Storing Cookies (See : http://ec.europa.eu/ipg/basics/legal/cookies/index_en.htm ) help us to bring you our services at overunity.com . If you use this website and our services you declare yourself okay with using cookies .More Infos here:
https://overunity.com/5553/privacy-policy/
If you do not agree with storing cookies, please LEAVE this website now. From the 25th of May 2018, every existing user has to accept the GDPR agreement at first login. If a user is unwilling to accept the GDPR, he should email us and request to erase his account. Many thanks for your understanding

User Menu

Custom Search

Author Topic: Mostly Permanent Magnet Motor with minimal Input Power  (Read 252425 times)

telecom

  • Hero Member
  • *****
  • Posts: 560
Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #195 on: August 19, 2014, 04:34:42 AM »
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

  • Hero Member
  • *****
  • Posts: 5365
Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #196 on: August 19, 2014, 05:00:03 AM »
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

  • Hero Member
  • *****
  • Posts: 560
Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #197 on: August 19, 2014, 05:21:24 AM »
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

  • Hero Member
  • *****
  • Posts: 5365
Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #198 on: August 19, 2014, 06:59:06 AM »
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

  • elite_member
  • Hero Member
  • ******
  • Posts: 3096
Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #199 on: August 20, 2014, 04:34:34 AM »
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

tinman

  • Hero Member
  • *****
  • Posts: 5365
Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #200 on: August 20, 2014, 05:37:26 AM »
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

  • Hero Member
  • *****
  • Posts: 5365
Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #201 on: August 20, 2014, 06:38:45 AM »
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

  • Hero Member
  • *****
  • Posts: 5365
Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #202 on: August 20, 2014, 07:39:56 AM »
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

  • Hero Member
  • *****
  • Posts: 5365
Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #203 on: August 20, 2014, 11:11:06 AM »
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

  • elite_member
  • Hero Member
  • ******
  • Posts: 3096
Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #204 on: August 20, 2014, 08:26:07 PM »
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

Khwartz

  • Hero Member
  • *****
  • Posts: 601
Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #205 on: August 24, 2014, 10:14:04 AM »
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

  • Hero Member
  • *****
  • Posts: 601
Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #206 on: August 24, 2014, 10:17:59 AM »
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

  • Hero Member
  • *****
  • Posts: 601
Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #207 on: August 24, 2014, 10:29:52 AM »
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

  • Hero Member
  • *****
  • Posts: 601
Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #208 on: August 24, 2014, 11:03:37 AM »
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 :D Caps are Very Great to provide accurate maths on energy delivering or input :)

tinman

  • Hero Member
  • *****
  • Posts: 5365
Re: Mostly Permanent Magnet Motor with minimal Input Power
« Reply #209 on: August 24, 2014, 11:46:22 AM »
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.