DUAL PERMANENT MAGNET ROTOR AXIAL FLUX ALTERNATOR GENERATOR

[ELEMAN]

This is Amazing  ElectricoMagnetic Motor Generator To Get "Mechanical" Set Up Voltage!  Features and Result Of Experiment!  ELEMAN MAGNET MOTOR, OverUnity is iT or Not... Motor Generator or a Toy... a Homemade Magnetic Motor ALMOST! Self Running or Quantum harmonic Oscillator Test the NEW design  2014...2015... https://www.youtube.com/watch?v=qnjfitCqeQQ&index=14&list=PLT7Zi_DRtP7fpdEHDlFIfaI_bOwZo5zyF

[gyulasun]

Hi Eleman,

Very nice setups you have been showing but the only thing I miss is putting a decent load resistor across your 330 uF output puffer capacitor.  Why do not you load the output at all??   You occasionally use a screwdriver to short the DC output this is a very rough load but not good for evaluating ouput power. 

For instance when the DC input is 11V at 60 mA and the DC output voltage is around 50.4 V, you short the capacitor, the input current jumps up to higher than 150-160 mA  (this is what the supply display is able to show) and RPM drops during the short circuit time. 
Please repeat the test with just the 11V at 60 mA DC input and when it is running, connect a resistor across the 330 uF capacitor, the resistor could be any value between say 420 Ohm to 510 Ohm, its rated wattage could be a few Watts like 4 or 5 W. 

Then watch 2 things: 

1) what value the DC output voltage drops to as per your DC voltage meter shows across the resistor?
2) what value the DC input current increses to as per the power supply meter shows?

This way you can estimate pretty well what the actual input and output power is, hence you have info on efficiency.
When you do not load your output voltage but just show how much the DC output voltage higher than the input DC voltage, then it is not a useful info at all, unfortunately.

Suppose you use a 420 Ohm load resistor across the 330 uF capacitor and suppose the DC output voltage drops from 50.4 V to say 40 V.  This would mean a heat dissipation of 40*40/420 = 3.8 Watts in the resistor.  And just suppose that the input current jumps to say 200 mA from the unloaded 60 mA draw, still at the 11 V input voltage, ok?  Now the input power would be 11*0.2 = 2.2 Watts.

So in this example I made up, you would have a COP = 3.8/2.2 = 1.72  which would be a fantastic result indeed. 
Understand the procedure?

Thanks,  Gyula