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Author Topic: Partnered Output Coils - Builders Group - Moderated!  (Read 20263 times)

Offline EMJunkie

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #105 on: June 12, 2020, 04:00:42 AM »
"If the directions of the two signals are such that opposite H-fields cancel and E-fields add, an apparently steady E-field will be created. The energy density of the fields remain as calculated above, but the value of the E-field will double from E / 2 to E."

An apparently steady E field will be created.

So how do we do this?

Mags


Mags, Newton told us, for every Action there is an Equal and Opposite Reaction - Right?

Floyd Sweet also said:

Quote


Electromagnetic induction with no measurable magnetic field is not new. It is well known that in the space surrounding a properly wound toroidal coil there is no magnetic field. This is due to the superposition of the fields.

However, when alternating current is surging through a transformer an electric field surrounds it. When we apply the principle of superposition to the vacuum triode it becomes more obvious how the device is in fact operating.




An Electric Field is formed when two Magnetic Fields Oppose, see the below graphic:

I said many years back, Newtons laws should be extended to: For every Action there is an equal and opposite Reaction and for every Reaction there is an equal and opposite Counter-Reaction.

Tinman got all excited and he agreed. Even quoting this later in his diagrams. Here.

You must think in units of Energy, and what that Energy is doing. M.M.F, is Energy, having a direct conversion to Joules. M.M.F also has a Magnitude and a Direction.

   1: Input - 1 unit in the Positive Direction.
   2: Output - 1 unit in the Negative Direction.
   3: Output - 1 unit in the Positive Direction.


So, the objective is, get 3 to oppose 2 and 2 will naturally oppose 1. This gives you a total M.M.F = 1 + -1 + 1 = 1.

Normally, you would see: M.M.F  = 1 + -1 = 0

So, to put it simply, you need a Third Force, an Asymmetrical Force, to add Energy to your System! The same as a Heat Pump, a Hydraulic Ram Pump and many other examples.

This energy can be Free, it can be the Counter-Reaction of your Reaction which Assists your Action!

Follow this post Here, and questions let me know.

A Magnetic Resonance can be found, when this is found, your Output will be maximum! By following these simple Rules and applying them as so, you will the Above unity results. Many have shown success using this method. Partzman is one Here.

Using the method I show, and have shown for nearly a decade, Reducing the Lenz's Effect in a Transformer! Almost to Zero and sometimes even Negative, power comes back on the Primary! Using an Asymmetrical Transformer, the one I have shown for nearly a decade now.

Best wishes, stay safe and well in these dire times,
   Chris Sykes

Free Energy | searching for free energy and discussing free energy


Offline bistander

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #106 on: June 12, 2020, 06:07:06 AM »
... M.M.F, is Energy, having a direct conversion to Joules. ...
   Chris Sykes

Hi EMJunkie,

So if I have an electromagnet energized with 750 AT (Ampere-Turns), how many Joules is that?

Thanks,
bi

Offline EMJunkie

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #107 on: June 12, 2020, 07:46:25 AM »
Hi EMJunkie,

So if I have an electromagnet energized with 750 AT (Ampere-Turns), how many Joules is that?

Thanks,
bi



Bistander,

I truly don't have time for your silly little games! If your'e not going to be serious, and do the homework yourself, then you can not be helped.

You choose what path you want to take, what you want to gain from the exchanges here. If necessary, if you keep acting like a juvenile, I will just put you on my ignore list and report you to Stefan!

M.M.F is directly convertible to AT Ampere Turns per meter: FMMF = NI AT/m or FMMF = 2 W / Φ, where W is the energy in joules, and Φ is the total flux going through part of a magnetic circuit.

1 Ampere is 1 Coulomb / 1 Second = Q / t

I am sure, if you are serious, you can do the rest, as Q is Charge, and 1 second gives the Energy.

I get tired of foolish people, the mouth is bigger than their Brain!

Best wishes,
   Chris Sykes

Free Energy | searching for free energy and discussing free energy

Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #107 on: June 12, 2020, 07:46:25 AM »
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Offline Smudge

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #108 on: June 12, 2020, 04:27:21 PM »
Forgive me for butting in but I think I can shed some light on the confusion going on here.  Magneto Motive Force (your symbol FMMM) is not Ampere Turns per meter.  MMF is just Ampere Turns, always was, always will be.  For a solenoid Ampere Turns per meter (or MMF per meter) is a valid expression for the linear density of the MMF along the solenoid.  It so happens that for a long solenoid having just a thin layer of winding (like a single layer) the internal magnetic field intensity H at its center point has a value equal to the linear MMF density (Ampere Turns per meter).  Thus H = NI/L AT/m where L is the solenoid length.  The presence of the magnetic field H means that there is also the magnetic induction field B present, related to H by the permeability.  The magnetic field there has an energy density BH/2 Joules per cubic meter.  If we multiply that energy density by the volume of the solenoid we get the stored energy W in Joules.  That multiplication includes length L and area A, so we can allocate the area to B and the length to H to get W = (BA)(HL)/2BA is the flux Φ through the coil while HL is the total MMF or Ampere Turns supplied to the coil (your FMMM).  We then get for the magnetic energy W = ΦFMMM/2.  Thus your FMMM = 2 W/Φ is correct for a solenoid where W is the energy stored in its magnetic field and Φ is the total flux through the solenoid (not flux going through just part of a magnetic circuit.) 

Smudge 

Offline bistander

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #109 on: June 12, 2020, 04:59:46 PM »
Forgive me for butting in but I think I can shed some light on the confusion going on here.  Magneto Motive Force (your symbol FMMM) is not Ampere Turns per meter.  MMF is just Ampere Turns, always was, always will be.  For a solenoid Ampere Turns per meter (or MMF per meter) is a valid expression for the linear density of the MMF along the solenoid.  It so happens that for a long solenoid having just a thin layer of winding (like a single layer) the internal magnetic field intensity H at its center point has a value equal to the linear MMF density (Ampere Turns per meter).  Thus H = NI/L AT/m where L is the solenoid length.  The presence of the magnetic field H means that there is also the magnetic induction field B present, related to H by the permeability.  The magnetic field there has an energy density BH/2 Joules per cubic meter.  If we multiply that energy density by the volume of the solenoid we get the stored energy W in Joules.  That multiplication includes length L and area A, so we can allocate the area to B and the length to H to get W = (BA)(HL)/2BA is the flux Φ through the coil while HL is the total MMF or Ampere Turns supplied to the coil (your FMMM).  We then get for the magnetic energy W = ΦFMMM/2.  Thus your FMMM = 2 W/Φ is correct for a solenoid where W is the energy stored in its magnetic field and Φ is the total flux through the solenoid (not flux going through just part of a magnetic circuit.) 

Smudge

Thank you Smudge,
So I don't see how Magneto Motive Force (AT) is Energy (Joules). And EMJunkie must have a different definition of "direct conversion", which to me implies a simple factor (constant) like '25.4mm / inch', not involving another variable like flux.

Regards,
bi

Free Energy | searching for free energy and discussing free energy

Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #109 on: June 12, 2020, 04:59:46 PM »
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Offline lancaIV

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #110 on: June 12, 2020, 07:08:08 PM »
bistander,but you can look fo another view :
J/T for Joule per Tesla
If you know the Tesla/Gauss/Oersted magnetomotive attraction( or repulsion) force from a permanent magnet and you unwrap this pm with a emf coil with winding turns you can experiment how many Ampere current turns
in relationship to the winding turns number you will need to a. neutralize b. amplify c. diminuate this permanent magnet force !
This is what the Flynn brothers,Dr.Pavel Imris  and Dr.med Keith Kenyon(R.I.P.)  over the last decades have shown and as information exposed !


Offline Smudge

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #111 on: June 12, 2020, 08:10:05 PM »
bistander,but you can look fo another view :
J/T for Joule per Tesla
If you know the Tesla/Gauss/Oersted magnetomotive attraction( or repulsion) force from a permanent magnet and you unwrap this pm with a emf coil with winding turns you can experiment how many Ampere current turns
in relationship to the winding turns number you will need to a. neutralize b. amplify c. diminuate this permanent magnet force !
This is what the Flynn brothers,Dr.Pavel Imris  and Dr.med Keith Kenyon(R.I.P.)  over the last decades have shown and as information exposed !
For a cylindrical bar magnet you don't need to experiment.  If you know the remanant B (in Tesla) for the magnet material then you can get the magnetization M from M = B/munought.  The dimensions of M are amps per meter and that gives you the winding detail you need.  You wind N turns over the length L of the magnet and drive it with I amps such that NI/L = M.  That will double the magnet's field for one direction of current and null it to zero for the other direction.  Needs a lot of ampere turns to do this though.

Smudge

Free Energy | searching for free energy and discussing free energy

Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #111 on: June 12, 2020, 08:10:05 PM »
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Offline EMJunkie

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #112 on: June 12, 2020, 11:16:02 PM »


Bistander: Did you even partially understand what Smudge wrote?


The magnetic field there has an energy density BH/2 Joules per cubic meter.  If we multiply that energy density by the volume of the solenoid we get the stored energy W in Joules.




http://info.ee.surrey.ac.uk/Workshop/advice/coils/terms.html


Girls, in case you thought you were on your high horse and here to try to look pretty:

M.M.F is directly convertible to Energy in Joules. lancaIV is right and correct, you are off on some weirdo anti Science Antifa movement!

As serious Energy Researchers, you should all know this basic Science!

Best wishes, stay safe and well in these dire times,
   Chris Sykes

@lancaIV - Thank You for your sensible, readable and short useful post!

P.S: I did miss the = sign: FMMF = NI = AT/m. 1 Oersted = 79.57747154594 ampere-turn/meter or Ampere per meter.

Quote

An ampere-turn per meter (AT/m) is another name, now obsolete, of the SI derived unit of magnetic field strength, which is also called magnetic field intensity or H-field. “Turn” in ampere-turn refers to the winding number of an electrical conductor comprising a solenoid or inductor.


Offline bistander

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #113 on: June 13, 2020, 12:21:58 AM »

Bistander: Did you even partially understand what Smudge wrote?



http://info.ee.surrey.ac.uk/Workshop/advice/coils/terms.html


Girls, in case you thought you were on your high horse and here to try to look pretty:

M.M.F is directly convertible to Energy in Joules. lancaIV is right and correct, you are off on some weirdo anti Science Antifa movement!

As serious Energy Researchers, you should all know this basic Science!

Best wishes, stay safe and well in these dire times,
   Chris Sykes

@lancaIV - Thank You for your sensible, readable and short useful post!

P.S: I did miss the = sign: FMMF = NI = AT/m. 1 Oersted = 79.57747154594 ampere-turn/meter or Ampere per meter.

Hi EMJunkie,
You ask:
"Bistander: Did you even partially understand what Smudge wrote?"

Yes, I understand all of it. I don't think you do. But I don't insult you. I merely point out legitimate differences so readers can judge for themselves. I can learn, can you?

I see you were able to edit or change your post before I had a chance to quote it. I guess you can learn. Good for you.

Regards,
bi

Free Energy | searching for free energy and discussing free energy

Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #113 on: June 13, 2020, 12:21:58 AM »
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Offline EMJunkie

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #114 on: June 13, 2020, 12:40:34 AM »
Hi EMJunkie,
You ask:
"Bistander: Did you even partially understand what Smudge wrote?"

Yes, I understand all of it. I don't think you do. But I don't insult you. I merely point out legitimate differences so readers can judge for themselves. I can learn, can you?

I see you were able to edit or change your post before I had a chance to quote it. I guess you can learn. Good for you.

Regards,
bi




A truly hypocritical post on your part Bistander!

Another way to look at this, I2R Losses, what are they, why do we look at this in an Inductor, whats applied across and Inductor before we can measure I2R Losses?

Floyd Sweet was very aware of the I2R Losses and his statements show his thinking:

Quote from: Floyd Sweet link="http://www.hyiq.org/Downloads/The%20Space-Flux%20Coupled%20Alternator.pdf"

Why the field of a magnet is not the property of the magnet: First the electromagnet – it takes power from a source to initiate and bring to steady state the field of the magnet. Once the field is stabilised and the exciting current is no longer changing, no further power is needed from the source. The only power required is that needed to support the I2R losses due to the ohmic resistance of the conductor comprising the coil of the magnet. This loss appears as heat.

Now we have a magnetic field, a potential source of energy in existence without support of the source of power to the coil. True, the moving charges through the copper conductor are accompanied by a magnetic field, also true this field requires no power from the source. As stated, the only power is that supporting the I2R losses. Then the field due to the moving charges is not a property of the current drawn from the source but a property of incoherent energy quanta in the surrounding space interacting coherently with fields produced by moving charges on the electrons in motion through the coil.



What is: I2R?

Very simple, Ohms Law: I2R = P or Power, Joules per second. Watt Seconds. Turns N having an Impedance, ΩjΩ. Simple, again proving my point! With Clarity!

Again, as serious Energy Researchers, you should all know this basic Science! Shouldn't you?

Best wishes, stay safe and well in these dire times,
   Chris Sykes

Offline bistander

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #115 on: June 13, 2020, 01:50:10 AM »



A truly hypocritical post on your part Bistander!

Another way to look at this, I2R Losses, what are they, why do we look at this in an Inductor, whats applied across and Inductor before we can measure I2R Losses?

Floyd Sweet was very aware of the I2R Losses and his statements show his thinking:


What is: I2R?

Very simple, Ohms Law: I2R = P or Power, Joules per second. Watt Seconds. Simple, again proving my point! With Clarity!

Best wishes, stay safe and well in these dire times,
   Chris Sykes

Hi EMJunkie,

Interesting. But I still take issue with your statement that MMF is energy.

Regards,
bi

Free Energy | searching for free energy and discussing free energy

Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #115 on: June 13, 2020, 01:50:10 AM »
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Offline bistander

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #116 on: June 13, 2020, 02:09:32 AM »



A truly hypocritical post on your part Bistander!

Another way to look at this, I2R Losses, what are they, why do we look at this in an Inductor, whats applied across and Inductor before we can measure I2R Losses?

Floyd Sweet was very aware of the I2R Losses and his statements show his thinking:


What is: I2R?

Very simple, Ohms Law: I2R = P or Power, Joules per second. Watt Seconds. Turns N having an Impedance, ΩjΩ. Simple, again proving my point! With Clarity!

Again, as serious Energy Researchers, you should all know this basic Science! Shouldn't you?

Best wishes, stay safe and well in these dire times,
   Chris Sykes

Hi EMJunkie,
Again you edited, the bold highlighted sentence being added, my bolding.

Another thing: you say "I2R = P or Power, Joules per second. Watt Seconds."

This, to me, implies you think "Joules per second" is equivalent to "Watt Seconds". Clearly they are not.

I've found that folks who confuse units are most often those that lack comprehension of the fundamentals.

Regards,
bi

Offline EMJunkie

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #117 on: June 13, 2020, 02:29:05 AM »
Hi EMJunkie,
Again you edited, the bold highlighted sentence being added, my bolding.

Another thing: you say "I2R = P or Power, Joules per second. Watt Seconds."

This, to me, implies you think "Joules per second" is equivalent to "Watt Seconds". Clearly they are not.

I've found that folks who confuse units are most often those that lack comprehension of the fundamentals.

Regards,
bi



Bistander - Man alive, what are you on?

Quote from: Wikipedia link="https://en.wikipedia.org/wiki/Watt"


The watt (symbol: W) is a unit of power. In the International System of Units (SI) it is defined as a derived unit of 1 joule per second, and is used to quantify the rate of energy transfer.





Quote from: Wikipedia link="https://en.wikipedia.org/wiki/Joule"


The joule (symbol: J) is a derived unit of energy in the International System of Units. It is equal to the energy transferred to (or work done on) an object when a force of one newton acts on that object in the direction of the force's motion through a distance of one metre (1 newton metre or N⋅m). It is also the energy dissipated as heat when an electric current of one ampere passes through a resistance of one ohm for one second.




Beg your pardon? Sorry EMJ, is that what I hear?

Man alive, no wonder you guys are SO FAR BEHIND! My Great Grandchildren even know this and they aint even born yet!!!

Best wishes, stay safe and well in these dire times,
   Chris Sykes


Offline bistander

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #118 on: June 13, 2020, 03:09:41 AM »


Bistander - Man alive, what are you on?




Beg your pardon? Sorry EMJ, is that what I hear?

Man alive, no wonder you guys are SO FAR BEHIND! My Great Grandchildren even know this and they aint even born yet!!!

Best wishes, stay safe and well in these dire times,
   Chris Sykes

Sorry EMJ,
But "Joules per second" is not equal to "Watt Seconds".

Regards,
bi

Offline EMJunkie

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Re: Partnered Output Coils - Builders Group - Moderated!
« Reply #119 on: June 13, 2020, 03:23:48 AM »
Sorry EMJ,
But "Joules per second" is not equal to "Watt Seconds".

Regards,
bi




You are saying, unequivocally, One Watt Second is NOT equal to 1 Joule per Second? Is this what you are saying? Because that's exactly what I am reading from your post!  :o

If this is what you are saying, then please don't ever post on my Threads again!  ::)



Quote from: traditionaloven.com/tutorials link="https://www.traditionaloven.com/tutorials/"


The joules unit number 1.00 J converts to 1 W·sec, one watt second. It is the EQUAL energy value of 1 watt second but in the joules energy unit alternative.




Best wishes, stay safe and well in these dire times,
   Chris Sykes

 

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