Thane Heins BI-TOROID TRANSFORMER

[ramset]

                                 PDF1

1] It's something that is hard to eyeball, and chances are the waveform does
not line up with the graticule on his scope. I am assuming that he measured this with a ruler??? I am not sure here, are there better ways to do this? It's a critical measurement because it affects his input power calculation.

2] So the OuterO is getting conflicting induced flux that cancels itself out. In other words, if the number of turns in the two secondary coils and the load resistors are perfectly balanced, then the OuterO does nothing because of the flux cancellation.

3] Therefore as a general statement, the OuterO flux ring is not serving any useful purpose. The setup would work just as well if you only had the Figure8 flux path with the three coils.

4] This would be telling you that almost no magnetic flux is flowing through the OuterO because of the flux cancellation problem as indicated in the previous posting. If you see a very low-level AC voltage, you should be able to tweak the self-cancellation to make it disappear almost completely. The easiest way to do this would be to add or subtract one or two turns to one of the secondary coils. You should be able to tweak the flux self-cancellation so that the AC voltage almost completely disappears. No AC voltage on the flux sensor coil means no flux. So what does this mean for any over unity with Thane's new configuration? Well, if you prove that the OuterO does not do anything of value,

5] When you drive a transformer hard into saturation, you waste a lot of power heating the core and the primary rather than transferring power to the load.

6] He claims it is all reactive. I doubt that or he would not have burned up the driving transformer. Current input rises rapidly when you approach saturation, loading on the secondary backs you off this part of the BH curve. He sees input power drop a bit when he connects the load. This is normal for a saturated transformer, as the load takes the transformer out of saturation a bit.

[ramset]

                               PDF2

7]To repeat my point, he is failing to note the flux cancellation problem that I described a couple of postings ago.

8]IT'S WRONG ANALISE IN VIDEO. 1/ WHEN USE BOTH SECONDARY COIL THEY CANCAL EACH OTHER SO THE ONLY WAY THAT FLUX GO IS IN PRIMER COLI LIKE A STANDART TTRANSFORMER. NO ENERGY WIN 2/ WHEN USE ONE SECONDARY COIL ON LOAD THERE IS NO CANCAL FLUX BUT PRIMERY COIL FLUX HAS E EASY TO GO THE SECONDAR COIL UNLOAD SO THE VOLTAGE ON SENDARY DROP MUCH. AND SO DON'T HAVE WIN ENERGY.

9]In this case almost all of the flux generated by the primary coil flows through the left (S1) flux path because it offers no resistance to the flux flow. This looks like a big inductor. Only a tiny amount of flux flows through the right (S2) flux path and this puts 1.09 watts through the light bulb load. From the perspective of the mains power, the load looks like a big inductor (the primary coil with the left flux path) and small separate resistive load.

10]In this case almost all of the flux generated by the primary coil flows through the left (S1) flux path because it offers very little resistance to the flux flow. I say "very little resistance" in this case as opposed to "no resistance" in Case 1. The fact that the secondary coil is short-circuited means that it looks like another inductor driving a small resistance, which is the resistance of the wire itself. That accounts for the "very little resistance" in the left flux path. In this case there are two coils (the primary and the S1 coil) so this ends up looking like a giant inductor that's driving a very small load.

11]So compared to Case 1, for Case2 a bit more flux can flow through the right (S2) flux path because of the very small resistance on the left (S1) flux path. That's why a bit more power flows through the light bulb, 1.65 watts.
11a]Thane thinks the power factor shows a 90 degree phase shift on his scope, but I suspect it's not quite 90 degrees but he can't see it.

11b]In summary, for this clip his setup in both cases looks like a giant inductive load plus a tiny resistive load. That's why his power factor is almost zero. The reactive part of the load is dissipating considerable power through the wire resistance. The fact that he is lighting up a light bulb is almost incidental. The problems associated with the asymmetrical flux paths for this particular experiment mean that the setup fails to efficiently transfer the mains power to the light bulb load. Instead the setup is acting more like a pure inductance through the left (S1) path than anything else.

11c]as well as exploring the issue relating to the canceling fluxes mentioned in post #13.

                                    overdrives

12]As anyone who has used a scope knows, you can position the current waveform exactly between the voltage traces by adjusting the vertical position for that channel. This would fool the viewer into thinking the zero power factor 90 degree phase difference is maintained. You could also add some delay to one channel so that when the PF appears to be approaching 90 degrees out, it is actually moving closer to unity.

[ramset]

                                 PDF3
13]but what Thane is not telling you is that the power input increases along with a nearly doubling of power factor.

13a]The burning of drive transformers was the dead giveaway.

14]No one can duplicate all details of Thanes setup.

15]saturated transformer,

16]Thane needs to connect a Watt meter to the input of his driving transformer.

17]Anyone skilled in the art knows that as long as there is resistance in the primary, the power factor can never be zero.

18]Watts input vs. Watts output is the proof of the pudding.

19]Too bad Thane does not read/post here. But I doubt he would stick around long if he did. I am amazed at the length of time he has retained funding. Good luck to him, but I'm afraid he's in the same boat as Steorn...one that won't float forever

20]Watts input vs. Watts output is the proof of the pudding. Doesn't it strike anyone odd that such a huge transformer arrangement has difficulty lighting a tiny lamp with a maximum of less than two watts of power? Thane needs to connect a Watt meter to the input of his driving transformer. Then you will see that he is drawing 150 Watts to get 1.6 Watts into his light bulb. After all he did burn up that 150 VA transformer and was well on the way to burning up the larger transformer he uses to drive his primary. Thane claims the power factor is zero. Anyone skilled in the art knows that as long as there is resistance in the primary, the power factor can never be zero.

21]The sad part of all this, is that it seems no one at OU has what it takes to pose the questions to Thane directly. Are you going to call him out? You are right, Thane must know what's really going on. What does that say about Thane?

22]It is ok to use amps times volts on the output if the load is purely resistive.

23]At the end of the day one must face reality and compute power input vs. power output.

                         TEST DATA LINK
              http://www.overunity.com/index.php?action=downloads

[maw2432]

Hmm..  I was wondering about the burning of the drive transformers also.

I guess no closed loop can be demonstrated.   

[CRANKYpants]

HERE ARE SOME OF THE EDITS. GOING TO THAT WEBSITE FOR EDUCATION IS NOT WISE SINCE IT "THE BLIND  LEADING THE BLIND" THERE, AS IS GETTING INTO A DEBATE.

CHEERS
T

1] It's something that is hard to eyeball, and chances are the waveform does
not line up with the graticule on his scope. I am assuming that he measured this with a ruler??? I am not sure here, are there better ways to do this? It's a critical measurement because it affects his input power calculation.

EXPAND THE X AXIS SCOPE WINDOW UNTIL 1/2 SINE WAVE IS SHOWN EVENLY OVER 40 X AXIS CROSS HATCHES. IN MY CASE 42. DIVIDE 180 DEGREES / 42 = 4.3 DEGREES PER CROSS HATCH. POSITION EITHER THE CURRENT OR VOLTAGE SINE WAVE EVENLY AND THEN COUNT THE NUMBER OF HATCHES (ON LOAD) THAT THE MIDDLE SINE WAVE DEVIATES.

I.E.
IF THE PURPLE LINE IS SMACK DAB IN THE MIDDLE OF THE TWO GREEN LINES THE PHASE ANGLE DIFFERENTIAL IS 90 DEGREES, WHICH REPRESENTS A POWER FACTOR OF 0.0 IF THE PURPLE LINE IS 5 HATCHES AWAY FROM THE CENTRE THEN THE PHASE ANGLE WOULD BE 90 - 5 = 85, COS 85 = 0.087 PF.

IN OUR CASE ON THE VIDEOS THE PHASE ANGLE IS ABOUT 1 DEGREE, WHICH REPRESENTS A PF OF 0.017.

RED FLAG: ANYONE WHO DOES NOT KNOW HOW TO USE OR READ PF ON A SCOPE SHOULD LEARN - SINCE IT IS THE FOUNDATION OF POWER ENGINEERING 101.

2] So the OuterO is getting conflicting induced flux that cancels itself out. In other words, if the number of turns in the two secondary coils and the load resistors are perfectly balanced, then the OuterO does nothing because of the flux cancellation.

WRONG
YOU CANNOT USE ONE MAGNETIC FIELD TO CANCEL OUT ANOTHER! THIS IS TRANSFORMER 101 AS WELL. IF THIS WERE SO EVEN A CONVENTIONAL TRANSFORMER WOULD NOT WORK BECAUSE BOTH PRIMARY AND SECONDARY FLUXES ARE MOVING IN OPPOSITE DIRECTIONS INSIDE THE PRIMARY CORE.

RED FLAG: A FORUM IS ONLY AS GOOD AS ITS ABILITY TO PROVIDE CORRECT INFORMATION TO THE READERS AND TO "SELF CORRECT" ITSELF WHEN A MISTAKE IS PRESENTED. SINCE NO ONE PICKED UP ON THIS AND CORRECTED IT - EVERYTHING ELSE THAT FOLLOWS IS ALSO IN ERROR.

3] Therefore as a general statement, the OuterO flux ring is not serving any useful purpose. The setup would work just as well if you only had the Figure8 flux path with the three coils.

WRONG NOT TRUE AT ALL.

4] This would be telling you that almost no magnetic flux is flowing through the OuterO because of the flux cancellation problem as indicated in the previous posting. If you see a very low-level AC voltage, you should be able to tweak the self-cancellation to make it disappear almost completely. The easiest way to do this would be to add or subtract one or two turns to one of the secondary coils. You should be able to tweak the flux self-cancellation so that the AC voltage almost completely disappears. No AC voltage on the flux sensor coil means no flux. So what does this mean for any over unity with Thane's new configuration? Well, if you prove that the OuterO does not do anything of value,

WRONG A VOLT METER READS NET FLUX VARIATIONS / TIME. A VOLTMETER IN THIS SCENARIO CAN ONLY READ FLUX FLOWING IN ONE DIRECTION. IF TWO IDENTICAL FLUX MAGNITUDES ARE FLOWING THROUGH A SENSOR COIL ON ONE CORE SECTION THE VOLTAGE WOULD BE 0 VOLTS EVEN THOUGH NET FLUX INSIDE THE COIL WOULD BE AT MAXIMUM. 

RED FLAG: KNOWING HOW A VOLTMETER WORKS AND HOW TO READ A VOLTMETER IS ELECTRICITY 101. ANYONE WHO DOES NOT KNOW HOW TO DO THIS SHOULD LEARN.

5] When you drive a transformer hard into saturation, you waste a lot of power heating the core and the primary rather than transferring power to the load.

WHO SAID ANYTHING ABOUT HARD SATURATION? THE LATEST VIDEO WAS OBVIOUS TO ANYONE OF PERCEPTION THAT THE PRIMARY RELUCTANCE ONLY HAS TO BE INCREASED ENOUGH (BY INCREASING PRIMARY FLUX LEVELS) TO HAVE A HIGHER RELUCTANCE THAN THE SECONDARY FLUX PATH.

6] He claims it is all reactive. I doubt that or he would not have burned up the driving transformer. Current input rises rapidly when you approach saturation, loading on the secondary backs you off this part of the BH curve. He sees input power drop a bit when he connects the load. This is normal for a saturated transformer, as the load takes the transformer out of saturation a bit.

THE PRIMARY IS NOT SATURATED - IT IS IN THE INITIAL STAGES OF SATURATION.