Skycollection's "Pentafilar Pancake" inductively coupled "Overunity Potential".

[skycollection 1]

Thanks Gyula, this project is 100% interesting and in the following days I will do some experiments as you describe in your comments, with 1K resistors in the current outputs and see what results it gives through those resistors, thank you very much for your support which will be very useful in the future.

[skycollection 1]

Hi Gyula, I did the experiment by placing a 2.2K resistor in the current output with the following result, from 66 volts the voltage drops to 40 volts through that resistance, the rotor did not stop, it did not affect its rotation, I don´t know how to interpret this result, maybe it is necessary to try other resistance values, what do you think ...? With a resistance IK the rotor slow down with only 4.5 volts.

[gyulasun]

Dear Jorge, 

Many thanks for the tests! I assume you connected the 2.2 kOhm resistor in parallel with the + and - outputs of the R rectifier diode bridge, right?  i.e. to the same pins where you discharged the puffer capacitors in the video. 

So in this case the 40 V DC across a 2.2 kOhm dissipates 40 x 40 / 2200 = 0.72 Watt power. If you use a second 2.2 kOhm resistor in parallel with the other + and - outputs of the other R rectifier bridge, then it will also dissipate about 0.72 W power. 

Question is whether the input current increases from the 0.23 A (at the 12 V input) when you connect the first 2.2 kOhm resistor and then the second 2.2 kOhm in parallel with the outputs? 

This would be good to know, especially because you wrote the rotor did not stop, the 2.2 k did not affect its rotation. 

You mentioned that with 1 kOhm load in parallel with the output, the rotor slows down and the 66 V drops to 4.5 V. Question would be how the brightness of the two LED bulbs running from the BACKEMF outputs changes during the 1 k and during the 2.2 kOhm loads?   

Thanks
Gyula

[skycollection 1]

Ok thanks, i have a new test now with both diode rectifiers: 

NUMBER ONE RESISTANCE 2.2K= 66 VOLTS DROPS TO 38.7 VOLTS
NUMBER TWO RESISTANCE 2.2K= 66 VOLTS DROPS TO 35.3 VOLTS

INPUT VOLTAGE WITHOUT RESISTANCES 12 VOLTS X 0.23 = 2.76 W

INPUT VOLTAGE WITH ONE RESISTANCE 2.2K BOTH SIDES  12 VOLTS X 0.30 = 3.6 W

LEDS BULBS BOTH BEMF OFF
what does this result mean?

[gyulasun]

Hi Jorge, 

Thanks for the new tests. To understand the results, first some calculations:
power dissipated in one of the 2.2 k resistors is 38.7 x 38.7 / 2200 = 0.68 Watt
power dissipated in the other  2.2 k resistor   is 35.3 x 35.3 / 2200 = 0.566 Watt
Summing these two numbers gives 1.246 Watts output power.

Because you experienced the LED bulbs across the BEMF outputs go off when the 2.2 k resistors are used, this means the following:

The two 2.2 kOhm resistors impose already such a high load for both bifilar coils that the peak amplitude of the BEMF is not enough any more to operate the 120 V LED bulbs. You know there is a minimum input voltage specified for LED bulbs by the manufacturer and in your setup the peak voltage of the BEMF was so far enough to light up both LED bulbs when the 2.2 k loads were not present. 

You could use for instance 3.3 kOhm or 4.7 kOhm load resistors instead of the 2.2 kOhm, so the LED bulbs may remain lit at a reduced brightness,  this depends on the trade-off how much part of the total BEMF you use up by the resistors and the rest of the total BEMF may light up the LED bulbs, albeit at a reduced brightness.  Let me note (although you may know also) that LED bulbs have a nonlinear voltage-current characteristic, this means that near their threshold ON voltage range a smal change in the voltage across them causes a high current change hence brigthness change. This is why we cannot really estimate actual power a LED bulb may consume by our naked eye. 

You know that the total output power comes from two sources: the rotor magnets induce current in the coils (normal Faraday motional induction) and the input current from the 12 V source charges up the coils and this latter appears at the BEMF outputs when the transistors switch this current off and the coils can release their stored energy. 

This is how I see this. 

Greetings
Gyula