Claimed OU circuit of Rosemary Ainslie

[Rosemary Ainslie]

I'm not sure what you mean Rose. What am I missing?

.99

Poynty - don't ask me to go into the details.  You know as well as I do where your circuit varies from the original.  And it's now 3.05 am here and I simply can't concentrate.  But if you want to talk about it I'm game.  It'll be a nice distraction from the emails I have to get through.

[poynt99]

Fundamentally, the circuits are the same.

The only physical differences are: Fuzzy has a hand-wound resistor, he is using a battery supply, and a breadboard circuit.

For purposes of power in vs. thermal heat out measurements, it does not matter if the supply is a battery or lab supply type. The COP>17 claim is based on more energy dissipated in the load compared to the energy supplied by the source. If required, I do have two 12V batteries lined up. I will need to assess their "health" beforehand though, as this is an unknown at the moment.

The notion of the hand-wound resistor having special properties is still up for debate.

The breadboard circuit I am willing to except some small differences.

So that leaves really only the resistor construction as the major difference for the tests we are doing. We agreed (I thought) that I would obtain some results using my resistor and you would rate my results compared to Fuzzy's #5.

I am having to leave this "rating" up to you Rose, because frankly I do not know what you are using as a basis for these ratings. All I can do is go by what Fuzzy has posted and your comments with each new test run, and the only difference I see between the several tests he has done, is the amplitude of the Drain spike, and also the amplitude of the bump on the battery voltage. If this is not it, I'd very much like to know the metric you are actually using for comparing one test result to another.

Incidentally, are you aware that this battery voltage bump on his commercial OTS resistor test was actually 30% higher than he got with his hand-wound resistor? The Drain voltage was lower though, and that would be expected imo.

.99

[Rosemary Ainslie]

Hi Poynty.  I take it you're not about to discuss anything.  And I'm way too tired to type anything more.  I'll try and deal with your points later.  But glad of the points.  Just sorry we can't actually talk it over.

[fuzzytomcat]

Quote from: poynt99 on Today at 02:02:05 AM
I'm not sure what you mean Rose. What am I missing?

.99

Poynty - don't ask me to go into the details.  You know as well as I do where your circuit varies from the original.  And it's now 3.05 am here and I simply can't concentrate.  But if you want to talk about it I'm game.  It'll be a nice distraction from the emails I have to get through.

The problem is the "Rosemary Ainslie COP> 17 Heater Circuit" has some distinct differences I would think any amateur that can read a schematic can tell .... and the modifications plus total refusal to use the correct items illustrated in the original or now modified "Quantum" October 2002 article and submit test result from a apparatus that is in no way like the original circuit as shown in "FIGURE 1 - DUT CIRCUIT" with even going so far to the earth grounding on the negative rail to eliminate any and all circuit effects using power supplies instead of batteries the incorrect 10 ohm load resistor ....etc. ..... etc. .....

Of course there are only 3 actual oscilloscope wave form images and only 2 modified data sheets not even the originals have been shown here by "poynt99" ...... to the 80 or more unmodified original images and data sheets that has been posted by myself at "Energetic Forum" for complete transparency of my testing "NOTHING HIDDEN" using all components as close as possible to original circuit.

I do at least have respect for poynt99 as he is the only member here at Over Unity to even have the "BALLS" to attempt a replication while others are only loud mouthed couch potatoes that don't know and have any clue the difference between the two schematics and test set ups .... giving there 2 cents and claiming some type of "FAKE" victory when the battle is just about to start.

Also the lack of my response for any additional testing is because as it is well known even by the nay sayers that I do not have in my possession none of the testing equipment that I had previously ...... but this will be resolved with the 100% full capabilities again in 1 to 2 weeks time. 

     

[allcanadian]

@fuzzytomcat
Thank you for the clarification on the circuit diagrams, I just stepped back into the fray, I mean thread--again and was not aware of the circuit discrepancies. In a perfect world we could say they achieve the same effects but this is not the case here. Consider the fact that if you attach two wires to a battery and these wires remain separated or open circuit that the opposite charges will extend to the ends of the wires, they are always present. What you have is a capacitor, the wires are the plates with a small surface area and a very wide spacing thus the capacitance or electric field between the two is small. When dealing with an inductive discharge having fast rise/fall times and higher voltages this capacitive effect increases drastically. Take a NE2 neon and touch a single lead to the circuit to see why Poynt99's circuit cannot be grounded and there can be no capacitors. Again, these discharge currents do not need conductors or closed circuits to transfer energy, they will discharge to the (-) or (+) terminal of a battery through a single wire, they will discharge completely to any ground connection, they are not conventional currents and they do not act like them. Consider why in the quantum circuit the source is in a closed loop with the resistive inductor and other than the isolated gate conductor on the mosfet there is only one external conductor leading from the working circuit--- and where this conductor goes? this is a discharge path because discharge currents seek greater surface area to find equilibrium by lowering it's potential. You can consider a discharge current (the spike) as a unidirectional impolar current in a sense(it will discharge to any source polarity (+) or (-)or ground). This is why when you approach the circuit with your hand the operating frequency charges, the discharge has created a large electric field around the whole circuit to extend it's surface area and find equilibrium. In this case the circuit as a whole is one plate of a capacitor and the space surrounding the circuit is induced with an opposite polarity--the other plate of the capacitor. To conclude you cannot treat this circuit as having conventional current, the same rules simply do not apply.
Regards
AC