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Author Topic: Mathematical Analysis of an Ideal ZED  (Read 746783 times)

mrwayne

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Re: Mathematical Analysis of an Ideal ZED
« Reply #330 on: March 03, 2014, 05:35:17 PM »
Eout - Ein =  extracted energy,  and of course with full cycle

do you have a self running machine ?

With Respect - I am sharing what I am not contractually bound to reserve.

Our Contract gives our Benefactor exclusive rights to first utilization of the manufactured models and their absolute first public demonstration.

Previous to that contract, we allowed a Skeptic to video our early model.

The Link has been posted.

Thank you.




MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #331 on: March 03, 2014, 08:54:13 PM »
The Ideal Analysis gets the calcs in your math right. Well Done.

...  Wall of pointless blather ...

Wayne Wayne Travis
President
Zydro Energy, LLC
Mr.Wayne@ZydroEnergy.com
Ah another wall of BS text from the shameless huckster Wayne Travis.  One might wonder why oh why is Wayne so anxious to try and deflect the analysis that shows that even under idealized conditions, the ZED is less useful than a brick.

Maybe the lying huckster wasn't completely honest with us about his investment situation.

Maybe the lying huckster has some fish he is trying to reel in and he doesn't want them to see the empty bag he is trying to trade them for their cash.

Note that in his entire wall of text, Wayne Travis cannot refute anything in the analysis.  Note that in the entire wall of text Wayne Travis fails to bring in any actual new information that supports his false claims of getting free energy by cyclically lifting and dropping weights.

mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #332 on: March 03, 2014, 09:00:34 PM »
One might wonder why oh why is Wayne so anxious to try and deflect the analysis that shows that even under idealized conditions, the ZED is less useful than a brick.

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.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #333 on: March 03, 2014, 09:02:12 PM »
To All;

... Wall of text ...

Since the inception of the NET Energy producing system - We worked hard to develop the systems, business, contact, and friendships.

I have put up with the nonsense from TK and his bandwagon now for two years for this reason.

Now we are hiring, we have Turn Key funding. We have selected 18 of our start up of 27 members.

We are hiring now, and we have temporary offices while our world class LEED building is being built.

.............................

We have Patent application pending in almost every county - some granted - some being exercised.

If you are interested in Clean and Green energy future - that has the only Alternative ON DEMAND CAPABILITY:

Welcome, our door has been open.

Wayne Travis

ZydroEnergy.com
Those are just more lies from you Wayne Travis.  You do not now have and have never had a device or a method to generate net energy by cyclically lifting and dropping weights as you claim.  The analysis of the "ideal ZED", the device model that you so glowingly approved, shows that the device is completely useless.  You can post as many walls of text as you want.  What you cannot do is get that ridiculous contraption to even match the efficiency of a brick.  Each time one goes from State 2 to State 3, a transition that performs no useful outside work, one loses 28% of the internally stored energy which must be replaced in order to return to State 2 to set-up for the next futile cycle.

mrwayne

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Re: Mathematical Analysis of an Ideal ZED
« Reply #334 on: March 03, 2014, 09:07:28 PM »
Those are just more lies from you Wayne Travis.  You do not now have and have never had a device or a method to generate net energy by cyclically lifting and dropping weights as you claim.  The analysis of the "ideal ZED", the device model that you so glowingly approved, shows that the device is completely useless.  You can post as many walls of text as you want.  What you cannot do is get that ridiculous contraption to even match the efficiency of a brick.  Each time one goes from State 2 to State 3, a transition that performs no useful outside work, one loses 28% of the internally stored energy which must be replaced in order to return to State 2 to set-up for the next futile cycle.

I am sorry MarkE,

You seemed to miss this post. I will post it again.

MarkE,

The Ideal Analysis gets the calcs in your math right. Well Done.

I will act as you were never told the states, again, again and again - for sake of time.

So lets get your states set up, and in order.

...........

Conditions

At least three layers and Pod each ZED:

ZED A Sunk remaining head due to riser weight and any added weight - ZED A  will be the receiving ZED,

ZED B is at the end of delivering a load and in the raised position - and was not allowed to Bob up after the load was removed.

.................

State one - Start with sunk - still head remaining - equal to the weight of the risers - and any additional load. (additional load is sometimes used to reduce time by reducing expansion and contraction during cycles)
p.s. Adding weight is counter intuitive - most people assume adding weight induces losses

Lesson to be learned - trying to achieve Ideal usage results in self determined conservative process.

The next state is post free flow - this is where the other ZED A and B have equalized between the stroked ZED and the sunk ZED. No riser movement in either ZED - only fluid and pressure.

Note: Free flow results in equalized pressure - but not equalized volume.

The next State is changing from Free flow too "precharge"

Full precharge is the end of the state between free flow and enough buoyancy to nuetralize the determined load and no riser movement either ZED.

The process to get to the full precharge state - two inputs are utilized :

One - the continued consumption of pressure from the ZED B - and the hydro assist.

The hydro assist adds enough pressure - that when combined with the exhuast from the other ZED - reaches load neutrality (buoyancy). This is full precharge for ZED A.

Note: ZED B will not sink until the stored head has dropped below nuetrality of the risers and any added weight.

The Hydro Assist continues to be combined with the Pressure from ZED B - the input cost is the differance between the sinking ZED pressure and the stroking pressure required.

The next state is the Production Stroke of ZED A. ZED A stroking and ZED B sunk is the first half of a Dual ZED cycle - the process repeats in the other direction - notice I did not say reverses.

.................

To understand Stroke - you must determine both the proper load and the proper stroke.

The proper load is the lift safely available at the determined end of stroke.

Iterations are helpful..... I will give you a rule of thumb - Do not make the stroke longer than 1/11 the height of the ZED.
(another counter intuitive - short stroke is a more efficient process)

Use your baseline calculator already prepared to determine what the load is at that height - and that is a good load - presuming riser weight and any added weight has already been considered.

.........................

Unlike the states Mark described - the precharge and stroke is only released into the other ZED - not bobbed up or consumed as production.

The transfer of the precharge and Stroke is made mechanically more efficient as Webby described and posted two of our methods.

but you do not need to add those improvements to find the outcome.


.......................

Last notes - when the full precharge is reached - any additional volume input into the ZED A results in production - so once precharge is hit - no consumption of the previous pressure occurs - the ZED B hits bottom at the end of the production stroke on ZED A.

In simple observation - the true cost of a stroke half cycle is all of the Hydro assist - which is also the stroking Pv ZED A, minus the sinking ZED B Pv, and then repeat for a full cycle.

The production cycle is both ZEDS having produced once and combined.

A full cycle is a return of ZED A to "Sunk.   

MarkE - if you do understand these States - you should be able to see how we transfer two sets of PV left and right - not consuming that value and truely reducing the total input cost - the remaining input cost is the hydro assist.

Lastly - the Hydro Assist can be a external input - or powered by the Production leaving excess. When you determine the cost of the Hydro Assist versus the production - you will understand why I have been so patient.

The Excess or Net per half cycle is no more than the value between the Pv sinking and the production - Not magical - but free.

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.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #335 on: March 03, 2014, 09:15:00 PM »
MarkE, please look at the ZED in State 3.  There is still a positive water head on the pod and each riser!  Therefore there is still much buoyant Force to be resolved.  Ego, the ZED could NOT come to rest in this condition.  It must stroke further if unrestrained.  This proves that the ASSUMPTION that Energy in is equal to Energy out was wrong.  Unless you have some other way to resolve the remaining buoyant Force without more stroke?
Mondrasek, would you like to bet?  The system as presented is physically and mathematically correct.  Why?  Because of the venting necessary to create State 1, the pod and the risers are stable as you stipulated as a requirement for the State 1 condition.  You will recall that I went to pains to discuss with you how the system would be brought to State 1. 

In State 2 3108 cir mm2*mm was added to the system.  Unrestrained that volume lifts the entire cross-section by a distance equal to the volume divided by the cross-section as shown.  A rising tide does indeed lift all boats.

If you object to State 3, then you must object to your own stipulations for State 1.  At the end of State 1 the massless materials:  The "air", the pod, and the risers are all in effect simply floating on the surface of one pool of water.  State 2 holds those masses restrained while water is pumped in.  State 3 releases those masses so that the water can equalize.  They rise, gaining no work because they are massless and the previously lifted water falls, losing a great deal of the energy added during State 2.



MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #336 on: March 03, 2014, 09:18:03 PM »
I realize you did not mean this seriously;

But you are right regarding the Set up pressures - the first time you introduce the differential density fluids - and set them up to the sunk and stroked position - that requires external input.

That input never leaves the system - and is continually recycled.

SO check in never check out - exactly

Wayne
Your piece of junk machine is a sink for energy.  The special features of:  A serpentine hydraulic piston corrupted by pockets of air, and with an outer venting column all take a useful machine:  a hydraulic piston and turn it into a lossy piece of junk.  This is of course contrary to your often stated lies that those very features enable you to cheat gravity.

mrwayne

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Re: Mathematical Analysis of an Ideal ZED
« Reply #337 on: March 03, 2014, 09:23:53 PM »
Ah another wall of BS text from the shameless huckster Wayne Travis.  One might wonder why oh why is Wayne so anxious to try and deflect the analysis that shows that even under idealized conditions, the ZED is less useful than a brick.

Maybe the lying huckster wasn't completely honest with us about his investment situation.

Maybe the lying huckster has some fish he is trying to reel in and he doesn't want them to see the empty bag he is trying to trade them for their cash.

Note that in his entire wall of text, Wayne Travis cannot refute anything in the analysis.  Note that in the entire wall of text Wayne Travis fails to bring in any actual new information that supports his false claims of getting free energy by cyclically lifting and dropping weights.

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.

......

I know, our ZED system is tough to wrap your head around for some, you are very close - don't give up.

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.

That is not dropping a rock twice.

Wayne

mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #338 on: March 03, 2014, 09:24:58 PM »
Mondrasek, would you like to bet?  The system as presented is physically and mathematically correct.  Why?  Because of the venting necessary to create State 1, the pod and the risers are stable as you stipulated as a requirement for the State 1 condition.  You will recall that I went to pains to discuss with you how the system would be brought to State 1. 

In State 2 3108 cir mm2*mm was added to the system.  Unrestrained that volume lifts the entire cross-section by a distance equal to the volume divided by the cross-section as shown.  A rising tide does indeed lift all boats.

If you object to State 3, then you must object to your own stipulations for State 1.  At the end of State 1 the massless materials:  The "air", the pod, and the risers are all in effect simply floating on the surface of one pool of water.  State 2 holds those masses restrained while water is pumped in.  State 3 releases those masses so that the water can equalize.  They rise, gaining no work because they are massless and the previously lifted water falls, losing a great deal of the energy added during State 2.

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?

mrwayne

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Re: Mathematical Analysis of an Ideal ZED
« Reply #339 on: March 03, 2014, 09:30:26 PM »
Your piece of junk machine is a sink for energy.  The special features of:  A serpentine hydraulic piston corrupted by pockets of air, and with an outer venting column all take a useful machine:  a hydraulic piston and turn it into a lossy piece of junk.  This is of course contrary to your often stated lies that those very features enable you to cheat gravity.

MarkE

Lots of losses - and bigger gains...

The ratio to losses is too high with a single layer ZED, We have agreed with you.

The ratio to losses does not increase at the same rate of output increase as you add layers.

Monderask and Larry detailed that.

But go to three layers - smile

Reuse the differential pressures, with a dual system and WOW.

DOn't give up!

Wayne




MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #340 on: March 03, 2014, 09:31:15 PM »
MarkE, I was going to draw your State 3 and dimension the remaining heads on the pod and each riser that need to be resolved.  But I ran into an error with your stated water height in AR2: 49.651mm.  That is obviously not correct.  Could you provide the correct value?  And you should try calculating the buoyant Forces that remain on the pod and each riser that still need to be resolved.  The ZED cannot remain in the position you show in State 3 unless restrained.  It has more Energy that needs to be released due to the still remaining buoyant Forces.

I would have liked to just present my own diagrams again, but I see you calculated the rise based on Volume in = Volume out.  This is another error since there is a third Volume of air that is interacting with the system by the nature of the outer annulus being open to the atmosphere.

I calculated my lift distance based on the ASSUMPTION that Energy in = Energy out, not by simple volumes.  That results in a stroke that should be 1.9094mm.  But the results are similar in that the system could not come to rest at that larger lift distance either.  There is still 31.828 grams of buoyant lift force at that larger lift distance.  So again, the lift would have to be even further to resolve the remaining buoyant Forces.

And FWIF, no iterations need to be performed for this simple analysis.  The iterations would be needed (for me at least) to find the final resting state of the charged ZED.  That state requires that the sum of all the internal buoyant Forces be zero.  That is definitely not the case in your State 3, nor in the one I calculated via an Energy Balance approach.
If you opened the spreadsheet, you will see that was a transcription error.  The value posted was the height of the air from the top of the innermost ringwall down to the surface of the water.  The correct value: 11.349mm is simply that value subtracted from the ring wall height. 

Once more:  Under the stipulation that you set that the system is stable, unrestrained in State 1, it is similarly stable unrestrained in State 3.

The materials are incompressible.  Did you get that?  They are incompressible.  One more time:  They are incompressible.  By definition their volumes cannot and do not change.  Fluid can be pumped in through the various inlets when those valves are open.  Fluid can be released through vents as you stipulated in State1.   Once the valves are closed, fluid may be forced into or drawn from AR7.

Your assumption that energy would not be lost is false for the reasons that I have already proven.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #341 on: March 03, 2014, 09:34:01 PM »
DOn't know what you are referring to?

The LOAD - is the energy extracted, with every half cycle?

We use Hydraulic production as the load - because it can be used to both apply Hydro assist and rotational output to a generator (thru a hydraulic motor).

Thanks
Gee, Mr. Wayne, where is the load bank?

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #342 on: March 03, 2014, 09:40:52 PM »
Two years ago - one of the TK "likes" said something Similar - meant to be an insult:

Why not just continue hooking the ZEDs together to forever..

In reality - the layering system works in much the same effect - but with reduced Capital cost and reduced foot print.

Net is the Product - two Six Layer ZEDs can be optimized to put out the same Net production as four three layer systems, in roughly the same foot print.

............

So it is a business decision - and a structural mechanical (Cost) limitation - to continue to up size each system.

Wayne
Ah more lies from Wayne Travis.  As can be seen from the analysis of the "ideal ZED", and LarryC's spreadsheet, one has two basically two choices:  make each successive annular ring  narrower and narrower to hold constant area per ring, or watch as the change in water height in the outer rings and therefore the stroke converges towards zero.  The ultimate ZED as opposed to the "ideal ZED" is a device with just one riser, and just one pod, where each are very, very wide, and the vertical stroke approaches zero.  In the limit, such a system approximates but never quite matches the efficiency of a brick.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #343 on: March 03, 2014, 09:42:50 PM »
With Respect - I am sharing what I am not contractually bound to reserve.

Our Contract gives our Benefactor exclusive rights to first utilization of the manufactured models and their absolute first public demonstration.

Previous to that contract, we allowed a Skeptic to video our early model.

The Link has been posted.

Thank you.
Once again the lying huckster Wayne Travis appeals to claims that he has something magic behind the curtain.  No worries Wayne.  Bubba is anxiously keeping his magic something for you with him in his cell.

mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #344 on: March 03, 2014, 09:43:15 PM »
If you opened the spreadsheet, you will see that was a transcription error.  The value posted was the height of the air from the top of the innermost ringwall down to the surface of the water.  The correct value: 11.349mm is simply that value subtracted from the ring wall height. 

I knew it was a simple typo.  Sorry I did not notice that you posted a spreadsheet from which I could have found the correct value.

Once more:  Under the stipulation that you set that the system is stable, unrestrained in State 1, it is similarly stable unrestrained in State 3.

I can't see how?  In State 1 there are no unresolved buoyant Forces.  The pod is in no water, and there is zero water head between the ID and OD surfaces of any riser.  This is clearly not the case in State 3 where the pod and risers are all affected by buoyant Forces that are not zero.

The materials are incompressible.  Did you get that?  They are incompressible.  One more time:  They are incompressible

I have never said or thought that they were compressible!  Compressibility is not at issue.  Unresolved buoyant Forces are the issue.