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Author Topic: Thane Heins BI-TOROID TRANSFORMER  (Read 471092 times)

lumen

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Re: Thane Heins BI-TOROID TRANSFORMER
« Reply #150 on: October 24, 2010, 06:24:34 PM »
Lumen I think it's mainly because of the 3d aspect. A 2d model is already a big job on the cpu, so you can imagine the strain of a 3d model. What I find too bad is the lack of software that simulates an actual running transformer design where voltage induction is part of the simulation. That would save a lot of time.

Broli
I setup to center coil at 120v with 200 turns and .8 ohms resistance, and run the frames @ .001 second steps up to .019 seconds using a 60hz coil input . The animated gifs that I uploaded show the core "B" flux for one entire cycle.
It seems to work very well, providing there are no operator errors.
The entire calculation only takes about two minutes with a mesh size of .01"



CRANKYpants

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Re: Thane Heins BI-TOROID TRANSFORMER
« Reply #151 on: October 24, 2010, 06:46:01 PM »
GOOD LUCK!  :P
CHEERS
T

BTW - THIS IS WHAT WE DO HERE ON THE WEEKEND! http://www.youtube.com/watch?v=tfC5oYcknew

BITT COMPUTER MODELLING TEST ITEMS:

1. Make the Primary and (net) Secondary with the same wire gauge and the same number of turns so it will be a 1:1 transformer,
i.e. Primary = 100 turns, S1 = 50 turns, S2 = 50 turns.
2. TEST: S1 + S2 (net)NO LOAD voltage = primary INPUT voltage.
3. Primary flux should be evenly distributed through NO LOAD S1 and S2.
4. Note and record Primary Current and Power Factor NO LOAD baseline.
5. Place S1 and S2 ON 100 ohm LOAD and note if Primary Current or Power Factor changes.
6. SOFTWARE TEST:
Remove S2 from load and note S1 load voltage which must = 0 volts because S2’s flux path route now represents a lower reluctance route than S1’s ON LOAD high impedance route.
7. S2’s NO LOAD voltage must = Primary INPUT voltage because S2 is getting all the Primary flux.
8. REPEAT with 50 ohm, 25 ohm, 10 ohm, 1 ohm and shorted Secondaries.
9. If Baseline NO LOAD Current or PF change when placed ON LOAD increase Secondary Outer Core Area to reduce reluctance until there is NO CHANGE from NO LOAD Baseline to ON LOAD shorted Seconaries.
10. Replace Secondary Outer Core with HIGH PERFORMANCE Low Reluctance Permalloy of Superpermalloy etc. And note performance advantage (if any).

TEST NOTES: Test Date: __________________

1. Number of Turns: Primary = _________ turns, S1 = __________, S2 =__________ turns.
2. Primary Input Voltage: Primary = ______ Volts, S1 = ______, S2 = ______ Volts.
3. S1 NO LOAD Voltage = _____ V, S1 NO LOAD Voltage = _____ V,
Flux Distribution: ________________ (even - uneven).
4. NO LOAD Baseline: Primary Current = ______ Amps, Primary Power Factor = ______.
5. S1 & S2 ON LOAD (100 ohms):
Primary Current = ______ Amps, Current change % = ______%, Power Factor = ______,
PF change % = ______%.
6. S2 NO LOAD: S1 ON LOAD Voltage = ______ Volts. (Must = 0 Volts).
7. S2 NO LOAD Voltage: S2 = ______ Volts. (Should = Primary Input Voltage).
8. S1, S2 50 ohm load:
Primary Current = ______ Amps, Current change % = ______%, Power Factor = ______,
PF change % = ______%.
S1, S2 25 ohm load:
Primary Current = ______ Amps, Current change % = ______%, Power Factor = ______,
PF change % = ______%.
S1, S2 10 ohm load:
Primary Current = ______ Amps, Current change % = ______%, Power Factor = ______,
PF change % = ______%.
S1, S2 1 ohm load:
Primary Current = ______ Amps, Current change % = ______%, Power Factor = ______,
PF change % = ______%.
S1, S2 (shorted) load:
Primary Current = ______ Amps, Current change % = ______%, Power Factor = ______,
PF change % = ______%.
9. New Secondary Core Area Increase: _______ %.
S1, S2 (shorted) load:
Primary Current = ______ Amps, Current change % = ______%, Power Factor = ______,
PF change % = ______%. (NOTE; if primary current and PF do NOT change with a short-circuit load they won’t change with lesser loads either).
10. New Secondary Core Material: ___________. Primary Current = ______ Amps,
Current change % = ______%, Power Factor = ______, PF change % = ______%.


« Last Edit: October 24, 2010, 07:07:21 PM by CRANKYpants »

ramset

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Re: Thane Heins BI-TOROID TRANSFORMER
« Reply #152 on: October 24, 2010, 07:35:11 PM »
Thane
AHHH fun with the Kiddypooos!!
So Thats not to different  from my trip to the "Bronx" last week?

Thats Poindexter [MY lab assistant} he loves it when I forget to put my helmet back on!!
{little Rascal]

Test Data link

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

« Last Edit: October 24, 2010, 08:32:27 PM by ramset »

lumen

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Re: Thane Heins BI-TOROID TRANSFORMER
« Reply #153 on: October 25, 2010, 03:31:59 AM »
Broli,
I think you may be right about the calculation times. After checking and finding the mesh settings of .01" were not operational, I changed them to a more realistic setting of .2". Even at the new setting of .2" the calculations take about 20 minutes. The good thing is now the results are more of what one might expect to see.
As MR "T's" test procedures go, I only wish I was good enough to setup the program to perform those tests. This is some powerful program, but you need to write formulas to return the results in volts and current based off things like core energy or the core "B" field intensity.
So I think at this point I am still a bit away form any software solution until I can put all the pieces together.

The "B" field vector plots at least now, look real and small changes to the core have small effects also, so I think I'm on the right track anyhow.



 

teslaalset

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Re: Thane Heins BI-TOROID TRANSFORMER
« Reply #154 on: October 25, 2010, 10:04:45 AM »
Guys, just a quick update where I am with my FEMM simulations.

In my earlier simulations I applied "pure iron" from the default materials library.
However, that is solid iron. So, I made a change to use laminated "pure iron".

I found a nice reference model that can serve as example how to do the detailed simulations:
http://www.femm.info/wiki/MyTransformer .
This example uses a combination of FEMM and Octave scripting, which is new for me, but it looks usable to get all currents and voltages out of simulations. It's a lot of work though, so it will take some time.

Looking at this model I found that the BITT has a different basic model.
Mainly the M (couple factor, or mutual inductance) is the difficulty here.
In my view M is depending on the saturation state of Core1.
So, in the BITT, within a full AC sinus cycle, M will vary.
I will first think of a good first attempt and show it here later.
Maybe you have some ideas you could share here.

Below the simple model details of an ordinary transformer in the given example:
« Last Edit: October 25, 2010, 11:27:42 AM by teslaalset »

broli

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Re: Thane Heins BI-TOROID TRANSFORMER
« Reply #155 on: October 25, 2010, 10:49:14 AM »
I just got a reply back on the price inquiry for the below design.

This is their offer:

piece price: 2.47€/pc.
material included (standard steel)
programming cost: 0€
*****************************
Total cost: 98.8€

To be honest I'm quite surprised at the low price. It could be even lower if I provided them my own material being true silicon steel for instance.

SchubertReijiMaigo

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Re: Thane Heins BI-TOROID TRANSFORMER
« Reply #156 on: October 25, 2010, 12:23:05 PM »
@ Teslaalset: This is normal your M is varying, in any transfomer generally B is linked to the voltage and H with with amperage...
Do you remenber, I have already try to explain the reluctance is directly linked to permeability, and permeability is directly linked to B/H formula...
The big problem is the most of magnetic material are highly non linear, that's why I propose to use a very small air gap or linear magnetic material (like nanoperm ?)... The core 1 is more induced or saturated than core 2 and we have varying voltage and B... It seems they are (for a very short time of the AC wave) we have some back coupling between Core2 to Core1. The main dificulty in the BTT is to preserve the "magnetic diode" in the whole time of the full sine wave and in any situation... I have read something interesting about transfomers: they are slightly no amplitude flux difference when the transfomer is loaded or not. Like Thane said the current draw in the primary is provoqued by the Power Factor. I hope everyonne understand this, transfomer are in nature very critical, I post some image to understand this. 2 months ago I knew nothing about Magnetics and Transfomers my original hobby is Pianist... but I have realized magnetic engineering is very subtil specially with OU device and new phenomena... :)
One of those pictures below was found here: http://www.feryster.pl/polski/nanoperm.php?lang=en it seems they have some linear core...

Good luck for all !!!  :)

Edit: In the Jean Louis Naudin website he have make some experiment with the nanoperm core look the linear permeability of the u=30000 core !!!
Beautiful !!!
Source: http://jnaudin.free.fr/2SGen/index.htm
« Last Edit: October 25, 2010, 01:46:29 PM by SchubertReijiMaigo »

ramset

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Re: Thane Heins BI-TOROID TRANSFORMER
« Reply #157 on: October 25, 2010, 06:20:46 PM »
Mr.T
In an effort to keep this civil,as well as advance the understanding of what the BTT is actually doing!
There is A very nice Man over here

http://www.overunityresearch.com/index.php?topic=261.msg6066

Called ION
That has a request?

Chet

teslaalset

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Re: Thane Heins BI-TOROID TRANSFORMER
« Reply #158 on: October 25, 2010, 08:50:15 PM »
Mr.T
In an effort to keep this civil,as well as advance the understanding of what the BTT is actually doing!
There is A very nice Man over here

http://www.overunityresearch.com/index.php?topic=261.msg6066

Called ION
That has a request?

Chet


Good point Chet.
I believe ION has some interesting observations that we need some response on.

Never the less, this principle is still worth looking at.
Maybe less saturation, maybe add an air gap will at least reveal the principle is working at the cost of some efficiency, I would be happy with 300%...
I'll continue to build the model and look forward to some comments from Thane on this.

CRANKYpants

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Re: Thane Heins BI-TOROID TRANSFORMER
« Reply #159 on: October 26, 2010, 02:38:54 AM »
Mr.T
In an effort to keep this civil,as well as advance the understanding of what the BTT is actually doing!
There is A very nice Man over here

http://www.overunityresearch.com/index.php?topic=261.msg6066

Called ION
That has a request?

Chet

CAN SOMEONE PLEASE CUT AND PASTE THE ERRORS (IN RED) SO WE CAN GO OVER THEM ONE BY ONE?

THANKS
T

ramset

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Re: Thane Heins BI-TOROID TRANSFORMER
« Reply #160 on: October 26, 2010, 03:01:22 AM »
                                 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

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Re: Thane Heins BI-TOROID TRANSFORMER
« Reply #161 on: October 26, 2010, 03:15:45 AM »
                               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

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Re: Thane Heins BI-TOROID TRANSFORMER
« Reply #162 on: October 26, 2010, 03:29:09 AM »
                                 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

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Re: Thane Heins BI-TOROID TRANSFORMER
« Reply #163 on: October 26, 2010, 12:06:36 PM »
Hmm..  I was wondering about the burning of the drive transformers also. ???

I guess no closed loop can be demonstrated.   :'(

CRANKYpants

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Re: Thane Heins BI-TOROID TRANSFORMER
« Reply #164 on: October 26, 2010, 01:45:45 PM »
HERE ARE SOME OF THE EDITS. GOING TO THAT WEBSITE FOR EDUCATION IS NOT WISE SINCE IT "THE BLIND  8) LEADING THE BLIND" THERE, AS IS GETTING INTO A DEBATE.

CHEERS
T


Quote
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.

Quote
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.
 
Quote
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.
 
Quote
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.
 
Quote
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.