Capacitive Sink Hole

[dragon]

Well...What do you gain out of having impulses of immense size traveling through a bank of resistive light bulbs? Do resistive loads really dissipate energy by conversion? or are we looking at things a bit askew. What exactly is the mechanism transforming amperage to heat and light?

If resistances dissipate (I hold that they do not) then as the pulse enters the resistive load, it will not come back out. It will be converted into light and heat, and the pulse cannot possibly exit the resistance, since it was all converted..This to me does not seem plausible.

Also, this entire circuit is at any one point unbalanced! this is a delay line, not a circuit in the traditional sense. There will never be equal amperage between any two points. And as to the ease, If you know about electrical length, wavelength, resonant frequency, and delay lines it should not be hard, especially since the circuit should be symmetrical.

I find this all quite interesting, some of the experiments I've put together would suggest your correct that resistances don't dissipate, at least not in the conventional sense. While studying alternator design I found that ohms law depicts a circuit that is 50% efficient which would actually make that statement quite obvious to most that it simply couldn't be true.... But... your talking about a single pass electrical circuit where energy flows in, throught the circuit and out again. If the circuit is energized and recirculated you get multiple passes of the same energy. If it flows through the same resistance then ohms law becomes somewhat questionable. The heat/light is the same yet the energy input is considerably lower.

I guess the way I envision the process would be you put 1 amp in and 1 amp is returned the only difference is the potential is altered. Since it's easier to manipulate voltage within a system it would make sense to replenish voltage and use that same amp for something else, recycle it as it were...

I don't claim to have the answer, I'm only going on some of the oddities I've seen during experimentation and unanswered questions arising from the same.

Your drawing sure does look alot like my ring cirucit..... mine has 48 diodes (LED's) in it though... very interesting indeed...
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[Armagdn03]

I guess the way I envision the process would be you put 1 amp in and 1 amp is returned the only difference is the potential is altered.

May I alter your statement???

"1 WATT in and one WATT out, only difference is the ratio between Potential and Kinetic energy."

[dragon]

May I alter your statement???

"1 WATT in and one WATT out, only difference is the ratio between Potential and Kinetic energy."

Ok, I can accept that... Ratio being the key. So we need to create an electronic equivilant of a tsunami or rogue wave. Then, devise a way to extract the peaks and leave the rest alone as to not disturb the original "swing".
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[Harvey]

The empircal evidence would indicate that the current is never consumed but instead it is the voltage that is consumed during dissipation.

Therefore, we can expect that once the Electrical Soliton leaves the resistance, the amplitude will be diminished even though the current is conserved. However, where it passes through an impedance that has no resistive properties, then neither voltage nor current is consumed and the action is considered a transformation of apparent power.

So let us suppose that our impedance here is an inductor. And let us suppose that the inductive reactance is such that it is seen as an impedance to the electrical soliton wave propagating through the conductor or wave guide.

When the wave encounters the impedance, some of the energy will respond as a reflected wave while some is used to charge the magnetic field of the inductor. The energy leaving the inductor will be reduced by the reflected amount. Upon reaching he diode, supposing the timing and phasing are properly accounted for, the reflected wave will be reflected again, this time off the diode and the full circle energy will be added to it to recreate the original wave. At this time a new pulse is added in and the original wave is bolstered and the cycle repeats until the amplitude reaches a point that causes destructive failure in the components or the injection site is unable to add more energy.


So the question arises as to whether current is reflected, voltage or both. Is the Soliton simply a moving wave of potential energy, static voltage propagating across the conductor? Or could there be bidirectional current (AC) flow occurring between the injection point and the impedance while a pulsed unidirectional flow (DC) occurs between the impedance and the diode?

Antenna systems tell us that it is the latter and that real power can be dissipated in such a system being converted to RF EM radiation. When a wave is reflected in a receiving antenna it certainly can produce a gain when properly phased. But when things are allowed to collide we get a standing wave.


Can current be transformed into voltage using a transformer and then be applied to dissipate the energy

For example, what if all the current running through the mains in your house was pushed through a step up transformer of say 100A x 1.2V primary to a 1A 120v secondary ? Would the meter detect the 120W extraction? Or would the extra 1.2V drop across the primary simply reduce the power elsewhere in the circuit? It would seem that while the voltage is consumed across a given load, it really is simply one form of energy being traded for another and in this case it is electrical power for whatever the load converting it to. We simply don't have a simple means (that I am aware of) that can take 100A of current and a zero volt drop and produce an output. But if that primary, was a superconductor . . .

Sorry for the digression.

[baroutologos]

http://i210.photobucket.com/albums/b...3/IMG_4949.jpg

3) We can replace the spherical top load with one side of a parallel plate capacitor. The plate connected to the Tesla transformer with rectifier will induce an equal and opposite charge onto the second plate. But what will this induced charge look like? In the “zoomed” in section, we can see that there will be a charge distribution within the second plate. This does not equalize within itself because it is held in place by the first plate. Equally the first plate and second plate cannot equalize.

hello there,

I found the whole concept rather interesting and a bit amusing. By the way, according my experience and understanding that cannot work as stated.

A capacitative sinkhole, maybe the holy grail of free enrgy. Kapanadze device from a solo ground sucks and pushes electrons in/out of earth hence achieving (assumed its truth) vast amounts of energy. Tesla envisioned that in his article about the problem of increasing human energy needs.

So it is a noble goal to pursue. Regarding your design, armagedon, assuming that somehow you managed your spark-gap to work as a diode (i have seen that in a video - magnetic diode!) you have the problem of charging the one capacitor plate to a - and the top plate to +.
This does not happen. I have seen that if current cannot circulate, the whole capacitor plates behave as a single terminal capacitance or a "sphere".

So if the capacitance plates have say, 300pf capacity, by not connecting the other to a normal ground, you will end up with a 10pf (eg) capacity for both bearing the same potential towards ground. (both + or -)

Tesla suggested some ways in Colorado Spring notes for boosting one-terminal capacitance (pp 88-89) but could not so far to replicate them neither i am sure they are supposed to work as said there.

I suggest before this thread goes 100 pages on theorizing few decisive experiment has to be conducted.


ps: Regarding the other thread concerning the eccentric transformer theory, i think i have informed you that have found it false. Clever thinking though

[Armagdn03]

hello there,



So if the capacitance plates have say, 300pf capacity, by not connecting the other to a normal ground, you will end up with a 10pf (eg) capacity for both bearing the same potential towards ground. (both + or -)

You are correct about this, however I NEVER stated that you would calculate for the full 300pf of the capacitor (example) I believe this may have been a mistaken inference by some readers, and I should have been more careful, however in totality, we are only dealing with free space capacitance through this idea. The idea of the grounding, or the ariel is not to magically charge a capacitor to its capacity as an energy gain mechanism, rather a method of allowing extraction of free electrons.

What is the most important concept with this thread however is the delay line ring circuit described more on page 2, operating with single solitons rather than wave trains.

[Armagdn03]

Because there is a distinct lack of Microwave Engineers on this forum, I decided to elicit the help of a microwave engineering forum

Here was my post and the best reply.

I had a question which is perfectly suited for this forum, and all you brilliant engineers out there who are better schooled than I!

I want to design a Ring circuit. The ring will be a resonant wave guide much like described by Tischer, F.J. "Resonance properties of ring circuits"

In this particular ring circuit, there is a resonant ring waveguide, with a directional coupler attached to introduce the signal. The directional coupler has a quarter wavelength spacing between the slits, creating a quarter wavelength interference pattern within the ring. The purpose of this is to double the wave propagating in one direction, and eliminate it propagating in the other direction. This way you have a unidirectional flow of energy around the ring circuit which will continually build on itself creating a sort of power multiplying device like a condenser.

Here is where I want to differ, and what also brings me to my question.

Say I have a ring circuit, with a directional device (similar to diode) inserted into it. Now I want to send a soliton around the ring. (a soliton is a solitary wave traveling in a restricted path, similar to the water hammer effect in plumbing)

YouTube - Collision de solitons hydrodynamiques (example of soliton)

now, without losses, the soliton should travel trapped in the ring circuit round and round.

But what happens if we stick a purely resistive impedance into its path???

The soliton will begin traversing the ring circuit down the transmission line with its characteristic impedance, then hit the resistive impedance. I understand that it will partially enter the impedance, and partially be reflected, depending the parameters.

This applet describes what happens to an alternating current hiting an impedance in a very similar way.

BesserNet Reflectometer

If you set the reflection coefficient to 0.5, half the wave is admitted through the impedance, and half is reflected. Now I am not trying to do a continuous train of waves, rather a single shot down the transmission line in one direction, so not exactly the same but similar.

In this applet, it looks like nothing is "lost" to the impedance as the wave travels through. This would lead me to believe that it is not resistive. If however the impedance is purely resistive, what will happen to the soliton???

One person has suggested that the amplitude (voltage) of the wave (soliton) will decrease but current will remain constant. The voltage will be "consumed" by the impedance, and less energy will exit the impedance than entered it.

I would assume if this is correct, and say we still have our reflection coefficient set to 0.5, then half the soliton will be reflected by the resistive impedance, half will traverse the impedance, and will exit with less total energy due to the resistive impedance converting some of the energy through joule heating. so energy reflected + joule heating + energy exiting the resistive impedance = total energy we started with.

does this sound like a reasonable take on what is happening???

Or is there a Bernoulli like effect happening in the impedance. I could see the energy entering the impedance undergoing a change where the ohmic resistance is the fulcrum on which the ratio of amperage to voltage pivots. a higher impedance, admits less (higher reflection coefficient) but what is admitted undergoes a change following the conservation of energy just like in Bernoulli's principle. Where the energy per unit volume before the impedance is equal to the energy per unit volume through the impedance, but the ratio of the voltage to amperage changes.

So these are the two plausible explanations, of which I do not know which is correct!

I hope someone out there has an enlightened point of view!!!

That was my post, here is the best answer...

Quote:
so energy reflected + joule heating + energy exiting the resistive impedance = total energy we started with.


This part is true. There is one error though previously in your explanation. The reflection coefficient is a voltage reflection coefficient and power and energy go as V^2. In order to reflect and transmit half the power the reflection coefficient needs to be 0.707 [or sqrt(0.5)].

The wave impedance by definition defines the ratio of voltage to current (Z=V/I) so that there is no "fulcrum" effect. The voltage and current take an equal hit.

In order for the resistive portion to reflect and transmit equal power the resistive portion would have to be very thin. (Although I suppose you could adjust the reflective part lower to account for the transmission loss through the resistive media.)

I don't understand why you want to add resistance when you can do the same thing in a lossless sense using L's and C's (meaning strictly imaginary impedances).

According to this individual, the current does not stay constant as Harvey has suggested. The reality of it is that both current and voltage will take a hit from the joule heating. This still seems fuzzy to me, and I will have to think on it.

I did not understand fully the following and have asked the engineer to elaborate....

"In order for the resistive portion to reflect and transmit equal power the resistive portion would have to be very thin. (Although I suppose you could adjust the reflective part lower to account for the transmission loss through the resistive media.)"

Will keep you all updated.

[Armagdn03]

here are a few of the helpful repplies I have gotten from the Microwaves101 forum.

It has to do with skin depth which is explained very well on this site. Essentially a resistive medium acts like a poor conductor. After propagating through a resistive region the power in the wave is reduced by 8.7 dB after only one skin depth.

So suppose the resistive region is one skin depth thick. The power making it through is -8.7 dB of the incident power (neglecting the reflection at the boundary leaving the region for simplicity but it should be accounted for). For equal power in the transmitted and reflected wave the return loss must be 8.7 dB which corresponds to a voltage relfection coefficient of 0.37 (way below the original 0.707). The power loss in the transmitted signal is from joule heating.

Quote:
I have gone over many many pages of information trying to find answers to these questions, however without the guidance of someone in the know it can be an exhaustive chase!


You might find some insightful information at: Emerson & Cuming Microwave Products

They make materials (both dielectric and magnetic) made to absorb RF and microwave energy. There are many good technical documents there.

and here is another good one

Hi armagdn03,
What you are describing is a Traveling Wave Ring Resonator.
You have added a few unnecessary parts but that is what you want.

The circuit was developed during WW2 in order to test high power components when the real power source was not available.
Power multiplications of 50 were practical.

My reference is
Microwave Filters, Impedance-Matching networks, and Coupling structures

G. Matthaei
L. Young
E.M.T. Jones

Jack K

Thank you microwaves 101 for this helpful information. If you have never visited their site, it is one of the best internet resources i have ever come across!

[Harvey]

When the engineer suggests that the current will take a hit, he is referring to the entire circuit. Adding a resistor in your circuit reduces the current for the entire circuit.

I apologize for not being more clear with my post. My intention was to stress that the current leading into , inside and trailing out of a resistor is the same. I did not mean to give the impression that adding a resistor into a circuit would not change the current in that circuit. Sorry for being unclear on that

As regards the engineer discussing "Imaginary Impedance", his phraseology can be very misleading. Inductors and Capacitors place very "Real" impedance on a circuit in the sense that your source and output stages will be affected by that impedance. However, in the equations (see Electrical impedance - Wikipedia, the free encyclopedia) you will find the variable 'j' . This is the square root of -1, and is considered an imaginary number and those impedance equations that use it, typically refer to the 'real" part as resistive and the "imaginary" part as reactive. So while the engineer is quite correct in his use of the term, it requires a bit of background to separate what the average person calls real and imaginary from what engineers call real and imaginary

The end result of this, I think, is that we get back to the two transformers back to back as I showed before - that is for electrical purposes where the Soliton is a voltage wave on the wire. That will give a Bernoulli effect.

But to just add an impedance in the line, a simple inductor would suffice - no Bernoulli effect, but it will reflect and pass at the same time with very little resistive losses.

[Hope]

Thank you for pointing me to this discussion topic! The quarter length quark is interesting. Has anyone started using flat coils and plate collectors,
ALU/BIS/ALU(Coils)/BIS/ALU(Coils)/BIS/ALU i think those plates only filter what is collected, since they collect some many bands of energy they MUST it would seem also then FILTER what is not helpful out or at least what is useful taken out.

[Hope]

Agreed with this theory!

Thanks again for this link to your discussion. It seems very plausible that the energy gained is as you say. You have my attention.

[Armagdn03]

@ hope,

May I ask which thread linked you to this one?

[Armagdn03]

When the engineer suggests that the current will take a hit, he is referring to the entire circuit. Adding a resistor in your circuit reduces the current for the entire circuit.

I apologize for not being more clear with my post. My intention was to stress that the current leading into , inside and trailing out of a resistor is the same. I did not mean to give the impression that adding a resistor into a circuit would not change the current in that circuit. Sorry for being unclear on that

As regards the engineer discussing "Imaginary Impedance", his phraseology can be very misleading. Inductors and Capacitors place very "Real" impedance on a circuit in the sense that your source and output stages will be affected by that impedance. However, in the equations (see Electrical impedance - Wikipedia, the free encyclopedia) you will find the variable 'j' . This is the square root of -1, and is considered an imaginary number and those impedance equations that use it, typically refer to the 'real" part as resistive and the "imaginary" part as reactive. So while the engineer is quite correct in his use of the term, it requires a bit of background to separate what the average person calls real and imaginary from what engineers call real and imaginary

The end result of this, I think, is that we get back to the two transformers back to back as I showed before - that is for electrical purposes where the Soliton is a voltage wave on the wire. That will give a Bernoulli effect.

But to just add an impedance in the line, a simple inductor would suffice - no Bernoulli effect, but it will reflect and pass at the same time with very little resistive losses.

Thanks Harvy,

I am sure many here knew what he was talking about when he said "replace with "Cs and Ls" but for those following along, this is still a good refresher since impedance can be bewildering to those simply familiar with ohms law and unfamiliar with imaginary numbers.

as to what has been discussed, In order to have a better grasp of this, i dont want to think in terms of lowering the amperage through the whole circuit because no two points in the circuit are going to be the same as current will not be uniform throughout. To say that it will lower it through the whole circuit, is averaging things out over a period of time, but I want to understand the instantaneous.

There are two types of solitons to consider...

One is the "water hammer" analogy and potential wave, and the other is the more akin to the you tube video of the water wave in a restricted channel.



With regards to the later....What is interesting is that there will not be an AC component to the wave being propagated, at least not in the current reversal sense. The soliton is created by releasing a tension, and allowing no reversal. Notice the water wave guide analogy shown in the previous youtube video. A standing column of water is released, which causes a raised potential AND current. This is different from the water pipe analogy causing water hammer, which is a compression wave.

so, perhaps a pertinent experiment (though extremely difficult to setup and quantify) Is to have a capacitive element with energy (X) discharge through an inductive element, with a resistive element inserted at the end. Then have another capacitive element at the other end. This is essentially a one way street. Discharge X joules, through the inductor, through the impedance, to the second capacitive element. When the energy is contained within the inductive element, there should be no charge on either capacitive element. when the inductor discharges, it should discharge into the second capacitive element. Does the action of passing the energy through the resistive element cause less energy to be stored on the second capacitive element than if there were no resistive element present?

[Armagdn03]

another interesting application to this is...Right hand circuitry and Antigravity.


If we consider the B field in this to be traveling around the circumference of the ring, then the A field will be perpendicular to that. (imagine a doughnut, the B field goes around the doughnut, the A field (potential field) goes through the hole). Since there is not a current reversal with respect to local zero (ground, local equi-potential) the A field does Not reverse like a normal resonant solenoid or any other resonant structure where there is a current reversal. This means that you have a resonantly created standing A field potential, similar to an ionic lifter, however created in a much much more creative, and efficient way.

[dragon]

another interesting application to this is...Right hand circuitry and Antigravity.


If we consider the B field in this to be traveling around the circumference of the ring, then the A field will be perpendicular to that. (imagine a doughnut, the B field goes around the doughnut, the A field (potential field) goes through the hole). Since there is not a current reversal with respect to local zero (ground, local equi-potential) the A field does Not reverse like a normal resonant solenoid or any other resonant structure where there is a current reversal. This means that you have a resonantly created standing A field potential, similar to an ionic lifter, however created in a much much more creative, and efficient way.

I've played with the A field somewhat in several experiments and it's potential is pretty amazing. Here is a good site that explains it a little more... SCIENCE HOBBYIST: Right Angle Circuitry .

I've noticed the amperage in a connecting heavy copper link through the "hole" of a toroid is quite high and that it seems to have a polarity, at least measuring it with conventional meters. Quite interesting indeed !
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