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Transmission Is Through Earth - Experimental Proof


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#21 dR-Green

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Posted 12 August 2013 - 07:29 PM

Yes that's true, the coil (wire) itself even radiates to a certain extent so I'm not sure that it will ever be possible to suppress it completely anyway. I posted the general (crude) field "shape" diagram on the other forum before, but I don't think I showed the secondary alone. In this case I mean in order to deliberately maximise the radiation, like you would use a small point to encourage a high voltage breakout as opposed to a larger surface in an effort to suppress and "contain" it.

 

For the record and future reference I'm not a physicist so I talk about these sort of things in basic and "practical" terms excluding the intricate details, people will notice the more specific things in their own experiments.

 

Thanks for the references.


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#22 jimm

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Posted 12 August 2013 - 08:03 PM

I'm not that stupid :P Pure water is more interesting. I don't know what the DC resistance would be through 2 metres of earth but I think safe to say "big".

 

I can't make any claims of 95% efficiency because I have not observed it. Then again I don't have RF watt meters (to put) on the output or input. But it's not a default or "automatic" result. That is, you don't just set up a transmitter and receiver and you suddenly have 95% efficiency. In this case the transmitter isn't optimised at all, and I hadn't tuned the receiver properly (using the proper tuning procedure), the receiver was just tuned to the signal source via the earth and balanced for primary voltage reading and audible signal level for the purpose of putting it in the video. With proper grounding and optimised coils who knows. Frankly I wouldn't have expected my setup to have worked at 2 metres given how crude the whole thing was and how little power I was using. I was putting copper pipes in the ground at random distances because my "conservative" efforts were turning out to be a waste of time because they all worked, I had to start trying random greater distances to try and make it NOT work.

 

Yes, Tesla did at first think of transmitting the energy through the air, but as far as I know he completely abandoned that idea, or at least adapted it. Even in my own experiments I've "discovered" (observed the effect before reading about it) something that I'm sure Tesla would have observed and been very excited by, causing him to adapt his ideas. By playing with the "wireless" around the top terminal one may start to wonder about more possibilities, as the human body appears to be quite conductive as a transmission medium allowing one to power loads at a greater distance from the coil, because the body is acting as a conductive medium in between the coil and the load. The thought then occurs, "what if I were to touch the coil terminal, and had infinitely long arms..." Just saying, ideas and opinions are adapted over time through experimenting with it and learning new things.

 

How does the Marconi type propagation argument stand when the output is observed to change based solely on the earth terminal? It shouldn't make any difference.

 

What would you consider to be reasonable power levels and distances? In the video I was using approx 23mW which basically means the setup would have to be relatively close range. But by increasing the power and distance, one may then argue that it's only working because of the higher power. So what is one to do? Go inside a cave? There aren't any that I know of around here, but there are plenty of mountains to stand behind. But then the signal is bouncing off the atmosphere :wacko: So what kind of power level doesn't supposedly bounce off the atmosphere, and so on?

No cave required, just the inverse square law that will make it impossible to induce the power.

I wouldn't care if you bounced the signal off the moon if you got 95% efficiency!!!! 

 

 Unless you have a TMT with the parameters mentioned earlier, You won't even be close ! ( I'm not sure if the original worked as advertized either) You have a scope, and a known load so you can calculate RF rms (.707Vp)^2/R in and out compare the the two for % efficiency. However , Useful measurements will be difficult if not impossible at such low power levels.

If you want to prove that you have done something, you have to have the data and experiential objectivity to back it up.

 

Read the TMT patents where Tesla describes the important aspects of the transmitter and why. There is nothing to debate, it's in black and white.


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#23 Robert

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Posted 12 August 2013 - 08:43 PM

"Unless you have a TMT with the parameters mentioned earlier, You won't even be close ! ( I'm not sure if the original worked as advertized either) You have a scope, and a known load so you can calculate RF rms (.707Vp)^2/R in and out compare the the two for % efficiency. However , Useful measurements will be difficult if not impossible at such low power levels.

If you want to prove that you have done something, you have to have the data and experiential objectivity to back it up."

 

Hey Jimm,

 

You seem to have a great amount of certainty in the absence of any evidence that you have built anything. I'll accept what you say* when you present either a small demonstration or some worked math.

 

* And only then, pertaining to the thing demonstrated.


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#24 dR-Green

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Posted 12 August 2013 - 09:05 PM

You have a scope, and a known load so you can calculate RF rms (.707Vp)^2/R in and out compare the the two for % efficiency. However , Useful measurements will be difficult if not impossible at such low power levels.

 

The power output meter should go between the transmitter coil and earth. I don't think the voltage measured across a resistance here will mean anything, and it would affect the coil's operation anyway. Hence I need an RF watt meter that's designed for the job, not modern "equivalents" and calculations. The power supplied to the transmitter is somewhat irrelevant because it's the output power and received power that's of interest. The actual power output of the transmitter is still unknown, I have only calculated the supplied power based on the primary voltage so far.


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#25 jimm

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Posted 12 August 2013 - 09:29 PM

"Unless you have a TMT with the parameters mentioned earlier, You won't even be close ! ( I'm not sure if the original worked as advertized either) You have a scope, and a known load so you can calculate RF rms (.707Vp)^2/R in and out compare the the two for % efficiency. However , Useful measurements will be difficult if not impossible at such low power levels.

If you want to prove that you have done something, you have to have the data and experiential objectivity to back it up."

 

Hey Jimm,

 

You seem to have a great amount of certainty in the absence of any evidence that you have built anything. I'll accept what you say* when you present either a small demonstration or some worked math.

 

* And only then, pertaining to the thing demonstrated.

I've played with this stuff quite a bit 10 years ago. I know some things, If you don't accept what I say , no problem. Prove me wrong. Only people who know these devices through experimentation can appreciate my point of view. anyway. You're new so you have to "pay your dues" before you understand. EXPERIMENT and DISCOVER.


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#26 jimm

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Posted 12 August 2013 - 09:44 PM

The power output meter should go between the transmitter coil and earth. I don't think the voltage measured across a resistance here will mean anything, and it would affect the coil's operation anyway. Hence I need an RF watt meter that's designed for the job, not modern "equivalents" and calculations. The power supplied to the transmitter is somewhat irrelevant because it's the output power and received power that's of interest. The actual power output of the transmitter is still unknown, I have only calculated the supplied power based on the primary voltage so far.

Using the formula I posted will suffice for both.  Put the load across the receiver coil secondary as Tesla indicated where it belongs! first check the open circuit voltage. try resistances until it's 50% of open circuit voltage. That is the maximum power point. plug the resistor value into formula with the voltage and that's all you need. Take the ratio of the two results and that is the % recovered power. Yes it will load the coil, but that's the point...how much of a load can it supply? Any load on any coupled tuned circuit will reflect that load onto the other winding. Radio or Tesla, it all works the same way.

You can do this!


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#27 dR-Green

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Posted 12 August 2013 - 10:03 PM

Using the formula I posted will suffice for both.  Put the load across the receiver coil secondary as Tesla indicated where it belongs! first check the open circuit voltage. try resistances until it's 50% of open circuit voltage. That is the maximum power point. plug the resistor value into formula with the voltage and that's all you need. Take the ratio of the two results and that is the % recovered power. Yes it will load the coil, but that's the point...how much of a load can it supply? Any load on any coupled tuned circuit will reflect that load onto the other winding. Radio or Tesla, it all works the same way.

You can do this!

 

That's already the plan for the receiver, but I'm not convinced in any way at all that a resistance on the transmitter output will be a good thing for the setup. I want something that's "invisible", not a definite resistance.


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#28 jimm

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Posted 12 August 2013 - 10:45 PM

That's already the plan for the receiver, but I'm not convinced in any way at all that a resistance on the transmitter output will be a good thing for the setup. I want something that's "invisible", not a definite resistance.

Agreed . I was talking about the receiver only. The scope will show the voltage on the collector, you know or can find the current input.  Another way is simply the V*I  DC input to the transmitter, so comparing the actual draw on the supply compared to the the recovered will take system losses into account by default.


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