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Overunity Machines Forum



Solid State Orbo System

Started by Groundloop, January 06, 2010, 12:21:24 PM

Previous topic - Next topic

0 Members and 10 Guests are viewing this topic.

gyulasun

Hi Bruce,

I think you can consider your 6 parallel coils as one "secondary" coil of a hypotetical transformer, with 28V AC output voltage and the DC resistance of this 'coil' is about 49 Ohm.

Now if you would use a 49 Ohm load resistor (any wattage) in parallel with your 'coil' output, and your measurements are correct, then you should find about 28/2=14V AC output voltage across the load resistor (i.e. at the 'coil' output) because the inner coil 49 Ohm resistance constitutes a voltage divider with your outside 49 Ohm load resistor, hence the output voltage is halved.
Try to do this and if this is ok, you may go further on. For AC voltage measurement, use the same meter what you used for measuring the 28V output measurement.

rgds, Gyula

gyulasun

Hi Bruce,

I highlighted your text what is the culprit....

Quote from: Bruce_TPU on March 07, 2010, 04:54:05 PM
....
If I were to take that voltage of 28 for each inductor at 254 ohms, that would give me a current of 0.11024 per coil, or .66144 I with all 6 coils in parallel.

With an I of .66144 this gives me a total power output of 18.52 watts.

But where is the current?  Is it still stored in the inductor?  Because I am not seeing .66144 of current.  I see the voltage.  I do the math and see the resistance.  I read the resistance on the meter.  But I am not seeing the current on the meter.
...

You cannot calculate current like that because the inner DC resistance of any one coil is 254 Ohms and if you short circuit such coil with a piece of wire, while there is 28V induced voltage across this coil, then your total 28V is dissipated in the inner 254 Ohm resistor. (To be more precise the inductive reactance of the coil comes in series with the 254 Ohm copper resistance to establish an series RL circuit, the inductive reactance is XL=2*pi*f*L  where f is the AC output frequency and L is inductance of the coil in question.  So the total impedance the shorted current 'sees' is Z=sqr(R2+X2L )

rgds, Gyula

Bruce_TPU

Quote from: gyulasun on March 07, 2010, 05:37:29 PM
Hi Bruce,

I think you can consider your 6 parallel coils as one "secondary" coil of a hypotetical transformer, with 28V AC output voltage and the DC resistance of this 'coil' is about 49 Ohm.

Now if you would use a 49 Ohm load resistor (any wattage) in parallel with your 'coil' output, and your measurements are correct, then you should find about 28/2=14V AC output voltage across the load resistor (i.e. at the 'coil' output) because the inner coil 49 Ohm resistance constitutes a voltage divider with your outside 49 Ohm load resistor, hence the output voltage is halved.
Try to do this and if this is ok, you may go further on. For AC voltage measurement, use the same meter what you used for measuring the 28V output measurement.

rgds, Gyula

Hi Gyula,

I have tried this, and the result is an output of 1.7 volts ac, so what does that now tell us?  It should have been about 12 or 13 volts.  I am running now about 25 or 26 volts steady.

I knew if no one else could figure this out, you would...LOL

Thanks!

Bruce
1.  Lindsay's Stack TPU Posted Picture.  All Wound CCW  Collectors three turns and HORIZONTAL, not vertical.

2.  3 Tube amps, sending three frequency's, each having two signals, one in-phase & one inverted 180 deg, opposing signals in each collector (via control wires). 

3.  Collector is Magnetic Loop Antenna, made of lamp chord wire, wound flat.  Inside loop is antenna, outside loop is for output.  First collector is tuned via tuned tank, to the fundamental.  Second collector is tuned tank to the second harmonic (component).  Third collector is tuned tank to the third harmonic (component)  Frequency is determined by taking the circumference frequency, reducing the size by .88 inches.  Divide this frequency by 1000, and you have your second harmonic.  Divide this by 2 and you have your fundamental.  Multiply that by 3 and you have your third harmonic component.  Tune the collectors to each of these.  Input the fundamental and two modulation frequencies, made to create replicas of the fundamental, second harmonic and the third.

4.  The three frequency's circulating in the collectors, both in phase and inverted, begin to create hundreds of thousands of created frequency's, via intermodulation, that subtract to the fundamental and its harmonics.  This is called "Catalyst".

5.  The three AC PURE sine signals, travel through the amplification stage, Nonlinear, producing the second harmonic and third.  (distortion)

6.  These signals then travel the control coils, are rectified by a full wave bridge, and then sent into the output outer loop as all positive pulsed DC.  This then becomes the output and "collects" the current.

P.S.  The Kicks are harmonic distortion with passive intermodulation.  Can't see it without a spectrum analyzer, normally unless trained to see it on a scope.

Bruce_TPU

Evening ALL,

Earlier today, I had a bit of frustration.  I had taken the core of soft iron laminations out of the center and tested as I posted earlier.

When putting it back, I could not get more than 5 volts.  Then I thought to use a compass on the soft iron laminations and realized they had been lightly magnatized, North down, and South pole up.  I found this interesting, and when reinserted correctly (as per the unit's liking), the voltage was back up.

I am personally convinced that it is "transforming" the input power created from the toroids and magnets, within the center core and increasing voltage at the sake of the current.  This is my theory for now.

I am looking forward to getting my electronic parts in.  The whole Ideas was to optimize as much as possible before bringing in the "big gun" which is the electronic high speed switching.

I feel that redoing the the circuit to run on 0.45 watts using a D-Cell is the next step.  And then go from there.

Cheers,

Bruce
1.  Lindsay's Stack TPU Posted Picture.  All Wound CCW  Collectors three turns and HORIZONTAL, not vertical.

2.  3 Tube amps, sending three frequency's, each having two signals, one in-phase & one inverted 180 deg, opposing signals in each collector (via control wires). 

3.  Collector is Magnetic Loop Antenna, made of lamp chord wire, wound flat.  Inside loop is antenna, outside loop is for output.  First collector is tuned via tuned tank, to the fundamental.  Second collector is tuned tank to the second harmonic (component).  Third collector is tuned tank to the third harmonic (component)  Frequency is determined by taking the circumference frequency, reducing the size by .88 inches.  Divide this frequency by 1000, and you have your second harmonic.  Divide this by 2 and you have your fundamental.  Multiply that by 3 and you have your third harmonic component.  Tune the collectors to each of these.  Input the fundamental and two modulation frequencies, made to create replicas of the fundamental, second harmonic and the third.

4.  The three frequency's circulating in the collectors, both in phase and inverted, begin to create hundreds of thousands of created frequency's, via intermodulation, that subtract to the fundamental and its harmonics.  This is called "Catalyst".

5.  The three AC PURE sine signals, travel through the amplification stage, Nonlinear, producing the second harmonic and third.  (distortion)

6.  These signals then travel the control coils, are rectified by a full wave bridge, and then sent into the output outer loop as all positive pulsed DC.  This then becomes the output and "collects" the current.

P.S.  The Kicks are harmonic distortion with passive intermodulation.  Can't see it without a spectrum analyzer, normally unless trained to see it on a scope.

aaron5120

Quote from: Bruce_TPU on March 07, 2010, 04:54:05 PM
Hello All,

I have been trying to find my math mistake all day, but can not.  Yet it does not feel like 16 watts of power when lighting up various items.

Each Brooks coil is 254 Ohms.  This is a fact.
All 6 are in parallel.  This is a fact.
Using the formula 254/6 gives a resistance of 42.3, and I am at 49.3 reading on the meter.  This is a fact.

Voltage remains relatively the same in each inductor, so I am producing 28 volts at output.  This is a fact.

If I were to take that voltage of 28 for each inductor at 254 ohms, that would give me a current of 0.11024 per coil, or .66144 I with all 6 coils in parallel.

With an I of .66144 this gives me a total power output of 18.52 watts.

But where is the current?  Is it still stored in the inductor?  Because I am not seeing .66144 of current.  I see the voltage.  I do the math and see the resistance.  I read the resistance on the meter.  But I am not seeing the current on the meter.

So, I am seeing far less current then ohms law tells me should be there, or is it all being stored in the magnetic fields of the inductors?  Any math assistance would be of help. 

Thank you,

Bruce

EDIT:
I have removed the iron lamination core and now the voltage is at 5.2 volts on the main output, with very dim lights indeed.
Bruce,
Sorry to hear about your frustration in measuring the output power. Please allow me to give you my point of view in regard to your problem.
I think it is due to phase factor of the V and I in the output. Since it is in AC and also it involves with many inductive components, there is a phase factor issue in your output: the voltage is not in phase with the current. If your correct the phase factor with a capacitor bank, maybe you can see the power. Also, the digital meters are no good for measuring AC above 60Hzs, because they are designed for measuring the Grid V and Amps.
If I am wrong, please forgive my ignorance. Anyway, please keep up your good work, We are pending on your great experments.