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### Author Topic: Mathematical Analysis of an Ideal ZED  (Read 658627 times)

#### mondrasek

• Hero Member
• Posts: 1301
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #675 on: March 06, 2014, 10:46:18 PM »
Another way to look at this is to calculate the Energy in the column of water that is on the piston.  That Energy is dissipated as the piston rises and the water spills out.  So that Energy must be equal to the Energy that is released when the ZED is allowed to rise.  That equation also resolves to 0.5*Fstart*S.

#### minnie

• Hero Member
• Posts: 1244
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #676 on: March 06, 2014, 11:33:10 PM »

Yes, you can't build any pressure in this because it will spill,so you have to let the big riser rise.
Buoyancy is really pressure,so the s.g. of the fluid providing the head is important and the
pressure acting on the largest riser is also important, these two should be what goes in and
what comes out. True the different densities of the internal fluid mix will have some effect
due to their weight, but I can't see it making much difference to the output.
John

#### Magluvin

• Hero Member
• Posts: 5886
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #677 on: March 07, 2014, 01:36:15 AM »
Pentane is flammable, volatile, toxic and carcinogenic probably. Sodium polytungstate? Man, you are weird. How about using galinstan and kerosene?

Or just a big spring and a rock.

Would using the lightest workable liquid vs. air, cause us to use even more input due to the lighter liquid would have a lot more weight added to the system? More weight to move around, and more in before we get anything out. As compared to air.

Was thinking also, being the air is compressible, initial input would be less loaded at the moment, rather than an abrupt requirement of 'full' input(surge) just to get things going. Just thinking. The total amount of possible excess input to begin a rise, as compared to an incompressible air substitute, might not be so much  in because of its compressibility. The incompressible air substitute definitely creates a much closer weight difference between it and the water, so more needs to be input than can be output because we are taking away the weight difference of the water, as compared to using air.

Fill a balloon with the oil and one with air and see which rises to the top of the pool from the bottom first.  So in the system being discussed, more oil would have to be moved than the amount of air mass, equals more loss to the system. Heck, might as well just use all water, if we dont care about the the object of changing the levels if weighted liquid in linked separate chambers as being discussed. In that case, then yes, buy a hand jack.

Also, once that air is up to pressure, that pressure contains the energy that put it there. So if we dont try to capture that after the riser is at bottom, then we just wasted it stupidly if it is just released foolishly without applying some use to it at release.  And beyond that, arrange the system to not release or use that compressed air that is at the pressure level of just before the riser is about to rise. Then we eliminated initial loss of compressing to the point of initial rise, again and again. We only have to go through 'that' loss 1 time. Pressure activated check valve?  Like a zener diode, wont let the current flow if below say 12v, for a 12v zener.

Mags

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #678 on: March 07, 2014, 01:38:00 AM »
MarkE.  I apologize for leaving out the word "integral" again.  It is a bad habit.

I think I saw this integral reduction shared by you.  It is the one we are using for the Energy in the water introduced during the change from State 1 to State 2.  That integral of F*ds is 0.5(Pstart-Pend)*V for the special case where the start or end Pressure is zero.  Is that correct?
For the special case of zero to HEND, or HSTART to zero, the work performed filling or emptying a column is:  E = 0.5*PEND*VEND filling, and 0.5*PSTART*VSTART emptying.  In all other cases the integral result is more complicated.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #679 on: March 07, 2014, 01:54:01 AM »
Another way to look at this is to calculate the Energy in the column of water that is on the piston.  That Energy is dissipated as the piston rises and the water spills out.  So that Energy must be equal to the Energy that is released when the ZED is allowed to rise.  That equation also resolves to 0.5*Fstart*S.
For the circumstance that you set-up that is correct:

So you get a movement S = FSTART/KRATE 0.30078N / 64.649N/m =  0.004653m and an output energy of
0.5*0.30078N*0.004653m   0.700mJ.

The energy expended, is the loss of potential energy in the annular columns as they move from the State 2 condition to the State 3 condition with the riser at 2.492mm lift:  The internal energy goes from 2.560mJ to 1.536mJ which equals 1.024mJ internal energy loss.

The efficiency is:  0.700mJ/1.024mJ = 68.3%.

This is the tyranny of N*(X/N)2.  Whenever you directly transfer energy between two potential energy stores you run into the N*(X/N)2problem.  In order to avoid the problem you have to retain N=1.0 and still transfer all of the energy.  You can only do that by converting the form of the energy from potential to kinetic.  That is how a pendulum works.  The insipidly stupid ZED design cannot do that.  It translates potential from one place to another.  There is even the "equalization phase" that HER/Zydro say is necessary.  Every time you take energy from one potential store and redistribute it to more stores than the original without first converting the form of the energy, you suffer losses.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #680 on: March 07, 2014, 02:00:01 AM »

Hi,
if you imagine a see-saw, equally weighted, try taking say one eighth of the weight from
one side. The now lighter side will stubbornly rise, that's what will happen if you try
The pod can be discounted because it's just Archimedes, and the rest of the thing will
just behave like any ordinary telescopic ram, if you have massless, incompressible air.
Anyone who seems to be getting more out than is put in has seriously got to hunt for
glitches in their work.
Anyone proving that you really can get free energy out of this sort of device is going to
become a very famous person!
I'm not saying that it can't be done, but it sure is one hell of a task
John.
I am saying that it cannot be done.  Gravity is a conservative field.  No amount of tinkering with:  weights, sloshing liquid volumes, compressed gas volumes, or other mechanical devices addresses that fundamental behavior of gravity.  Wayne Travis talks out of his hat when he claims as he does on the HER/Zydro web site that "Our Technology produces clean energy Mechanically, by altering the once believed conservative field of gravity - allowing s to supply endless and abundant clean Energy."  The statement is a bald-faced lie.  HER/Zydro have no evidence and never had any evidence that the claim was ever true.  They cannot demonstrate any means within their possession to alter the conservative nature of gravity.  "Slapstick" was yet another very entertaining work of fiction by Vonnegut.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #681 on: March 07, 2014, 02:09:17 AM »
MarkE,

I was wondering why you have od squared of the outside of the risers as the area,, I think that is what is meant in the spreadsheet?
I calculate circular areas for many things in the spreadsheet, including the surface areas of the riser top surfaces identified as: RiserNODArea, where N is 1, 2, or 3, and each of the annular rings, ring walls, and the riser walls. You will see a note in column D that the areas are calculated as circular areas, IE the area that a square with the same width as the diameter would occupy.  This is merely a convenience for calculations where the ratio of the true area to the circular area:  pi/4 ends up common to a number of terms.  Consequently, that ratio is just rolled up in a constant when needed, or not used at all when it would appear in both the numerator and the denominator of some relationship.  Avoiding inclusion of irrational numbers like pi into equations when it cancels out reduces numerical error.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #682 on: March 07, 2014, 02:17:58 AM »
Would using the lightest workable liquid vs. air, cause us to use even more input due to the liighter liquid would have a lot more weight added to the system? More weight to move around, and more in before we get anything out. As compared to air.

Was thinking also, being the air is compressible, initial input would be less loaded at the moment, rather than an abrupt requirement of 'full' input(surge) just to get things going. Just thinking. The total amount of possible excess input to begin a rise, as compared to an incompressible air substitute, might not be so much  in because of its compressibility. The incompressible air substitute definitely creates a much closer weight difference between it and the water, so more needs to be input than can be output because we are taking away the weight difference of the water, as compared to using air.

Fill a balloon with the oil and one with air and see which rises to the top of the pool from the bottom first.  So in the system being discussed, more oil would have to be moved than the amount of air mass, equals more loss to the system. Heck, might as well just use all water, if we dont care about the the object of changing the levels if weighted liquid in linked separate chambers as being discussed. In that case, then yes, buy a hand jack.

Also, once that air is up to pressure, that pressure contains the energy that put it there. So if we dont try to capture that after the riser is at bottom, then we just wasted it stupidly if it is just released foolishly without applying some use to it at release.  And beyond that, arrange the system to not release or use that compressed air that is at the pressure level of just before the riser is about to rise. Then we eliminated initial loss of compressing to the point of initial rise, again and again. We only have to go through 'that' loss 1 time. Pressure activated check valve?  Like a zener diode, wont let the current flow if below say 12v, for a 12v zener.

Mags
The best case for maximum force generation in a given volume is a very dense working fluid in place of water and a massless, incompressible fluid in place of the air.  The best case for efficiency is to replace both fluids with a massless, incompressible fluid.

The real air being compressible creates pumping losses, making the already stupid contraption even worse than the "ideal ZED".  More work has to be exerted changing the volume because it goes into thermal losses compressing / decompressing the air.

See the discussion where Mondrasek came up with a clever way to only lose 31% of the energy each cycle.  His scheme identically loads the riser with just the mass that the riser can lift at any moment less a little tiny bit that we ignored in the analysis to ensure that it does rise.  It still loses 31%.

#### Magluvin

• Hero Member
• Posts: 5886
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #683 on: March 07, 2014, 03:41:34 AM »
The best case for maximum force generation in a given volume is a very dense working fluid in place of water and a massless, incompressible fluid in place of the air.  The best case for efficiency is to replace both fluids with a massless, incompressible fluid.

The real air being compressible creates pumping losses, making the already stupid contraption even worse than the "ideal ZED".  More work has to be exerted changing the volume because it goes into thermal losses compressing / decompressing the air.

See the discussion where Mondrasek came up with a clever way to only lose 31% of the energy each cycle.  His scheme identically loads the riser with just the mass that the riser can lift at any moment less a little tiny bit that we ignored in the analysis to ensure that it does rise.  It still loses 31%.

"The best case for efficiency is to replace both fluids with a massless, incompressible fluid.  "

So your saying it is no more than a hydrolic jack with a squishy piston.  And to improve on that, replace the piston with a rigid one and say aero gel for fluid.  Dont remember if it is compressible. lol, Im so beat, I should not be trying to think and post. late nights. You know.

Mags

#### TkMhMefan

• Newbie
• Posts: 2
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #684 on: March 07, 2014, 03:43:11 AM »
Mags

Why didn't you jump in sooner   I'm not sure what your motive is on this topic.  Yep i'm a noob and you drew me out.

--R

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #685 on: March 07, 2014, 04:02:27 AM »
"The best case for efficiency is to replace both fluids with a massless, incompressible fluid.  "

So your saying it is no more than a hydrolic jack with a squishy piston.  And to improve on that, replace the piston with a rigid one and say aero gel for fluid.  Dont remember if it is compressible. lol, Im so beat, I should not be trying to think and post. late nights. You know.

Mags
It is a hydraulic lift jack with polluted hydraulic fluid.  The first energy efficiency improvement is to remove the pollutant "air".  If you still want high force for low fluid volume, use a large diameter insert.  That will reduce the amount of fluid that you have to store.  It will not reduce the amount of fluid required to obtain a given amount of lift against a particular maximum load.

#### MileHigh

• Hero Member
• Posts: 7600
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #686 on: March 07, 2014, 06:25:09 AM »
Just one more time for review.  Quoting Wayne:

Quote

Try using this process - and you will see that we do not release the "spring" as you describe.

I will post it again.

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

...........

Conditions   At least three layers 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.

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

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.

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.

I submit to all of you that the above quote is nonsensical gibberish.  I can't understand it not because I am dumb, but because it is nonsensical gibberish.

I challenge anybody that claims they understand what he is saying to restate what he is saying in terms of cycles and energy.  In my opinion, if no one takes up the challenge, that represents a tacit admission that you agree with my evaluation of Wayne's statements.  It's another smoking gun indicating that Wayne will never produce anything.  He is just a cash burner and it's other people's cash.

MileHigh

#### mrwayne

• Hero Member
• Posts: 975
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #687 on: March 07, 2014, 06:30:14 AM »

I can't understand it not because I am dumb, but because it is nonsensical gibberish.
MileHigh

ENough said...

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #688 on: March 07, 2014, 06:33:38 AM »
Just nitpicking

If you do not have air, or its equivalent, in the system, there is no lifting at all,, there is no seal to create pressure, no pressure no work.
...

That's right:  The best ZED is no ZED at all.  First remove the pollutant: air.  Then seal off AR7, them remove the pod.  Then you have the best ZED ever.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #689 on: March 07, 2014, 06:36:59 AM »
ENough said...
Hey there Wayne, how's your technical cadre doing with the spring that ate Cincinnati?   Have your minions figured out any talking points that can make the "ideal ZED" at least seem like it does anything more than a 1000X smaller, ~\$1 spring?