Partnered Output Coils - Free Energy

[MarkE]

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 V_SUPPLY²/R_WINDING.  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]

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]

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]

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]

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 V_SUPPLY²/R_WINDING.  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.