Mostly Permanent Magnet Motor with minimal Input Power

[Khwartz]

If a spring is not an energy source,but only an energy storage device,how will it make the output higher?
Think of the weight as an inductor,and the spring as a capacitor. All you need to do then is hit the right frequency of the tank circuit to gain resonance. Once resonance is achieved,we will get a large amplitude-in this case,the 1/2kg weight being lifted that 2cm twice per minute

To be within test guidlines,the 2cm lift would have to start from the point at which the spring is not compressed-just off the surface from which it will contact as the weight drops.If the spring is to compress 1cm,then the total travel of Luc's slide(coils travel) on his device would need to be 3cm.

Why the spring would allow the weight to be lifted more efficiently than if it wasnt there?.

The answer is simple. If there were no spring,and the weight was just allowed to hit the surface as it fell to a resting state,then energy is disipated from the system as noise/vibration. The spring removes this loss,and stores that energy that would have been disipated as noise/vibration,and returns it back to the system. So the spring wont actually distort the output,but it will make it higher due to less loss in the system.

Very Good and Accurate analysis! 

[Khwartz]

That is correct-the spring stores the energy,and then returns it back to the system.This is in relation to this thread in that a system be designed and understood,so as Luc can achieve maximum efficiency from his DUT.A correct understanding as to how your system opperates, where losses may occur, what those losses are,and how to remove those losses,is the best way to achieve the results you are after,and make your DUT the most efficient it can be. Although the spring itself will have losses (vibrational/noise),it will increase the efficiency of the DUT,as those losses in the spring are not as much as they would be without it.

All in all,we agree that the spring will increase the efficiency of the DUT.

Very Correct!

[Khwartz]

That sounds about right Luc. The less current your coil draw's,the higher the backEMF voltage will be in your coil-and of course,the higher the backEMF voltage,the lower the current draw.This is the very same in my little DC motor,and in most electric motors i know of.The more load you place on the motor,the less the BackEMF voltage in the inductors will be,so up go's the current to try and maintain the forward voltage within the inductor.

I dont know if that little motor would have 500g's of torque in it,but we can make some smoke and try lol.
Will go do it now.

Please,  dear Tin, don't use "torque" when you speak about "tangential force". The first is in N.m  ("Newton.meter" or "kilogram.force.meter" if you want) and the second in N only ("Newton" or "kilogram.force"). This is confusing and may lead to erroneously concepts and ways in experiments and calculations them :/

[Khwartz]

Would be good to have some test perameters right about now.
So we could do it like this.
It takes 1 joule of energy to lift 1kg of mass 10cm high.
If we are setting the weight to 1/2kg(500g's),with a lift hight of 2cm,then we need 100millijoules.
A cap that has say 5000uf would then need 6.325volts in it to give us our 100mJ.Or if we have a 10000 uf cap,we would need only 4.47 volts in it to give us our 100mJ.

So Luc,if you can lift that 500g weight by 2cm using a 10000uf cap with only 4.47 volts in it-you have hit unity. If you need only 4.46 volts in the cap to do it,you are OU.

Dear TinMan, I think you're on the Very Right path in term of Correct Methodology for o.u. checking Caps are Very Great to provide accurate maths on energy delivering or input

[tinman]

Please,  dear Tin, don't use "torque" when you speak about "tangential force". The first is in N.m  ("Newton.meter" or "kilogram.force.meter" if you want) and the second in N only ("Newton" or "kilogram.force"). This is confusing and may lead to erroneously concepts and ways in experiments and calculations them :/

The torque of an electric motor is measured exactly as i did it.We are measuring static rotor torque.Torque is a measure of rotational or "twisting" force. I can dig up my torque dial if you like,and show you that the results will be exactly the same for the P/in we used.