# open-source-energy.org

## Open - Source - Research => Open-Source Research => HHO / Browns Gas / Hydroxy / Stan Meyer => Topic started by: Heuristicobfuscation on October 18th, 2014, 10:47 AM

Title: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on October 18th, 2014, 10:47 AM
The higher the voltage the higher the resistance has to be!

If we are aiming to increase the voltage potential accros the Cell then the higher the voltage is the higher the resistance has to be
to maintaine the same watt or power consumpiton other wise if the resistance goes down the power we are consuming goes up.!

check tabel attached.

volts = √watts × ohms

Doesnt this make sense? This is kinda counter intuitive isnt it?

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Lynx on October 18th, 2014, 12:38 PM
Actually, these quantities do go hand in hand, just as they should!

With the WFC hitting high enough voltage, dissociation of the water molecules takes place, which is the working theory anyway with regards to Meyer's WFC tech, so instead of getting/maintaining traditional water electrolysis, hydrogen fracturing is the way to go, splitting water into hydrogen & oxygen using high voltage/low current as opposed to the "traditional" way of applying electrolysis to the water using low voltage/high current.

Apparently the last thing that should take place in the (Meyer) cycle is a sudden inrush of high current, which is detected by the control circuit which in turn pauses the pulsetrain hitting the cell, giving the water time to "recover" inbetween the cycles and also to replenish the cavity in the cell which now momentarily has become somewhat depleted of water due to the effective gasproduction.

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on October 18th, 2014, 08:38 PM
What determines the Amp Draw in a given circuit?
is it not the load?

why is it our capacitors are acting as a load and consuming so much current?
capacitors are not suposed to consume curent.

here is the key.
limit current.

how?

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on October 18th, 2014, 08:51 PM
A capacitor will not charge to its voltage limit unless the source voltage reaches that limit.

Example...a 12v capacitor will not charge with a 2v constant charge.
we need a 12v constant charge supply.

how can we expect to charge a capacitor with same voltage pulse?

ok maybe thats were stan meyer mentions the electron bounce phenomenon.
so the electron gets elongated form its orbit and bounces back with kinetic energy in form of voltage.

this couses the higher voltage spike needed to charge the water beyond tranformer action?
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on October 18th, 2014, 08:57 PM
How is the voltage going to climb into infinity if the resistance doesnt climb to infinity? :huh: :idea: :huh:
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Matt Watts on October 19th, 2014, 12:16 AM
Quote from Heuristicobfuscation on October 18th, 2014, 08:57 PM
How is the voltage going to climb into infinity if the resistance doesnt climb to infinity? :huh: :idea: :huh:
The amperage drops to zero, but only when the voltage is at its peak.  And when amperage is zero, resistance has no meaning.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Gunther Rattay on October 19th, 2014, 01:51 AM
Quote from Lynx on October 18th, 2014, 12:38 PM
Actually, these quantities do go hand in hand, just as they should!

With the WFC hitting high enough voltage, dissociation of the water molecules takes place, which is the working theory anyway with regards to Meyer's WFC tech, so instead of getting/maintaining traditional water electrolysis, hydrogen fracturing is the way to go, splitting water into hydrogen & oxygen using high voltage/low current as opposed to the "traditional" way of applying electrolysis to the water using low voltage/high current.

Apparently the last thing that should take place in the (Meyer) cycle is a sudden inrush of high current, which is detected by the control circuit which in turn pauses the pulsetrain hitting the cell, giving the water time to "recover" inbetween the cycles and also to replenish the cavity in the cell which now momentarily has become somewhat depleted of water due to the effective gasproduction.

If you want to check that gating theory out you can use my pre-buit application for stuff like that:
Quote from Gunther Rattay on September 18th, 2014, 11:58 AM
nav, with this setup you can control the gating and test your theory (at 13:50 min):

(for a full screen display click on YouTube bottom right hand corner)
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on October 19th, 2014, 07:18 AM
Quote from Matt Watts on October 19th, 2014, 12:16 AM
The amperage drops to zero, but only when the voltage is at its peak.  And when amperage is zero, resistance has no meaning.
Matt thanks for your comment...

I agree that the amps  need to fall as the voltage rises.... but I also belive the resistance has to Rise!  (this sounds counter intuitive) Plese check out the math..the power formula tells  us the resistance needs to rise...

Power consumption need to stay the same for  any given voltage rise!
If the power goes up when we apply higher voltages then we are using brute force electrolosis.

If voltage rises amperage need to fall!

wattage needs to stay the same! not  rise...

which  means  according to (volts = √watts × ohms) Resistance needs to Rise!   Please see attached spreadsheet....
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Matt Watts on October 19th, 2014, 11:00 AM
Quote from Heuristicobfuscation on October 19th, 2014, 07:18 AM
which  means  according to (volts = √watts × ohms) Resistance needs to Rise!   Please see attached spreadsheet....
And it does, in the form of impedance, not resistance.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on October 19th, 2014, 06:23 PM
Quote from Matt Watts on October 19th, 2014, 11:00 AM
And it does, in the form of impedance, not resistance.
Thanks again Matt,

I respectfully disagree. :P

Resistance is a must!  bare with me..
we may be out of context in regard to each others tought.
your probably thinking of the impedance required in inductive coupling to Water Cell?

Im refering to the Water itself.. Stan Meyer said that the water becomes part of the circuit in the form of resistance!

water as a dielectric! water as a resistor!

If the water property loses the ability to resist current then we have conventional electrolosysis.
The power formula tells us that in order for the voltage to rise for a given watt, the resistance has to rise.

here is the problem, contaminates in water  create dead short condition. voltage rises and resitance  falls cousing watts to go up.
(by the way this is something i learnd here in the forums probably in previus chat we had)

how in the world are we to increase the  water resistance?
so far in my personal experiments i havent been able to charge a water capacitor to significant levels using distilled water.

this tells me that there are so many parts per billion or million of contaminates present in the water bath. dont know the details of how many can creat the dead short condition. causing paralel path of current with in the inside of the cell that  shorts out and drains the voltage.

but i know enough that its not holding a significant charge for a significant period of time. therefore how can i possible increase the voltage if the capacitor cant even hold a significant charge without discharging to under 2 volts?

I  am not exploiting the real dielectric characteristic of the water.

In my case i need to fiqure out a way to either filter out the contaminates or clean the cells or purchase better quality distilled water.

This is the problem i see with water. Water is acting more like a semiconductor than a resistor or dielectric material.

If we cant predict the resistance of water then how can we expect it to not become more conductive?
If the water becomes more conductive then how can i control the power bieng feed into it?

Resistance is what regulates the Power in watts bieng consumed by the cell. V = √W × Ω
The more Resistance we have in the cell the less power we will consume!
The more Resistance we have in the cell the higher the voltage potential across the plates.
The more Resistance we have in the cell the less current will flow in water bath.
The more Resistance we have in the cell the better we can match external circuit impedance.

With Resistance we will not have a Dead Short Condition in water bath :exclamation:

I would be dead wrong if i did not recognize the importance of total ciruit impedance, but i would be swiming against the
current avalache produced by the dead short if i did not recognize the importance of  water dielectric.

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Matt Watts on October 19th, 2014, 09:06 PM
Okay, let me hop over to your side of the fence for a moment.

I want a high voltage, a low amperage, a low wattage and high resistance.  Okay, so far so good.

Lets pick some ballpark numbers to shoot for and see what we are up against.

How about 20000 volts drawing only 3 watts of power.  Sound reasonable?

So we have:  V = √W × Ω , lets re-arrange to:

Ω = V / √W  or in our case:  Ω = 20000 / √3

Ω = 11547

Seems to me 11.5k Ω isn't out of reach.  If this were mega-ohms instead of kilo-ohms, we would have our work cut out for us.

Now if we use chlorinated tap water, with tubes like Stan's, 11.5k Ω might be a bit hard to come by.  BUT...

Let's imagine though this resistance isn't what we see when everything is turned off, instead this resistance is what we see when the unit is powered up and producing gas.  We have bubbles on the tubes, we have Exclusion Zones near the tube surfaces and we are pulsing instead of just pouring the juice in continuously.  So in reality, we are using an average of 3 watts; our burst power over small intervals of time could be a lot higher.

Seems to me, you just made a case for what things should look like when the VIC is working properly.  I think the magic happens getting everything started.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: ~Russ on October 20th, 2014, 07:39 AM
some good thoughts flowing here.

keep thinking. keep posting.

~Russ
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Webmug on October 20th, 2014, 12:26 PM
Quote from Matt Watts on October 19th, 2014, 09:06 PM
Okay, let me hop over to your side of the fence for a moment.

I want a high voltage, a low amperage, a low wattage and high resistance.  Okay, so far so good.

Lets pick some ballpark numbers to shoot for and see what we are up against.

How about 20000 volts drawing only 3 watts of power.  Sound reasonable?

So we have:  V = √W × Ω , lets re-arrange to:

Ω = V / √W  or in our case:  Ω = 20000 / √3

Ω = 11547

Seems to me 11.5k Ω isn't out of reach.  If this were mega-ohms instead of kilo-ohms, we would have our work cut out for us.

Now if we use chlorinated tap water, with tubes like Stan's, 11.5k Ω might be a bit hard to come by.  BUT...

Let's imagine though this resistance isn't what we see when everything is turned off, instead this resistance is what we see when the unit is powered up and producing gas.  We have bubbles on the tubes, we have Exclusion Zones near the tube surfaces and we are pulsing instead of just pouring the juice in continuously.  So in reality, we are using an average of 3 watts; our burst power over small intervals of time could be a lot higher.

Seems to me, you just made a case for what things should look like when the VIC is working properly.  I think the magic happens getting everything started.
So now we know why Stan went to the Injector design with high ohm coils. Reverse the problem! :cool:

~webmug
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on October 20th, 2014, 07:25 PM
Thank you all for your feedback.. This has helped me allot in trying to understand this beautiful enigma.

I would like to share with you guys the following.

In his very first memo that is on the book WFC technical brief pg 1-1 he says the following.

"Water now becomes part of the Voltage Intensifier Circuit in the form of "Resistance"
See Picture attached.

If Water conducts electricity due to contaminates how can it act like a Resistor?
According to Stan is it still part of the Voltage Intensifier Circuit?

How can it possibly be part of the intensification if the contaminates are
hindering its dielectric performance on the contrary its  de-intensifying the circuit.

Also on the following too pages that is pg 1-3 of Memo 420
See Attached pic

He says the following:

"Voltage across the inductor and the capacitor are theoretically infinite.
However, physical constraints of the components and circuit interaction prevent the voltage from reaching infinity."

I would like to point at the following phrase.  "physical constraints"

What would physically constrain the voltage from climbing?

Yes the water we are using has contaminates even the distilled water!
Those contaminates are reducing the resistance of water.

I don’t know if Stan Meyer filtered his water. :huh:
I don’t know how his VIC worked as he claimed with any kind of water. :huh:
Actually there is allot I don’t know.. :huh:

But I do know the following..

The Math does not lie :exclamation:

And so far it keeps telling me that  Resistance needs to Rise!

Ok now here goes the good stuff and I’m hoping you guys can prove me wrong...

Some of you will say and quote Stan mentioning that using the external circuit one can "tune" into the contaminates
Changing the external frequency and resistance!

Ok this is my response and please help me with this one:

Does the External "Tuning" Increase  the internal water resistance?

If it doesn’t then I’m sorry but you are using conventional electrolysis. :imsorry:
I personally don’t think it will affect the resistance but i could be wrong.

Please, please, prove me wrong show me that you can affect the internal resistance
because the alternative is much more complicated. :facepalm:
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Ravenous Emu on October 21st, 2014, 06:03 AM
Quote from Heuristicobfuscation on October 20th, 2014, 07:25 PM
"Voltage across the inductor and the capacitor are theoretically infinite.
However, physical constraints of the components and circuit interaction prevent the voltage from reaching infinity."
http://www.electronics-tutorials.ws/accircuits/series-resonance.html
Resonant Voltage Rise or "Q - Magnification".

He's saying you'll have losses in the "electronics".
I.E. The whole circuit.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: firepinto on October 21st, 2014, 06:08 PM
There was a water filter built into the Dunebuggy on the 11 cell system.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on October 21st, 2014, 08:06 PM
Quote from firepinto on October 21st, 2014, 06:08 PM
There was a water filter built into the Dunebuggy on the 11 cell system.
Thanks for your comment!

Funny ive seen this picture before and never paid attention to it. :exclamation:
This picture probably answers one of my  questions.
Did stan meyer ever used a filter? .. aparently he did for the 11 cell system!

it actually says on the drawings "Entraped positve charged contaminates" and  Entraped 'Negative contaminates"
interesting using "electrostatic filter"

Ok  great i wont sleep tonight lol...going to hit the books again.
How in the world does that "Electrostatic filter" work? :huh:

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Lynx on October 21st, 2014, 11:33 PM
Quote from Heuristicobfuscation on October 21st, 2014, 08:06 PM
How in the world does that "Electrostatic filter" work? :huh:
My guess is that it works much the way an electrostatic air cleaner works, the debris/contaminants gets negatively charged and "sticks" to the positive electrode and out comes the clean/filtered water.

(http://cdn4.explainthatstuff.com/electrostaticsmokeprecipitator.gif)

I guess you'd have to clean the filter every now and then, but I can live with that ;-)
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: brettly on October 22nd, 2014, 12:32 AM
the previous link by ravenous emu on series resonance, also mentions that at resonance there is zero voltage drop across the resonating portion of circuit ( capacitor (wfc)/inductor(choke)), which supports the use of a light globe across the choke(s) which goes out when resonance is hit.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: firepinto on October 22nd, 2014, 06:27 AM
Quote from Heuristicobfuscation on October 21st, 2014, 08:06 PM
Thanks for your comment!

Funny ive seen this picture before and never paid attention to it. :exclamation:
This picture probably answers one of my  questions.
Did stan meyer ever used a filter? .. aparently he did for the 11 cell system!

it actually says on the drawings "Entraped positve charged contaminates" and  Entraped 'Negative contaminates"
interesting using "electrostatic filter"

Ok  great i wont sleep tonight lol...going to hit the books again.
How in the world does that "Electrostatic filter" work? :huh:
I'm not sure, but I have one to test with. :D Six SS fender washers, four SS nuts, one piece of plastic cut from a jar lid, one 2 inch long 1/4-20 Nylon screw and nut, and some ring terminals.  A couple crude tests I did before my MOSFETS blew up did show some charging.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: ~Russ on November 3rd, 2014, 01:55 PM
Quote from brettly on October 22nd, 2014, 12:32 AM
the previous link by ravenous emu on series resonance, also mentions that at resonance there is zero voltage drop across the resonating portion of circuit ( capacitor (wfc)/inductor(choke)), which supports the use of a light globe across the choke(s) which goes out when resonance is hit.
this would be a good test...

~Russ
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on November 3rd, 2014, 05:21 PM
Quote from ~Russ on November 3rd, 2014, 01:55 PM
this would be a good test...

~Russ
Good Idea!

will have to try. and seems easy enough..

Thnks
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on November 15th, 2014, 04:52 PM
This is how we can increase the  RESISTANCE OF WATER ! :idea:  yes we can increase the resistance!   volts = √watts × ohms

The more I Study Water the more it behaves like a Semiconducor!

So instead of treating it like a dielectric. I will start to treat it like a Semicondutor.

Here is the plan...

#1 how can one change the characterisitics in a semicoductor?
in the industry this process is called "Doping"

#2 So I will Start "Doping" Water.

Becouse of internal parrallel path and "Ez Zone" there exist leakage that the capacitor does not maintiane charge for longer periods.
by doping the dielectric we can extend the charge capacty of water...

The question is what material can we add  to the water that does not involve chemical reaction?

Here is the answer and run with it yes fly:

Option number #1   Silica  Dioxide Carefull not to breath.
Option number #2   Crushed Quartz crystal
Option number #3   Crushed Glass, fine powder.

What do they have in common?
[{(sand...Silicone)  Same material used in semicondutors!}]
Yes
They all have a resistive carachteristic to current and do not bond chemicaly to the water.

if we are to Dope water then we can saturate the water with these elements.
The more we Dope the water with resistive elements the more we impede  current flow due to parralel path!

Ok  lets think about that....

if we can manage to have the water maintaine the charge for longer periods of time then the charge rate will be significantly increased.

dont know about you guys but those of us who have spent countless hours  experimenting we are constantly trying to charge the  water :exclamation: thats the name of the game.

problem is as soon as you disconect the power the amount of charge is significantly reduced!
By Doping the water the charge should no longer significanlty be reduced due to constant dielectric element introduced in mixture.

We have a constant strugle with the VIC... and that is we have to pulse the voltage in an atempt to catch the charge before it de-energizes .

Im hoping that with the introduction of "Doping" we can increase the capacity of charge in the WFC.

What do you think?

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Enrg4life on November 15th, 2014, 07:24 PM
Do we need to increase resistance in the water or the resistance between the cathode and the anode. I say this because when the cell is producing bubbles the bubble them selves would increase the resistance between the anode and cathode which should allow the voltage to go higher
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on November 17th, 2014, 03:31 PM
Quote from Enrg4life on November 15th, 2014, 07:24 PM
Do we need to increase resistance in the water or the resistance between the cathode and the anode. I say this because when the cell is producing bubbles the bubble them selves would increase the resistance between the anode and cathode which should allow the voltage to go higher
What im proposing is increasing the Resistance of Water.
Yes I think you are right as far as the bubbles cousing increased resistance...it will probably rise...not sure to what point...

but this is the problem.. Bubbles are dynamic and moving meaning the resistance will ultimately fall once power is switched off.

No longer will we have steady  resistance. This is whats happening with the cell . if you apply electrical power to water it will charge then once power is off it will discharge significantly..

Why doesn’t it maintain charge like any regular capacitor?
I have done experiments were the capacitor charges then discharges into an led then a small motor.
but the charge level is not near where it should be.

As it is we are constantly "Doping" the water weather we realize it or not... :thinking:
Some add electrolytes to make it more conductive!
Some use tap water under the assumption that it will not conduct not taking into account that the water utility company are already "Doping" the water with chemicals.
Others use pure distilled water. yet the WFC doesn’t seem to manage to maintain the charge..

when using distilled water its as if when we apply voltage the cell opposes or resist the current yet when power is of the cell changes characteristics and internally discharges that energy.

Just as when we add chemicals to increase the conductivity of water why not add elements that does the opposite?

What I’m saying is instead of "Doping" the water to conduct current lets use the process of "Doping" to resist current!

Stan Meyer always mentioned that we should not add chemicals in the water to make it more conductive. on the contrary he wanted to eliminate the "Dead Short" condition.
so with out adding conductive chemicals the water is still internally acting like a dead short when power is off.

Stan Meyer never mentioned that we couldn’t add resistive elements in water to suppress the dead short condition. :idea:

Ok so here it is in a nutshell. By adding resistive impurity’s in the water ('Doping"). This will ultimately start to impede internal parallel path to complex groups of molecular crystals. This lattice or crystals, forms groups of  "charged networks"  within the water bath.

Because of freedom of movement within the water bath these charged networks will seek each other and exchange charge condition.
in effect neutralizing the charge.  This is why I believe that when power is applied it acts like a dielectric and due to special property of
water when we disconnect the power. The charge  mysteriously start to disappear.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: ~Russ on November 17th, 2014, 08:21 PM
FYI,

i have a nice system to play with... the EDM...

read this, this is what the water is in the system.

http://edmtodaymagazine.com/AAweb2_2010/TechTips2010/TechArt2010/Regeneration_Summer_2010.pdf

now what would happen if it put this water in my cell?

Deionized
water,

would it help achieve those high voltages? i beleave yes... should i try it?

~Russ
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Matt Watts on November 17th, 2014, 09:05 PM
Quote from ~Russ on November 17th, 2014, 08:21 PM
now what would happen if it put this water in my cell?

Deionized
water,

would it help achieve those high voltages? i beleave yes... should i try it?
Fully deionized water shouldn't conduct any electricity as its H and OH ions are in perfect balance.  But it won't stay that way, it will start to absorb CO2 from the air and the pH will change.  It may also pick up contaminates from the cell, so I would wash down those cells with this water several times prior to adding it for testing.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on November 18th, 2014, 03:49 PM
Quote from ~Russ on November 17th, 2014, 08:21 PM
FYI,

i have a nice system to play with... the EDM...

read this, this is what the water is in the system.

http://edmtodaymagazine.com/AAweb2_2010/TechTips2010/TechArt2010/Regeneration_Summer_2010.pdf

now what would happen if it put this water in my cell?

Deionized
water,

would it help achieve those high voltages? i beleave yes... should i try it?

~Russ
Something to put on your list of things to try lol...
thats a good idea. if you have the deionizer its defenitly worth a try.
Matt mentioned something that has affected my test and that is the cross contamination from previus experiments.

Funny i have a deionizer i use for work for cleaning PV arrays and havent used it. Im gonna have to take a look at it. It comes with a
digital impurity tester. hmm.

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on November 18th, 2014, 04:01 PM
Quote from Heuristicobfuscation on November 15th, 2014, 04:52 PM
This is how we can increase the  RESISTANCE OF WATER ! :idea:  yes we can increase the resistance!   volts = √watts × ohms

The more I Study Water the more it behaves like a Semiconducor!

So instead of treating it like a dielectric. I will start to treat it like a Semicondutor.

Here is the plan...

#1 how can one change the characterisitics in a semicoductor?
in the industry this process is called "Doping"

#2 So I will Start "Doping" Water.

Becouse of internal parrallel path and "Ez Zone" there exist leakage that the capacitor does not maintiane charge for longer periods.
by doping the dielectric we can extend the charge capacty of water...

The question is what material can we add  to the water that does not involve chemical reaction?

Here is the answer and run with it yes fly:

Option number #1   Silica  Dioxide Carefull not to breath.
Option number #2   Crushed Quartz crystal
Option number #3   Crushed Glass, fine powder.

What do they have in common?
[{(sand...Silicone)  Same material used in semicondutors!}]
Yes
They all have a resistive carachteristic to current and do not bond chemicaly to the water.

if we are to Dope water then we can saturate the water with these elements.
The more we Dope the water with resistive elements the more we impede  current flow due to parralel path!

Ok  lets think about that....

if we can manage to have the water maintaine the charge for longer periods of time then the charge rate will be significantly increased.

dont know about you guys but those of us who have spent countless hours  experimenting we are constantly trying to charge the  water :exclamation: thats the name of the game.

problem is as soon as you disconect the power the amount of charge is significantly reduced!
By Doping the water the charge should no longer significanlty be reduced due to constant dielectric element introduced in mixture.

We have a constant strugle with the VIC... and that is we have to pulse the voltage in an atempt to catch the charge before it de-energizes .

Im hoping that with the introduction of "Doping" we can increase the capacity of charge in the WFC.

What do you think?
Ok so i couldent wait and had to try some preliminary experiments with what i had.. :excited:

went to local pet store and purchased 3 dollars worth of silica sand. unfortunatly it wasnt pure it was
coated with polymers.

Wasnt to happy with results .. didnt get the higher resistance values i was looking for. :(
of course it was durring lunch time so my phone was going off anywasy managed to get some numbers posted bellow.
I will be patience and order the right stuff and performe a better and controlled experiment next time.

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Breakzeitgeist on November 20th, 2014, 01:58 PM
Perhaps these are the droids your looking forhttp://m.youtube.com/watch?v=bXtO7McrRIY
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on November 26th, 2014, 11:59 AM
Just an observation i would like to post before i forget and hopefully some feedback.

probably some of you have experienced this effect..

when i use a MOT transformer to connect to the VIC. The following happens.

cell starts to charge and voltage slowly increases say about 20plus volts. Ive managed to charge the cell up to 40volts.
the amperage was about .11 amps.  there was significant bubbles. not huge amounts but significant.

now the wattage on the secondary side was 4.4 watts!.
This is the problem: on the primary side of the transformer i was consuming 15amps! at 30 to 40 volts on the variac!

now on other ocasions ive managed to bring the volatge up to about 70 volts with very little bubbles.. ive noticed that some
current is necesary for the effect to occur.

the high voltage produced by the MOT is consumed by the water the instant it senses it.
ive noticed this test easilhy by connecting a neon gas bulb accross cell. initial voltage presence will light it up then it will shut down.
if  somehow  we can  manage to maintiane the primary at low power consumpition then the voltage on secondary should rise on the cell.

not sure why the MOT was consuming so much power on the primary if the secondary was insignificant!  Any Ideas?
im pretty lost on this one. :thinking:

Also I did not want to push the transformer. but it felt as if the more voltage i feed it the more voltage would rise on the cell almost at an exponetial rate. problem was did not want to exede primary amp rating.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Dynodon on November 26th, 2014, 01:38 PM
I don't know if you posted the specs on your set up anywhere, but without knowing what you are using, it's hard to answer your questions. I can tell you this much, if you are using a variac to power a MOT and hook that up directly to your cell, it won't work. You can't use a variac to pulse a Meyers type VIC coil. You need a primary, secondary, blocking diode and two choke coils to even attempt Meyers work. You also need a frequency generator with two output. One for the main frequency and the other for the gating. The primary frequency must be a 50% duty cycle, and the gating needs to have an adjustable duty cycle.
Don
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on November 26th, 2014, 03:09 PM
Quote from Dynodon on November 26th, 2014, 01:38 PM
I don't know if you posted the specs on your set up anywhere, but without knowing what you are using, it's hard to answer your questions. I can tell you this much, if you are using a variac to power a MOT and hook that up directly to your cell, it won't work. You can't use a variac to pulse a Meyers type VIC coil. You need a primary, secondary, blocking diode and two choke coils to even attempt Meyers work. You also need a frequency generator with two output. One for the main frequency and the other for the gating. The primary frequency must be a 50% duty cycle, and the gating needs to have an adjustable duty cycle.
Don
Thanks Dynodon

please see attached picture hopefully that will explain beter what i was doing.

I used the MOT to drive the inductive coils using a rectifier.
did not use a triger circuit to interupd the pulses coming from the transformer.
didnt want to disasembly the 8xa circuit i had for this test. Even dou i do realize
that the gating and frequency is very important. what i was trying to do was just simply
restric the current on the secondary via adding more and more inductace. I happend to have about 20 inductors that
never got used for single bulb flourescent fixtures laying around. these inductors bobbin come wraped in ferrite core.
and gage is realy small forget what size. anyways works well at limiting consideerabley the current. this would allow the
voltage to rise on the cell. but the more voltage would rise the more current the MOT was consuming on the Primary.

What im still puzzled about is that the primary on the transformer is consuming alot more energy than the secondary.

this has been the only setup that I have been able to significantly increase the voltage and restrict the amperage on the secondary.
problem is the primary power consumption increases while the secondary power decreases.

The more inductors I add in series the less amps flow to the cell and voltage goes up but the more amps the primary consumes.

Any ideas on why the primary is consuming more?
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Matt Watts on November 26th, 2014, 05:26 PM
It's a bit pricey I know, but the performance far and away exceeds what you will ever get from a MOT:

http://www.antekinc.com/as-15t950-1500va-950v-transformer/

I used this toroidal transformer as a substitution for a MOT with my plasma spark ignition system and the results are stunning.  Much lower power draw, especially at idle.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Dynodon on November 26th, 2014, 06:00 PM
Heur, I see that you are driving the MOT with AC voltage, and the secondary side is pretty much like Stan's. When using the 8xa circuit, your power from the variac is run through a full bridge rectifier. Have you tried that? Is the MOT fully stock? By that I mean , did you modify it? Some people remove a some of the current limiting part from it, so there is only the primary and secondary windings. Doing this I think makes it very likely to shock the hell out of you or even kill you.

I would say that the high current you are seeing when you raise the voltage may be do to the out of resonance condition you are working at. Because you are not able to change the frequency from the variac from the 50hz or 60hz your using driving the circuit at, you aren't able to tune into the LC circuit you have. Each choke (inductance) and cell (capacitance) will resonate at a given frequency. So you are asking the primary side of the MOT to operate at a frequency it isn't happy at. If you were to size the cell and chokes to resonate at the 50hz or 60hz, then I would expect you would see the input power to the MOT drop. That's just my thoughts, I may be wrong. But this set up with the chokes and cell are meant to be used in resonance. You don't have the means to tune into resonance with a fixed frequency input.
Don
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on November 26th, 2014, 08:13 PM
Quote from Dynodon on November 26th, 2014, 06:00 PM
Heur, I see that you are driving the MOT with AC voltage, and the secondary side is pretty much like Stan's. When using the 8xa circuit, your power from the variac is run through a full bridge rectifier. Have you tried that? Is the MOT fully stock? By that I mean , did you modify it? Some people remove a some of the current limiting part from it, so there is only the primary and secondary windings. Doing this I think makes it very likely to shock the hell out of you or even kill you.

I would say that the high current you are seeing when you raise the voltage may be do to the out of resonance condition you are working at. Because you are not able to change the frequency from the variac from the 50hz or 60hz your using driving the circuit at, you aren't able to tune into the LC circuit you have. Each choke (inductance) and cell (capacitance) will resonate at a given frequency. So you are asking the primary side of the MOT to operate at a frequency it isn't happy at. If you were to size the cell and chokes to resonate at the 50hz or 60hz, then I would expect you would see the input power to the MOT drop. That's just my thoughts, I may be wrong. But this set up with the chokes and cell are meant to be used in resonance. You don't have the means to tune into resonance with a fixed frequency input.
Don
Thanks for the feed back.

Yes i have tried using a full wave bridge rectifier on the primary side of the  MOT..
and I dont recomend it. :thumbdown:. The MOT does not like dc pulses it does  :nono: not function properly all sorts of bad stuff starts to happen.
Burned up one variac, burned up a PWM  and I forget how many bridge rectifiers i burned befrore i gave up trying to use DC pulses on the MOT lol.  In short MOT's will not function with DC Pulses on the primary.

The MOT that im using I havent modified in any way. just used it strait off the microwave oven.
just havent used the capacitor that it came with.

probably the only other option i have is to just add the trigering circuit like you had mentioned thats included in the 8xa.
Controling the secondary frequency i can test and see if my primary drops or goes up in amps.

Still trying to analyze why more power on the primary...
something i forgot to mention dont know if its that important but
when i was taking the primary reading on one leg the amps would be lets say at 8amps and the other leg would show 3 amps.

now that was little wierd becouse i can understand that when there are multiple lines or phases.
but in this scenario ther is only one line. and the "neutral" wire should have the same balance load as the "hot" or line wire.

in reality the "neutral" wire is realy only "neutral" when two diff phase lines cancel themselves while consuming same load then the  common wire becomes the "neutral".

this is not the case in this application :huh: why then the difference in load?

same amperage should be flowing in both the hot and "neutral" wire.

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on November 26th, 2014, 08:21 PM
Quote from Matt Watts on November 26th, 2014, 05:26 PM
It's a bit pricey I know, but the performance far and away exceeds what you will ever get from a MOT:

http://www.antekinc.com/as-15t950-1500va-950v-transformer/

I used this toroidal transformer as a substitution for a MOT with my plasma spark ignition system and the results are stunning.  Much lower power draw, especially at idle.
NIce! I  like the transformer.
ive used similar toroid transformer.
the one i have is a step down transformer and ive used it in reverse...
works realy good actually better than the MOT. but limited in primary current due to wire size.

hmm... maybe i should document the results and post it. using the low voltage transformer in reverse vrs the MOT.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Gunther Rattay on November 27th, 2014, 06:13 AM
Quote from Heuristicobfuscation on November 26th, 2014, 08:13 PM
...
Still trying to analyze why more power on the primary...
something i forgot to mention dont know if its that important but
when i was taking the primary reading on one leg the amps would be lets say at 8amps and the other leg would show 3 amps.

now that was little wierd becouse i can understand that when there are multiple lines or phases.
but in this scenario ther is only one line. and the "neutral" wire should have the same balance load as the "hot" or line wire.

in reality the "neutral" wire is realy only "neutral" when two diff phase lines cancel themselves while consuming same load then the  common wire becomes the "neutral".

this is not the case in this application :huh: why then the difference in load?

same amperage should be flowing in both the hot and "neutral" wire.
whenever something weird happens like you are experiencing you should triple-check your measurements.
often  our meters or the way we use them trick us and of course always the grounding aspect can be a source for trouble at measurement.
whenever you can use a scope for voltage and amp and wattage measurements and calculation because true-RMS meters can cheat you if your´r dealing with weird frequency mixes.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Dynodon on November 27th, 2014, 07:12 AM
Heur, Have you tried taking those current readings with the secondary of the MOT left unconnected? Maybe it's the load you have connected with the one diode that is causing the effects your seeing in the current. I also second what Gunther said. Double check you measurements.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: kenssurplus on November 27th, 2014, 10:07 AM
I know this doesn't apply to the main topic of the thread, but I hope that I might help Heuristicobfuscation obtain higher voltages without greater amperage input.

Hector Perez Torres disclosed the RE – amp in Doug Konzen's EVGRAY Yahoo group.
The concept is very simple. It uses a ferroresonant transformer and taps off of the ferroresonant winding and ferroresonant capacitor. It then steps that voltage down from 600 V AC through a MOT to 18 V for battery charging.
Although I did not achieve over unity, it did allow me to utilize a resonant condition, and learn how to extract power from it.

I was able to short-circuit the output without adversely affecting the components.
I was also able to send the power over long distance cables with little attenuation.

If you use the same concept, of a ferroresonant transformer feeding your MOT, you may achieve high voltage without increasing the amperage.

I have not tested the RE – amp with the MOT in normal mode, only in step down reverse mode.

Here is the original post:
Quote
Message number 25810

Hi All,

I have been short on power for the last little while so I haven't posted much.
While I was in the desperation mode, I decided to build the RE-AMP that H
detailed.

Most everything is the same as Hector's schematic diagram except for the caps
are slightly different:
Ferroresonant cap same at 2 uf 660 vac.
second series cap at 1 uf 4.5 kv instead of .86 uf 2000 vac (all I could find on
hand)
third low voltage cap was the difficult one to find, so I used 120 MFD 450 VDC
POLARIZED caps back to back to make them unpolarized.
which equates to about 60 uf 450 vac (I think).

Anyway, the circuit is powered by a 325 watt invertor from a battery stack.
The output into the batteries through FWBR was about 1/2 amp @ 12 volts.
That was kindof low I thought, so I placed more of these back to back caps in
paralell with the output cap to boost up the MDF value.
As I did this the output amps would rise slightly each time I added a new pair
of back to back caps.
When I fanally ran out of caps to put on, I had paralelled 7 pairs of 120 MFD
450 vdc caps which should equate to about 420 MFD 450 vac (anyone correct me if
I am worng here).

Output at this level of caps tuning was 1.75 amps @ 12 volts. I used a 2 amps
100 mv shunt on the invertor input and read 325 mv.
So, if I understand all the conversions correctly I have 78 watts into the
invertor and 21 watts back into the battery.

I suppose I could build another binary cap bank to use here but I don't think
that I have access to the correct sized MFD oil caps.

At this point it looks as though this experiment has hit a brick wall as the
output is not even close to being ou.

Any thoughts?

Ken
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on November 28th, 2014, 08:33 PM
See video bellow...
update on restricting current in VIC network via Inductance..

As I was adding more and more inductance. The current does go down significantly!
for me it was a record... to the point that my meter could not register the low amper consumed.

This system allows the Voltage to pass on to the cell.
On the VIC side I had a 500volt Potential.

A reading on the actual Cell yielded very little voltage compared to what was being applied.
Thats due to water doing its usual thing, consuming voltage and converting it to current.

In this test I just wanted to confirm that by simply adding more and more Inductance that we can further restrict the
VIC network current to allow voltage to "pass."...and so far it worked..
The next phase is to stimulate the voltage to "rise"

To get the voltage to rise a test will be performed where the secondary of the transformer on the VIC side will be trigered and gated.
This will alllow to change the frequency on the cell and also will allow via the gate for the cell to react.

Different frequency and gating settings will be documented and tested against power consumption and voltage potential or rise on the cell itself.

any ideas let me know

Thanks,

At 500 volts

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Dynodon on November 29th, 2014, 09:03 AM
First off, I'd like to say it's good to see that you are experimenting with this research into Stan Meyers work. Now I'm going to make some comments on what I have seen in your video. Now this is only meant to be helpful criticism.
There are many problems with the way your trying to go about this. The first problem is that you are mixing up more than one of Stan's projects together. A lot of experimenters have done the same thing. From what I see, you are trying to attempt to replicate Meyers Resonant system.
You have many problems with your approach.
The first problem is that your using the demo tube cell for your test cell. The demo tube cell was only used with the alternator set up. It looks like you are only using one tube, which is good, but it is way too long. The resonant cell only used 3-4 inch long tube sets. Now the latest idea is that we need to use several tube sets of the 3-4 inch long six to ten sets, and that they need to be wired in series not parallel. Also the tube set must be encased in delrin and isolated  from each other but still submerged in the same water bath.
Second item is the use of the variac. The variac was used with the plate demo cell, as you are trying to use it. The variac can be used as your power supply to the MOT, but it would require that you use full bridge rectifier and enough capacitors to filter out the dc ripple, to get a clean straight dc power. This dc power would then need to be run through a frequency generator and be pulsed to the MOT. Then you will be doing it right. This will allow you to tune into resonance.
Third item. If you try to use your current set up with the optocoupler circuit triggering an SCR, it won't work either. That set up will only create a gate signal if it is only run at a frequency slower than the 50/60hz that your variac is running at. If you pulse it faster than that, it will only look like a 50/60hz signal with a higher frequency pulses riding on top of it. It doesn't do anything. You could use a circuit similar to the SCR to pulse the variac power, but it can't be an SCR. You will need a mosfet or a transistor like a tip120. An SCR can't be turned on and off with the trigger pin only. The power that flows through a SCR once switched on will stay on as long as the power flows through it. If the trigger pulse is sent to the trigger pin, it will turn on. If the trigger pulse to the pin is turned off, the SCR will stay on. It will only turn off when the power going in to the SCR is turn off, it then turn off the SCR. So once the SCR gets a trigger, it won't turn off until power is removed from the input.
So if you want to continue your work into the resonant cavity system, you will need to correct these issues, to even get close to going in the right direction.
Don
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on November 30th, 2014, 01:54 PM
Quote from Dynodon on November 29th, 2014, 09:03 AM
First off, I'd like to say it's good to see that you are experimenting with this research into Stan Meyers work. Now I'm going to make some comments on what I have seen in your video. Now this is only meant to be helpful criticism.
There are many problems with the way your trying to go about this. The first problem is that you are mixing up more than one of Stan's projects together. A lot of experimenters have done the same thing. From what I see, you are trying to attempt to replicate Meyers Resonant system.
You have many problems with your approach.
The first problem is that your using the demo tube cell for your test cell. The demo tube cell was only used with the alternator set up. It looks like you are only using one tube, which is good, but it is way too long. The resonant cell only used 3-4 inch long tube sets. Now the latest idea is that we need to use several tube sets of the 3-4 inch long six to ten sets, and that they need to be wired in series not parallel. Also the tube set must be encased in delrin and isolated  from each other but still submerged in the same water bath.
Second item is the use of the variac. The variac was used with the plate demo cell, as you are trying to use it. The variac can be used as your power supply to the MOT, but it would require that you use full bridge rectifier and enough capacitors to filter out the dc ripple, to get a clean straight dc power. This dc power would then need to be run through a frequency generator and be pulsed to the MOT. Then you will be doing it right. This will allow you to tune into resonance.
Third item. If you try to use your current set up with the optocoupler circuit triggering an SCR, it won't work either. That set up will only create a gate signal if it is only run at a frequency slower than the 50/60hz that your variac is running at. If you pulse it faster than that, it will only look like a 50/60hz signal with a higher frequency pulses riding on top of it. It doesn't do anything. You could use a circuit similar to the SCR to pulse the variac power, but it can't be an SCR. You will need a mosfet or a transistor like a tip120. An SCR can't be turned on and off with the trigger pin only. The power that flows through a SCR once switched on will stay on as long as the power flows through it. If the trigger pulse is sent to the trigger pin, it will turn on. If the trigger pulse to the pin is turned off, the SCR will stay on. It will only turn off when the power going in to the SCR is turn off, it then turn off the SCR. So once the SCR gets a trigger, it won't turn off until power is removed from the input.
So if you want to continue your work into the resonant cavity system, you will need to correct these issues, to even get close to going in the right direction.
Don
Thanks for your observation.

the approach that im using is based on the tools and knowledge that i have.

I do agree that a "perfect" replica would be the Ideal scenario.
But i am realistic with my intelectual and practical limitations.
at the moment I dont have the capability to bring togherther all the disipliines requreed to achive
an exact prototype.

The approach i have choosen is to work with what i have (ie knowledge and tools). and document as much as possible.

Just embarking in this proces I have gained knowledge that i did not have before and slowly still learning lots of practical
applications.

Example:

In  trying to apply the volts = √watts × ohms formula
I posted the video above which was an attempt not to exacly replicatte the stan meyer resonant action rahter it was to test how much current
the Inductance network can actually resist.

So in the context of (volts = √watts × ohms formula) the network impedance was working like it should have.
the problem has always been not the network but the internal water resistance.  Which is what this post is trying to
investigate.

Thanks again for your input.

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Breakzeitgeist on December 1st, 2014, 08:16 PM
Quote from Heuristicobfuscation on November 26th, 2014, 11:59 AM
Just an observation i would like to post before i forget and hopefully some feedback.

probably some of you have experienced this effect..

when i use a MOT transformer to connect to the VIC. The following happens.

cell starts to charge and voltage slowly increases say about 20plus volts. Ive managed to charge the cell up to 40volts.
the amperage was about .11 amps.  there was significant bubbles. not huge amounts but significant.

now the wattage on the secondary side was 4.4 watts!.
This is the problem: on the primary side of the transformer i was consuming 15amps! at 30 to 40 volts on the variac!

now on other ocasions ive managed to bring the volatge up to about 70 volts with very little bubbles.. ive noticed that some
current is necesary for the effect to occur.

the high voltage produced by the MOT is consumed by the water the instant it senses it.
ive noticed this test easilhy by connecting a neon gas bulb accross cell. initial voltage presence will light it up then it will shut down.
if  somehow  we can  manage to maintiane the primary at low power consumpition then the voltage on secondary should rise on the cell.

not sure why the MOT was consuming so much power on the primary if the secondary was insignificant!  Any Ideas?
im pretty lost on this one. :thinking:

Also I did not want to push the transformer. but it felt as if the more voltage i feed it the more voltage would rise on the cell almost at an exponetial rate. problem was did not want to exede primary amp rating.
try placing the microwave oven capacitor across the primary
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Breakzeitgeist on December 1st, 2014, 08:30 PM
By doing this it will allow the voltage to lead the way
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Breakzeitgeist on December 1st, 2014, 08:31 PM
Let me know what your results are and thoughts on your observations
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Dynodon on December 2nd, 2014, 07:19 AM
Heur, again it's good to see you or anyone else experimenting. That's more than most people do here or anywhere else. That's the best way to learn what doesn't work and what does. The whole key to getting this project to work is LC resonance. Without it, you will never get high voltage across the water. The only true way to get to that is with an adjustable frequency generator that you can use to tune into it. Once resonance is hit, the resistance goes high as does the voltage. It's part of the process on it's own. The high resistance is a byproduct of resonance.
Again, it doesn't hurt to experiment, but in the end, you will need to follow Stan's path to get there.
Good luck and happy hunting
Don
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Breakzeitgeist on December 2nd, 2014, 02:09 PM
I agree don but still put the cap across the primary and see what happens. If it at least takes care of what you were talking Let us know your findings
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on December 2nd, 2014, 04:47 PM
Quote from Breakzeitgeist on December 2nd, 2014, 02:09 PM
I agree don but still put the cap across the primary and see what happens. If it at least takes care of what you were talking Let us know your findings
thanks,

yes thats a good idea.. I will put the cap in MOT primary side.. easy enough to test.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: ~Russ on December 2nd, 2014, 08:14 PM
good work going on here, use what tools you have and publish the result, that's what this place is about! you learn and teach along the way! keep it up!

~Russ
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Cycle on March 30th, 2015, 05:32 PM
Quote from Heuristicobfuscation on November 15th, 2014, 04:52 PM
The more I Study Water the more it behaves like a Semiconducor!

So instead of treating it like a dielectric. I will start to treat it like a Semicondutor.

Here is the plan...

#1 how can one change the characterisitics in a semicoductor?
in the industry this process is called "Doping"

#2 So I will Start "Doping" Water.

Becouse of internal parrallel path and "Ez Zone" there exist leakage that the capacitor does not maintiane charge for longer periods.
by doping the dielectric we can extend the charge capacty of water...

The question is what material can we add  to the water that does not involve chemical reaction?

Here is the answer and run with it yes fly:

Option number #1   Silica  Dioxide Carefull not to breath.
Option number #2   Crushed Quartz crystal
Option number #3   Crushed Glass, fine powder.

What do they have in common?
[{(sand...Silicone)  Same material used in semicondutors!}]
Yes
They all have a resistive carachteristic to current and do not bond chemicaly to the water.
Putting silica in water results in silicic acid.

https://en.wikipedia.org/wiki/Silicic_acid
"Silicic acid is the general name for a family of chemical compounds containing the element silicon attached to oxide and hydroxyl groups."

So the moment you put current to the water and start generating hydroxyl ions (OH-), you'll be generating silicic acid, which will carry the current, destroying your hopes of high voltage, low current dissociation.

Pure water acts as a dielectric because the molecules are highly polar. In fact, the water molecule is so polar that it'll rotate in an electric field to align itself with that electric field. Use that fact to align the water molecules properly to make dissociating them easier, then hit them with some other form of energy (x-rays would be the most efficient, given the resonant frequency of the water molecule is in the hard x-ray range) at such an angle that you have a wider cross-section of energy absorption (higher barn).

You can hit it with sub-harmonic frequencies, but between each "pulse" of that sub-harmonic, the proton will 'spin down' a bit... so the lower the sub-harmonic frequency you use, the harder it is to keep the proton 'spun up' to the point that it deprotonates. X-ray frequency will spin it up so quickly that it'll deprotonate immediately, with no spin-down.

"Guenther and Holzapfel irradiated water with X-rays in contact with a large free volume in a vacuum system and found large continuing yields of hydrogen gas."

How to generate the x-rays? Well, you can "amplify" the light from an LED until it's in the x-ray range:
Miniaturized high-speed modulated x-ray source(https://www.google.com/patents/US20140044239)
"A miniaturized high-speed modulated X-ray source (MXS) device and a method for rapidly and arbitrarily varying with time the output X-ray photon intensities and energies."

That's an x-ray source that essentially is an electron "amplifier", taking the light from an LED and adding energy to it until it's in the x-ray range. The good thing about this device is you can tune its wavelength from 120 eV (10.332 nm) to 120 KeV (0.010332 nm).

From my E-F posts, reverse engineering the resonant frequency of water:
======================================
http://www.energeticforum.com/water-fuel/19877-molecular-distances-corresponding-frequencies.html

According to:
Hydrogen Bonding And Orbital Models(http://pages.swcp.com/~jmw-mcw/Hydrogen%20Bonding%20and%20Orbital%20Models.htm)
In ambient atmosphere the O—O in the water dimer is 2.985 angstrom (calculated by JMW); the short O—H bond is 0.948 angstrom and the long bond is 2.037 angstrom.

That's .2985 nm, .0948 nm and .2037 nm.

FREQUENCY & WAVELENGTH CALCULATOR(http://www.1728.org/freqwave.htm)
2.037 angstrom corresponds to a frequency of 1.4717e+18 Hz or 1.4717351890034363270 ExaHertz. X-ray range.

2.985 angstrom corresponds to a frequency of 1.0043e+18 Hz or 1.0043298425460636160 ExaHertz. X-ray range.

0.948 angstrom corresponds to a frequency of 3.1624e+18 Hz or 3.1623677004219407360 ExaHertz. X-ray range.
======================================

Keep in mind that you'll want to target the short O-H bond, since the long O-H bond is so close in frequency to the inter-molecule O--O dimer bond that hitting it without stressing the dimer bond is near impossible. And if you stress that dimer bond, you also make it harder to cleave the O-H bond.

Also keep in mind that as you stress the short O-H bond, you'll need to 'blue-shift' your frequency (increase the frequency) to account for the change in O-H bond length as you stress it. I think that's why Meyer's waveform stepped up like it did.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on March 31st, 2015, 07:56 PM
Quote from Cycle on March 30th, 2015, 05:32 PM
Putting silica in water results in silicic acid.

https://en.wikipedia.org/wiki/Silicic_acid
"Silicic acid is the general name for a family of chemical compounds containing the element silicon attached to oxide and hydroxyl groups."

So the moment you put current to the water and start generating hydroxyl ions (OH-), you'll be generating silicic acid, which will carry the current, destroying your hopes of high voltage, low current dissociation.

Pure water acts as a dielectric because the molecules are highly polar. In fact, the water molecule is so polar that it'll rotate in an electric field to align itself with that electric field. Use that fact to align the water molecules properly to make dissociating them easier, then hit them with some other form of energy (x-rays would be the most efficient, given the resonant frequency of the water molecule is in the hard x-ray range) at such an angle that you have a wider cross-section of energy absorption (higher barn).

You can hit it with sub-harmonic frequencies, but between each "pulse" of that sub-harmonic, the proton will 'spin down' a bit... so the lower the sub-harmonic frequency you use, the harder it is to keep the proton 'spun up' to the point that it deprotonates. X-ray frequency will spin it up so quickly that it'll deprotonate immediately, with no spin-down.

"Guenther and Holzapfel irradiated water with X-rays in contact with a large free volume in a vacuum system and found large continuing yields of hydrogen gas."

How to generate the x-rays? Well, you can "amplify" the light from an LED until it's in the x-ray range:
Miniaturized high-speed modulated x-ray source(https://www.google.com/patents/US20140044239)
"A miniaturized high-speed modulated X-ray source (MXS) device and a method for rapidly and arbitrarily varying with time the output X-ray photon intensities and energies."

That's an x-ray source that essentially is an electron "amplifier", taking the light from an LED and adding energy to it until it's in the x-ray range. The good thing about this device is you can tune its wavelength from 120 eV (10.332 nm) to 120 KeV (0.010332 nm).

From my E-F posts, reverse engineering the resonant frequency of water:
======================================
http://www.energeticforum.com/water-fuel/19877-molecular-distances-corresponding-frequencies.html

According to:
Hydrogen Bonding And Orbital Models(http://pages.swcp.com/~jmw-mcw/Hydrogen%20Bonding%20and%20Orbital%20Models.htm)
In ambient atmosphere the O—O in the water dimer is 2.985 angstrom (calculated by JMW); the short O—H bond is 0.948 angstrom and the long bond is 2.037 angstrom.

That's .2985 nm, .0948 nm and .2037 nm.

FREQUENCY & WAVELENGTH CALCULATOR(http://www.1728.org/freqwave.htm)
2.037 angstrom corresponds to a frequency of 1.4717e+18 Hz or 1.4717351890034363270 ExaHertz. X-ray range.

2.985 angstrom corresponds to a frequency of 1.0043e+18 Hz or 1.0043298425460636160 ExaHertz. X-ray range.

0.948 angstrom corresponds to a frequency of 3.1624e+18 Hz or 3.1623677004219407360 ExaHertz. X-ray range.
======================================

Keep in mind that you'll want to target the short O-H bond, since the long O-H bond is so close in frequency to the inter-molecule O--O dimer bond that hitting it without stressing the dimer bond is near impossible. And if you stress that dimer bond, you also make it harder to cleave the O-H bond.

Also keep in mind that as you stress the short O-H bond, you'll need to 'blue-shift' your frequency (increase the frequency) to account for the change in O-H bond length as you stress it. I think that's why Meyer's waveform stepped up like it did.

Silica dioxide was a sugestion to increase resistance altough Silica acid is whats converts to Silica dioxide.... not the other way around.. according to same wiki source...
although if it did , then would the quartz crystall be affected same way?  what about the glass?

ok so you mentioned that pure water acts like a dielectric..(i used to think this way)

Yes it "acts like a dielectric" then it decides its no longer going to "resist" and starts to conduct....
this is problem we are always runing into...
ive purchaced multiple diff brands of pure distilled water..countless days, cheking notes and experimenting .... and always  get same results.

water has amazing ability to covert voltage...into current...

This is my argument with water acting more like a semiconductor rather than a dielectric...
I dont agree with the interpretations that water is the ultimate dielectric.

we know that in the industry Distilled water is considered  a fairly ok dielectric. this is why its not used  in capacitors.
becouse it does not hold the charge for long periods.......

if the water did hold a charge for long periods of time this would mean that the resistiance does not  change......
that would be perfect... but not the case. resistance is constantly changning.  constantly .......

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Gunther Rattay on April 1st, 2015, 12:49 AM
so the water stores charge like a capaciter and additionally reacts like a non-linear resistor.
observing the voltage/current/time dependency gives answers to the question how water interacts with a transformer coil configuration pulsed system.

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on April 1st, 2015, 08:15 PM
Quote from Gunther Rattay on April 1st, 2015, 12:49 AM
so the water stores charge like a capaciter and additionally reacts like a non-linear resistor.
observing the voltage/current/time dependency gives answers to the question how water interacts with a transformer coil configuration pulsed system.
yes..

#1 so resistance is present...

#2 upon application of power  voltage starts to rise with current...

#3 Then resistance drops  and current rises as voltage falls.

Here are my two cents on this...
I agree with most that resonance is important.. but i dont tink thats what we are realy after... to me resonance makes what we are
looking for operate at  max efficiency..
Is it running at max efficieny? then resonance probably has something to do with it.

So what are we realy after?

there are two objectives..

#1 Limit Current...

#2 charge capacitor with step charge phenomenon....

ok so number one is managable to a degreee. see link bellow on inductive network... works great the more inductance the
less current...

here is problem with number two..

A Capacitor canot charge any higher than applyied voltage!

and we are dealing with the most complicated capacitor setup possible in the world.
that wont hold the charge applied.....will lose its resistance...and wont care if you apply 12volts or 6,000volts
it will drop it back to 1.5volts...and watever charge it does maintane will convert to current and consume internaly
producing bubbles.

All of that and we havent even raised the voltage beyond applyied secondary!
that according to stan meyers should go beyond applyied voltage .. with this "step charge phenomenon"

This is my second thought on this.... before we can acheive "Step charge Phenomenon"
we first have to maintaine a managable amount of time interval of applyied  charge

Before the voltage drops...we need to switch off the circuit!

Maybe once we can manage that, then ..we could try and fiqure out the true "Step charge Phenomenon" " were voltage climbs to Infinity!"    At witch point water resistance also has to rise to infinity!

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Matt Watts on April 1st, 2015, 08:36 PM
Okay, let me ask this...

Has anyone connected a fairly large high voltage capacitor to the coil, creating a tank circuit, then with a diode in series, connect the WFC across the capacitor?  The idea here is to have your tuned resonant circuit and piggy-back the water cell (with diode) on to the capacitor.  If your tuned circuit is good enough (high Q), voltages should go way up and the cell shouldn't be able to knock it down much, it's basically just along for the ride.

Now, if you do simulations of this, you should see some very high amperages in the tank circuit.  Whether these high amperages cross over to the WFC or not, I can only speculate.  If the voltage and amperage are 90 degrees out of phase, it really doesn't matter how much amperage is there, any resistance the WFC has, cannot have an effect on the tank circuit.  It's somewhat a game of proportions.  Let's say the WFC has a calculated capacitance of 10nF and our tank capacitor is 1uF (100 times larger).  So we find a suitable inductor to resonant with that drops the 1st harmonic frequency around 25kHz.  What can we expect this to do?  Has anyone tried it?
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on July 28th, 2015, 07:29 PM
Here is an idea...

Have you tried freezing your water into ice?

method is as follows. .

#1 ground one of the plates... disconect the anode.

#2 freeze the water.
# 3  disconect earth ground and apply charge potential using complete circuit.

first step will allow alignment of molecule without current consumption.
second step will freeze the water molecule pre aligned in respective polarity.
third step....well molecule will not be able to turn or move due to frozen water structure
the only movable particle will be the electron.

in normal water bath the molecules move all over the place creating all sorts of ions.
in frozen water bath the molecules are stuck!  but the electrons are free.

this method will surely "limit current"..........  no ions no current; no standard electrolysis.

Ok... here is problem number one...

obviosly ther will be no particle "bombartment or impact" aiding with kinetic energy to weaken the molecule.

here is problem number two..

ahhh.... I find it difficult the molecule will be able to "elongate"

so .
for this to work an external force will have to be greater than the solidified crystal matrix that was formed.
said force can be a high voltage potential...

if the static lines can overcome the shielding effect of the bond geometry....
then im assuming the molecule will split... and stay stuck within the bond or spontaneously expand  into gas bypassing the liquid state.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on September 21st, 2015, 06:08 PM
V = √W × Ω  Higher voltage requires Higher resistance!

The statement above is very true...and in this post that I started some time ago many toughts
have gone into it..

honestly was expecting to be  challenged on this argument... that is of resistance having to rise!

but ive come to a realization concerning the formula.....

voltage is independant of resistance!
you dont need resistance to have voltage...

but you do need voltage to measure resistance!

yes the "formula" is a tool of measurement..
and like all tools there is always a right tool for the job.
onfurtunatly this formula is not the right tool for the job.

what is the job?

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: massive on September 22nd, 2015, 01:20 AM
a fly back transformer does not rely on winding ratio , it stores energy in air gap and transfers from primary to secondary . 2 coils isolated from each other , also called a swinging choke .
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on September 26th, 2015, 03:48 PM
Is the following statements true or false?

#1 voltage is independant of resistance!
#2 you dont need resistance to have voltage...

#3 but you do need voltage to measure resistance.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Matt Watts on September 26th, 2015, 05:31 PM
Are we talking just in general or specific cases of DC or high frequency RF through a transmission line?

When I hear the term resistance, I typically refer to the DC case and the term impedance to the AC one.  Regardless, I think there is a tight relationship there which forces us to better understand what DC actually is.

In the case of the infinite line below, there is no actual resistor components, but the circuit behaves as though there is.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: massive on September 27th, 2015, 12:20 AM
Resistance is for passive components .     with non linear resistance , negative resistance , voltage and current are differentials .
with water its hard to tie it down because water is self ionizing .  filling it with electrolyte brings it more into line with ohms law but with straight water its behaving non linear , being self ionizing its behaving more like an 'active' component
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on September 27th, 2015, 05:30 AM
Quote from massive on September 27th, 2015, 12:20 AM
Resistance is for passive components .     with non linear resistance , negative resistance , voltage and current are differentials .
with water its hard to tie it down because water is self ionizing .  filling it with electrolyte brings it more into line with ohms law but with straight water its behaving non linear , being self ionizing its behaving more like an 'active' component
Yes agreed...

that is problem we are facing... water doing what it wants...
eating up the voltage converting it to amps...

According to the formula  V = √W × Ω      if we apply  say 1000volts... the Resistance should rise! if we manage to restrict amps..

So how is the  resistance increased in water?

that has been the question ive been pursuing in this post...

But lately  ive come to the realization that this formula may not apply in all cases....
we may not need the resistance to rise at all... it may just be that voltage does not need resistance in the conversion to current.

example... all closed circuits have a source of power....in which in its path always returns to source...

This is not the case with static electricity...........
lightning does not return force back to source!    //   rather its an interchange of charge.....
But this is not the case with conventional closed loop circuits!

This is what ive been thinking.....
It could be that Stan Meyers was using the VIC/EEC as a(Pulsed DC to Static Generator converter).

This makes lots more sense to me because upon "switch off " or "gate time"  of the  VIC the EEC would take over and
Attract charge in water thereby reducing amounts of ions in it...possibly allowing voltage to accumulate rather than
allowing ion charge to avalanche..

This is a hybrid circuit........a circuit that goes from current flow to Static charge!

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Gunther Rattay on September 27th, 2015, 05:49 AM
Quote from massive on September 27th, 2015, 12:20 AM
Resistance is for passive components .     with non linear resistance , negative resistance , voltage and current are differentials .
with water its hard to tie it down because water is self ionizing .  filling it with electrolyte brings it more into line with ohms law but with straight water its behaving non linear , being self ionizing its behaving more like an 'active' component
Good discussion, massive!

Now add in to the non linear behaviour of water those special dynamics taking place in a bifilar wound transformer coil stack enabling the core material to build cooper pairs of electrons as Bill Alek describes here http://auroratek.us/SCIENCE.html.

Imagine what dynamics take place in such a non-trivial system. So we need lots of measurements of single parts of the system before a one-ín-all application can work properly:

voltage and current, phase shift measurements in several parts of the system, maybe magnetic flux measurements etc.

who thinks that this one can be solved with backyard stuff?

If the idea of cooper pairs is correct the transformer itself is no longer a closed system because accessing extra energy from a universal field by quantum level tunneling effect.

william alek demonstrates his bifilar transformer, it gets cold at shortcut condition and delivers 160% efficiency

Stan Meyer extremely understated the complexity of his system and that way he cheated most of us. Maybe that´s the reason why most activities in that field have failed or ended so far ... :hedidit:
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Gunther Rattay on September 27th, 2015, 06:14 AM
Quote from Matt Watts on September 26th, 2015, 05:31 PM
Are we talking just in general or specific cases of DC or high frequency RF through a transmission line?

When I hear the term resistance, I typically refer to the DC case and the term impedance to the AC one.  Regardless, I think there is a tight relationship there which forces us to better understand what DC actually is.

In the case of the infinite line below, there is no actual resistor components, but the circuit behaves as though there is.
Matt,

this one is the exact configuration Stan Meyer has set up:

(http://open-source-energy.org/?action=dlattach;topic=2119.0;attach=11783;image)

Inserting an AC pulsing source (transformer secondary) thru his diode replacing the battery and taking into account that all those capacitors are stray capacitances of the upper and lower path chokes and taking into account that destilled water up to 100 µs has no resistive component we see that the system itself protects from current flow at correct pulse length and resonance.
... of course there is also mutual inductance in the system so that we have serial and parallel LC components at oscillation enabling complex AC dynamics behind the diode.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Gunther Rattay on September 27th, 2015, 06:30 AM
Quote from Matt Watts on April 1st, 2015, 08:36 PM
Okay, let me ask this...

Has anyone connected a fairly large high voltage capacitor to the coil, creating a tank circuit, then with a diode in series, connect the WFC across the capacitor?  The idea here is to have your tuned resonant circuit and piggy-back the water cell (with diode) on to the capacitor.  If your tuned circuit is good enough (high Q), voltages should go way up and the cell shouldn't be able to knock it down much, it's basically just along for the ride.

Now, if you do simulations of this, you should see some very high amperages in the tank circuit.  Whether these high amperages cross over to the WFC or not, I can only speculate.  If the voltage and amperage are 90 degrees out of phase, it really doesn't matter how much amperage is there, any resistance the WFC has, cannot have an effect on the tank circuit.  It's somewhat a game of proportions.  Let's say the WFC has a calculated capacitance of 10nF and our tank capacitor is 1uF (100 times larger).  So we find a suitable inductor to resonant with that drops the 1st harmonic frequency around 25kHz.  What can we expect this to do?  Has anyone tried it?
tbd,

should work similar by paralleling a large hv capacitor to the cell ...
but in both cases the pulsing is a forced oscillation on the wfc while the cell itself taking part in the oscillation dynamics can add in it´s specific harmonic modes.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Matt Watts on September 27th, 2015, 07:03 AM
Quote from Gunther Rattay on September 27th, 2015, 06:30 AM
should work similar by paralleling a large hv capacitor to the cell ...
but in both cases the pulsing is a forced oscillation on the wfc while the cell itself taking part in the oscillation dynamics can add in it´s specific harmonic modes.
This coupled with an electrostatic effect (Heuristicobfuscation mentions) from the surrounding environment appears to be the exact same principal utilized in the Ruslan/Akula device (and probably others).

Let me just pose a simple observation made from some RV batteries...

Recently I had to replace several sets of batteries used in RVs.  Reason?  The end cells nearest the negative terminals had run dry of water, permanently damaging the batteries.  I ask myself, so why are these particular cells charging at a rate faster than the other cells depleting the water?  It can't be a simple matter of Ohms Law here since all the cells in the battery are receiving the same identical charging current.  The only answer I can come up with is the grounded body of the RV is somehow providing additional electrostatic charge to this directly connected end cell.  Now wouldn't it be nice to figure out exactly how this is happening and optimize this process on demand...
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: massive on September 27th, 2015, 12:43 PM
personally I dont like dealing with std transformer construction . Im into FBT , HV destruction hobby stuff .  with FBT primary energy is stored in the ferrite , the sec is invisible so theres no opposition . pri current stops sudden then sec diode is fwd bias and stored energy is released .
I plain dont like silicon steel etc cores or function .
non conductive cores rule!

"If the idea of cooper pairs is correct the transformer itself is no longer a closed system because accessing extra energy from a universal field by quantum level tunneling effect"

^^^^^ Tunnel diodes deal in negative resistance ,crazy as it sounds this 'Area' of electronics produces power , this Aurora tek is a cleaver dude . some guys can figure stuff out .

atm Im looking into negative resistance , Ive only ever reached for components to build circuits but recently my interest has been in NR

I see it as unreasonable to demand more energy out of a circuit , theres has to be a way in for energy from outside .
also using the primary power source can only result in depletion

as for voltage across water , the only example we have in nature is lightning , Ive been looking into that for years .
there was the "chemical explosion" theory of the 1800s .
theory being the HV field ionized water molecules , stripped electrons added to neg charge , ionized air created channel , HV breakdown of air , arc begins .....boom!   , electrons  shoot to earth.

modern theory isnt too far off the same thing except water self ionizing and H2 isnt mentioned .  NOR do they mention that ozone is highly explosive  .... which just adds to the "chemical explosion" theory.

anyway theres a water molecule suspended in mid air with no ground contact / isolated , self ionizing .  does voltage rise across the molecule ? idk
I only assume if it is balanced then it has equal charge

another thing..... the return lightning strike from ground to cloud in the opposite direction . is it a flow of electrons?
is it positive charges

I am also interested in GROUND as an electron source . Stubblefield did amazing things even though I was not there to witness any of it , the concept sounds reasonable .(also Hermann Plausons work)

paralleling is another thing Ive stuffed around with , theoretically a parallel R should lower the R of water ..... yea nup!  :0)
^^^ its the non linear + linear Resistance thing again
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on October 11th, 2015, 10:35 AM
Quote from Heuristicobfuscation on July 28th, 2015, 07:29 PM
Here is an idea...

Have you tried freezing your water into ice?

method is as follows. .

#1 ground one of the plates... disconect the anode.

#2 freeze the water.
# 3  disconect earth ground and apply charge potential using complete circuit.

first step will allow alignment of molecule without current consumption.
second step will freeze the water molecule pre aligned in respective polarity.
third step....well molecule will not be able to turn or move due to frozen water structure
the only movable particle will be the electron.

in normal water bath the molecules move all over the place creating all sorts of ions.
in frozen water bath the molecules are stuck!  but the electrons are free.

this method will surely "limit current"..........  no ions no current; no standard electrolysis.

Ok... here is problem number one...

obviosly ther will be no particle "bombartment or impact" aiding with kinetic energy to weaken the molecule.

here is problem number two..

ahhh.... I find it difficult the molecule will be able to "elongate"

so .
for this to work an external force will have to be greater than the solidified crystal matrix that was formed.
said force can be a high voltage potential...

if the static lines can overcome the shielding effect of the bond geometry....
then im assuming the molecule will split... and stay stuck within the bond or spontaneously expand  into gas bypassing the liquid state.
check this video ... Thomas Kin actually tested this idea... very good results..!  what does it prove?  #1 that in solid crystal structure it will act more like a resistor.

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Cycle on October 20th, 2015, 01:57 PM
Quote from Heuristicobfuscation on March 31st, 2015, 07:56 PM

Silica dioxide was a sugestion to increase resistance altough Silica acid is whats converts to Silica dioxide.... not the other way around.. according to same wiki source...
If you're using frequency to try to split water, try adding acetone to the water. For some unknown reason, the acetone increases the water's resistance to high frequency electricity far more than it should. Careful you don't have any sparks, though.

Note that while acetone does increase the resistance of water to high frequency electricity, it doesn't affect the electrolysis of the water. I think that might be part of the reason Stan Meyer was able to use low enough power into his WFC that splitting the water took less energy than what he got back out of it when running the engine to power the car. Nothing to back that up, of course, just a hunch.
Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Heuristicobfuscation on December 12th, 2015, 02:39 PM
Quote from Heuristicobfuscation on September 27th, 2015, 05:30 AM
Yes agreed...

that is problem we are facing... water doing what it wants...
eating up the voltage converting it to amps...

According to the formula  V = √W × Ω      if we apply  say 1000volts... the Resistance should rise! if we manage to restrict amps..

So how is the  resistance increased in water?

that has been the question ive been pursuing in this post...

But lately  ive come to the realization that this formula may not apply in all cases....
we may not need the resistance to rise at all... it may just be that voltage does not need resistance in the conversion to current.

example... all closed circuits have a source of power....in which in its path always returns to source...

This is not the case with static electricity...........
lightning does not return force back to source!    //   rather its an interchange of charge.....
But this is not the case with conventional closed loop circuits!

This is what ive been thinking.....
It could be that Stan Meyers was using the VIC/EEC as a(Pulsed DC to Static Generator converter).

This makes lots more sense to me because upon "switch off " or "gate time"  of the  VIC the EEC would take over and
Attract charge in water thereby reducing amounts of ions in it...possibly allowing voltage to accumulate rather than
allowing ion charge to avalanche..

This is a hybrid circuit........a circuit that goes from current flow to Static charge!
This video shows simple concept of water attraction to charge.      molecule are definitely aligning up to polarity.just as Stans mentions....

we dont want the water to discharge...
we dont want the water to conduct current.
we dont want "Catastrohic failure" of dielectric property of water.
we dont want such a high voltage potential that arcing occurs within the capacitor.

Rather:

we want the water to "align".   (like in video bellow)
we want the water to "elongate"  (that is) increase energy source of aligment..
we want the water molecule  to  "switch off" or give up its electrons...(this is were the EEC comes in) link bellow.

http://open-source-energy.org/?topic=2425.msg34436#msg34436

once the  covalent bond or internal charge no longer exist...the molecule no longer sticks togherter. (This is "Switch off")

were does all the static potential go?
has it been converterd to kinetic?

if it has then we failed to restrict the amps flow.

if up to this point we have managed to restrict amp flow in cell.
Then a nonviolent separation has occurred withing the water molecule.

At this point we have created free floating electrons........ what to do with all this electrons?  Extract them.... (EEC) use as DC power.

practicaly how can we achieve this?
This is what I think..
we need two ingrideients..

#1... constant source of Electrostatic diff of potential.
#2.. a means to extract the free floating electrons produced by the elongation.

so for the first one... I belive this is the purposse of the VIC... a Electromagnetic to Static converter.
for the second one we need a true nonreplicated (EEC) Electron Extraction Circuit) (nobody has replicated this..at least not in public)

Title: Re: V = √W × Ω Higher voltage requires Higher resistance!
Post by: Cycle on January 25th, 2016, 01:19 PM
I just had a thought... what if one were to provide two negative plates separated at the angle of the two hydrogens, and a single positive plate oppositely positioned to those two negative plates. This would align the water molecules pretty precisely, better than only two plates could.

Then, we hit the water with the high frequency from the hydrogen side, from two different angles, at the necessary angles and frequencies (because remember, we'd have to use two different frequencies since each O-H bond is a different length, and we'd have to shift the frequency down a bit as the O-H bonds become stressed to keep adding energy to them) to slice the hydrogens off the oxygen with maximum efficiency?

Obviously, since we're dealing with angles, there'll be a "sweet spot" in the electrolizing container in which the angles of the voltage aligning the water molecules, and the angles of the frequency emitters which we're using to hit the O-H bonds, are just right.

For tube-within-tube designs, I believe it'd be better to go negative on the outside tube (so you'd have your water molecules aligned hydrogen molecules facing outward), positive on the inside, but that makes it difficult to hit the O-H bonds at the right angle to cleave the molecules, since you can't really fit separate frequency emitters in between the tubes at the correct angles.