Demonstration Cell - Tubular Cluster Array

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No I don't see what you mean about the diodes, I have all six diodes in my alternator but i am not using the negative from the diodes, i am just going straight from the center of the stator to the cell like in the diagram, so i bypass the negative diodes.

I did the test, here are the results:

Cavity = 1 in * 4 inch * 3.14 = 12.56 Cubic Inch = 205.8 Cubic Centimeters

Salt Water
5 psi in 16 seconds

Rain Water
1.3 psi in 30 seconds
2.6 psi in 60 seconds
4 psi in 90 seconds
5 psi in 113 seconds

Tap Water
1.9 psi in 30 seconds
3.7 psi in 60 seconds
5 psi in 81 seconds

5 psi @ 205.8 cc  Cavity = 115 cc of gas produced

Salt Water = 431 cc/min
Rain Water = 61 cc/min
Tap Water = 85 cc/min

All tests performed with 10 volts on the Variac, no other electrical measurements taken.
This is with Four 3" Tubes

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From Internet;

Imagine a cylinder 20 inches high with a radius of 5 inches. To find the radius squared, multiply that number by itself, so 5 times 5 equals 25. Now multiply Pi by the radius squared: 3.14 times 25 inches equals 78.5 inches. Next, multiply 78.5 by the height (20 inches). The product is 1,570. Therefore, the volume of this cylinder equals 1,570 cubic inches.

(1 * 3.75 * 3.14 = 11.775)
Your Radius Squared, 1.875 * 1.875 =  3.515625
3.14 * 3.515625 = 11.0390625
11.0390625 * 1" = a volume of 11.0390625
The Volume of the Above cylinder is 11 Cubic Inches. 11 Cubic inches is 180.2 Ml or CC.
You need 1k cc for a 1 liter cavity, not 180 cc's.

Converting Cubic inches into Milliliters, Example. (Volume Multiplied by 16.387064.)

11 Cubic inches * 16.387064 is 180.257704 cc's of volume.

61 Cubic inches * 16.387064 is 999.610904 cc's or ML of volume.

Assuming your Container is 4" in Od, this would leave you with 3.75 ID, If your wall was 1/8th Thick. Just take your time to find your 1 liter cavity.

I'll post a schematic.

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Crayola Schematic

You guy's are onto stan like Flys on Stank! Proud to see this being deeply looked into.

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Hrm,,, I wish everything back then was as Clear to me as it is clear to me now. Keep testing, id like to play with my tubes and alternator again just for fun but I cant.

I used 6 diodes to get 7 Psi in one minute, stan used 3, Now that brings out some questions also, But, if you remember in his video, He did say he Made a Tweak, i wonder what that tweak was, it was also recorded to have 12 volts across his cell, Hrm, something Does not sound right about this at all. 

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My Inside Radius is 2 inch

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what i dont get is why would stan say that as voltage continues to rise amps drop  when hitting resonance..  he also emphasized on how more winds on the choke would allow for a higher production.. so you say you are only hitting 12.5 volts how much gas will 30 volts at 4.4 amps produce?

when he says more winds more volts does he mean that you have more resistance  and the power supply can be of higher potential to supply the  load? so there for your intial power generation can be like 20,000 volts and the choke has the resistivity to only allow 12.5 volts at 4.4 amps into the cell? almost like making a minature power grid.. the primary and secondary wires are cosidered the high end like line power, while the chokes are minature step down?

but this observation wouldnt conclude with stans statement of being able to have a cavity in such a resonant that it would turn water into gases faster then water could enter the tube. stated in one of his videos.

@ warp.

i like your perspective brought forward about how a pressurized gas reacts to gating..

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Here are some pictures of my cell taken with the camera on my laptop.


(http://img.photobucket.com/albums/v81/bigbuba/Photo61.jpg)

(http://img.photobucket.com/albums/v81/bigbuba/Photo62.jpg)

(http://img.photobucket.com/albums/v81/bigbuba/Photo63.jpg)

(http://img.photobucket.com/albums/v81/bigbuba/Photo64.jpg)

(http://img.photobucket.com/albums/v81/bigbuba/Photo65.jpg)

(http://img.photobucket.com/albums/v81/bigbuba/Photo66.jpg)

(http://img.photobucket.com/albums/v81/bigbuba/Photo67.jpg)

(http://img.photobucket.com/albums/v81/bigbuba/Photo68.jpg)

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Alumina Quenching Circuits - 0.016"

1 hole
2 holes
4 holes

(http://img.photobucket.com/albums/v81/bigbuba/Picture13-1.png)

The one I'm holding is a 4 hole, but it's too hard to get it to focus, the holes are so tiny!

I got these a while ago, but have been too busy to build the nozzles and test them, and then equip the cell at both outputs. Within 2 weeks I will do this.