at least the lift force of the displaced volume of water MUST be considered.

You have not identified any errors in the R4 spreadsheet.  Your spreadsheet double counts values.  Your spreadsheet is wrong.  If you want to understand your mistakes see the drawing below.