Author Topic: 3 Questions  (Read 2459 times)

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3 Questions
« on: February 07, 2013, 13:42:53 pm »
If you were not able to achieve water fuel gas on demand using low watt, than why you didn't used more watts?

Are you sure that you were applying the energy on the water?

Do you know how to apply more energy to the water to make fuel gas on demand?

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    • water structure and science
Re: 3 Questions
« Reply #1 on: February 07, 2013, 23:59:49 pm »
I did...
I think so...
Dito...


 :)

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Re: 3 Questions
« Reply #2 on: February 16, 2013, 01:34:51 am »
I think that basically meyer used this kind of flyback effect,,, in such way as to charge up the water molecules energy levels at specific frequency which causes and maintain molecular ringing..

I asked that because i'm sure that many don't have a clue of how an inductor or a capacitor woks, and are using huge number of turns and small amount of amps so un matched in impedance that just dissipate energy at the switching... not in the water...

I used meyer values applied to the formulas and got reliable results. I mean 40watts is enough to send 5khz pulses that charge the capacitor to 20kv and beyond hundreds times per second... and with 30 turns or so if done the right way!

This gives parameters for experimenting...

Water fuel on demand is on the way...

I believe i found the way for saving the switches... using a series capacitor wit the primary...
« Last Edit: February 16, 2013, 15:29:05 pm by sebosfato »

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Re: 3 Questions
« Reply #3 on: February 16, 2013, 19:51:24 pm »
I think that basically meyer used this kind of flyback effect,,, in such way as to charge up the water molecules energy levels at specific frequency which causes and maintain molecular ringing..

I asked that because i'm sure that many don't have a clue of how an inductor or a capacitor woks, and are using huge number of turns and small amount of amps so un matched in impedance that just dissipate energy at the switching... not in the water...

I used meyer values applied to the formulas and got reliable results. I mean 40watts is enough to send 5khz pulses that charge the capacitor to 20kv and beyond hundreds times per second... and with 30 turns or so if done the right way!

This gives parameters for experimenting...

Water fuel on demand is on the way...

I believe i found the way for saving the switches... using a series capacitor wit the primary...

I assume you mean 30 turns on the chokes.  And by my own experiments I am in full agreement.

Please explain "I believe i found the way for saving the switches... using a series capacitor wit the primary"

TS
« Last Edit: February 16, 2013, 21:22:16 pm by timeshell »

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Re: 3 Questions
« Reply #4 on: February 17, 2013, 17:33:11 pm »
I mean the main problem with the switches is the high voltage developed in the primary which sums with the source voltage to destroy the switch. When you try to reduce this voltage by dissipating means the voltage output of the secondary will be proportionally reduced too. This mean that the energy stored should go all into the water... not on the switch to destroy it.

If you add a capacitor in series with the primary having a diode from drain to the capacitor connection with the diode from source positive.

than add a switch in parallel with this diode to discharge the capacitor into the primary to allow it to reverse its polarity and allow more energy to go into the resonance in the primary.

This capacitor saves the day since it isolate the source and the coil and switch.... so the potential is allowed to rise elsewhere other than the switch... making this energy also available to the circuit. Being the capacitor already charged from pulse on the primary than will discharge into it at specific time constant... given by the component sizes..

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Re: 3 Questions
« Reply #5 on: February 17, 2013, 18:51:13 pm »
What is this switch that you are talking about?

TS

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Re: 3 Questions
« Reply #6 on: February 17, 2013, 21:39:18 pm »
i think he mean =mosfet

thanks
geenee

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Re: 3 Questions
« Reply #7 on: February 18, 2013, 04:22:02 am »
yes correct!

I'm talking about the moment where the switch opens and primary voltage reverts polarity thereto summing with the source, destroying the switch...
If simply a diode is connected across the primary than the primary will be shorted during pulse off, which works but prevents the voltage to rise...
if instead a capacitor is added in series with the primary, than the diode will charge this capacitor to a greater voltage but its inverse relative to the source so it won-t burn the switch... i mean a diode will be able to protect it..


The basic circuit to prove step charging is

dc   d1    c1             L1
+---->!-----! !-----><><><><><
           _!_                               !
       d2  /\                                !
              !-------------------------!
                               Q1 ___---
                                _! !___--------->.G