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### Author Topic: Confirming the Delayed Lenz Effect  (Read 819828 times)

#### gyulasun

• Hero Member
• Posts: 4144
##### Re: Confirming the Delayed Lenz Effect
« Reply #300 on: October 16, 2011, 08:05:03 PM »
Hi Luc,

Quote
....  I now understand that a higher voltage across the Shunt Resistor does not necessarily matter if the Phase (cos angle) stays the same. Do I have that correct?

I did not mean it that way, though what you say I think it also matters because an increasing shunt voltage means an increasing input current hence an increasing input power draw while the phase angle may remain the same.  I meant when your LC tank circuit was activated by connecting the 39uF cap, the input current got reduced (voltage drop across the shunt decreased) but the phase angle between the input voltage and input power has changed from the 82Â° to 34.5Â°, this caused a significant increase in input power.

Quote
....
The First Shot below is the no load centered Reference. We have exactly 4 squares between our rising and dropping voltage phase. Each square should be 22.5 degrees, so then one of the 5 divisions in each square should be 4.5 degrees. The Current is 3 divisions behind the Voltage (3 x 4.5 = 13.5) so 90 - 13.5 = 76.5 degrees.
So would you agree this is an accurate way to get Phase degree?

Yes it is more accurate now than I managed to estimate it but then the 76.5Â° is even further away from the 90Â° goal. Maybe you wanted to write 8 squares above instead of 4?

Quote
I still don't understand how you came up with the numbers above so I better let you do it. Hopefully in time I will be able to learn the equations

I used conventional AC power calculations whereby the phase angle between the AC current and voltage is considered and you multiply the rms values of current and voltage with the cosine value of the phase angle.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/powerac.html

I will revise the calculations though I just noticed Thane already gave a hint.

Gyula

#### gotoluc

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• Hero Member
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##### Re: Confirming the Delayed Lenz Effect
« Reply #301 on: October 16, 2011, 09:07:21 PM »
Hi Luc,

I did not mean it that way, though what you say I think it also matters because an increasing shunt voltage means an increasing input current hence an increasing input power draw while the phase angle may remain the same.  I meant when your LC tank circuit was activated by connecting the 39uF cap, the input current got reduced (voltage drop across the shunt decreased) but the phase angle between the input voltage and input power has changed from the 82Â° to 34.5Â°, this caused a significant increase in input power.

Okay I think I get it.

Yes it is more accurate now than I managed to estimate it but then the 76.5Â° is even further away from the 90Â° goal. Maybe you wanted to write 8 squares above instead of 4?

I edited it to say:
We have exactly 4 squares between each 90 degrees side of the rising and dropping voltage phase.

I will revise the calculations though I just noticed Thane already gave a hint.
Gyula

Please do the complete calculations. I will go over it on the phone with Thane to get a verbal explanation of how to do the calculations.

Thanks

Luc

#### gyulasun

• Hero Member
• Posts: 4144
##### Re: Confirming the Delayed Lenz Effect
« Reply #302 on: October 16, 2011, 09:53:51 PM »
Hi Luc,

Focusing on your third scope shot now, I assume there is no 39uF connected and CH3 and 4 show the output voltage across the 10 Ohm resistors, respectively.
CH1 is the voltage across the input current shunt and CH2 is the input voltage as before, right?
Now the phase angle between input current and voltage seems to be near 90Â° for me?  One full wave is 4 msec rounded down (248 Hz is 4.03 msec) and there is 1 msec for the 90Â° section. The 1 msec section is divided to 5 smaller divisions, this means 18Â° for any two neighbouring small divisions and I can see 4.5 small such divisions between current and voltage curves, giving a phase angle of 4.5 x 18Â°=81Â°

The input power now is Pin=7.16V*0.0177A*cos81Â°=0.0198W

Output power on one of the 10 Ohm load is Pout1=(0.208*0.208)/10= 0.00432W
Output power on the other 10 Ohm load is Pout2=(0.171*0.171)/10=0,00292W
Summing the two outputs gives 0.00725W

efficiency is (0.00725W/0.0198W)*100=36.6%

Now if you could expand the third scope shot to see more precisely the phase angle (I used an estimated 81Â°)  and what is more you could attain a phase shift much nearer to 90Â° degree between input current and voltage, then the real input power would be much less hence efficiency should improve.  The exact 90Â° phase shift would mean a fully reactive input power and no real input power (cos90Â°=0)  this is what Thane aims at.

Gyula

#### gotoluc

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##### Re: Confirming the Delayed Lenz Effect
« Reply #303 on: October 16, 2011, 10:32:47 PM »
Hi Luc,

Focusing on your third scope shot now, I assume there is no 39uF connected and CH3 and 4 show the output voltage across the 10 Ohm resistors, respectively.
CH1 is the voltage across the input current shunt and CH2 is the input voltage as before, right?
Now the phase angle between input current and voltage seems to be near 90Â° for me?  One full wave is 4 msec rounded down (248 Hz is 4.03 msec) and there is 1 msec for the 90Â° section. The 1 msec section is divided to 5 smaller divisions, this means 18Â° for any two neighbouring small divisions and I can see 4.5 small such divisions between current and voltage curves, giving a phase angle of 4.5 x 18Â°=81Â°

The input power now is Pin=7.16V*0.0177A*cos81Â°=0.0198W

Output power on one of the 10 Ohm load is Pout1=(0.208*0.208)/10= 0.00432W
Output power on the other 10 Ohm load is Pout2=(0.171*0.171)/10=0,00292W
Summing the two outputs gives 0.00725W

efficiency is (0.00725W/0.0198W)*100=36.6%

Now if you could expand the third scope shot to see more precisely the phase angle (I used an estimated 81Â°)  and what is more you could attain a phase shift much nearer to 90Â° degree between input current and voltage, then the real input power would be much less hence efficiency should improve.  The exact 90Â° phase shift would mean a fully reactive input power and no real input power (cos90Â°=0)  this is what Thane aims at.

Gyula

HI Gyula,

yes, that is correct!... and no Capacitor used.

Third Scope shot is the same as the Second Scope Shot. Nothing has changed other then the Scopes Voltage and Time Divisions so you can see the complete picture.

So your 81 degrees estimate is not correct.

I have checked all the Transformers I have, even a high end Audio Power Amp Toroid. None of them are 90 degrees out of phase idle (no load). The Toroid is at 40.5 degrees with no load.
So I don't know why we have to be at 90 degrees to get a winner. If we start with 40.5 and we connect a load and it stays at 40.5 then is the Power on the load not coming from Reactive power?

Luc

#### gyulasun

• Hero Member
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##### Re: Confirming the Delayed Lenz Effect
« Reply #304 on: October 16, 2011, 11:40:53 PM »
Luc,  sorry for the misunderstanding....

Pin=7.16*0,0177*cos76.5=0.02958W

efficiency (0.00725W/0.02958W)*100=24.5%

You wrote: I found that it's only @248Hz that there is Zero Phase Shift when under Load. See Second Shot below. Frequencies above 248Hz the Phase slowly goes up and below 248Hz Phase slowly drops.

My question is where do you mean exactly the zero phase shift happens at the 248Hz frequency?

According to Thane, a 90Â° phase shift is the goal between the input current and voltage. This is a situation when input power is fully reactive and in case this happens at the 60 Hz mains frequency than utility WattHour meters in most homes would not measure the load's consumption.  Of course reactive current would still load the mains and this reactive current should still be supplied by the utility providers.
Considering your question on the 40.5Â° phase angle: it does not represent a fully reactive power like it would when the phase angle were at or very near to 90Â° angle.

Phase angle, Phi = arctan(XL/R)   so if you have R=120 Ohm and a coil's inductive reactance XL=150 Ohm at a given AC frequency, than the phase angle between their current and voltage is arctan(150/120)=51.34Â°  arctan function is available in Windows built-in scientific calculator and click on Inv icon inside it when wish to take arctan. (the 120 Ohm is the coil's DC resistance)  So in this example we have both a reactive and a real (heat) dissipation for this coil. If you had a coil which would have say XL=800 Ohm at an AC frequency and it would have only 1 Ohm DC copper resistance, its phase angle would be arctan(800/1)=89.92Â° so pretty close to 90Â° an almost ideal coil.

This means that to get a phase angle very near to 90Â° the L/R for a coil has to be a high value, meaning a low resistive part. By manipulating the frequency to find this situation may help but I guess it brings in capacitive part of the coil windings (like for a MOT secondary) and an unpredictable nonlinear core behaviour at the higher than manufactured frequencies, so all in all here is where you have to ask Thane on any further such questions. I really mean this, without any bias.

Gyula

#### gotoluc

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##### Re: Confirming the Delayed Lenz Effect
« Reply #305 on: October 17, 2011, 02:16:22 AM »
You wrote: I found that it's only @248Hz that there is Zero Phase Shift when under Load. See Second Shot below. Frequencies above 248Hz the Phase slowly goes up and below 248Hz Phase slowly drops.

My question is where do you mean exactly the zero phase shift happens at the 248Hz frequency?

Humm ... lets try it this way. If I connect the Primary of the Transformer I made to the output of my Signal Generator to find the Frequencies that when I connect the load to the Secondary it does not consume any extra current.
This happens at 248Hz. At lower Frequencies it slowly starts to consume current and at higher Frequencies it slowly starts to reduce current consumption.

Now when doing the same as above but using the Scope to only look at Phase ( scope adjusted like fist and second shot) I can see at 248Hz there is Zero Phase change when adding the load or removing it.
Now, if I start to lower the Frequency it slowly starts to shift in a negative way (reducing phase degree)and if I increase the Frequencies it slowly start to shifts in a positive way (adding phase degree)
This is what I mean with a Zero Phase Shift Frequency.

I hope you understand as I don't know how to explain it any better. I would have to do a video if you don't get it.

According to Thane, a 90Â° phase shift is the goal between the input current and voltage. This is a situation when input power is fully reactive and in case this happens at the 60 Hz mains frequency than utility WattHour meters in most homes would not measure the load's consumption.  Of course reactive current would still load the mains and this reactive current should still be supplied by the utility providers.

I agree to the above but I don't think it's possible for a transformer Primary to not consume any current, so I don't think you can start at 90 degrees Phase Shift. That is what I've been trying to say. However, from my tests I do see that it's possible to build a Transformer in a way that when the Secondary is on load it will not shift the Primary Phase to a lower degree. This I have proven with my tests with special core. Please correct me if I'm wrong but as far as I am concerned there are no low Impedance Transformers that you can take off the shelf and have these results.

Considering your question on the 40.5Â° phase angle: it does not represent a fully reactive power like it would when the phase angle were at or very near to 90Â° angle.

I agree but like I said above, I don't think a full 90 degree shift could be achieved. I hope I'm wrong but I have not seen it yet.

Phase angle, Phi = arctan(XL/R)   so if you have R=120 Ohm and a coil's inductive reactance XL=150 Ohm at a given AC frequency, than the phase angle between their current and voltage is arctan(150/120)=51.34Â°  arctan function is available in Windows built-in scientific calculator and click on Inv icon inside it when wish to take arctan. (the 120 Ohm is the coil's DC resistance)  So in this example we have both a reactive and a real (heat) dissipation for this coil. If you had a coil which would have say XL=800 Ohm at an AC frequency and it would have only 1 Ohm DC copper resistance, its phase angle would be arctan(800/1)=89.92Â° so pretty close to 90Â° an almost ideal coil.

This arctan calculation sounds very interesting. At what Frequency do you think you would get 800 Ohms Reactance?

Could you calculate heat dissipation (losses) in my Primary @248Hz Sine Wave with DC R of 17.2 Ohms and L of 220mH

Luc
« Last Edit: October 17, 2011, 04:20:10 AM by gotoluc »

#### gyulasun

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##### Re: Confirming the Delayed Lenz Effect
« Reply #306 on: October 17, 2011, 11:52:16 AM »
Hi Luc,

Ok, now it is clear how you meant the zero phase shift occuring i.e. not occuring...

Quote
I agree to the above but I don't think it's possible for a transformer Primary to not consume any current, so I don't think you can start at 90 degrees Phase Shift. That is what I've been trying to say. However, from my tests I do see that it's possible to build a Transformer in a way that when the Secondary is on load it will not shift the Primary Phase to a lower degree. This I have proven with my tests with special core. Please correct me if I'm wrong but as far as I am concerned there are no low Impedance Transformers that you can take off the shelf and have these results.

Yes, a transformer will always consume at least reactive current and the less copper and core loss a transformer has got, the closer the phase shift would be to the ideal 90Â° phase shift. When you start loading it, surely you can find a frequency where this phase shift would not change due to the load but the same reactive current is still consumed by the primary coil. I do not know if such a low impedance transformer exists off the shelf, the frequency you test it is always a question and what purpose a specific transformer was manufactured for.

Quote
This arctan calculation sounds very interesting. At what Frequency do you think you would get 800 Ohms Reactance?

Could you calculate heat dissipation (losses) in my Primary @248Hz Sine Wave with DC R of 17.2 Ohms and L of 220mH

Well, the formula to calculate the frequency where a coil can have 800 Ohm inductive reactance is f=XL/(2pi*L)  i.e. f=800/(6.28*L)

For your 220mH coil the frequency where you would have 800 Ohm inductive reactance would be f=800/(6.28*0.22)=579 Hz  However, the 17.2 Ohm resistance would "ruin" the phase angle a little:  arctan800/17.2=88.76Â°phase angle would be involved so an even higher frequency should be found... At 9 kHz the 220mH coil has an XL of 124344 Ohm reactance this gives a phase angle of 89.99Â° with its own 17.2 Ohm wire resistance...

From your third scope shot the current via your 220mH primary is 0.0177A this is what is established by the series 10 Ohm shunt and the inductive impedance of this coil at 248 Hz. This current dissipates PL=0.0177*0.0177*17.2=0.00538W power i.e 5.38 mW loss in the coil.

Gyula

#### Overunityguide

• Full Member
• Posts: 107
##### Re: Confirming the Delayed Lenz Effect
« Reply #307 on: October 17, 2011, 01:46:52 PM »
Yes, a transformer will always consume at least reactive current and the less copper and core loss a transformer has got, the closer the phase shift would be to the ideal 90Â° phase shift. When you start loading it, surely you can find a frequency where this phase shift would not change due to the load but the same reactive current is still consumed by the primary coil. I do not know if such a low impedance transformer exists off the shelf, the frequency you test it is always a question and what purpose a specific transformer was manufactured for.

@Gyula

The Primary coil resistance is 1.8 Ohms in my MOT transformer experiment...
This is exactly why I am using a MOT transformer to do this kind of experiments.

My MOT High Freq. (Delayed Lenz) Experiment:

And Yes Reactive Power will be consumed, but in my opinion this is a nonissue, because we always can power factor correct this.

With Kind Regards, Overunityguide

#### gyulasun

• Hero Member
• Posts: 4144
##### Re: Confirming the Delayed Lenz Effect
« Reply #308 on: October 17, 2011, 03:53:10 PM »
...
The Primary coil resistance is 1.8 Ohms in my MOT transformer experiment...
This is exactly why I am using a MOT transformer to do this kind of experiments.

Yes, high power mains transformers in the half to the 1-2 kWatt range have primary coils made from thick wire, so copper resistance is not the main problem.  However their core laminations can have upper frequency limits  that may prevent them to get meaningful power output because losses above some hundred Hertz or in the very low kHz range become prohibitive. Perhaps the so called Hypersil cores do not suffer from  as much frequency limitation as the laminations with normal 4% Si content.

And I assume your MOT's primary may serve as a "coupling coil" to the very high impedance secondary coil.

Quote
And Yes Reactive Power will be consumed, but in my opinion this is a nonissue, because we always can power factor correct this.

Yes this is what would be the most welcome tests and looking forward to seeing such tests where really useful output power can be received at the output and the big question is: at what input power price?

Thanks,  Gyula

#### DeepCut

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• Posts: 640
##### Re: Confirming the Delayed Lenz Effect
« Reply #309 on: October 17, 2011, 05:22:36 PM »
Hello Everyone,

The coils employed in this prototype are 4.5 ohms, 16 gauge bi-filar wound series connected with M1 core laminations and create acceleration at 1800 RPM with a 10 ohm light bulb. Each coil can produce 50 Watts or more and the magnets are 90 lb pulling weight. They create so much torque and acceleration that two set screws on each rotor were not enough to keep them secured to the drive shaft and they had to be returned to the machinist to have key-ways installed. Even now the air gap on each side is about 1/2 an inch. When properly balanced with three rotors and offset cores the cogging torque is virtually zero and the core "cost" was very low - which is reduced as speed increases anyway and is NOT an issue.

I will post the test data when I find it to end this discussion (which is a waste of time BTW) because all generators have coils and cores with some hysterisis losses inherent in them but not all generators accelerate when a load is applied which is the REAL issue.

Cheers
Thane

WOW Thane,

i'm just catching up on bits of this thread.

4.5 ohms @ 16-gauge is 9.5 pounds of wire, crikey !!!

No wonder you're getting 50 watts

I've got hold of a MOT so i may join you chaps on that journey soon.

In the meantime i've been watching ALL of Thane's youtube videos (all 62 of them !) :

It's a very sound and logical journey, it's very interesting seeing the concept go through it's stages of development and i recommend that everyone views the lot of them.

If you use FireFox and want to download them all in one go then i'd recommend this plugin :

Best to all,

DC.

#### CRANKYpants

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• Posts: 1062
##### Re: Confirming the Delayed Lenz Effect
« Reply #310 on: October 17, 2011, 06:47:16 PM »
WOW Thane,

In the meantime i've been watching ALL of Thane's youtube videos (all 62 of them !) :

It's a very sound and logical journey, it's very interesting seeing the concept go through it's stages of development and i recommend that everyone views the lot of them.

Best to all,
DC.

HEY DC,

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CC

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<jackiesyrett@rogers.com>
« Last Edit: October 17, 2011, 07:07:50 PM by CRANKYpants »

#### gotoluc

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##### Re: Confirming the Delayed Lenz Effect
« Reply #311 on: October 17, 2011, 06:49:18 PM »
Hi Luc,

Ok, now it is clear how you meant the zero phase shift occuring i.e. not occuring...

Yes, a transformer will always consume at least reactive current and the less copper and core loss a transformer has got, the closer the phase shift would be to the ideal 90Â° phase shift. When you start loading it, surely you can find a frequency where this phase shift would not change due to the load but the same reactive current is still consumed by the primary coil. I do not know if such a low impedance transformer exists off the shelf, the frequency you test it is always a question and what purpose a specific transformer was manufactured for.

Well, the formula to calculate the frequency where a coil can have 800 Ohm inductive reactance is f=XL/(2pi*L)  i.e. f=800/(6.28*L)

For your 220mH coil the frequency where you would have 800 Ohm inductive reactance would be f=800/(6.28*0.22)=579 Hz  However, the 17.2 Ohm resistance would "ruin" the phase angle a little:  arctan800/17.2=88.76Â°phase angle would be involved so an even higher frequency should be found... At 9 kHz the 220mH coil has an XL of 124344 Ohm reactance this gives a phase angle of 89.99Â° with its own 17.2 Ohm wire resistance...

From your third scope shot the current via your 220mH primary is 0.0177A this is what is established by the series 10 Ohm shunt and the inductive impedance of this coil at 248 Hz. This current dissipates PL=0.0177*0.0177*17.2=0.00538W power i.e 5.38 mW loss in the coil.

Gyula

Thanks Gyula for all the calculations and suggestions.

Always lots to learn

Luc

#### CRANKYpants

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##### Re: Confirming the Delayed Lenz Effect
« Reply #312 on: October 17, 2011, 06:58:39 PM »
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CC
-------- Original Message --------
Subject: HOW TO AVOID OIL WAR # 3 with IRAN
From: <thaneh@potentialdifference.ca>
Date: Sat, October 15, 2011 9:10 am
To: thaneh@potentialdifference.ca
Cc: "Bill Moore" <editor@evworld.com>, "Mike Brace"<techeditor@evworld.com>, "JÃ©rÃ´me_DANGMANN" <jerome.mog@free.fr>,
phil.petsinis@gm.com, "Hearsch,Dan" <Dan.Hearsch@ricardo.com>,"Bailo,Carla (NTCNA)" <BailoC@NRD.NISSAN-USA.COM>, "éƒ½ç•™ å…¸å­"
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« Last Edit: October 17, 2011, 07:26:29 PM by CRANKYpants »

#### CRANKYpants

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##### Re: Confirming the Delayed Lenz Effect
« Reply #313 on: October 17, 2011, 07:19:07 PM »

FUNNY THING, MAINSTREAM SCIENCE PEOPLE DON'T USE TERMS LIKE OVERUNITY OR VIOLATING THE LAW OF CONSERVATION OF ENERGY THEY USE TERMS LIKE...

NET GAIN

DO YOU THINK WE CAN CONVINCE STEFAN TO CHANGE THE OU FORUM'S NAME TO THE NET GAIN FORUM?

CHEERS
T

ps - is it just me or does that wall plug look like this  ?