When the magnet is at TDC the inductance of the coil is 0.1mH. Assuming we have a 30V power supply that can supply 5A then we choose R = 6 Ohm so the charge time is 66µs. Assuming we use neo magnets and our C core gets perhaps saturated 10° before and after TDC. So we give ourselves 10° of charge time. Using our time constant this leads to a constant rpm of 2500 RPM. After this 10° the influence of the magnet on the core starts to diminish, but also meanwhile current is used as joule heating. Let's assume that 45° later the magnet has no more effect on the core and the coil can be discharged. First let's calculate the joule heating energy.
given period = 45°, since rpm = 2500 time is then 300µs. Joule heat enegy = V*I*t = 0.045J.
Notice that I ignored inductive energy input and just put it as a "loss" in joule heating, I'm that confident

So finally now we can discharge our inductor without any external influence. I took the final inductance to be 0.1H and the load 333.33 ohM. First of all lets calculate the energy of the inductor.
E=0.5*I^2*L=0.5*25*0.1 = 1.25J.
Electric wise the energy out / energy in = 27 or 2700%.
Since this needs again time to discharge. 4*L/R = 1.2ms. From our 2500 rpm rotation that is equal to 180° of rotation. So it will take 180° of rotation to discharge our high inductive coil through the 333.33 Ohm load.
And to finish it all let's look at the power generation. We know it rotates at 2500 rpm and that every rotation steals 0.045J and gains 1.25J. In average watt values this results in 1.87 Watt of loss and 52.08 Watt gain.
And that's just one example of high energy gain from this system. Many parameters can be adjusted to gain any desired cop. I might make a javascript calculator that gives these results automatic given a certain parameters.
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