Author Topic: The VIC with resonant cavity project by Steve..  (Read 92737 times)

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Re: The VIC with resonant cavity project by Steve..
« Reply #328 on: October 30, 2016, 16:18:21 pm »
Steve and others, I am giving out a lot of information on how Stan's fuel cell works over at RWG forum You guy's may want to go check it out.

http://open-source-energy.org/?topic=2785.0

Thank you Ronny!
I appreciate your information very much.
I have also troubles to get registrated on rwg forum.
Not sure why.
Can you maybe publish some of yr content here?
Sofar, i can only read without pictures etc at rwg....

Good work my friend!
Steve

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Re: The VIC with resonant cavity project by Steve..
« Reply #329 on: October 31, 2016, 03:09:29 am »
Steve and others, I am giving out a lot of information on how Stan's fuel cell works over at RWG forum You guy's may want to go check it out.

http://open-source-energy.org/?topic=2785.0

Thank you Ronny!
I appreciate your information very much.



I have also troubles to get registrated on rwg forum.
Not sure why.
Can you maybe publish some of yr content here?
Sofar, i can only read without pictures etc at rwg....

Good work my friend!
Steve

Heres the picture Ron colored Steve. I added the  remarks 3A -3F  to it tryn to satisfy a curiosity ,not finished yet trying to show a relationship to conductivity.The second pic is a version of the original Ron colored,I haven't found the same one yet.The 3rd pic is part of where I referenced 3A-3F.
« Last Edit: October 31, 2016, 07:14:12 am by Newguy »

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Re: The VIC with resonant cavity project by Steve..
« Reply #330 on: October 31, 2016, 11:06:29 am »
A lot of people are having trouble with impedance matching. I worked out Stan's Vic for example for everyone.

Using Stan's Vic and the numbers Don gave us as and example, I will attempt to show how to impedance match it all.
Question is what is the purpose of Impedance matching?
The answer is Watts in must equal Watts out.

Let's start with the Primary, I have already show it has 10 ohms of impedance in it and how it is calculated.

Line(Primary) side=10 ohms
12volts/10ohms=1.2amps
1.2amps*12volts=14.4watts

Next we use a transformer (Amplifier) to match the Load side.
we need to know the total resistance of the load side.
Secondary side= 72.4+76.7+70.1+Re78.54+11.5=310 ohms

Now that we have a total resistance of the line side of 10ohms
and a total resistance of the load side of 310ohms

Next we take the 310ohms and 10ohms and use this formula to get the turn ratio.
Ns/Np=sqrt Zs/Zp   sqrt (310/10)=5.567
So we need a turn ratio of 5.567 to 1

We know our line voltage is 12volts We can times this by the turn ration of 5.567 which is =66.8 Load Voltage
Now we have our load voltage.
Next we calculate the load watts
using formula (66.8 ^2)/310ohms= 14.39 watts

That's how you do it

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Re: The VIC with resonant cavity project by Steve..
« Reply #331 on: November 04, 2016, 09:28:46 am »
A lot of people are having trouble with impedance matching. I worked out Stan's Vic for example for everyone.

Using Stan's Vic and the numbers Don gave us as and example, I will attempt to show how to impedance match it all.
Question is what is the purpose of Impedance matching?
The answer is Watts in must equal Watts out.

Let's start with the Primary, I have already show it has 10 ohms of impedance in it and how it is calculated.

Line(Primary) side=10 ohms
12volts/10ohms=1.2amps
1.2amps*12volts=14.4watts

Next we use a transformer (Amplifier) to match the Load side.
we need to know the total resistance of the load side.
Secondary side= 72.4+76.7+70.1+Re78.54+11.5=310 ohms

Now that we have a total resistance of the line side of 10ohms
and a total resistance of the load side of 310ohms

Next we take the 310ohms and 10ohms and use this formula to get the turn ratio.
Ns/Np=sqrt Zs/Zp   sqrt (310/10)=5.567
So we need a turn ratio of 5.567 to 1

We know our line voltage is 12volts We can times this by the turn ration of 5.567 which is =66.8 Load Voltage
Now we have our load voltage.
Next we calculate the load watts
using formula (66.8 ^2)/310ohms= 14.39 watts

That's how you do it

Dear Ronny,

Thank you for your post and Explanation.
This calculation will be handy for the persons who didnt know how to do this.

To add, i think, we should mention as well that XC and XL must match to come to these numbers of resistance.
XC and XL only match at a certain frequency. (resonance). In resonance, the fysical resistance of the coils and cell  are left over.

You also wrote on the other forum on how you start the gas production, when you have the setup complete.
I found it very interesting.
Starting the cell is like plain dc electrolysis, till water is removed and is replaced by gas.
In that state resonance seems to possible. Not in the first electrolysis state.
Can you please explain this a bit more?

Cheers!
Steve