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The way to understand this setup is to start off by envisioning Tesla's

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example

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or analogy of how his Wardenclyffe was intended to operate in which he has drawn

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the picture of the Earth and a hand is pumping He's operating a pump,

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pumping energy in a linear reciprocal fashion into the earth which is acting as

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a spherical capacitor.

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Now Wardenclyffe Tower, the top of the tower was virtual ground.

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He couldn't afford to waste any energy out the top

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so there are no lightning bolts coming out of the top.

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He did that in Colorado Springs and he learned through experience that that's

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the wrong way to do things.

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It has to be a virtual ground because all the energy is going through the earth

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Because underneath Wardenclyffe Tower is an extensive array of an antenna,

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an aerial.

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In the old days of analog television reception,

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you know,

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like a tree,

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a Christmas tree,

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buried in the bedrock of the earth is where he's sending all his energy.

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He's pumping the earth.

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Now, my circuit has three loops.

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And one linear relationship.

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Each one represents a different aspect of electricity.

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The linear represents the electrostatic,

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the dielectric, and the three loops represent the magnetic.

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Now, the loop on the top that is L5 coil going through the R11 resistor,

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if I'm not mistaken,

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and ground,

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That was intended to be a load or a ballast or throttle or preferably a load

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if it could be.

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Of course, I took out the other side, so now it would have to be the load.

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And that's why I initially started to measure it when I first started my dialogue

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with you or recently when I took out the other side.

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And that's a kind of a self-short.

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And it's a pair and they need each other.

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The resistor R11 exterior to the L5 coil.

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Now that's necessary to retain the energy and to remember it.

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That's why it's the load.

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Because it's also why I call it a ballast.

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It's an inductive ballast in conjunction in series with a resistive ballast.

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And that's why all my resistors are 1 ohm.

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Because that's the null point, the fulcrum point, between too low and too high.

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You know, on a graph paper, XY axis, we have zero as the null point.

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But when it comes to resistances, one is the null point, the fulcrum point.

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And it acts as a superb ballast whenever it's required in different areas of a circuit

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to

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smooth out the operation so that the simulator has less trouble figuring everything out.

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And if the simulator has less trouble,

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if the simulator is happy,

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then conservation of energy is happy because that's where the difficulty lies.

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The simulator has to comport with conservation of energy so

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that everything balances out to zero.

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And because my circuits are so weird,

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It's sometimes difficult for that to occur,

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and it's my own fault because I don't know what I'm doing.

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So my job is to make it easy on the simulator,

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which means it becomes less stressful for the circuit if it were built,

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and it operates more happily, very quickly.

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It doesn't take so long for the (simulator's approximation) engine to figure things out.

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So all the resistances are raised above your normal solder joint resistance,

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which would be like 1 milli-ohm or 100 micro-ohms and instead is raised to

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a full ohm.

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And they act as ballasts.

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So the top loop involving L5 and R11 is a ballast in itself in its entirety

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because it's going to retain the memory of the current that is being passed to it by

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the L3 coil.

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Otherwise, nothing would be retained in memory and it wouldn't be able to build up.

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It has to remember in order to build up that humongous amount

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of voltage that's rising inside of the neon bulb.

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In order for that to rise inside of that,

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the current has to be retained in the L5 coil and R11 resistor.

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Now, the loop on the right involving, I believe it's L1, if I'm not mistaken, 10 Henrys,

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Oh, and by the way, the load coil can be anything I found.

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It really doesn't seem to matter, I don't think.

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It looked that way when I tried different values.

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And to some degree, it doesn't matter how the L1 coil is sized.

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But the load coil that remembers L5 is a 25 gauge.

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It's a standard gauge, or in the range, in the ballpark.

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But the L3 and the L1 are different.

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They're a 10 gauge, and they're not...

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They're intended to not create a voltage difference between their terminals.

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So they're both behaving the same way.

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They're trying to keep their terminals equal in voltage.

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But they are trying to retain a tiny memory of current.

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And they do it by being there.

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Now the L1 coil is the pump.

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And...

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It shorts to itself through two ballasts, two resistors of 1 ohm each.

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And

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it creates an activity of magnetism that the rest of the circuit doesn't see

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because it loops to itself.

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All it sees is what's translated,

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and that's an electrostatic linear relationship that goes through

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the small capacitor of 100 femtofarads,

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And then the next loop in the center,

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which is the L3 coil, the two resistors of 1 ohm each, and the neon bulb.

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And then to the other side,

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the linear continues through another 100 femtofarad capacitor, and then to ground.

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And that's the pump that's pumping the Earth.

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But we're not trying to do anything to the Earth per se.

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We're trying to do something to the L5 coil up above.

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Now, the L3 coil is also low resistance,

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10-gauge wire, same as the L1, and it matters what it is.

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It's a throttle of sorts.

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I haven't tested it out enough to become totally familiar with it,

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but it has to be at 2 Henrys in order to Allow the energy to rise and sustain.

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So when the voltage rises inside the neon bulb to 10 to the 17th power, it plateaus.

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And everything else exterior to that plateaus.

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But if I pick a different value, it tends to explode.

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And I'm sure somewhere in that variation above or below 2 Henry's of L3...

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It will go comatose.

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But most of the values I tried, it wants to gradually explode.

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And it's nice that the circuit takes a long time to explode.

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Because then we have a chance to study it.

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We have a chance to possibly regulate it.

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And now it's self-regulated at putting out, I think it was 500 watts at the L5 coil.

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Now it's a jumpy, jerky waveform.

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Probably no good, but for having...

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A heating situation in which maybe you'll heat water, I suppose, boil water.

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I don't know.

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I guess you can boil water at 500 watts if the water, if you wait long enough, maybe it'll boil.

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In any case, let's see.

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So the purpose of the L3 coil is to do similarly to the L1 coil,

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only in this case it's in

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parallel, in series, depending on how you want to phrase it.

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Yeah, it's in parallel with the neon bulb.

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And so the neon bulb, we want its terminals to be equal in voltage.

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Yet, we want to retain a memory of current which the L3 coil will

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succeed at doing.

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Now, we go back to the L1 coil.

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What is the L1 coil of 10 Henry's is doing?

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It's creating...

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Two directions of current simultaneous to each other because it has such depressed resistance.

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And they leave at the same time.

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That means we're pumping the neon bulb.

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Well, firstly,

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we're pumping,

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excuse me,

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the two capacitors on either side of the neon bulb at each terminal of

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the two terminals in the neon bulb.

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Using a capacitance that's so low, the capacitor doesn't want to hold on to it.

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It immediately expels it, gets rid of it, and passes it to its environment.

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It can't go back to the L1 coil necessarily, so it moves on.

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It wants to move on in the direction of ground.

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And so in order to go to ground, it has to go through the neon bulb interior.

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And by doing so,

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now what we're doing is we're pumping both We're pumping the neon bulb

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from both directions through both of its terminals in opposite directions

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towards its interior with a current surge that meets in

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the middle like two bulls butting heads, two rams butting heads.

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And that's why the voltage is zero, difference between the two terminals of the neon bulb.

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Yet, we've got current there of 65 amps.

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Because L3 is holding on to that and sharing it in parallel with the neon bulb.

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And so its internal voltage is able to rise because it has a capacitor in there

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which is a functional capacitor.

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That's why they don't sell capacitors less than 1 picofarad

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because they're non-functional below 1 picofarad.

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But the capacitor inside the sub-file of the neon bulb Is functional, minimally functional.

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It can't be maximal because then it will be able to dump readily.

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So it has some retentive ability,

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unlike the 100 femtofarad capacitors on either side, outside the neon bulb.

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Inside, it can retain it, and it does.

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And that's why...

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I arranged the circuit by trial and error.

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I learned how to do it because the two switches that are opposed

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to each other inside the neon bulb have to occur at the same time in order

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to facilitate this headbutting of the two rams,

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the two (symbolic) bulls in the interior of the neon bulb.

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And, of course,

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the resistance that's in there helps to differentiate the internal voltage so

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that it can rise.

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So it helps the capacitor do its job.

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Remember,

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this is an electrostatic linear lineup between the loop,

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the L1 coil loop on the right and the ground on the left.

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And the L3 and the L5 coil are facilitating this,

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but this is altogether a linear relationship between ground on the left and

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the L1 coil on the right.

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And so that's why it's a capacitive situation going on along with a resistor

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to create this, to help this head-butting situation.

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So we literally have a tank circuit because we have the L3 coil on the outside and

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then we have a (macro) capacitor,

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a minimally functional capacitor and (macro) resistor on the inside (of the neon bulb) creating a tank circuit.

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But it's not your normal tank activity.

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It's not oscillating in any way other than a standing wave.

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And engineers, you know, like to argue with me about this.

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But we have evidence that it is a standing wave because the voltage

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at either terminal of the neon bulb is equal for all intents and purposes.

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It's zero difference.

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That means it's a standing wave because we have current.

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When we have current and zero voltage difference, we have a standing wave.

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And that's when we get that we have a standing wave because that's the net result.

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It's not the cause.

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It's the effect.

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The cause are two waves (of current) happening simultaneously in opposite directions,

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butting heads within the neon bulb and within the L3 coil and within the L1 coil.

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In all three locations simultaneously, it's all happening at the same time.

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Only in the L5 coil do we actually get a directional flow of

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a singular dominant vector so that it can retain,

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it can remember what's happening and magnetically build up a memory while

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the capacitor inside the neon bulb Is building up the retention of an electrostatic

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or a dielectric memory to counterpoise the magnetic memory that's happening in

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the L5 coil.

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So does any of this make sense?

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I hope so.