Electron Reversing Device

[tinman]

Hi TK
Well i live on the coast of western Australia,so we dont quit hit 50c.
But we often get to 45c in summer,but when you live with it all your life-you get use to it.
Then on the other end of the scale,in winter we can often go below 0c.
No snow here,just ice.

This dosnt realy have anything to do with the circuit,but it was something else i found interesting reguarding a BPC.
And i also put forth a question in reguard to reading the current flow in both directions,useing two diodes in opposite directions off the positive input.Then hooking 1 DMM to each diode to measure the current flow in both diections.We then add the two amounts together to get our total current amount.

http://www.youtube.com/watch?v=AmR_Sqw7wv0

[ramset]

TM
Quote
But the thing i need you to understand is this-in no way do i claim any OU from this circuit.
What i am doing is sharing something i found to be interesting.

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Well apparently you are not alone in your "curious" observation,I personally feel your worth your weight in Gold to this community !

Sometimes Providence brings Men together to ponder a problem or  curiousity,These moments can be Fleeting or be part of Much bigger "events".

It is This "Event" we work towards.And what you do here brings us Closer!
you set a good example .[as always as far as I can tell?]

Thank you
Chet
PS
I also hope you do not suffer from that heat wave .

[poynt99]

And i also put forth a question in reguard to reading the current flow in both directions,useing two diodes in opposite directions off the positive input.Then hooking 1 DMM to each diode to measure the current flow in both diections.
http://www.youtube.com/watch?v=AmR_Sqw7wv0

Apparently you did not see this post tinman:
http://www.overunity.com/13234/welcome-to-understanding-overunity/msg350306/#msg350306

I have reproduced the diagram here for you.

We then add the two amounts together to get our total current amount.

This is incorrect. The correct procedure to obtain the average current supplied by the FG is to take the absolute value of each reading (convert the negative current to positive), add them together, then divide by 2. As follows:

I_FG(avg) = (|Ipos| + |Ineg|)/2

[poynt99]

Ok to answer some question's
First up,i make no claims of any type,other than what i see is something i havnt seen befor.

Relax mates, I've not alluded that you have made any claims.

So a question of my own would be-if the current is changing direction(as it is) then why do the meters not read in AC insted of DC?
At a guess i would think this could be due to the wave shape itself,in that the meters must see a sign wave and not a square wave to read in AC mode.

By all means try it on ACmA. You won't get an accurate reading unless your meter is true rms, but you will get a much more meaningful reading than what you have on DCmA.

Now the r3,150uf cap and diode are not for any operational purpose of the circuit.As mentioned in one of the video's,it is for a voltage reference point only.This tells me how much voltage the circuit is running on.The diode loss must also be added to that voltage>in my case it is +.5 volts for those diodes.

I don't quite understand this. Why can't you measure across the FG output?

Although the FG may be useing so many miliamps at 12 volt's,that is not what the circuit in question is useing.

If you were to measure the average battery current, this would give you a pretty good idea what the circuit in question is using, less 5mA or so for the FG itself.

You should post the original schematic for the FG. It appears you deleted a line or two that is required to show proper operation.

[tinman]

Apparently you did not see this post tinman:
http://www.overunity.com/13234/welcome-to-understanding-overunity/msg350306/#msg350306

I have reproduced the diagram here for you.
This is incorrect. The correct procedure to obtain the average current supplied by the FG is to take the absolute value of each reading (convert the negative current to positive), add them together, then divide by 2. As follows:

I_FG(avg) = (|Ipos| + |Ineg|)/2

Well i guess the only answer i have as to why i missed that post is that i thought anything related to this circuit would be posted here on the thread about the circuit.
But looking at your diagram,it seems that we agree on how to measure each flow of current.
But yes,i did overlook the fact that it must be devided by two for a full cycle current average.
With the meters hooked like i was going to hook them up,then we would not have to convert negative current to positive current.If you flip your bottom meter around(reverse the polarities)then both meters will read a positive current flow.
In reguards to your question in your next post-I don't quite understand this. Why can't you measure across the FG output?
Well my meters just wont read a voltage that is going high then low,so i collect the peak to peak voltage and store it in the 150uf cap,so as i have a stable voltage to read.
I then placed my scope across the output of the FG output,and make sure the cap is correct.But we must remember to add the voltage loss from the diode aswell to that cap voltage-and this will vary slightly depending on what diodes you use ofcourse.

I would also like to say thanks for the help and input you have given so far-its been a great help