The bifilar pancake coil at its resonant frequency

[tinman]

@Tinselkoala and Milehigh,

Message successfully sent to Stephan Hartiberlin this afternoon at 4:15 P.M. CT.

It's time for anyone else who disapproves of these "Disinformation Agents" to let their feelings be heard by Stephan.

I think you may have shot yourself in the foot synchro,as i too will be sending an email to Stefan.

I think you need to go back to that junk forum !EF!-and stay there,as they thrive on rubbish like your's,as it can make AM and PL a heap of money,through sales of books full of secrets.

All i have seen on this thread,is multitudes of unfounded claims and insults from you

Brad

[ARMCORTEX]

@Tinselkoala,

You two "Fossil Fuel Lobby Sock Puppets" degraded John Bedini the same way. That's how you both got booted off Aaron's Forum.

I have also been banned from those sites, those sites are for Low IQ incompetent people that talk to themselves and enjoy wasting time.

Tinselkoala is a man with dirty fingers because of all he builds, the knobs of his oscilloscope are damaged because he has used it so much.

You on the other hand are a little whiny baby, nobody is stopping you from building anything. Go and prove TK wrong nobody is stopping you, TK is not tying your hands up, you are free as a bird to go and do whatever you want.

Its funny to see people use this excuse... Tk is stopping me from achieving over unity... Because he is arguing with me on a forum... Poor me... All this bad stuff is happening because of oil companies ... Boo hoo woo  . Everybody who disagrees with overunity  should be banned because once those people are banned overunity  will happen

[MileHigh]

I cant agree with what you say about the capacitance. Just because it is minute does not mean it does not function. The only difference between a very tiny capacitance or a larger capacitance, in a resonant LC, is the freq of the resonance and the time it takes to ring off once input is taken away. The larger capacitance should ring longer, but each cycle takes longer to complete than the tiny cap.

If we were to discharge an inductor into a cap, the same inductive charge for each cap, the larger cap will have a lower voltage than the smaller cap. None the less, the tiny cap will take all that the inductor can give, just like the large cap will, just that the smaller cap will be at a higher voltage than the larger cap, of which levels out the playing field as to which cap is more significant than the other

By saying that the capacitance of a bifi coil is so insignificant that it cant be noticed or even compare to a larger cap is nonsense. The tiny capacitance, no matter how small can hold just as much energy as a larger cap. The problem with realization of that is the difference in voltage for each respectively.   Sure you can have 2 caps 100uf and 1uf and compare. But if both are rated and limited to 16v, then yes, the larger cap can and will be a more significant charge at the same voltage level for each. Maybe some people have that stuck in their heads and thats just how it is. But lets use a 1uf at 1kv in comparison to the 100uf at 16v. Now the amount of energy in each is so much closer.

Mags

Okay, so let's use Conrad's numbers for his series bifilar pancake coil from his second clip and make some inferences.

His numbers:  34 microhenries, 49 picofarads of self-capacitance and a 3.9 megahertz self-resonant frequency.

We know that air typically breaks down at 20 kilovolts per centimeter.  Let's assume the spacing between turns is one millimeter and we will simply ballpark the breakdown though the insulation as 4 kilovolts.  In other words, we are doubling the air breakdown to factor in the insulation.

Max energy stored in the self-capacitance (at peak voltage):  E = 1/2 * 49 pF * (4 kV)^2 =  392 microjoules.

How much current does that correspond to (at peak current):  i = sqrt ((2*E)/L) =  4.8 amps

So the RMS current through the self-resonating coil would be on the order of 4.8/1.414 = 3.39 amp RMS

So if we assume a 10-ohm coil the power dissipated would be on the order of 3.39^2 x 10 = 114 watts.

Obviously it is not realistic to have a hand-sized self-resonating pancake coil dissipating about 114 watts.  So let's scale everything down to 10 watts dissipation.  So the RMS current has to be about one amp and the peak current 1.414 amps.  So the inductive energy becomes 34 microjoules.  That translates into a peak capacitance voltage of 1178 volts.

That seems pretty realistic.  You have a fairly hot pancake coil self-resonating at 2356 volts peak to peak, storing 34 microjoules of energy and you are feeding about 10 watts of continuous power into it by some means to keep it resonating at 3.9 megahertz.

So the questions must be asked:  SO WHAT, and WHY?

Why would I want to store a measly 34 mircojoules of energy in a resonant cavity blazing away at 3.9 megahertz when I have to continuously pump 10 watts into the series bifilar pancake coil to sustain the resonance?

What precisely is that going to do for me?

[synchro1]

No
The calculated amp hours a battery will deliver,is based around a !usable! voltage value--which in most cases is between 10.5 volts and 13.2 volts for a 12 volt battery-not just 12 volts.

So,for every hour your load draws 5 amp's from that battery,the energy delivered to the load will be less and less every hour.


Brad

@Tinman,

Please answer the question:

What amount of power would be required to generate 1 Tesla of magnetic flux density in an "Inductor Coil" with a measured inductance of 1 Henry?

I don't want to re-hear why the answer of "1 Ampere Hour" that I supplied is wrong; I want the to hear the right one if indeed it's wrong. Milehigh's opinion that the question is meaningless is unacceptable; If you want lynch me over this, so-be-it.

[MileHigh]

Also let's use Conrad's numbers again to look at the series bifilar coil a different way:

34 microhenries, 49 picofarads of self-capacitance

Let's just say that a reasonable peak-to-peak voltage for the self-resonating coil is 400 volts.

So the capacitive energy is 0.98 microjoules.

Now, I am going to make an apples and oranges comparison here.  Let's assume the inductor is pulsed at a low duty cycle in some kind of pulse circuit such that the average power dissipation from the 10-ohm coil is low and not an issue, and the peak current through the coil is 400 milliamperes.

So the inductive energy is 2.72 mircojoules.

Well, in this apples and oranges comparison, the two energy levels are pretty comparable and it certainly doesn't look like a gnat sitting on the back of an elephant.  However, there is one caveat.  If the coil is functioning in some kind of pulse circuit the self-capacitance is not really there, there is no real self-resonance taking place.  In that context you have what looks like a gnat sitting on the back of an elephant.

Finally, one last kick at the napkin.

If the self-resonance is say 40 volts peak-to-peak, then the energy stored in the self-capacitance is 0.0098 microjoules.  And the energy in the coll at 400 milliamperes is 2.72 microjoules.

In this apples and oranges comparison the inductive energy is 277 times the capacitive energy.  So more like a big racoon on the back on an elephant than a gnat on the back of an elephant.

Note how everything is in microjoules territory?  Here is an experiment for the "daring" - and it is an exercise in number crunching.  Find the right capacitance value for 10 microjoules of energy to charge the capacitor up to 10 volts.  Then touch the two terminals of the capacitor to your tongue and see what 10 microjoules feels like.  I am not even sure you will be able to feel the discharge.

What purpose does the series bifilar pancake or regular series bifilar coil serve?  I have asked this question many times and never got an answer.

MileHigh