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Author Topic: what happens between a teslacoil the battery the incadescent bulb and the ground  (Read 13029 times)

Offline tinman

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For those that want to know how a transmission line works,and how power is delivered to a load via 1 wire.
The second conductor(wire) in the case of experiments carried out here in this thread,and else where,is the ground(ground plane) it self.

https://www.youtube.com/watch?v=ozeYaikI11g


Brad

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Offline TeslaScientific

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Because if current could be delivered to a load via one single wire,then that current should be able to be delivered to that load,regardless of where that load is placed along that single wire.

But your statement above say's  that one wire cannot light the bulb unless--Quote:When you put a terminating capacitance AFTER the filament, now the energy flows THROUGH the filament, and it lights.

That is correct. There needs to be some form of plate after the bulb in order for it to light--that is the conductor acting as one capacitor plate.

The fact that you need anything after the bulb,in order for the bulb to light,is proof beyond doubt that it is capacitive coupling.

Nature it self shows us this with lightning.
One of the plates is the ground,and the other is the clouds. When the potential is high enough,the discharge arc takes place-over fare greater distances than you can demonstrate with your systems.

So once again,you need some form of plate or antenna !after! the bulb,in order for the bulb to light,and so,the bulb is being lit due to capacitive coupling,and not by current flowing back and forth along a single conductor.

Whether you want to believe it or not,in order for current to flow,there must be a path for it to follow.

Please read,so as you may understand what is happening,and stray from giving wrong information.
I have highlighted the part you are missing.

https://en.wikipedia.org/wiki/Single-wire_transmission_line

At the end of the 19th century, Nikola Tesla demonstrated that by using an electrical network tuned to resonance it was possible to transmit electric power using only a single conductor, with no need for a return wire. This was spoken of as the "transmission of electrical energy through one wire without return".[1][2]

In 1891, 1892, and 1893 demonstration lectures with electrical oscillators before the AIEE at Columbia College, N.Y.C., the IEE, London, the Franklin Institute, Philadelphia, and National Electric Light Association, St. Louis, it was shown that electric motors and single-terminal incandescent lamps can be operated through a single conductor without a return wire. Although apparently lacking a complete circuit, such a topology effectively obtains a return circuit by virtue of the load's self-capacitance and parasitic capacitance

So once again--the bulb lights because of the capacitive coupling.


Brad

Capacitive coupling is the coupling of energy through a capacitance. The bulb is connected by a conductor, not a capacitance.

It makes all the difference where the load is connected along the wire, because you are dealing with high frequencies. You can connect the bulb back to a different point along the same wire in a short circuit and the bulb will still light, because there's a potential difference across the filament, and there isn't a short circuit there at all at those frequencies. That's all it needs. Tesla demonstrated that experiment in public to illustrate the existence of standing waves along the length of a conductor, corresponding to the frequency.

How is one side of a bulb terminal considered a "load"? Why should it light? What is the current supposed to do in this situation? A filament lights because the energy flows through it, so when the filament is made to be a part of the transmission line rather than the end of it, energy flows through it, and it lights. The same energy is flowing, or rather oscillating, along the entire length of the wire, but the wire isn't a bulb filament so you can't see it. Again, put a variable capacitor across the bulb and you will be able to control the brightness through adjusting the capacitance. So again, the energy is not received through capacitive coupling nor does it close any circuit back to the source. The capacitance merely facilitates a sufficient difference in potential across the filament as an isolated part of the system. The bigger the capacitance, the bigger the potential difference, the brighter the bulb. That's all there is to it.

Unfortunately Wikipedia doesn't know any more than the best expert does. And clearly none of the editors have ever bothered to repeat Tesla's experiments because they are quite happy to apply their own explanations which explain nothing. What you say is valid within the bounds of what it's applicable to, but it's most certainly not applicable in this situation. It's a capacitance to SPACE, not back to the power source. Also in the articles the Wikipedia page references, Tesla is speaking of his own design of single terminal bulbs consisting of a single piece of carbon, not a filament which power must flow through.

Quote from: Nikola Tesla
we may consider the inside surface of the bulb as one coating of a condenser, the air and other objects surrounding the bulb forming the other coating.

http://www.tfcbooks.com/tesla/1891-05-20.htm

This is the part that you are missing:

http://www.youtube.com/watch?v=DovunOxlY1k

Offline TeslaScientific

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For those that want to know how a transmission line works,and how power is delivered to a load via 1 wire.
The second conductor(wire) in the case of experiments carried out here in this thread,and else where,is the ground(ground plane) it self.

https://www.youtube.com/watch?v=ozeYaikI11g


Brad

The "ground plane" is the ONLY conductor. That is the wireless = single wire transmission system. Forget the air.

Also, it's quite easy to measure faster than light propagation, it's all about which path the energy takes. It's right there in Colorado Springs Notes. Take the extra coil conductor length and resonant frequency, and see what you come up with...

Offline Magluvin

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Ive always thought of it as electrons being compressed onto one end and depleted at the other end, capacitance in the air or not. Like say if we have a 12v car battery, I would bet that there are an excess of electrons in the outer exposed - lead terminal and a depletion of electrons in the + lead terminal, meaning the lead terminal mass.  If we have a high voltage, high freq nst with an open secondary, I believe that the electrons in the sec are being compressed and decompressed from one end of the winding to the other at the freq of operation. With that, if I connect 1 lead of the sec to ground, and the swing of compression is in the direction of the ground, then those compressed electrons have a place to go and not be only compressed in the end of the winding, whether there be capacitance in the air or not. I will say that the capacitance in the air can help release more from the winding to the ground, but I think that even without or in the absence of the air capacitance that the compression alone in the open circuit could cause currents to flow through the bulb.  In a way, we could look at the earth as a huge reservoir that can accept and give electrons easily as with what we are doing wouldnt put that much of a charge to affect the reservoir charge in any significant way that the exchange would be impeded much. And when we hold the light bulb at the base and 1 wire from the coil on the other bulb terminal, our body acts as a similar reservoir, only not so large, but large enough to take and give electrons, especially at high freq.

The capacitance can help more electron transfer as shown in the vid below, but I do believe that it is not absolutely necessary at high freq hv potential.

Like I keep hearing that the capacitance of a bifi is so insignificant and only parasitic, yet when it comes to these things, the minute capacitance involved seem to be a big part of the explanation for the 1 wire deal.  Capacitance will always be at play. But like I said, I really think there is compression and decompression within the sec of the tc, and even the tiny nst that drives it, of which I use it in the vid to charge the caps via AV plug in the vid.

But, we cant really eliminate the cap effects to really test it all.  But I dont find it hard to believe that if we attach 1 end of the hf hv sec to a chunk of copper that the copper chunk wouldnt give and take electrons being compressed and decompressed on that end of the sec.

And the pics below show the diff between the bulb lighting 1 wire to the bottom and then the top of the sec of the Lil Tc. I am the reservoir.

https://www.youtube.com/watch?v=TR3WmK3qrws



Mags

Offline TeslaScientific

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There are a few things that you may or may not know so I'll say them anyway:

It's a natural requirement/condition of 1/4 wave resonance that the current lags the voltage by 90 degrees = 1/4 cycle of the wave = the length of the resonator. Due to the reflections that occur, a standing wave forms, and the current and potential are distributed at opposite ends of the coil. Maximum current is at the ground end, and maximum potential is at the top/free end, 90 degrees or 1/4 wavelength apart, physically, in space. "Free" being the end with the least capacitance.

When one end of the coil is connected to earth or a big capacity, the free end becomes sensitive to parasitic/external capacitance, which throws the coil out of resonance, and if it's too much then it will stop working altogether.

When both ends of a coil are free or have equal capacitance then it will become 1/2 wave resonant. Through increasing the capacity on one end, the coil can effectively be "biased", and the highest potential will be forced to the end with the least capacity, which presents the greatest opportunity to escape from the system. But this will be the end that's most sensitive to external capacitance.

As some of my pictures show, the bulbs are brightest when they are in series with the capacitance terminal on the top end of the coil, because the energy has to flow from the coil to the terminal and back again in order to resonate, and that's where the highest potential is. But I could never hold the bulb in my hand and light it from the top terminal, because the capacitance of my body would throw the coil completely out of resonance, and so in order to do that I would need to retune it to work while my body is connected to it.

On the other hand, when there's a larger capacitance on the ground end of the coil, which includes the primary and its inductance and the capacitance of the circuit etc, the coil is tuned to its natural resonant frequency, and so there's no amount of capacitance that I can add that would affect its resonant frequency, so I can easily hold the bulb to the terminal and light it in my hand without having to retune the coil, because I'm taking the output from the end that ALREADY has the bigger capacity, and the coil is already tuned to output energy. The free end is doing what it's supposed to do - remaining free to resonate with the maximum potential, and the ground end is outputting the energy.

So I would say that for these reasons, your bulb is brighter at one end than the other. If the primary was placed along the centre of the coil, then assuming everything is mirrored, then the bulb should be equally bright (or not) at both ends.

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Offline tinman

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There are a few things that you may or may not know so I'll say them anyway:

It's a natural requirement/condition of 1/4 wave resonance that the current lags the voltage by 90 degrees = 1/4 cycle of the wave = the length of the resonator. Due to the reflections that occur, a standing wave forms, and the current and potential are distributed at opposite ends of the coil. Maximum current is at the ground end, and maximum potential is at the top/free end, 90 degrees or 1/4 wavelength apart, physically, in space. "Free" being the end with the least capacitance.

When one end of the coil is connected to earth or a big capacity, the free end becomes sensitive to parasitic/external capacitance, which throws the coil out of resonance, and if it's too much then it will stop working altogether.

When both ends of a coil are free or have equal capacitance then it will become 1/2 wave resonant. Through increasing the capacity on one end, the coil can effectively be "biased", and the highest potential will be forced to the end with the least capacity, which presents the greatest opportunity to escape from the system. But this will be the end that's most sensitive to external capacitance.

As some of my pictures show, the bulbs are brightest when they are in series with the capacitance terminal on the top end of the coil, because the energy has to flow from the coil to the terminal and back again in order to resonate, and that's where the highest potential is. But I could never hold the bulb in my hand and light it from the top terminal, because the capacitance of my body would throw the coil completely out of resonance, and so in order to do that I would need to retune it to work while my body is connected to it.

On the other hand, when there's a larger capacitance on the ground end of the coil, which includes the primary and its inductance and the capacitance of the circuit etc, the coil is tuned to its natural resonant frequency, and so there's no amount of capacitance that I can add that would affect its resonant frequency, so I can easily hold the bulb to the terminal and light it in my hand without having to retune the coil, because I'm taking the output from the end that ALREADY has the bigger capacity, and the coil is already tuned to output energy. The free end is doing what it's supposed to do - remaining free to resonate with the maximum potential, and the ground end is outputting the energy.

So I would say that for these reasons, your bulb is brighter at one end than the other. If the primary was placed along the centre of the coil, then assuming everything is mirrored, then the bulb should be equally bright (or not) at both ends.

Is your device grounded?--as in,do you have a ground spike driven into the ground,that is connected to the negative side of your circuit.


Brad

Offline TeslaScientific

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Is your device grounded?--as in,do you have a ground spike driven into the ground,that is connected to the negative side of your circuit.


Brad

It depends on the experiment. It's not earthed when it's used as a single wire source in specific situations, but it must be earthed to transmit.

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Offline gotoluc

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Hi everyone,

I've been fortunate to obtain an original TinselKoil made by TinselKoala himself

The below is the first test video of many to come.
I have come up with what I believe to be an accurate input power measurement circuit and an estimated output power comparative method which is (for now) visually done.

Link to video: https://youtu.be/vPpgHSN9jPc

Regards

Luc

Offline woopy

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Hi everyone,

I've been fortunate to obtain an original TinselKoil made by TinselKoala himself

The below is the first test video of many to come.
I have come up with what I believe to be an accurate input power measurement circuit and an estimated output power comparative method which is (for now) visually done.

Link to video: https://youtu.be/vPpgHSN9jPc

many thank's Luc and TK

really intersting first results

Iam looking forward for the next video

Laurent


Offline Magluvin

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Hi everyone,

I've been fortunate to obtain an original TinselKoil made by TinselKoala himself

The below is the first test video of many to come.
I have come up with what I believe to be an accurate input power measurement circuit and an estimated output power comparative method which is (for now) visually done.

Link to video: https://youtu.be/vPpgHSN9jPc

Regards

Luc


Very cool. 

So then probably the next step would be to add more receivers around the transmitter and see what happens. Would the transmitter start to show loading on its input? Would a second receiver take away from the first or would 8 of them light up just as much as 1 at the same distance?

Could the receiver get a third receiver going, while the 1st receiver is driven from the transmitter? Put the 2nd receiver on the opposite side of the first receiver, all 3 inline, transmitter, 1st receiver then second receiver.

Mags

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Offline woopy

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Hi Luc

I have seen at minute 8.50 in your video, that you have already prepared a separated bulb on the table.
 
Did you try the original experiment of TK and connect the bulb between the ground and one of the battery terminal (+or- ) to see the difference between the battery receiver (if it is a receiver and if it works at such big distance) and the second coil receiver.

Thank's

Laurent

Offline gotoluc

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Very cool. 

So then probably the next step would be to add more receivers around the transmitter and see what happens. Would the transmitter start to show loading on its input? Would a second receiver take away from the first or would 8 of them light up just as much as 1 at the same distance?

Could the receiver get a third receiver going, while the 1st receiver is driven from the transmitter? Put the 2nd receiver on the opposite side of the first receiver, all 3 inline, transmitter, 1st receiver then second receiver.

Mags


Thanks Mags


Yes, that next step is exactly what I did!... but since the transmitting Tesla coil was getting hot probably due to the high resistance of the 33 AWG wire used, I decided to wind 3 new identical coils using 24 AWG wire. So one transmitter and two receivers at opposite sides. I did the test yesterday and unfortunately the power is divided when I connect the second receiver coil but the input power consumption is much less then my first video demo but not OU.


I'll do an update video in the next few days to demonstrate the differences


Regards


Luc

Offline gotoluc

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Hi Luc

I have seen at minute 8.50 in your video, that you have already prepared a separated bulb on the table.
 
Did you try the original experiment of TK and connect the bulb between the ground and one of the battery terminal (+or- ) to see the difference between the battery receiver (if it is a receiver and if it works at such big distance) and the second coil receiver.

Thank's

Laurent

Bonjour Laurent,

I had the bulb ready but forgot to demonstrate it in the video demo. If I hold the bulb with my fingers and touch a battery terminal the bulb lights dimly (about half) but lights brighter if I touch the circuit side. So the circuit reflection still makes it back to the batteries (through distance and filter network) probably through the wires but a little less then what TinselKoala demonstrated.
The receiver coil bulbs also dim when I connect the bulb to the batteries and the current goes up as well.

Regards

Luc

Offline woopy

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hi Luc

yes it seems that this energy can travel everywhere , through wires  , capacitors ,   measuring instrument ......and air .....and perhaps other materials.

What about installing a wood plate ( or any other material )  shield between the emitter coil and the receiver coil when you have the best distance efficiency to see if it affect the results ?

Laurent

Offline Magluvin

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I just built a subwoofer box to test a speaker that Pioneer makes ans it has an unusually high sensitivity level(96db@1w and my sub box program when all the other parameters are plugged in reveals 97db) compared to just about every subwoofer out there, and at least in the price range. Ebay $85   TS-W3003d4

It is a ported enclosure tuned to 25hz. In car you can hear almost to 15hz. It can handle a clean 1kw, claim is 800w and on the bench im giving it 150w per 4ohm coil.

Testing in the shop, 1000sq ft at 45hz the ceiling tile frames make a lot of noise. At 25-28hz the large garage door rattles and you can feel the vibration of the door.

So Im thinking on the same lines. If I were to have 10 garage doors that basically are resonating at the same freq, would they all be just as noisy and would they all have the same vibration level as the one door I have? Or, would adding the doors reduce the pressures on the first door and we get less and less as we add them. My shop is 50x20ft and the sub is in the middle of the shop length but nearer to on side wall. Im pretty sure there wouldnt be any increase in input if there were more doors added, but Im kind of betting on adding doors would affect the amount of energy they each would receive as they simply have to have an affect on the pressures in the outer perimeter of where the doors would be, as in those large panels at resonance must be affecting the other doors in a negative way since they would all be in phase.

It is very possible that sound waves in air are going to be different than the electromagnetic waves you are displaying. But Im working on a small scale with a 6.5in woofer ported to a low freq and see if I make a duplicate box without a speaker to see if the driver affects the passive box and at what if any output can possibly enhance the sound.  Like if I built the speaker drive box in the middle of 8 duplicates with all ports in the same side of the structure.

Anyway, can hardly wait to see your results as it will be very interesting. ;)

Mags

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