This is what I read from another post [which deals with internal resistance from a electrical-mechanical energy conversion] which would suggest that such a formula would more or so be derived from empirical results rather than formulating a purely mathematical construct (makes sense due to the nature of mechanical energy)
https://www.physicsforums.com/showthread.php?t=297491
This is what was stated;
"Suppose the height of initial release of the mass minus the height of detachment is h. Then if the mass falls freely (without flywheel), the total kinetic energy of the mass is mgh = (1/2)mv
2. But in the case where it is attached to the flywheel, its final velocity at point of detachment is v
1. So its energy at release is now (1/2)mv
12. So the missing energy is (1/2)m[v
2 - v
12]. This was all transferred to the flywheel.
If the flywheel were frictionless and the generator perfect, all the flywheel energy would be converted to electric power. All inertial energy remaining in the flywheel after release of the mass would also be converted to electric power. To maximize the total electric energy, the objective then is to make the final velocity of the mass v
1 as small as possible. If the flywheel were massive and v
1 were small, then the electrical energy would be nearly mgh. If mgh were expressed in joules, then mgh/3600 would be the energy in watt-hours, or in volt amp hours. "
So I would suppose that the relationship of magnetism and electricity isn't direct, in the sense that the factors that I listed in my post from above would deal with imperfections and internal resistance which would differ from the conversions of electricity-heat or electricity-chemical because these ones are more direct.