Pulsing A Bifilar Coil And Collecting From It And Various Ways Of Going About It

[TinselKoala]

Mags, when you break the series connection between the Files of a Tesla Bifilar coil, it is no longer a TBF coil. It has become a Spirinductpacitor. (tm TKLabs). When you reconnect the series connection to make it a TBF coil again, everything changes. You can't measure the one type and then expect its values to apply to the other type.

Get out your capacitor meter and measure the capacitance of the fully connected TBF coil. I doubt if you can get any reading at all from the capacitor meter. Does this mean it has no capacitance? No, of course not, because we know that it does. The way to measure the capacitance of the TBF coil, in contrast to the Spirinductpacitor, is to measure the resonant frequency and then calculate the capacitance.

After all, this is how many L C meters do it, they just do it internally where you can't see it happening.

[Magluvin]

Mags, when you break the series connection between the Files of a Tesla Bifilar coil, it is no longer a TBF coil. It has become a Spirinductpacitor. (tm TKLabs). When you reconnect the series connection to make it a TBF coil again, everything changes. You can't measure the one type and then expect its values to apply to the other type.

Get out your capacitor meter and measure the capacitance of the fully connected TBF coil. I doubt if you can get any reading at all from the capacitor meter. Does this mean it has no capacitance? No, of course not, because we know that it does. The way to measure the capacitance of the TBF coil, in contrast to the Spirinductpacitor, is to measure the resonant frequency and then calculate the capacitance.

After all, this is how many L C meters do it, they just do it internally where you can't see it happening.

Im, yes, revisiting the possibility of something special in the bifilar coil. Been thinking clearer than I have in the past couple years, for reasons Ill explain in a thread I made here on cell phones and headaches... So some ideas are coming out on this that I never thought of before and Im going to lay it all out here...

The idea that we can determine the res freq of a bifi by calculating the measured separate winding capacitance and the inductance of just 1 winding is in the least, a tool. For some that may be valuable.

Just like the theory I had on the cap to cap deal that the 50% loss is not due to resistance, and resistance only changes the time it takes to level out the charge, and then I proved it, without a cap on the bench, I have that kinda going on right now with this coil. So Im going with it.

That cap to cap 50% loss just 'really' didnt seem right to me, even before I thought more on the possibility of 'if' we could count the electrons being taken from the + of a cap and shoving them into the = of the cap, we could determine the voltage pressure in the cap by that count alone. So is the resistance eating up electrons in the transfer, or would the count be the same in an 'ideal' situation?  And Im kinda at that point in the back of my mind that is saying, something isnt right here with this bifi coil according to what others are claiming, as in there is nothing special there.  Not saying anyone is misconstruing info, just saying there are some things to do that I havnt really seen nor thought of before with this. So thats where this thread is going.    Ill just post it all over here at OU. No interest in this over there really.

Mags

[Jimboot]

Augment the wave with an impulse.   

like a tether ball?

[SkyWatcher123]

Hi all, Hi magluvin, had a thought about the 50-50 capacitor deal.
It is progressively, acting as two batteries in parallel.
The receiving capacitor, is incrementally acting like another voltage source, much like a coil being induced by a magnet.
So, our primary feed capacitor is fighting more and more over time against the receiving capacitors voltage source.
So i agree, it is not due to ohmic resistance.
I wonder if some arrangement can be made in the 50-50 capacitor setup, that would mimic two same polarity magnetic flux sources, entering a ferromagnetic core.
As long as the core is not saturated, it will accept the flux and not fight against the incoming flux.
We need the receiving capacitor to stop fighting against the feeding capacitor.
While writing this, i just had an interesting idea for a rotating generator.
peace love light

[Magluvin]

Hi all, Hi magluvin, had a thought about the 50-50 capacitor deal.
It is progressively, acting as two batteries in parallel.
The receiving capacitor, is incrementally acting like another voltage source, much like a coil being induced by a magnet.
So, our primary feed capacitor is fighting more and more over time against the receiving capacitors voltage source.
So i agree, it is not due to ohmic resistance.
I wonder if some arrangement can be made in the 50-50 capacitor setup, that would mimic two same polarity magnetic flux sources, entering a ferromagnetic core.
As long as the core is not saturated, it will accept the flux and not fight against the incoming flux.
We need the receiving capacitor to stop fighting against the feeding capacitor.
While writing this, i just had an interesting idea for a rotating generator.
peace love light

Im not saying there is free energy there in any way. The discussion was whether the 50% energy loss in doing so was due to heat in resistance, no matter the value above 0ohm, or my argument that the resistance heat is not the cause of the 50% loss. The claim on the resistance loss side was that if the caps had no resistance, ideal, then the end result would be 7.07v in each cap from 1 10v cap.  But that is not the case. If we we able to count the electrons, just like we could count water mass/volume in buckets, then just by that count alone we could determine the voltage in the cap, every time.  So resistance doesnt take electrons from the system, it only slows them down in their travels. The ideal caps would still have 5v each as the count of electrons would be the same as a real world circuit.

So where is the loss?  Well in real world with resistance, if you wanted to generate heat, then good, you used that supposed 50% loss and it is now not a loss. But if there was no resistance and no heat, then where did we lose 50%?? I say we lost is stupidly by not making any use of the current during the transfer from cap to cap.

Now if it were ideal again and we used an inductor with say a 0v drop diode, then the transfer would be a complete transfer, the 10v cap would end at 0v and the 0v cap would end at 10v.... We used the current in the cap to cap transfer to energize the inductor(flywheel) and it uses that lost 50% to pump every last bit from the source cap to the receiving cap.  But we can also have that situation in the real world, and minus the diode voltage drop and some loss(not 50% at all) in resistance only due to voltage division in the circuit( of which we dont have in the actual cap to cap mostly due to only 2 components other than resistance).

Anyway, thats that.

Mags