Partnered Output Coils - Free Energy

[verpies]

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]

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]

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/R_WINDING*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. 

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]

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]

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.

I believe that we had the discussion on the energy stored within a PM,and it was very little.

Yes it is small.

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.