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Overunity Machines Forum



STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM

Started by PaulLowrance, December 04, 2009, 09:13:07 AM

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LarryC

#3405
June 19, 2010, 08:18:35 PM
Quote from: Omnibus on June 19, 2010, 07:34:20 PM
Did you read that carefully, Larry? I guess not. So, I'll repeat it: in an RL circuit the power which is dissipated is only that due to the dissipative (Ohmic, active) resistance. that is, I^2R where R is the dissipative (Ohmic, active) resistance having nothing to do with inductance L. The part of the power delivered to the inductance is not dissipated. Again -- the part of the power delivered to the inductance is not dissipated when integrating over a full cycle. Can this come across to you or you need more explanation? Here in the discussion at hand we are only concerned with the dissipated power. Only the power dissipated is the output power (the power delivered to the inductance is not dissipated and therefore is not of interest to us in this discussion). Therefore, because the only dissipated power is I^2R where R is the dissipative (Ohmic, active) resistance having nothing to do with inductance or L, that's what we calculate as output power in our case. Get it?

Omnibus stated:
'Did you read that carefully, Larry? I guess not. So, I'll repeat it: in an RL circuit the power which is dissipated is only that due to the dissipative (Ohmic, active) resistance. that is, I^2R where R is the dissipative (Ohmic, active) resistance having nothing to do with inductance L.'

Larry states:
Yes, I did read carefully, but somehow what you say you read is totally different from the page that I posted. Total disinformation. Where did you get that statement from, you must be delusional? Can anyone here find what Omnibus stated in my page listing?

I've included the page listed again below to shows the actual original statement: 'In an RL Circuit, only a portion of the input power is dissipated. The part delivered to the inductance is returned to the source each time the magnetic field around the inductance collapes.'

Regards, Larry

IceStorm

#3406
June 19, 2010, 08:30:20 PM
LarryC , I admire your patience but i still think its a waste of time, Omni want to stick to his view even if we know that's not the correct one. At the end he will come to the same conclusion , will just take more time.

Best Regards,
IceStorm

Omnibus

#3407
June 19, 2010, 08:36:50 PM
Quote from: LarryC on June 19, 2010, 08:18:35 PM
Omnibus stated:
'Did you read that carefully, Larry? I guess not. So, I'll repeat it: in an RL circuit the power which is dissipated is only that due to the dissipative (Ohmic, active) resistance. that is, I^2R where R is the dissipative (Ohmic, active) resistance having nothing to do with inductance L.'

Larry states:
Yes, I did read carefully, but somehow what you say you read is totally different from the page that I posted. Total disinformation. Where did you get that statement from, you must be delusional? Can anyone here find what Omnibus stated in my page listing?

I've included the page listed again below to shows the actual original statement: 'In an RL Circuit, only a portion of the input power is dissipated. The part delivered to the inductance is returned to the source each time the magnetic field around the inductance collapes.'

Regards, Larry

Of course. Do you understand what you just posted? Obviously not. Can't you understand that the power returned to the source is not dissipative power? Power delivered by the inductance and returned back to the source is not dissipative power. In this discussion we are only interested in dissipative power. Inductive power is not dissipative power. Only  power delivered to the active resistance (dissipative resistance, Ohmic resistance) is dissipative power. Only power causing Joule heating (Ohmic heating) is dissipative power. Only the power delivered to the active resistance, not to the inductance brings about Joule heating. Get it? Read first the scanned pages and try to understand them before posting them here.

Bruce_TPU

#3408
June 19, 2010, 09:32:29 PM
Hi Omnibus,

Thank you for the pictures of the primary, and of it rewound on your plastic bobbin.  Please humour me and give me a diameter measurement in millimeters...  Thank you!

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.

Omnibus

#3409
June 19, 2010, 11:08:52 PM
@Bruce_TPU,

That last coil is approx. 1.5cm in diameter and 1cm thick--425turns. I wonder what the awg of the wire is. You said 30awg which may be right. Will have to look around to see if I can order from somewhere similar or even thinner (if such exists at all). Also, you may try with the magnet wire from RadioShack. I'm playing right now with such. This is getting more and more interesting.
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