Mostly Permanent Magnet Motor with minimal Input Power

[tinman]

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

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.

[tinman]

I have completed the test Luc,and video is uploading now. I will post the results with the video. I am now wondering if that power consumption would go down if the motor was actually running?-that will be my next test i think. Now just need to find my torque meter lol.

[tinman]

So here is that second test Luc,and yes-more than double the power to achieve twice the pull force.

250g's-.888 volts@2.02 amps= 1.793 watts
500g's- 1.45 volts@3.18 amps= 4.8 watts

https://www.youtube.com/watch?v=VSvp5PtffUo&list=UUsLiBC2cL5GsZGLcj2rm-4w

[tinman]

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.

[gotoluc]

So here is that second test Luc,and yes-more than double the power to achieve twice the pull force.

250g's-.888 volts@2.02 amps= 1.793 watts
500g's- 1.45 volts@3.18 amps= 4.8 watts

https://www.youtube.com/watch?v=VSvp5PtffUo&list=UUsLiBC2cL5GsZGLcj2rm-4w

Thanks for doing this excellent tests Brad

This gives us a good idea that your motor does NOT produce much torque per watt since it needs about 10 times more watts to achieve the same 500g pull as mine.
If you have a larger DC motor it may be worth making the same 500g pull test and see how much it changes with the size of motor.
I know pull force is not everything but it's a good start since stall torque test are part of testing an electric motor.
See this link: http://lancet.mit.edu/motors/motors3.html
They say a motors maximum Torque is in stall position. You may not want to leave it for long in that position though!... best to give it small shots till you have it pulling the correct amount.

I'm still trying to get my latest prototype back to do more tests. I've pulled out my large magnets from storage and will start to design the new more powerful prototype to see how far we can push this puppy.

Thanks for taking the time to do the test and video.

Luc