V = √W × Ω Higher voltage requires Higher resistance!

Heuristicobfuscation

V = √W × Ω Higher voltage requires Higher resistance!
« 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?




Lynx

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #1, 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.

Read all about it, http://www.rexresearch.com/meyerhy/meyerhy.htm

Heuristicobfuscation

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #2, 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?








Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #3, 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?
Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #4, 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:

Matt Watts

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #5, 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.

Gunther Rattay

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #6, 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.

Read all about it, http://www.rexresearch.com/meyerhy/meyerhy.htm
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):


https://www.youtube.com/watch?v=LKIMpeXp42c#t=834
(for a full screen display click on YouTube bottom right hand corner)

Heuristicobfuscation

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #7, 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....

Matt Watts

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #8, 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.

Heuristicobfuscation

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #9, on October 19th, 2014, 06:23 PM »Last edited on October 19th, 2014, 06:32 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.


Matt Watts

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #10, on October 19th, 2014, 09:06 PM »Last edited on October 19th, 2014, 09:07 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.

~Russ

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #11, on October 20th, 2014, 07:39 AM »
some good thoughts flowing here.

keep thinking. keep posting.

~Russ

Webmug

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #12, 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

Heuristicobfuscation

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #13, 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? 

The answer: and please Wright this in your heart. The answer is Water! 

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:

Ravenous Emu

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #14, 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.

firepinto

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #15, on October 21st, 2014, 06:08 PM »
There was a water filter built into the Dunebuggy on the 11 cell system.

Heuristicobfuscation

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #16, 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:


Lynx

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #17, 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.



I guess you'd have to clean the filter every now and then, but I can live with that ;-)

brettly

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #18, 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.

firepinto

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #19, 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.

~Russ

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #20, 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

Heuristicobfuscation

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #21, 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
Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #22, 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?


Enrg4life

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #23, 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

Heuristicobfuscation

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #24, 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.