#### viktor94

##### The VIC need a higher resistance load?
« on August 2nd, 2012, 02:32 PM »
Hi, I recently read the "Stanley Meyer Explainded" thread over at overunity.com and came across the common problem that no one seems to be able to get a high voltage to develop across the WFC.

The suggested problem is that the WFC has too much of resistace (ohms), all of you experimenters are using a single WFC right? The VIC transformer dont put out enough amperage to drive the voltage very high with such a high resistive load (single WFC). Is ohms law applying? U = R * I? He is comparing it to match a speaker to an audio amp, a too low ohm speaker (ex 1 ohm) is going to burn up an amp rated for ex an 8 ohm speaker.

The solution would then according to the thread be to connect more WFC cells in series, the resistance would then increase and a higher voltage would be allowed to build up.

I dont know how Meyer himself wired the VIC´s to the actual cells but here is a picture that show 10 of the 11 cells hooked up in series:

http://www.overunity.com/7030/stanley-meyer-explained/240/#.UBrmG1JHHi4

Hope this can be to some help!:)

/viktor94

#### Webmug

##### RE: The VIC need a higher resistance load?
« Reply #1, on August 2nd, 2012, 02:58 PM »
Hi,

1 WFC had a VIC transformer connected to it. So 10 VICs were connected to 10 WFCs operating as independend systems.

Br,
Webmug

#### viktor94

##### RE: The VIC need a higher resistance load?
« Reply #2, on August 3rd, 2012, 01:03 AM »Last edited on August 3rd, 2012, 01:05 AM by viktor94
Thats what i thought first too, and although it makes sense im not shure it was connected in that way.

Look at the pictures,

The first one shows the VIC unit were you can count the VIC cards to 9.
[attachment=2038]

The second one shows the resonant cavity were you can count to 10 cells (the 11th seemed to be unconnected/taken out). It also shows all the ten cells connected in series.
[attachment=2039]
It doesent seem like there was enought VIC´s for all the WFC cells?

The third photo shows the two cables (not twenty) coming out from the bottom of the resonant cavity.
[attachment=2040]

I think that the system with one VIC to one WFC didnt work out, so he ended up with connecting all of the WFC in series to just one VIC card.

This is just what i think, has anyone else a theory/know how it was connected?

/viktor94

##### RE: The VIC need a higher resistance load?
« Reply #3, on August 3rd, 2012, 04:18 AM »
Everybody sais series series....cell.Take a closer look..if it was to be series it would be one wire on top and one at the bottom ,but in the pic is zig-zag,that's result in a ''short'' between anode and cathode.About H2opower he always says that series cels are beter but series resistance means more resistance for the transformer to handle with.The trafo could not handle 1cell resistance,think about 10 cells in series..more resistance....so i think..is BS..
Look at the att

#### viktor94

##### RE: The VIC need a higher resistance load?
« Reply #4, on August 3rd, 2012, 09:07 AM »
The cells are SERIES connected. When you wire capacitors, batteries etc in series positive (+) will be connected to minus (-). The "short" is supposet to happen. You are talking about hooking them up in parallel configuration.
[attachment=2044]
Capacitors in series.

As for that the increasing resistance would have a negative effect on the VIC´s performance, look at this:

OHMS law: U = I * R, or Voltage = Amperege * Resistance

If one cell for example has a resistance of 100 ohms and the VIC is capable of supplying 0.5A the maximum voltage that can develop across the WFC would be 50 v.

U = 0.5 * 100
U = 50 Volts

If you instead connected 10 WFC in series the total resistance would be 1000 ohms and the voltage that develop across the WFC would be 500 v. The VIC might have to get over this threshold to function best and allow the voltage to rise even higher (500+).

U = 0.5 * 1000
U = 500 Volts

The "conditioning coat" that we see on lawton and ravis (even on myers demonstration cell in some videos) might just do this (increase the cells resistance so that voltage can take over), it is a high resistance coat, not isolating but still resistance.

The resistance of one single WFC with distilled water might be enough to allow voltage to take over, but i think its worth a shot for those that have more than one cell to try out the concept of series WFC as its appears to be what meyer himself used.

/viktor94

#### Webmug

##### RE: The VIC need a higher resistance load?
« Reply #5, on August 3rd, 2012, 10:19 AM »
Quote from viktor94 on August 3rd, 2012, 09:07 AM
The cells are SERIES connected. When you wire capacitors, batteries etc in series positive (+) will be connected to minus (-). The "short" is supposet to happen. You are talking about hooking them up in parallel configuration.

Capacitors in series.

As for that the increasing resistance would have a negative effect on the VIC´s performance, look at this:

OHMS law: U = I * R, or Voltage = Amperege * Resistance

If one cell for example has a resistance of 100 ohms and the VIC is capable of supplying 0.5A the maximum voltage that can develop across the WFC would be 50 v.

U = 0.5 * 100
U = 50 Volts

If you instead connected 10 WFC in series the total resistance would be 1000 ohms and the voltage that develop across the WFC would be 500 v. The VIC might have to get over this threshold to function best and allow the voltage to rise even higher (500+).

U = 0.5 * 1000
U = 500 Volts

The "conditioning coat" that we see on lawton and ravis (even on myers demonstration cell in some videos) might just do this (increase the cells resistance so that voltage can take over), it is a high resistance coat, not isolating but still resistance.

The resistance of one single WFC with distilled water might be enough to allow voltage to take over, but i think its worth a shot for those that have more than one cell to try out the concept of series WFC as its appears to be what meyer himself used.

/viktor94
The VIC was designed to operate with a dead-short condition and prevent current flow.

Br,
Webmug

#### Gunther Rattay

##### RE: The VIC need a higher resistance load?
« Reply #6, on August 3rd, 2012, 01:12 PM »Last edited on August 4th, 2012, 04:12 AM by bussi04
Quote from Webmug on August 2nd, 2012, 02:58 PM
Hi,

1 WFC had a VIC transformer connected to it. So 10 VICs were connected to 10 WFCs operating as independend systems.

Br,
Webmug
The 2nd  pic shown above demonstrates a serially wired cell configuration. we must take into account that meyer has tried several different approaches as we do. taking into account that the whole meyer stuff has been hidden for 10 years it may be that only those parts not working correctly have been given back to the public.

so there will have been direct coupled wfcs  wired one by one and those in serial configuration.

#### Amsy

##### RE: The VIC need a higher resistance load?
« Reply #7, on August 5th, 2012, 08:34 AM »Last edited on August 5th, 2012, 01:12 PM by Amsy
Quote from viktor94 on August 3rd, 2012, 01:03 AM
Thats what i thought first too, and although it makes sense im not shure it was connected in that way.

Look at the pictures,

The first one shows the VIC unit were you can count the VIC cards to 9.

The second one shows the resonant cavity were you can count to 10 cells (the 11th seemed to be unconnected/taken out). It also shows all the ten cells connected in series.

It doesent seem like there was enought VIC´s for all the WFC cells?

The third photo shows the two cables (not twenty) coming out from the bottom of the resonant cavity.

I think that the system with one VIC to one WFC didnt work out, so he ended up with connecting all of the WFC in series to just one VIC card.

This is just what i think, has anyone else a theory/know how it was connected?

/viktor94
Hi,

Spontaneously I would say: Let´s try to find out where the two (red and black) cables were connected. Maybe we can ask the photograph.

On the other hand, I was thinking of something like that:

Car Batterie:
12V / 11 = 1,09V
12V / 10 = 1,2V
12V / 9 = 1,33V

Car alternator:
14V / 11 = 1,27V
14V / 10 = 1,4V
14V / 9 = 1,55V

The theoretical minimum voltage of electrolyses is 1,23V.

Maybe the WFC was supplied by the batterie and the VIC hit a high voltage field to each tube. (diode in the VIC to inhibit dead short of +12V). A high voltage field can lower the resistance of water in the tubes, so more gas can be produced. (tap water has high resistance).
Normally I would say, this only ends up in electrolyses, but this can be performed like in the water spark plug project. -->the high voltage ignites the plasma in the spark plug. Something like a eleictrical arc or discharge can be formed, and the water is splitted with higher effectivity. The isolated VIC transformer would be very important!

Edit: This would make only sense when all VIC are running in ON Mode.
Watched the patent again. It says that every VIC is connected seperated to one tube pair (to find their "resonance"). But there is no comment about a serial connection from tube to tube.
So if the system of one tube pair is highly sensitive to frequencys etc., it would not make sense to conncect the tube pairs together.

Maybe only one VIC was connected to all.... but why to 10, not to 11?

#### geenee

##### RE: The VIC need a higher resistance load?
« Reply #8, on August 5th, 2012, 10:14 AM »Last edited on August 5th, 2012, 10:18 AM by geenee
11 resonant cavity(10 cells make HHO(series connected= 1 unit) and 1 cell make steam water).

but in control vic unit has 9 vic cards(resonant cavity), 1 steam resonator card and 1 vic card(gas processor).

i thinks "if i make hho gas then i will make a circuit at maximum hho output(not variable).but if i need variable output then i will add double circuit,triple circuit,4X circuit etc.that easy way and Stan use like that ,Stan has 10X level(more unit more Capabilities of amps(amps=magnetic flux)).maybe vic unit connect paralleled(9X unit+1x gas processor=10x).End of vic unit connect to resonant cavity 2 wire(pos and neg)."

just thought.
geenee