Hi Bill
Thanks for your reply,be sure I will listen to MrMags and your self,but I think you my be be misunderstanding me,what ever juice is taken from the rotor coil will have to be paid for,I have a simple rig that seems to show that the fixed coil is free as the rotor coil only sees an even more increasing flux,and that increase is free.
peter

Hi Bill
Just to add,you are right in saying the fixed coil will impose a load on the rotor,but in return,the rotor coil will give more juice,I think?
peter

Exactly.  And, according to the laws as I understand them, (I am no expert here) those values will balance out, minus any losses and drag which will/should/ keep it underunity.  In other words, the more you get out of the rotor, the harder it will be for the rotor to turn.  I know you are using the stator as a pick-up point but, the power comes from the rotating rotor as I see it.  These are the balance of forces as represented by the 2nd law of thermodynamics, such as it is.

One more disclaimer here, this is my understanding, it is not the end all, be all, of any evaluation of your particular device.

Bill

Hi Bill
Many thanks for you continuing replies.
As I see it you will only pay for the juice taken from the rotor coil,not a penny,not a penny less.
Taking it step by step.
Moving coil go's to fixed mag.normal genny.
If i were to move the fixed mag.towards the coil,more juice,but more drag on the rotor,still every thing normal,plus the effort in moving the mag.
Now,with the fixed coil loaded,same as above,but no effort in moving the mag.plus some juice.I wish I could see were I'm going wrong!!!
peter

Peter:

You may indeed not be wrong.  Let's see what MrMags has to say about this.  I am only telling you what I think with what I THINK I know.  Nothing definitive here on my part at all as I am still learning.  This is just my understanding is all, for what it is worth.

Bill