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Overunity Machines Forum



Oscillating sine wave LC tank magnet motor.

Started by synchro1, August 31, 2014, 09:26:50 AM

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synchro1


Here's a youtube comment from the "Old Scientist's" Bifilar tank video:

"Tuning the Tank circuit and sweeping up to 40 MHz to find the harmonics. The Bifilar seams not major influenced by the capacitor, at least not like a standard LC tank circuit. I also used high NF capacitors with no influence on the frequency response to resonance. It seams that the Bifilar coil does not participate in the dependence of capacitance. That would explain my good result in the radiant energy video with Bifilar coils where only the 1/4 wave coil was required to draw the energy".


These "Old Scientist" findings have persuaded me into believing the Reed Switch simple breaks the circuit when it goes normally open. The only thing left to do is to accelerate the magnet spinner up to the series bifilar's self resonant frequency, then disconnect the coil from the positive power pole of the battery. It has to be that simple and that's the end of it. No external capacitor needed.

synchro1

The finalized circuit has to be merely a series bifilar coil grounded on one side and open on the other where it acts as an antenna. The only other criterion is to match the dipole magnet spinner's R.P.M. to the series bifilar's self resonating frequency. That's all there can be to this sine wave motor. Naturally, we have to choose a way to power the magnet spinner to the threshold R.P.M.


My warning; The spinner goes ballistic on the unpowered sine wave, so make sure you take the safety precautions I outlined!



synchro1

@MarkE,


             You persisted in projecting your misunderstanding on me. I want to clear this up once and for all; The series bifilar tank circuit does not conform to to the  resonant formula for the standard LC circuit.


             The capacitance of the series bifilar coil is "Evenly Distributed". The external bifilar tank condenser is connected to the coil through two wires. The resistance of these wires acts as a bottle neck that results in losses that interfer with the "Capacative resonance" of the condenser and the bifilar coil. The "Bifilar Coil" can't see the external capacitor! 


               To sum it up: The "Bifilar Tank" is a pig with wings!     

MarkE

Quote from: synchro1 on September 04, 2014, 08:53:54 AM
@MarkE,


             You persisted in projecting your misunderstanding on me. I want to clear this up once and for all; The series bifilar tank circuit does not conform to to the  resonant formula for the standard LC circuit.


             The capacitance of the series bifilar coil is "Evenly Distributed". The external bifilar tank condenser is connected to the coil through two wires. The resistance of these wires acts as a bottle neck that results in losses that interfer with the "Capacative resonance" of the condenser and the bifilar coil. The "Bifilar Coil" can't see the external capacitor! 


               To sum it up: The "Bifilar Tank" is a pig with wings! 
Again: There is no such thing as "capacitive resonance".  I have explained this several times and yet you persist in using that meaningless term.  It takes capacitance and inductance to resonate.  Physical components such as capacitors and inductors all exhibit self resonances once connected with external wiring such that an inductance loop is closed.  This is true for a simple single loop of wire, or even a soda can.  In the 1960's and 1970's it was popular among microwave engineers to use beer cans as microwave resonators.  A bifilar coil, is no exception, and has parasitic capacitance that sets a self resonance like all the aforementioned devices and structures.  The resonance results from the combination of the parasitic capacitance acting in parallel with the inductance.  Connecting a larger value discrete capacitor across a coil of any kind creates a dominant pole pair at a lower frequency than the coil's self-resonant frequency.  The resistance in the circuit dissipates energy, which means for a larger resistance, a lower percentage of energy remains in storage each cycle than with a lower resistance.  IE the circuit has a lower Q.   

synchro1

@MarkE,


             Maybe you should go back and re-read the comment. The "Old Scientist" reports that the external capacitor does not have a major influence on the "Series Bifilar Coil" as you infer! There's an additive formula for the twin capacitors in the Colpitts oscillating circuit that won't work on the bifilar tank as you imply. You're making a mistake! Capacitance is evenly distributed in the series bifilar unlike the external capacitor that's wired across a divide of calculated resistance.


Quote from the "Old Scientist":


                            "It seams that the Bifilar coil does not participate in the dependence of (external) capacitance".

          Wiring a separate capacitor to a series bifilar coil does not behave additively like wiring two capacitors in parallel to a regular coil. The unequal distribution of capacitance between the bifilar coil and the external capacitor precludes the use of the standard formulas you're falsley trying to apply!