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Overunity Machines Forum



Electric motor coupled to a PMA

Started by returnedintime, January 28, 2013, 06:39:45 PM

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0 Members and 7 Guests are viewing this topic.

gyulasun

Hi,

Although the PMA can become cogless, unfortunately it does not mean it will represent a dragless load to your fan motor when you start loading the PMA with real (I mean heavier) load, to really take out 8-10 Ampers. Your PMA is an off the shelf device so that Lenz law surely manifests in it, unfortunately.

Gyula

returnedintime

Thank you for replying:
Could you please expand on your statement of..."Heavier Load"?
The electric motor operates @ 1075 rpm,....the PMA is designed for windmill generator,..it produces electricity at low speeds,..but it has been tested @950 rpms and will put out 12.7 amps
It is brushless,...the only restraint is the weight of the shaft and not the outter core,...but at 950 RPM,..it is nearly coggless.
I have no doubt that there will be a slight cog present,..but I do not believe it is enough to hinder the output.
The electric motor(Fan) Is @ 1075 rpm,......considering I figured there would be a slight drag,...that is why I anticipate the rpms on the Electric motor fan to round out @ 1000 rpms.
Electric motors have shown to continue running without burning up when drag is present. 

gotoluc

Quote from: returnedintime on January 28, 2013, 06:39:45 PM
am new here of course,..I have only one question,....I have been kicking this idea around for quite a while and I just want to know if you think it may work as I invisioned here.
I have a fan (electric motor) in a ac ground unit.
It operates at 1075 rpms @ 2.6 amps.
I was thinking about setting a PMA above it set on brackets, with the shafts lined up and coupled.
The pma will generate 12.7 amps@950 rpms and is nearly coggless.
The weight of the pma(including shaft)would hover a 1/2 inch above the electric motor shaft on a bracket, thereby, the weight of the PMA is not adding weight to the electric motor.
My thoughts are that the electric motor that operates @ 1075 rpms,...when coupled with the PMA,..the rpm would round off to about 1000 rpm.
My question is,..can this work without burning up the electric motor?
If this will not burn up the electric motor,...then even though it takes 2.6 amps to operate the electric motor from the grid,..I am gaining basicly 10.1 amps excess that can be attached to a grid-tie inverter.
Your thoughts would be appreciated..thankyou

Hi returnedintime,

it would be nice if it could work. However, you have not mentioned how many Volts your PMA puts out when it reaches 12.7 Amps at 950 rpm. It could be in the 12 volts range to keep a battery charged?... that would be about 150 Watts.

Now your fan motor probably works on 120 Volts AC and at 2.6 Amps it would come to a little over 300 Watts.

If you're low on cash I would not invest money in this, as many have tried this before. I wish it would be so simple :-\

Nothing wrong in thinking of alternative energy solutions but using off the shelf items chances it would work are next to none.

I wish you all the best in your search for a solution

Luc

gyulasun

Quote from: returnedintime on January 29, 2013, 06:59:54 PM
Thank you for replying:
Could you please expand on your statement of..."Heavier Load"?
The electric motor operates @ 1075 rpm,....the PMA is designed for windmill generator,..it produces electricity at low speeds,..but it has been tested @950 rpms and will put out 12.7 amps
It is brushless,...the only restraint is the weight of the shaft and not the outter core,...but at 950 RPM,..it is nearly coggless.
I have no doubt that there will be a slight cog present,..but I do not believe it is enough to hinder the output.
The electric motor(Fan) Is @ 1075 rpm,......considering I figured there would be a slight drag,...that is why I anticipate the rpms on the Electric motor fan to round out @ 1000 rpms.
Electric motors have shown to continue running without burning up when drag is present.

Hi,

On 'heavier load' I mean as follows:  you use say a 12 Ohm power resistor as a load across the output of your alternator, the current draw will be 1 Amper if the alternator output voltage is around 12V. Now suppose you replace the 12 Ohm resistor with a 2 Ohm power resistor,  this is what I call a heavier load because the current draw goes up to at least 6 Amper provided the alternator keeps its output voltage at 12V as earlier.   

Some more notes:  it would be good for you to feel the torque by your gloved hand on the shaft this PMA must receive from the wind or from any other prime mover when it rotates at 950 RPM and you take the 12.7 Amper out of it.  Will the fan motor be able maintain that much torque on its shaft when the PMA is tied to it?  The only correct answer you can get is to test this and see for yourself. I understand you do not belive us, this is natural.

rgds,  Gyula

tinman

Here is some thing you all may find quite interesting.
In the video below,a guy has a standard generator and a standard motor.
He show's the motor and coupled generator winding up,then disconecting the power-with the generator output disconected.
Then he show's the motor and generator winding up ,then disconecting the batteries and conecting the generator output to the motor.
Now one would think that if the motor and generator both have losses,then the thing should wind down to a stop quicker?
But watch the video.
One has to wonder just how hard it would actualy be to get this to work-because this standard motor-generator setup is very close already
http://www.youtube.com/watch?v=hl57y-c_bWc