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Re: Tesla Coil RF Transmitter

Original poster: Jim Lux <jimlux@xxxxxxxxxxxxx>

At 09:30 AM 9/20/2005, Tesla list wrote: 
Original poster: "Gary Peterson" <gary@xxxxxxxxxxxx>


Original poster: "Antonio Carlos M. de Queiroz" <acmdq@xxxxxxxxxx>

Interesting to see that some people was already at that time seeing that
something
was terribly wrong...

Tesla's analog is incorrect:

...
Tesla
    I will explain it by an analogue.
    Suppose that the earth were an elastic bag filled with water.  My
transmitter is equivalent to a pump.  I put it on a point of the globe,
and work my little piston so as to create a disturbance of that water.
 . . .

Tesla forgets that the "water" sent into the ground is charged, and so is
attracted back to the oppositely charged top terminal (and coil) of his
transmitter. It then does NOT spread through all the Earth, but stays
concentrated around the transmitter.

In Tesla's analogy the water is not charged, it IS the charge. In other words, the water represents Earth's background DC electrostatic potential, which is already spread throughout all the Earth. In operation, a grounded non-sparking Tesla coil transmitter creates a local disturbance in this charge. The disturbance manifests itself as an annular distortion of the background field around the transmitters ground terminal. At a point in time when a measurement of the e-field component of the EM field at the ground terminal shows zero volts above the background potential (see www.tfcbooks.com/special/images/system.gif), other measurements will show it rising in intensity until a point 1/4 wavelength (1/4 λ) away from the ground terminal is reached (axial projection). From there the e-field diminishes in intensity until, at 1/2 wavelength from the terminal, it again shows zero. At a measurement point approximately one wavelength away from the ground terminal, an induced e-field once again begins to emerge above the background field, again increasing in intensity until a second maxima is reached at 1 1/4 wavelength away from the oscillator. With a sufficiently powerful transmitter this phenomenon repeats itself over and over until the antipode is reached, at which point reflection takes place and the transmitted energy begins to travel back to its point of origin in the reverse direction.

Gary Peterson

A couple problems:
1) that wavelength thing... It assumes uniform propagation speed, which is hardly the case, particularly for propagation over a boundary.
2) The earth isn't a smooth uniform sphere, so the paths aren't symmetric
3) Sufficiently powerful transmitter doesn't have much to do with it, although that will affect the detectability.
4) There's no reflection at the antipodes (which doesn't really exist anyway, because of #1 and #2), the wave just keeps on going and eventually gets back to the start.

The latter effect is widely noted as "long path" propagation on radio transmissions. You get two copies of the transmitted signal, separated in time by the differential propagation delay for the short way and long way.

In any case, the propagation around the earth (at whatever frequency) is sufficiently non-uniform and lossy that you're not going to get any significant standing waves. No "ringing like a bell"