another small breakthrough on our NERD technology.

[Rosemary Ainslie]

Ok, Rosemarry please excuse me for asking these questions.

First, do I understand correctly that there is a current in these circuits that not only is against the direction applied by the battery and the collapse, but that it moves through diodes/transistors in the wrong direction without apparent damage?

Actually, that really is the only question I had.

I am interested in replicating one of these circuits, what is the most robust preferred embodiment and parts?

Also for what it's worth I certainly know part of how these types of circuits work and I think I probably know enough to make them produce more power if you are interested. (and I'm not just blowing smoke)

Thanks,
John

Hello John.  If you can open those files that I posted you should get all you need on this subject.  I'll go back and see if I can repost it.  You've more or less got the gist of the claim - but it's rather more comprehensive.  Anyway - hold fire.  I'll see if I can find it.

Kindest regards,
Rosemary

It took me forever to get back here.  Something's seriously wrong with this new system Harti. In any event John - here are those links.
Again - all the best.  Let us know if you do a replication.

http://www.overunityresearch.com/index.php?action=dlattach;topic=13.0;attach=6766

http://www.overunityresearch.com/index.php?action=dlattach;topic=13.0;attach=6767

[gravityblock]

True Electrical Units

Gravock

[gravityblock]

Definitions are funny things. While watts are generally considered to be a measure of instantaneous power, and watt hours or watt seconds are a measure of power over time (energy), it must be recognised that time is already factored into the definition of a watt. How so? Power is Voltage times Current. - Yeh Since charge is measured in coulombs and time is measured in seconds, 1 Ampère is the same as 1 Coulomb per second. That is, Current (Amps) is Coulombs per second, therefore Power is Voltage x (Coulombs per second -  for 1 second). So 1 watt can be the equivalent of 1 Volt x 1 Coulomb per pecond, for 1 second. Hmmm, no wonder confusion over power or energy measurement occurs Cheers from Hoptoad

By using the true electrical units, this hidden factor, which has been the author of confusion, is now clearly exposed.

q = kg.
A = m/s^2
Z = seconds

Energy, work, quantity of heat = Joule or qA^2 Z^2 while,
Power, radiant flux = Watt or qA^2 Z

Electromotive force, potential difference = Volt or qA
Electric Current = Amp or AZ

Electric Resistance = Ohm or q/Z
Electric Charge, quantity of E = Coulomb or AZ^2

Gravock

[poynt99]

Alternatively we could, perhaps, just stick to that earlier post of yours where you proposed to multiply the voltage across the load with the voltage across the battery to ascertain it's power?

WTF, I proposed no such thing. 

Your fundamental problem in all this is that you either don't have the capability to correctly understand and interpret these simple technical diagrams and the salient points being made about them, or you are intentionally trying to cloud them with your nonsense. Which is it?

.99

[Rosemary Ainslie]

WTF, I proposed no such thing. 

Your fundamental problem in all this is that you either don't have the capability to correctly understand and interpret these simple technical diagrams and the salient points being made about them, or you are intentionally trying to cloud them with your nonsense. Which is it?

.99

Golly Poynty Point,

You're parading all that exasperation again.  It's becoming seriously repetitive.  But if it helps you to pretend that you can barely tolerate my intellectual incompetence - my willful and deliberate obfuscation - feel free.  I suspect you'll need every trick in the book now.  You'll need to find cause for denying us our claim for your prize.  And you're hoping against hope that a scornful dismissal - may yet cut it.   But it's an unfortunate choice to refer to that curious post of yours.  I doubt it will survive scrutiny.  However.  Since you insist - then I, MOST reluctantly, MUST engage.  Let's see exactly where this argument poynts - no pun intended.       

Here's the post

The electric field across an electric power SOURCE is always in OPPOSITE polarity to the direction of current through the power source when the power source is supplying current in the circuit. Therefore when a power calculation is performed on the power source in such case, (V x I), the two possible scenarios are the following, either:

1) +V x -I, or
2) -V x +I.

In either case, the result of the product is a NEGATIVE value.

The electric field across an electric power LOAD is always in EQUAL polarity to the direction of current through the load when the load in the circuit is dissipating energy. Therefore when a power calculation is performed on the load, (V x I), the two possible scenarios are the following, either:

1) +V x +I, or
2) -V x -I.

In either case, the result of the product is a POSITIVE value.

Although outlined in the detailed analysis06, the simple example below illustrates these facts quite well also. Note the difference in the direction of current and potential difference across each component. 

.99

If I can manage the schematic I'll post it later.  Meanwhile.  What you STATE is that the 'electric field across an electric power SOURCE is always in OPPOSITE polarity to the direction of current through the power source when the power source is supplying current in the circuit.'  If - by this - you're proposing that the potential difference imposed on the circuit materials has an opposite polarity to the applied current flow from the source - then who's arguing?  Therefore -  for instance - if the battery is delivering a POSITIVE current flow - then the measured voltage across the load - the wires - and so on - will be NEGATIVE and vice verse. Again.  Who's arguing?

But then you state that therefore 'when a power calculation is performed on the power source in such case, (V x I), the two possible scenarios are the following, either: 1) +V x -I, or 2) -V x +I.  Golly.  We know that the watts (dare I use that term?) is determined by the product of the voltage across the battery supply - in the schematic example that you use -  and the direction of that current flow through the circuit (per second and so on).  AND.  You've agreed that IF the voltage measured across the circuit components is negative then you can put MONEY on it that the current flow is POSITIVE.  BUT YET?  With a 'flick of the wrist - so to speak - with a tan tan tara - with all the flamboyance of a magician - pulling the rabbit out the hat - you THEN propose that that current flow must be given a NEGATIVE VALUE. There it is.  As written in your first example.  1) +V x -I.

AND.  As if that's not enough!  You do it AGAIN - A SECOND TIME - to include a second option.  That - 2) - V x + I? What can I say?  What can any of us say?  Except that if this is a serious proposal then - you are grossly unaware of your own contradiction. OR. You've somehow managed to GROSSLY underestimate - what we both know - is my rather average intelligence. 

I put it to you Poynty Point - that the direction that current flows is ALWAYS consistent with the polarity of the applied voltage form its source. Therefore IF the voltage is positive then the current flow is positive.  And IF the voltage is negative then the current flow is negative.  And if the applied voltage across those circuit components is NOT negative when the flow is positive or correspondingly, if the applied voltage is NOT positive when the flow is negative - then we can all RETHINK the standard model.  It will mean that you have, indeed, discovered something that diametrically contradicts everything that we have all rather come to depend on.

Kindest as ever
Rosie Posie

added
'in the schematic example that you use'