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Re: just wondering (Schumann resonance)



Original poster: Ed Phillips <evp-at-pacbell-dot-net> 

"Original poster: "Steve Conner" <steve.conner-at-optosci-dot-com>

I have a couple of thoughts about the Schumann resonance.

As a radio ham I know about propagation and stuff. And I understand the
Schumann resonance is a standing wave caused by a radio wave that, in
wavelength terms, fits exactly once (or a whole number of times) round
the
earth. (please correct me if I'm wrong.)

Now, if my "standing wave" analogy is correct, I would imagine the
propagation path for the wave that forms the Schumann resonance is
pretty
lossy. When talking to very high powered foreign stations on the HF
band, I
could often hear a distinct "echo" on their signal, which is of course
caused by their radio waves reaching me twice, the short way round the
world, and the long way.

Anyway, from observations like these, I would estimate the loss for a
trip
round the world is over 100dB. I was working at 10s of MHz, it would of
course be lower at the 10s of Hz where the Schumann resonance happens.
But
even assuming the loss goes down in proportion to the frequency, it's
still
a 40dB loss (40dB= 1/10000) at 10Hz.

So how on earth you could ever excite a resonance in such a lossy
system,
using even the most powerful transmitter ever made, I don't know.

If these loss figures are ballpark, it blows all that "wireless death
ray"
cr*p out the water too: For instance, Tesla couldn't have caused the
Tunguska explosion ;) without causing an explosion 40dB bigger at his
transmitter site, and I think folks would have noticed New York missing
:)))

Steve C."

	You'll bring down the wrath of the gods with that note, but it's pretty
accurate and to the point, even the editorializing.  The Schumann
resonance bandwidths are so broad that the signals are hard to detect,
implying a very low Q and even higher losses than you speculate.

Ed