Bedini/Cole Bipolar switch (or how to drastically reduce your input!)

[Tehnoman]

Not through! That is the catch. I said exactly what I said, but - I didn't say that these were the same electrons. But battery doesn't care, whether these are same electrons or are not. The cap gets his energy and potential (voltage) only with help of charged particles (one side - electrons, other side - excess of electrons).

I edited your picture to ilustrate my thoughts.

[elias]

I haven't seen that paper, could you please upload it somewhere or at least advise on it's exact nomenclature number so that I can find it? Thx!

I can send them for you next week, because I have left them on my other computer. You'd better provide your email address, for this purpose.

These are two links, but you need to subscribe in order to download the papers:

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Elias

[elias]

It could be, as far as I can understand from information gathered here. But it is strange - RC circuits are quite well investigated, yet no such effects was detected with conventional voltmeter at cap side. Maybe this is more complicated phenomena than usual RC circuit stuff. It could depend on R (resistance) value - when observing RC circuit, we are using rather big R values, if compared with plain Cu-wire resistance, so we can observe cap charging process with ampermeter and voltmeter (commonly, osciliscope). And it could be, that in directly unobservable case (resistance is only wire) these effects start to show up.

Tehnoman

Some phenomenon is not so explainable by conventional means, such as where does half of the energy go IF your charge an Ideal capacitor WITHOUT any resistance (R=0)? I mean the energy of the Capacitor is always 0.5CV^2
and half of the energy is supposedly wasted in R, no matter what size R is, but ideally if you assume R = 0 then where does half of the energy go?

I will post one of the papers next week and it can be easily observed that IF Nothing gets wasted in R, then we have charged our capacitor WITHOUT consuming any energy, and the potential is simply TRANSFERRED to the capacitor. I have concluded that Electrical Current has nothing to do with charging a capacitor, and it is only the way we charge the capacitor is what makes us waste some heat.
The problem we have is to TRANSFER the potential on the battery to the capacitor WITHOUT drawing any current. This paper provides an experimental way to achieve this BUT I am sure that there must be better ways to block electron current. We have to change the way we look at such things significantly to be able to SEE better ways.

Elias

[elias]

Superconductivity

For example what happens if you charge a capacitor by using a superconductor?! There is no R to waste energy. Then the capacitor must be able to charge instantaneously without any energy loss.

[Tehnoman]

Tehnoman

Some phenomenon is not so explainable by conventional means, such as where does half of the energy go IF your charge an Ideal capacitor WITHOUT any resistance (R=0)? I mean the energy of the Capacitor is always 0.5CV^2
and half of the energy is supposedly wasted in R, no matter what size R is, but ideally if you assume R = 0 then where does half of the energy go?

As far as I remember, capacitor gets 0.5CV^2, and over resistance in heat is lost the same amount, that I was told at school. And from battery we got energy value of CV^2. If we assume that R=0, then majority heat is lost in power supply, unfortunately we don't have ideal power supplies or batteries with zero internal resistance. Not sure, whether it is the same 0.5CV^2, but that could be found out experimentally.

I will post one of the papers next week and it can be easily observed that IF Nothing gets wasted in R, then we have charged our capacitor WITHOUT consuming any energy, and the potential is simply TRANSFERRED to the capacitor. I have concluded that Electrical Current has nothing to do with charging a capacitor, and it is only the way we charge the capacitor is what makes us waste some heat.
The problem we have is to TRANSFER the potential on the battery to the capacitor WITHOUT drawing any current. This paper provides an experimental way to achieve this BUT I am sure that there must be better ways to block electron current. We have to change the way we look at such things significantly to be able to SEE better ways.

Elias

I'm opened in new ways and ideas. No offense. But a little criticism hasn't killed anyone. It could be, that there is something that is not explained by traditional science, yet people has been charging caps for long time, but I haven't heard a single statement, that they could somehow conserve or get energy from ZPF/environment. So I am little bit cautious about statements that is directly opposing observations made by engineers and scientists for centuries. Really looking forward to read one of these papers about this phenomena and experiments, that would prove it - and so understand were classic physics has gone wrong! Haven't made any tests myself, as I said, but will start doing them soon. And I guess that cap charging will be one of the first.

For example what happens if you charge a capacitor by using a superconductor?! There is no R to waste energy. Then the capacitor must be able to charge instantaneously without any energy loss.

Again, power supply/battery. Basically, at the initial cap charging point it is short circuit - empty cap has practically zero DC resistance, and so are superconductor. The current draw would be as high as can be supplied.

One more situation worth exploring would be charging cap from other cap. This would eliminate the heating in power supply/battery, because as far as I know, cap hasn't any significant internal discharge resistance, has it?

[lighty]

@Elias

I asked a friend who asked a friend so I got the papers regarding capacitor charging. Interesting read indeed.

[Maxwell]

cap charging energy

Let's not confuse the theory on paper with what happens in reality. We can draw a circuit with no resistance in it, but we can never actually construct it.

It is true that any power supply or battery will have internal resistance, to which we lose energy when current flows (if it gets warm, energy is lost). Any capacitor will have internal resistance and inductance (just google capacitor ESR or ESL) to which energy is lost.

On paper, when we apply a voltage source to a resistor and capacitor we will get the familiar RC time constant as the cap charges. The source will lose CV^2 joules of energy, the cap will end up with 0.5CV^2, and the other half lost to the resistor. The quirk, as mentioned be Elias, is that when we throw out the resistor (limit as R->0) we still lose half the energy but there is no explanation for where it went. The same thing happens when we charge a cap with another cap. Consider this quote from "Electrical Engineering: Principles and Applications" 2nd edition by Allan Hambley (pg 124):

"Thus, we see that the stored energy after the switch is closed is half of the value before the switch is closed. What happened to the missing energy? Usually, the answer to this question is that it is absorbed in the parasitic resistances. It is impossible to construct capacitors that do not have some parasitic effects. Even if we use superconductors for the wires and capacitor plates, there would be parasitic inductance. If we included the parasitic inductance in the circuit model, we would not have missing energy. To put it another way, a physical circuit that is modeled exactly by [a cap charging another cap] does not exist. Invariably, if we use a realistic model for an actual circuit, we can account for all of the energy."

So the conventional explanation for this quirk is that the theory is made for physical situations. Pose a situation that is unphysical, and we will get weird results. I think a better explanation is that it is a failure of lumped circuit analysis; if we construct the situation and analyze it with Maxwell's equations directly, I bet this quirk will disappear.

[Aaron]

what charges the cap?

IF electron current is needed to charge a cap...even if it isn't needed, no matter...lets just say a cap is charged up.

If we let it sit around...the voltage will leak down to a lower level....the more leaky the cap the faster the voltage drops on its own.

If current is supposed to charge the cap, does that mean current leaves the cap when it goes down by itself and if so, how is current or the electrons simply draining out of it?

In my opinion, the cap is leaking gas. The aetheric gas that makes up the heaviside flow made of the photon potential or voltage gas. A cap to me is like an aerosol can. You can pressurize it to its level and it will leak out down to its intended capacity. If it is short circuited...it goes to zero voltage...then there is a little vacuum that sucks the radiant potential back inside and the voltage goes back up a little.

A cap can both charge up itself open looped and go down by itself in open loop. Neither involve closed loop electron current.

[Aaron]

Eric Dollard Notes on Dielectricity & Capacitance

Read pages 26-29 of this document:
Eric Dollard Notes on Dielectricity and Capacitance.
Eric Dollard Notes (1986--1991)

"The misconception that capacitance is the accumulation of electrons has seriously distorted our view of dielectric phenomena."

Eric is probably considered the leading Tesla expert because he actually duplicated more of it than anyone else...from my understanding.

Therefore, his opinion is valuable in my opinion.

[gmeat]

Read pages 26-29 of this document:
Eric Dollard Notes on Dielectricity and Capacitance.
Eric Dollard Notes (1986--1991)

"The misconception that capacitance is the accumulation of electrons has seriously distorted our view of dielectric phenomena."

Eric is probably considered the leading Tesla expert because he actually duplicated more of it than anyone else...from my understanding.

Therefore, his opinion is valuable in my opinion.

Hi Aaron,

Thx for posting that link to some of Eric's writings.That was a very interesting read .


-Gary

[vzon17]

Peak switch for pulse charging from caps to bats or coils

I had this idea I am going to experiment with thought I would let you guys in on it. I was trying to think of a simple peak switch that would turn on when the voltage rises past a certain point. Well they make those and are commonly available and used as surge suppressors. the part is carlled a varistor. or MOS varistor. and what they do is conduct current when they reach the clamping voltage. Usually they are only hooked up to suppress surges and spikes in the power line. So I thought why not hokk one in searies with a capacitor so that when it reaches a cetain voltage it can discharge into a battery or a coil. it would save on having to make the bedini wheel with the pully and the cam with a switch to fire the cap into the battery. just hook a varistor in series with the cap and then when it gets up to the selected voltage it fires into the battery. you can get them with various clamping voltages and they have a super fast respons time. It may also be good for firing a coil with a fast rise time. I am just thinking at thins point and have not obtained one to test it.
Here is some information.

varistor: Definition and Much More from Answers.com

[elias]

IF electron current is needed to charge a cap...even if it isn't needed, no matter...lets just say a cap is charged up.

If we let it sit around...the voltage will leak down to a lower level....the more leaky the cap the faster the voltage drops on its own.

If current is supposed to charge the cap, does that mean current leaves the cap when it goes down by itself and if so, how is current or the electrons simply draining out of it?

In my opinion, the cap is leaking gas. The aetheric gas that makes up the heaviside flow made of the photon potential or voltage gas. A cap to me is like an aerosol can. You can pressurize it to its level and it will leak out down to its intended capacity. If it is short circuited...it goes to zero voltage...then there is a little vacuum that sucks the radiant potential back inside and the voltage goes back up a little.

A cap can both charge up itself open looped and go down by itself in open loop. Neither involve closed loop electron current.

This is very interesting of course, I have observed a capacitor of 330uF jump up about 15volts, when suddenly discharged from 200V.

[elias]

Hi

The second paper "Ideal Capacitance Circuits and Energy Conservation" has an interesting conclusion:

To state our conclusion in a general manner, so long as the
voltage across a capacitor is a continuous and differentiable
function of time, the energy delivery to the capacitor is complete
without the missing energy. The idealized circuits in
Fig. 1~a! and ~b! happen to be exceptions.

So, it is assumed that one cannot abruptly change the voltage across a capacitor, and if one does, no explanation is for it through conventional means.
So This is EXACTLY what we are trying to do with Bedini circuits isn't it?
Charging and discharging capacitors and batteries as abruptly as we can. (Sharp Gradients). This was also what gray was doing. And of course Tesla, which found out that abruptly charging and discharging not only doesn't fit the standard theories of energy loss, but it even results in energy gain.

And Maxwell, thanks for taking time to explaining this stuff it was really informative, but what if we could build these circuits and capacitors by using superconductor material, which is an ideal conductor with R = 0.

Elias

[Aaron]

Impulse

We want abrupt discharges...this is taking advantage of time energy. Sharp gradients violate conventional thermodynamics. This is IMPULSE technology. Read SECRETS OF COLD WAR TECHNOLOGY chapter 1 on Tesla by Gerry Vassilatos.

[Aaron]

1 Wire Capacitor Charging

I have a question, if I put from any type of circuit only a positive lead to a capacitor...can I make it charge without connecting the ground? I'm talking for example a 33,000uf 60v capacitor.

Will it charge if I connect only 1 lead?

According to conventional theory, will it charge? If so, why, if not why?

According to non-conventional theory, will it charge? If so, why, if not why?

I would appreciate any input.