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

~Russ

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #25, 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

Matt Watts

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #26, on November 17th, 2014, 09:05 PM »Last edited on November 18th, 2014, 06:04 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.

Heuristicobfuscation

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #27, 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.

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #28, 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.


Breakzeitgeist

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #29, on November 20th, 2014, 01:58 PM »
Perhaps these are the droids your looking forhttp://m.youtube.com/watch?v=bXtO7McrRIY

Heuristicobfuscation

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #30, 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.

Dynodon

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #31, 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

Heuristicobfuscation

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #32, 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?

Matt Watts

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #33, 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.

Dynodon

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #34, 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

Heuristicobfuscation

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #35, 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.









Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #36, 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.

Gunther Rattay

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #37, 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.

Dynodon

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #38, 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.

kenssurplus

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #39, on November 27th, 2014, 10:07 AM »Last edited on November 27th, 2014, 10:14 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

Heuristicobfuscation

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #40, 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,


https://www.youtube.com/watch?v=wwEUWcA8dJI#ws

At 500 volts 


Dynodon

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #41, 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

Heuristicobfuscation

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #42, 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.



 








Breakzeitgeist

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #43, 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
Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #44, on December 1st, 2014, 08:30 PM »
By doing this it will allow the voltage to lead the way
Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #45, on December 1st, 2014, 08:31 PM »
Let me know what your results are and thoughts on your observations

Dynodon

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #46, 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

Breakzeitgeist

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #47, 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

Heuristicobfuscation

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #48, 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.

~Russ

Re: V = √W × Ω Higher voltage requires Higher resistance!
« Reply #49, 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