Mostly Permanent Magnet Motor with minimal Input Power

[gotoluc]

There must be a certain amount of distance the weight needs to be lifted in a certain time frame no?

Luc

[tinman]

There must be a certain amount of distance the weight needs to be lifted in a certain time frame no?

Luc

@ Luc
Syncro is using metric horse power there.If you wish to use that,then the calculations are:

1 metric horse power =735.498 watts
To have one metric horse power,you must lift 75kg's 1 meter in 1 second.

Power= work/time
Lifting 1kg per second, converts 10 joules of energy to 10 watts of power.
It takes 10 joules of energy to lift 1kg 1 meter.
So to lift 1kg up 100mm,it would take 1 joule of energy.
As an example-10v in a 20000uf cap is 1 joule of energy.

What we need is a constant p/in(watt hours) to suspend say a 1kg mass. I think it can be done using the acelleration of G in this case. I will look into it.

[Khwartz]

@tinman

Your perfectly right in your calculations, Brad.

Just at the end , if you mean by "watt hours": "watts times hours", I remember you it is no more a power but an energy

Good idea to use gravitation but then only 1 phase of the cycle will be needed as excitation, right? And we would need to take in account that only half the time period the consumption will occur but that gravity will work the second part of the cycle; isn't it?

Theoretically, the pulling work made to lift the crank would be equal to the work made by the gravity in the second phase of the cycle. I am not sure that the system would be better if only excited on 1 half of the cycle, but experiments should tell.

@gotoluc

Please, be careful with your fingers and skin hands when you'll handle these so powerful neodymium magnets ^_^

Yes, if we can get o.u. in the straight design, a rotary system would be A Must!

Thanks for keeping going and sharing, Luc

[tinman]

@tinman

Your perfectly right in your calculations, Brad.

Just at the end , if you mean by "watt hours": "watts times hours", I remember you it is no more a power but an energy

Good idea to use gravitation but then only 1 phase of the cycle will be needed as excitation, right? And we would need to take in account that only half the time period the consumption will occur but that gravity will work the second part of the cycle; isn't it?

Theoretically, the pulling work made to lift the crank would be equal to the work made by the gravity in the second phase of the cycle. I am not sure that the system would be better if only excited on 1 half of the cycle, but experiments should tell.

@gotoluc

Please, be careful with your fingers and skin hands when you'll handle these so powerful neodymium magnets ^_^

Yes, if we can get o.u. in the straight design, a rotary system would be A Must!

Thanks for keeping going and sharing, Luc

yes-not sure why i put watt hour's,should just be watts.
So would we use 1/2 G in this case? 4.45m*/s

So as 1 watt /4.9x1kg=204mW.
A pull force of 1kg @ 204mW of power?.

[gotoluc]

Hi everyone,

I got my 1" cube mostly magnet motor back and have put together a new test which I feel should be a fairer comparison.
In this test I use a large 20VDC 400 RPM Permanent Magnet Motor since it's torque should be very higher and a better comparison to the torque of my 3xM design.

My reasoning for this is the 3xM design is a back and forth solenoid action which has a limit of how many times it can efficiently oscillate per second. At this time it seems to do well with 6 to 10Hz. So if we added a crank shaft to convert it to rotation it should turn in the 360 to 600 RPM range.
So using the above motor seems to be a more ideal comparison. However, keep in mind that the 3xM is only 1" wide compared to the DC motor rotor which is about 6" wide. So obviously surface area is an important factor for high torque hence my new 8" wide super build.

Link to video:  https://www.youtube.com/watch?v=cS5pmKBruuU

Luc