Stanley Meyer Explained

[h20power]

Do not worry about that as the figure 6-1 VIC transformer is not to be built unless you are making the injectors as they will not work on the WFC's. Plus I already go over just what bidirectional means and all it is something that is cross wrapped, nothing more. Please I know you mean to help but your making this hard for people to understand. The figure 6-1 VIC Matrix Circuit has a lot more rules that go along with than that of figure 3-23. These people want results not confusion, okay? Please I got this, okay?
In the 8xa circuit the 2 H inductor is a guess that seems to work with the 3 inch overlap capacitors. But there is a way to calculate the exact size needed and that it to fully integrate the time constance (TC) to Meyer's work. For these are very leaky capacitors and the resistance and impedance totals in ohms go towards giving you enough time to be able to charge such a leaky capacitor. The Time Constant (TC) for a capacitor is TC = R x C, where R is resistance in ohms and C is capacitance in farads. This directly relates to what size the chokes need to be in order for the capacitors to be charged, and I am still working through the math as we speak. The inductor TC is also important TC = L / R. As I have been saying these VIC Matrix Circuits look deceptively simple but nothing could be farther from the truth of how they really work.

h2opower

[guruji]

That coil is to wrap three coils bi directional near each other?

[illuminati]

guruji

Email me dude 

bc109@hotmail.co.uk

Very easy to build this to try out.

[h20power]

http://www.youtube.com/user/Venturecaplaw#p/u/52/GXcxswDcUbI

http://www.youtube.com/user/Venturecaplaw#p/u/51/-vznuNkEBto

http://www.youtube.com/user/Venturecaplaw#p/u/54/HGgfOqlxWqU

http://www.youtube.com/user/Venturecaplaw#p/u/55/V6xa76IshbE



In these videos you will learn these two time constant (TC) formulas

Inductance TC = L / R

Capacitance TC = R x C

And it takes a total time of 5 TC's to fully charge or discharge a inductors current and a capacitors voltage.



Now since we are dealing with a leaky capacitor the time it take my electrode pair to discharge is:

5(3778.83 pf x 78.54 ohms) = 2.967893^-7 sec x 5 = 1.48395^-6 sec to fully discharge my capacitor. Not a lot of time, huh?

Now by adding in resistance you gain more time before the capacitor fully discharges.

Example if I add in 40k ohms, which would be all coils added up together, I'd get this: 5(3778.83 pf x 40078.54 ohms) = 7.5724995^-4 sec

And this takes place at the same time the inductor is dumping current into the capacitor as it losses current giving even more time before the capacitor fully discharges.


So now you can see that adding in resistance has two purposes, to prevent amp leakage, and to add more time to the capacitors fast discharge time so it isn't too fast. 

So the bifilar chokes will go to slow this leaky capacitors discharge time down and the resistance will go to slow this leaky capacitors discharge time down. Wow, I learn something new everyday!


We need to calculate the total resistance of the circuit, and due to the way the circuit works we only get to add one resonant value of a choke with this resistance.



So that gives us all resistances that are in-line with the capacitor plus the XL of the choke plus the Xc of the capacitor at the resonant frequency. This value is then added to the Time Constant so see if the choke size will give a time that is greater than the frequency time count per second. Remember the maximum pulse rate is always twice the resonant frequency, that means you divide the one pulse time by two.



Unfortunately I can't add all the resistance of the wires due to I don't know the resistance per foot data. I only know the 0.125 mm wire is around 24 ohms per foot. I will only have all the data on the chokes after it is built and I put a resistance meter to it to take a reading. So here we go .



Planned resonant frequency for me is 1.83k Hz which gives a single pulse time of 5.4645^-4 seconds divide that by two and I get 2.7322^-4 seconds. This is my charge time and it must be less than the capacitor drain time.



So starting with the secondary I will be using 263.6 ft of wire at 24 ohms per foot. Which gives me 6326.44 ohms of resistance.

The chokes @ 2 H gives me, 2 x pie x 1830 Hz x 2 H = 22996.46 ohms

The capacitor gives 1 / 2 x pie x 1830 Hz x 3778.83^-12 f = 23015.04 ohms plus 78.54 gives 23093.58 ohms



Now to use the TC = R x C formula

TC = (6326.44 + 22996.46 + 23093.58) x 3778.83^-12 = 1.9807^-4 seconds is for one time constant. So it is multiplied by 5 giving me 9.9036^-4 seconds for full discharge time of the capacitor.

This shows me that the 2 H choke will be enough for me to charge the capacitor since 2.7322 is less than 9.9036. With this size choke I will have a voltage drop just over 36.8% between charges so I will go to a larger choke size to improve those numbers.



Now at least all of you know how to calculate the value of your chokes so you get a working VIC Matrix Circuit the first time around, no more guessing! 


Enjoy!

h2opower

[guruji]

Hi illuminati I've emailed you. Another thing I built two cells and put them in parallel is this ok or in series bettet?
Thanks