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### Author Topic: Partnered Output Coils - Free Energy  (Read 3323194 times)

#### verpies

• Hero Member
• Posts: 3473
##### Re: Partnered Output Coils - Free Energy
« Reply #4875 on: July 19, 2015, 03:12:05 PM »
Ok,lets do it ass about.
We want to achieve say a 10 Lb pull force on that spring. With the ferrite block in place,we may need say 12v @ 2 amp's applied to our electromagnet to achieve such a force. We then replace that ferrite block with a PM of the same size,and we now find that we need only apply say 6V @ 1 amp to achieve the same pull force on that spring. So the work done against the spring is the same,but we need only apply half the energy to that electromagnet to achieve the same work done.
You are making a conceptual mistake by ignoring the time that it takes to reach this level of current.  I have written an entire article with diagrams about that issue here.  -  click on the blue links in that article to see  diagrams.

Here are some things to remember:
1) The final current flowing through a coil, does not represent the energy that was transferred to that coil's magnetic field (E=Li2)
2) The power of the final electric current flowing through a coil, does not represent the energy transferred to that coil's magnetic field.
3) Average current * Average voltage <> Average power
4) Power is not energy
5) Current is not energy
6) Voltage is not energy

Q: So what does the final current flowing through a coil represent?
A: It represents the power of the heater formed by the coil's resistance (according to P=i2R, because R is constant)

If you integrate the instantaneous products of amps through your coil and volts across your coil,  over time of the entire cycle, then I will have no qualms about the sensibility of your experiment.

P.S.
When I write that something is not energy, I do not mean that it is not a factor in an energy equation.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Partnered Output Coils - Free Energy
« Reply #4876 on: July 19, 2015, 03:16:17 PM »
What is electrical energy?
mechanical energy(the HP)is now converted into electrical energy(KW's)
My cars engine is now calculated in KW's,not HP<--which is a mechanical energy amount.
Please go back and reread what I have posted and look at the plots.  It should be obvious that the current vs time profiles, IE when multiplied by the constant supply voltage and integrated: the energy used establishing the magnetic field is different in all three cases.  The situation is similar to confusing the fuel your car uses going from 0-100km/h versus the fuel it consumes cruising at 100km/h over any given period of time.

#### tinman

• Hero Member
• Posts: 5363
##### Re: Partnered Output Coils - Free Energy
« Reply #4877 on: July 19, 2015, 03:20:19 PM »
Hopefully these timing diagrams will help make the situation more clear.  The energies that we are interested in are the: energy in the spring which you qualitatively have, and the energy needed to get more or less energy into the spring.  Over any significant period of time the energy that is going into the electromagnet is going to asymptotically approach:  V/RWINDING*T.  But once the magnetic field has been established (and the spring has stabilized) there is no more energy being added or removed from the spring.  The work that goes into the spring comes from establishing the magnetic field.  Because we don't have superconductors, we are then stuck feeding the I2R losses of the winding.

In the drawing below you can see that the forces are different using:  a ferrite keeper, or a PM aiding or bucking the electromagnet.  The current profile and therefore power and energy up to the point the magnetic field stabilizes is different for all three cases.  So: literally all you have done is to set up a situation with a stronger magnet, and it takes more energy to magnetize it.  There is a component of work that is also performed moving the PM into place.

Quote
The experiment does not measure the electrical energy used to magnetize, or the work performed placing the PM.

The same must apply for the work being done to place the ferrite keeper in position,and the energy used to make that ferrite keeper.
I believe that we had the discussion on the energy stored within a PM,and it was very little.
As far as using stored energy within the PM,then i refer to your reply email when i asked if there was any way to recover the stored energy within a PM.
Quote: You can do this by shattering the magnet while surrounded by a pick-up coil.

As we are not shattering the magnet,we can assume that we are not draining any energy from the magnet. This being the case,no energy that went into making the PM is being used as an additive of energy to the system,so the energy use to make the PM can be discarded.

#### Liberty

• Hero Member
• Posts: 524
##### Re: Partnered Output Coils - Free Energy
« Reply #4878 on: July 19, 2015, 03:20:35 PM »
Force is not energy.
I can place a wooden beam against a brick wall,and that beam will apply a force on that brick wall,but there is no energy being dissipated .

However, if force is "fashioned" to travel a distance and then be "reset" to travel the distance again over time, it performs work and is a useful form of energy.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Partnered Output Coils - Free Energy
« Reply #4879 on: July 19, 2015, 03:24:51 PM »
The same must apply for the work being done to place the ferrite keeper in position,and the energy used to make that ferrite keeper.
The energy to make the parts does not enter into the equation.  But the energy to put the pieces in place does, which for the ferrite is virtually zero.
Quote
I believe that we had the discussion on the energy stored within a PM,and it was very little.
Yes it is small.
Quote
As far as using stored energy within the PM,then i refer to your reply email when i asked if there was any way to recover the stored energy within a PM.
Quote: You can do this by shattering the magnet while surrounded by a pick-up coil.

As we are not shattering the magnet,we can assume that we are not draining any energy from the magnet. This being the case,no energy that went into making the PM is being used as an additive of energy to the system,so the energy use to make the PM can be discarded.
That is correct.  The energy to make the magnet is not a factor.  There is some energy exchange moving the magnet into place.

#### tinman

• Hero Member
• Posts: 5363
##### Re: Partnered Output Coils - Free Energy
« Reply #4880 on: July 19, 2015, 03:31:54 PM »
I have written an entire article with diagrams about that issue here.  -  click on the blue links in that article to see more diagrams.

Here are some things to remember:
1)  (E=Li2)
2) The power of the final electric current flowing through a coil, does not represent the energy transferred to that coil's magnetic field.
3) Average current * Average voltage <> Average power
4) Power is not energy
5) Current is not energy
6) Voltage is not energy

(according to i2R)

P.S.
When I write that something is not energy, I do not mean that it is not a factor in an energy equation.

Quote
The final current flowing through a coil, does not represent the energy that was transferred to that coil's magnetic field

And this relate to my experiment how?. Dose the PM's field change the way current flows through a coil when that coil is an air core electromagnet?.

Quote
You are making a conceptual mistake by ignoring the time that it takes to reach this level of current.

Time is not a factor in the test,as it is an applied DC current,and this current may be applied to the coil for as long as you like.The point of the test is to show the force applied against the spring with a set amount of power/electrical energy,what ever you like to call it on the day, flowing into the coil.

Quote
Q: So what does the final current flowing through a coil represent?
A: It represents the power of the heater formed by the coil's resistance

So it takes no power to create the magnetic field around the coil?
If it dose,then that should be added to your above answer.

#### Spilled Fluids

• Full Member
• Posts: 138
##### Re: Partnered Output Coils - Free Energy
« Reply #4881 on: July 19, 2015, 03:32:27 PM »
Just as i thought,you have no answer,and thus you hide behind babble.
I have shown you how a PM increases the work that can be done with a set amount of power.
No other material you can come up with will increase the work being done against that spring with a set amount of power as show in my simple experiment.

Like all other self acclaimed guru's,you fail at providing a simple answer,and the reason is-you have none.The electrical power remains the same,and yet the work done against the spring is increased simply by adding a PM into the system. You deny the outcome,and yet you have no argument against it.

The only babble here is coming from you and your bent thinking.

You can't show me an experiment that does not use some outside energy input to get a PM to do work so you launch off on another one of your diatribes.
I gave you a chance to redeem yourself but you came up with an epic fail.

#### tinman

• Hero Member
• Posts: 5363
##### Re: Partnered Output Coils - Free Energy
« Reply #4882 on: July 19, 2015, 03:34:21 PM »
The energy to make the parts does not enter into the equation.  But the energy to put the pieces in place does, which for the ferrite is virtually zero.Yes it is small.That is correct.  The energy to make the magnet is not a factor.  There is some energy exchange moving the magnet into place.
If the ferrite keeper and the PM's mass is the same,and they have to be moved the same distance to be put in place,then the energy required to do so is the same in both cases.

#### Spilled Fluids

• Full Member
• Posts: 138
##### Re: Partnered Output Coils - Free Energy
« Reply #4883 on: July 19, 2015, 03:36:45 PM »
Ok,lets do it ass about.
We want to achieve say a 10 Lb pull force on that spring. With the ferrite block in place,we may need say 12v @ 2 amp's applied to our electromagnet to achieve such a force. We then replace that ferrite block with a PM of the same size,and we now find that we need only apply say 6V @ 1 amp to achieve the same pull force on that spring. So the work done against the spring is the same,but we need only apply half the energy to that electromagnet to achieve the same work done.

Exactly, the electromagnet is doing the work not the PM, not the ferrite and nor would a plain iron bolt...or are you going to try telling us that a plain iron bolt can do work?

#### tinman

• Hero Member
• Posts: 5363
##### Re: Partnered Output Coils - Free Energy
« Reply #4884 on: July 19, 2015, 03:40:08 PM »
The only babble here is coming from you and your bent thinking.

You can't show me an experiment that does not use some outside energy input to get a PM to do work so you launch off on another one of your diatribes.
I gave you a chance to redeem yourself but you came up with an epic fail.
Dear mister !my rocks wont bounce!
I gave you an experiment to carry out that shows a PM doing useful work. It dosnt matter what situation that PM dose that work in,weather it is additive to an existing force or not.
You(like all others so far) have failed to prove me wrong,and in stead,you hide behind irrelevant excuses. It dose not matter how that PM dose the work,nor dose it matter in what situation it dose it-->the fact is,it dose useful work. Now stop hiding,and explain as to how it is not doing useful work in the experiment i gave you-->you are yet to answer.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Partnered Output Coils - Free Energy
« Reply #4885 on: July 19, 2015, 03:44:55 PM »
And this relate to my experiment how?. Dose the PM's field change the way current flows through a coil when that coil is an air core electromagnet?.

Time is not a factor in the test,as it is an applied DC current,and this current may be applied to the coil for as long as you like.The point of the test is to show the force applied against the spring with a set amount of power/electrical energy,what ever you like to call it on the day, flowing into the coil.

So it takes no power to create the magnetic field around the coil?
If it dose,then that should be added to your above answer.
Tinman if you drive your car at a constant speed for three hours, the fuel that you use will be some amount, say 24 liters.  That will completely obscure the gas you use to get going, whether you floored the gas to accelerate to speed as fast as you could or nursed the car to speed over a couple of minutes.  The same thing is happening here.  The running power once the magnetic field is set-up is just VSUPPLY2/RWINDING.  That is analagous to the specific fuel consumption of your car at cruise. The energy at that constant power level just integrates with time.  The energy that we are interested in is the energy needed to magnetize the magnetic circuit.  With the ferrite you get one current build-up versus time, with a PM bucking another, and a PM aiding a third.  Those current build-ups versus time multiplied by the constant supply voltage establish the magnetization energies.  They are different in all three cases.

#### tinman

• Hero Member
• Posts: 5363
##### Re: Partnered Output Coils - Free Energy
« Reply #4886 on: July 19, 2015, 03:44:59 PM »
Exactly, the electromagnet is doing the work not the PM, not the ferrite and nor would a plain iron bolt...or are you going to try telling us that a plain iron bolt can do work?
Do you know of equal and opposite forces?
If the electromagnet is applying a force on a iron bolt,then the iron bolt is applying an equal and opposite force against the electromagnet.

Now tell us why these forces should increase simply by replacing the iron bolt with a PM,without having to increase the energy applied to the coil?.If you cant answer that,i will answer it for you.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Partnered Output Coils - Free Energy
« Reply #4887 on: July 19, 2015, 03:46:57 PM »
If the ferrite keeper and the PM's mass is the same,and they have to be moved the same distance to be put in place,then the energy required to do so is the same in both cases.
No, not at all, because the PM's field will act on the soft iron in the electromagnet.  The ferrite's residual field is essentially zero and virtually no energy is exchanged in the magnetic circuit moving it into place.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Partnered Output Coils - Free Energy
« Reply #4888 on: July 19, 2015, 03:48:15 PM »
Do you know of equal and opposite forces?
If the electromagnet is applying a force on a iron bolt,then the iron bolt is applying an equal and opposite force against the electromagnet.

Now tell us why these forces should increase simply by replacing the iron bolt with a PM,without having to increase the energy applied to the coil?.If you cant answer that,i will answer it for you.
But the energy required to magnetize the coil does change.  This has now been explained to you more than half a dozen times by Verpies and myself.

#### tinman

• Hero Member
• Posts: 5363
##### Re: Partnered Output Coils - Free Energy
« Reply #4889 on: July 19, 2015, 03:50:39 PM »
Tinman if you drive your car at a constant speed for three hours, the fuel that you use will be some amount, say 24 liters.  That will completely obscure the gas you use to get going, whether you floored the gas to accelerate to speed as fast as you could or nursed the car to speed over a couple of minutes.  The same thing is happening here.  The running power once the magnetic field is set-up is just VSUPPLY2/RWINDING.  That is analagous to the specific fuel consumption of your car at cruise. The energy at that constant power level just integrates with time.  The energy that we are interested in is the energy needed to magnetize the magnetic circuit.  With the ferrite you get one current build-up versus time, with a PM bucking another, and a PM aiding a third.  Those current build-ups versus time multiplied by the constant supply voltage establish the magnetization energies.  They are different in all three cases.
the difference is more the car traveling down hill as apposed to up hill.
But before i go any further,we will wait for !self bouncing masses with elasticity! to answer my question.