Mostly Permanent Magnet Motor with minimal Input Power

[telecom]

No,the kinetic energy isnt being absorbed by the string,it is being transformed into vibrational energy via the string through the framework of the device,and then to the bench the device is mounted on,and finally to ground(earth). The spring stores the kinetic energy that would be normally transformed into vibrations/vibrational energy/sound,and returns it back to the system apon spring decompression. Energy is never absorbed,it's stored or transformed.

If you have a brick and drop it on top of the spring, the spring will first compress
to a certain length, after that it will straighten up and propel the brick to a height
which is smaller than the initial height due to the losses...
What this discussion has to do with the topic of the thread?

[tinman]

If you have a brick and drop it on top of the spring, the spring will first compress
to a certain length, after that it will straighten up and propel the brick to a height
which is smaller than the initial height due to the losses...
What this discussion has to do with the topic of the thread?

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.

[telecom]

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.

It may increase the efficiency of this particular setup, but may be irrelevant
to a real life application of the device.
In  real life it will be connected to a generator with a resistance in both parts of a stroke - forward and return. There will be no free fall like in a setup.

[tinman]

It may increase the efficiency of this particular setup, but may be irrelevant
to a real life application of the device.
In  real life it will be connected to a generator with a resistance in both parts of a stroke - forward and return. There will be no free fall like in a setup.

Indeed that is correct. But think of this-if Luc can raise that 1/2kg weight 2cm 2  times per second,he has a unity device.Thats a 100% efficient motor,and thats something that dosnt exist today. Every device i have tested so far,has no where that pull force that Luc shows with his device. I would hope that the high end builders here would try and build something that shows the efficiency of Luc's DUT-in fact,i make it a challenge to all here.I would spend more time on it,but im working on my Inertia drive at the same time,which i want to get finished.

[gotoluc]

Here is the pull force(or torque) from a standard small dc motor.

https://www.youtube.com/watch?v=pDLLphaRC-k&list=UUsLiBC2cL5GsZGLcj2rm-4w

Hi Brad,

I was giving your test more thought and remembered that the Watts consumed for a specific amount of grams pulled is not linear, well not in my device anyways.
Let me give you an example, if I can pull 500g with 0.43W and I reduce the input to pull 250g the watts are not divided in half, it's much less than that!... more like 1/3 or less. Unfortunately I don't have my latest test device with me but if I pulled 130g like you did it would probably use around 0.08W to do it.

If you can re-test your motor and raise the voltage till you achieve 500g of pull and calculate the watts used you may see what I mean.

Luc