Hydro Differential pressure exchange over unity system.

[MT]

Hi,
thank you for responses and posts, I'm learning from it.


My previous model for inverse travis effect (1inverse riser) showed no OU. Advice was to go multi riser system similarly as ZED, reuse mass etc


So I created the following model and would still call it 2riser. I think see3d would name it one riser but name is not important now.


Model is 2D but it can be mapped on a 3D model with cylinders.


I used following simplifications:
- air in below text should be replaced by icompresible weightless fluid lighter than water. I agree with parisd that air IS compressible but it just makes calculations more complex and since we can use icompresible fluid why not make things simpler if it helps and simplification is valid.
- risers have negligible weight
- compartements have negligible thin walls
- no losses


1. picture shows starting situation, risers are locked. There is inner compartment with the pod(1th riser) and 2 units of water. In the outer compartement containing 9 units of water on each side is floating 2nd riser. Air pressure above 2-riser but also air pressure in inner compartement are the same = 1atm. Under 2riser left side are 6 units of air and 6 on right side.
Total lift force of risers is 6 units. 2riser contributes 0 and the pod 6.


2. Precharging, risers are still locked - injecting 2 units of water in the inner compartment (at its bottom) raises water in 1 and 12 column by one.
Total lift force 16. The pod contributes 12 units and 2nd riser 4.


3. Stroke - risers are allowed to lift up one unit distance. To keep internal pressure the same(real ZED keeps pressure in inner compartment also the same 8psi), 10 extra water units need to be injected. Most of this water (6 units) is needed to fill space under rising pod.
Total lift force after stroke is 22. Inner pod contributes 18 and 2riser 4.


So work that can be done using this model equals lifting force caused by 16 volume units of displaced water over one unit distance. In fact lifting force increases from 16 to 22 over one unit distance. Water injected at bottom = 12 units.
Now I'm trying to answer the question: is the work done enough to inject 12 units of water at bottom and by this close the loop? Webby1 said some posts ago we can use builded pressure of water from stroke step and lead it to another 2 riser. I assume equalization would fill first 6 units in another 2riser. Question (restated) is then: is the work done enough to inject only 6 (not 12 units) of water and close the loop?


I have a feeling that it can but think at the same time that feelings have no place here, just facts.


thanks,
Marcel

[Red_Sunset]

But my own position is that I'd want to avoid people wasting their time on overunity concepts that don't work.
Until you can provide either solid experimental evidence or a plausible theory on why gravity acts in a non conservative way in this device then I think it is a reasonable stance to take.

Seamus & TK,
As watchdogs that qualify over-unity examples with a plausible theory that will waste no-body’s time,  see below ONE of the MINOR Zed OVER-UNITY aspects that should have woken up the dog that was watching !

I re-arranged some of the measurement details somewhat to make it stand out more. Reduced the pod area to 50% of the outer lift layer area. The idea is to illustrate Wayne's strategy &  logic towards over-unity, not by creating energy, but by reducing costs (governments could learn something here). The point here is not to explain where the energy is coming from and how this impacts the sacred laws and re-arranges what we new before. The point is that the logic flow is sound and it produces results.

Sample Specs
Total lifting area (risers + pod)= 26031cm2
Outer Layer 4 area = 5500 cm2
Control lift area (pod)  =  2750 cm2 (50% of outer layer area)
Layers = 4 + pod


Lift efficiency ratio = Total Lift Area/Pod area
                                 =  26031/2750
                                 = 9.46

Pod efficiency ratio = Pod area / Total Lift area
                                 =  2750/26031
                                 = 10.5%

Virtual displacement water for stroke 3” or 7.5cm
Virtual volume = Displacement volume  - Pod volume
                         = 195.23 â€" 20.62
                         = 174.6 Ltrs   (cm3 -> Ltr)
** The virtual water is non existing water that plays a role as real water as per standard Archimedes

Real  displacement water for stroke 3” or 7.5cm
Total Volume to rise 3” = Outer Lift area x stroke length
                                       = 5500 x 7.5
                                       = 41.2 Ltrs    (cm3 -> Ltr)
** The real displacement water must be provided at the appropriate pressures for each layer. This is offset in the energy balance. The water is effectively shared in the same way the air spaces are shared by all layers.  As a quick calculation, the balance of energies would be unity  (if we ignore the heads)

Actual used volume for for stroke 3” or 7.5cm ( using pod only)
Total Volume to rise 3” = Pod Layer area x stroke length
                                       = 2750 x 7.5 
                                       = 20.6 Ltrs    (cm3 -> Ltr)
** Here we use the pod area for stroke displacement water injection. Supported by sufficient balanced water in the U-bend for the stroke length. The pod area is 50% of the largest lift outer area. This reduces our water and input costs by 50%
                 
Energy used = 20.6 x 5.62 = 115.7 KgMtr   (8psi = 5.62 mtrs)
Potential energy gained by lifting  3000Kg for 7.5cm @ 8psi = 225KgMtr

NET EFFICIENCY =  225/115.7 = 144.5 %
OVERUNITY by 44.5%

Note: This is only one of the minor optimizations towards over-unity in the Travis Zed. The major optimization trump card is way more simple in its setup but because of it more ingenious and more beneficial 
Do not be fooled by the first sight and your first impressions of this set of inverted cooking pots. There is way more cooking going on here than you think.

Please Note: Figures should be accurate within a 5-10% margin either way.   The figures  listed are rudimentary and intended to show the general operational energy flow of the Zed device. No overheads incurred by mechanical or other losses have been included.

Regards, Michel

[LarryC]

I've corrected the SI calculation sections on the attached spreadsheets and showing the results from each in the first picture.

The last two results are from the System Rise water height calculator to show the force before a rise and after a three inch rise, they are the same. It is accomplished by increasing the water level between the pod and the pod retainer wall during the rise. The total force is different from the 3 Riser at the top because the Pod water level was reduced and replaced with air pressure to maintain the water head in the Risers. This has been explained many times, but some just keep implying that there is no force over distance values.

For the real anal, use the System Rise water height calculator and start with a rise value of .00001 and press the Adjustment button, wait 10 seconds for the number to quit spinning. Record the force results. Continue incrementing by .00001 until you reach 3 inches. Please report back in a month with your results. I'm sure you will remember next time before implying again.

Regards, Larry   

[TinselKoala]

Why can you mathematicians not answer the simple questions, without all this obfuscation?

You start at the bottom, precharged however. DON'T TELL ME THE DETAILS of the precharge, I will take your word for it AS LONG AS IT REMAINS IN THE SYSTEM AND IS THE SAME AT THE END OF THE CYCLE. You have some "load mass" sitting at some initial height and this mass and this height you will tell me.

You use an external piston and cylinder and plumbing to apply A KNOWN AMOUNT OF INPUT WORK by injecting your desired quantity of water into the chamber surrounding the pod. All I need to know here is the surface area of the piston, the stroke length, and the average force exerted on the piston to push it in for the stroke length.

You then measure the HEIGHT of the LIFTED WEIGHT sitting on top.

You have now completed ONE HALF CYCLE. The input work can be compared to the output work at this point, without all the complications introduced by Marcel and Larry. But you aren't done, because perhaps your precharge has boosted the output, so for TOTAL work balance you need to return to the initial state EXACTLY.  So now you compute the work in the second half of the cycle: Your lifted load mass becomes the "input" and your injection piston travel is now your "output".  If the system returns to the initial state on its own, just from removing whatever force you are holding the piston in, you are done. If you have to put in extra work, either by pulling out the piston or pressing down on the lifted weight, this counts as input in the total work calculation. If the input piston pushes out with more force than you took to put it in initially, averaged over the full stroke, then this difference counts as output work in the total work calculation.

Please use measurements from an actual system. Made-up numbers and complex calculations only serve to muddy the waters.

So, will one of you numbers crunchers please just give the values that will allow actual calculations of work in and work out, total?  This should be very easy to do, and the only time that it has actually been done, by webby some days ago, the calculation I performed on the given data indicated an "ou" value that was in the "noise floor" or the margin of error in the measurements.

The fact that there is so much reluctance to give the required numbers in an easily interpretable form indicates to me that there must be some reason for not doing so. Perhaps, using the measurements and calculations that I would like to see used, you don't get such big overunity numbers.

In fact Michel's given OU number is so embarrassingly great that he has to remind us, over and over, that losses aren't being included... as if losses in a wellconstructed system would prevent the OU from several layers from building up drastically.

So let's break it down again.
MrWayne has said that he has a simple, three layer system that is clearly overunity by itself. These are his own exact words. And I have asked over and over, how is this clear overunity measured IN THAT SYSTEM, and what is the ratio of input work to output work for this simple system that is clearly overunity by itself?

All that is needed to answer these questions properly is to set THAT SYSTEM up with an external piston and cylinder that can be used to apply a known input work, and do a few simple measurements of the work put in and the work gotten back out. Webby is nearly doing this with his system.... the two thousand dollar set of tennis ball tubes.... but not quite. In the only experiment he reported that is "almost" this simple, I got a numerical answer that was "positive" but not by much, in the noise floor of the measurements.

[fletcher]

Actual used volume for for stroke 3” or 7.5cm ( using pod only)

Total Volume to rise 3” = Pod Layer area x stroke length
                                       = 2750 x 7.5 
                                       = 20.6 Ltrs    (cm3 -> Ltr)

** Here we use the pod area for stroke displacement water injection. Supported by sufficient balanced water in the U-bend for the stroke length. The pod area is 50% of the largest lift outer area. This reduces our water and input costs by 50%
                 
Energy used = 20.6 x 5.62 = 115.7 KgMtr   (8psi = 5.62 mtrs)

Potential energy gained by lifting  3000Kg for 7.5cm @ 8psi = 225KgMtr

NET EFFICIENCY =  225/115.7 = 144.5 %

OVERUNITY by 44.5%

Note: This is only one of the minor optimizations towards over-unity in the Travis Zed. The major optimization trump card is way more simple in its setup but because of it more ingenious and more beneficial.

Regards, Michel

TK .. I agree with your objective approach to method, measurement & data collection 100%.

Just a side note about Mr Sunsets last contribution.

I get 194 % Net Efficiency based on his figures i.e. that's OverUnity by 94 % [give or take 5-10% margins of error & no losses] - re 144.5 %, but I don't want to seem trivial.

I also have a little problem in comprehension just near the end - he says, I think, that the systems input is 20.6 liters of fluid which raises the PSI from 0 to 8 PSI - then he uses a head justification of 5.62 m = 8 PSI.

AND ... 8 PSI is indeed 5.62 m thereabouts - so he seems to be saying, I think, that 20.6 liters [kg] of fluid is raised by a height of 5.62 m = 115.7 kgm = 1135.72 joules Input Cost ["Energy Used"].

Yep, that looks ok - the trouble I'm having is that 8 PSI at 5.62 m head means that 20.6 liters [kg] would only raise its PE by half that distance because Static Pressure = pgh = 1000 kg x 9.81 m/s^2 x 5.62 m = 55.132 KPas = 8 PSI [i.e. fluid depth 5.62 m], while an equivalent column of solid would have its CoG half that distance [i.e. 2.81 m] since density is uniform.

So was the potential raised 5.62 m or 2.81 m ? If 5.62 m is correct for the "Energy Used" calc the 20.6 liters would have to be raised 5.62 m to get 115.7 kgm or 1135.72 joules & this doesn't seem to match the head statement of 8 PSI at 5.62 m.

If the former even 194 % looks paltry.

Or maybe the hydrostatic paradox is coming home to roost - higher pressure means more work done to force a fluid into the system, especially if you want to force it in at a lower level & take it out again - I can only think that the major trump card lies in volume changes beneath the risers but then we all know pressure is scalar & pushes up & down & that's what the paradox is about ?

I hope you get straight answers this time, especially from the spreadsheet kings.