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Overunity Machines Forum



COP 20.00 (2000%) Times, Reactive Power Energy Source Generator,

Started by synchro1, May 07, 2014, 01:25:54 PM

Previous topic - Next topic

0 Members and 4 Guests are viewing this topic.


picowatt

With regard to the two screen captures posted a few pages back:

1.  The captures were made approx 1/2 hour apart, so one cannot be absolutely certain that adjustments to the circuit (variac, etc) were not made in between the two captures.

2.  If the right side capture is indeed the I and V seen by Rload, then Rload appears to be a 10 ohm resistor (if I am reading the captures correctly, hard to see the divisions)

3.  The left side capture seems consistent with the V trace being 125V pk to pk or so (reduced line voltage via the variac?)

4.  The peak V values on the left and right captures appear different, so either the V was adjusted in between captures or possibly the transformer is being used to reduce voltage to Rload.

5.  The right side captures are consistent with full wave rectification.  Can anyone discern if there is a FWB just behind the left most capacitor in any of the images? 

6.  Possibly the transformer primary is in series with AC line (in Harti's lamp position in his cap switching schematic) with the secondary rectified and applied to the load.  Alternately, possibly a full wave bridge is in that location with its DC out to the load resistor, and as others have speculated, the transformer used for isolation. 

7.  If someone were to take the time to accurately graph out the left side capture, the V trace could be partially reassembled from the I and W trace values.

Looks like it is time for some of the sim people to try a few things to see if they can replicate the waveforms based on best guesses.

Just a few thoughts...

PW

SchubertReijiMaigo

Look like don't work for me here...
I've made a simulation and don't seem to have excess.
The logic is like this.
0-90° --> charge in parallel.
90-180° --> discharge series.
then same thing but for the inverted polarity.
Putting cap to series/parallel doesn't give excess (this a well know fact).
When you charge cap 1/2 of energy is lost no matter what kind of resistance you use. It can be the parasitic resistance of the wire, capacitor ESR, or a bulb.
This is also true at discharge cycle 1/2 is again lost.
So for one unit of energy, 1/2 get "lost" and at discharge 1/2 of 1/2 get "lost" again.
So it give that 3/4 is lost/burned and only 1/4 return to source... 3/4+1/4=1 so energy is conserved in the whole process, I don't see unfortunately OU in this setup at least.
---------------------------------------------------------------------
Here the code:

$ 1 5.0E-6 13.654669808981877 50 5.0 50
w 240 288 240 208 0
w 240 208 288 208 0
r 288 208 384 208 0 100.0
s 384 208 496 208 0 0 false
w 496 256 496 208 0
w 496 256 432 256 0
r 496 256 496 288 0 0.01
c 496 288 496 352 0 1.0E-5 3.4950131883260247
r 496 416 496 448 0 0.01
c 496 448 496 528 0 1.0E-5 3.495013188326026
w 496 528 240 528 0
w 240 528 240 400 0
v 240 400 240 288 0 1 50.0 325.0 0.0 0.0 0.5
178 432 256 432 336 0 1 1.0E-9 1.0000000000157452E-5 0.05 1000000.0 1.0E-9 1000000.0
178 608 368 608 448 0 1 1.0E-9 1.0000000000157452E-5 0.05 1000000.0 1.0E-9 1000000.0
w 496 352 608 352 0
w 544 416 496 416 0
w 544 416 544 464 0
178 752 464 752 368 1 1 1.0E-9 -1.000000082740371E-5 0.05 1000000.0 1.0E-9 1000000.0
w 752 464 544 464 0
w 608 352 608 368 0
w 560 368 560 320 0
w 576 368 576 320 0
w 576 320 640 320 0
w 560 320 560 272 0
w 640 320 640 272 0
v 560 272 640 272 0 2 100.0 5.0 5.0 0.0 0.5
w 384 256 384 224 0
w 384 224 560 224 0
w 560 224 560 272 0
w 400 256 400 240 0
w 400 240 640 240 0
w 640 240 640 272 0
w 704 368 704 320 0
w 704 320 640 320 0
w 720 368 720 304 0
w 720 304 544 304 0
w 544 304 544 320 0
w 544 320 560 320 0
w 496 416 416 416 0
w 416 416 416 336 0
w 592 448 592 528 0
w 592 528 496 528 0
w 768 368 768 352 0
w 768 352 608 352 0
o 12 64 0 291 640.0 6.4 0 -1
o 2 64 0 291 320.0 6.4 1 -1
o 12 64 1 291 916.4449253911987 9.765625000000002E-255 2 -1
o 2 64 1 291 428.6034428745069 2.4414062500000007E-305 3 -1
o 12 64 0 290 559.9361855444511 1.3998404638611277 4 -1
o 26 64 0 35 10.0 9.765625E-5 5 -1

---------------------------------------------------------------------------------





e2matrix

Quote from: MarkE on July 13, 2014, 12:41:23 AM
In over three decades in the business I have never seen a circle with an X in it used for a light bulb.
Just another example of living inside the box.   After about 20 years of looking at alternative energy I assumed everyone recognized that as a light bulb who had interest in free energy.   It's so common in my experience that you don't want to know what I was thinking of saying about you for not knowing but I decided to chill   8) .   Surprised TK had any question though....   
BTW I don't think you'll ever see a multiplier with only 2 connections.   

picowatt

One more note.

As the left side capture is labeled as transformer output, it is possible the variac is driving the primary of the transformer and that the secondary (transformer output) is connected to the circuit.  This arrangement allows for both isolation and input V adjustment.

I must again address my concern, however, that the data collected from the transformer output is not necessarily going to reflect similar data measured directly at the AC line.  It is difficult to believe that anyone claiming to be able to draw real power from the AC line while returning the same or more power back to the AC line, would not include AC line measurement data.  A variac and that transformer must surely represent a fair amount of inductance between the AC line and the points in the circuit used for input power measurements.

I will remain skeptical until AC line input measurements are released that reflect what is happening directly at the AC line, and that also includes any power drawn from the driver circuitry.  The total AC line input power measured directly at the AC line, compared to that simultaneously made across Rload, would paint a more accurate picture.

PW