Magnetic energy pump OU motor

[Ergo]

I hope you are right. 
Good luck in your findings.

[wizkycho]

I hope you are right. 
Good luck in your findings.

I'm certanly wrong, but proper experiment is more and more likely to be right.

all the best

Wiz

[Ergo]

Things and formulas to have in mind when doing your experiments.
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Advices:
1) The torque force of your motor must be stronger than the Lenz law drag.
    The formula below shows you the Lenz drag you need to overcome to reach overunity.

2) The radius of the measured force is of utmost importance.
    You need heavy force at large radius to accomplish lots of work.

3) Rpm times Force = Hp.
    The more force you can get at high rpm, the more output you will have.
    In real life you will get half the stall torque at half the free rpm = optimum output.
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Explanations:
1) Force Hp Formula = (10ft-lbs x 2 x 3,14 x 200rpm ) / 33000 * 746 = 284W
    This tell's you it will take 10ft-lbs of torque at 200rpm to develop 284watts of power.
    If your motor consumes less than 284watts at 200rpm and 10ft-lbs loaded shaft, then you have reached OverUnity.
    You can rearrange the numbers to fit your type of motor and output.

2) When you measure the shaft force then consider that you have to recalculate
    the numbers to ft-lbs, this meaning 1 foot from the shaft centre.
    Simply put. If you measure 1kilo of force at 15cm distance from centre (like in your setup) then
    you have to divide the measured force by two and multiply by 0.445 before using the formula above.
    The divisions gives you 30cm distance = 1 foot and the 0.445 multiply converts kilo to pound.

3) In all motors only half the stall torque is accessible at half the free rpm.
    If your motor is free spinning at 400rpm and no shaft load, then you should load it down to 200rpm.
    This is where you will find the optimum output vs rpm mode. The torque is completely rpm dependent.
    The higher the rpm the lesser the torque, and the lower rpm the higher the torque. This is simple physics.
    See the formula below.

T  = Ts - (N * Ts / Nf)

T  = Torque at loaded rpm
N  = Loaded RPM
Ts = Stall torque              (in this case 20 ft-lbs)
Nf = Free spinning rpm     (in this case 400 rpm)

T  = 20 - (150 * 20 / 400) = 12,5 ft-lbs
Output = ((12,5*2*3,14*150)/33000)*746 = 266 watt <-- you can see that you get less out when loaded harder.

T  = 20 - (200 * 20 / 400) = 10 ft-lbs
Output = ((10*2*3,14*200)/33000)*746 = 284 watt <--Always max output at half the "free rpm". This is a physics law. Use it.

T  = 20 - (250 * 20 / 400) = 7,5 ft-lbs
Output = ((7,5*2*3,14*250)/33000)*746 = 266 watt  <-- you can see that you get less out when spinning faster.
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I hope you take your time and read this carefully until you understand the importance of my advices.
If you don't understand then please ask me to be more specific and I'll try my best to help you understand.
Or else you will just be reaching in the dark when performing your tests.

[wizkycho]

Hi !

many thanks for the input...

I'm aware that when messuring actuall power output I can expect FE (motor going over 100%) in certain range of RPM and of course certain
range of input (and pulsing at ceratin angles of rotor to make ON window where efficency of transfer of PM field to rotor is greatest).
And it would be quite a task....but once....

I didn't know that at half RPMs the power product is highest and this will help a lot - unless magnets will change power output peak appears at some other RPM value.
Anyhow I intended to messure all of the "spectrum" of RPMs - and all of the input power range as input....It is neccessery to define all of the spectrum so to use it as FE at first and then with some genarator (first some efficient classic one, and later allso FE type - toroid,faraday,lorentz, peripeteia... ) as OU.

Many thanks

all the best

Wiz
btw. (what you describe is not Lenz - but none the less)

[wizkycho]

hi all !

What I'm allso thinking about is another type of Magnetic Transistor - with mechanicall input
either another magnet or permeable material instead of coil (current-voltage) input,
and allso using balanced equilibrium in input rotor...haven't baked the idea throughly.
two rotors again - one equilibrilized for using very small input that interferes with magnetic resistance
on stator with Pmagnet and therefore much stronger Power output on another rotor.

This will be great improvement since coils here can not be wound in bucking or biffilar for faster switching and still temper properly with magnetic resistance.
(or can they?)..and there is resistance of coils.....

benefit: much higher COP, wider range where device works in FE

does it seems doable? can someone think of obviously working Magnetic Transistor (or pair - for balancing (in-out)) with small power mechanical input ?


Wiz