Re: 7.1Hz, how the heck did Tesla succeed?

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: William Beaty <billb@xxxxxxxxxx>

On Sat, 16 Jul 2005, Tesla list wrote:

> Original poster: Ed Phillips <evp@xxxxxxxxxxx>
>
> > "Was this low value for loss associated with a leakage vertically
> > across the atmosphere (which would be constant whether there were any
> > power receivers in use?)
>
> 	I'd have to look up his reference but he says that "only a few
> horsepower" are needed to excite his antenna; I take that to mean loss.

Ah, I remember that.  You're right, it would take WAY more than "a few
horsepower" to run Wardenclyffe if nobody was loading it.  But the Corum
paper said ~1 megawatt.  Was "few horsepower" about Wardenclyffe or about
some other unit?

> If you perform a calculation as to the capacitance to earth of the
> "conducting layer" he proposed to use (remember he says in his patents
> that his proposed transmission medium was conducting layer to receiver
> with earth return) you'll find that its so high that the reactive power
> it would store at any reasonable operating frequency would be hundreds
> of millions of "horsepower".  That's the stored energy; divide it by the
> claiamed loss and you get a very, very high Q.

Yep.  This is all about the number of times a wave goes around the Earth
before it falls to half it's original value.  If the frequency is moving
around and we didn't know this, we'd measure a broad spectral line and
assume that Q was low.


> 	How do they explain those low losses considering the exceedingly low
> conductivity of the ionosphere and the relatively high resistivity of
> the earth?

I'll have to find that paper again.   I think they based calculations on
Q value for a resonance, not on resistivity.


>
> > "On the other hand, there's that Corum paper that say that low Q on
> > the transmission end is not such a big issue, since it's analogous to
> > the corona leakage and insulator conductivity on a continent-wide power
> > grid.
>
> 	They can't make the issue go away that easily, no matter what they
> say.  Remember that Tesla's scheme of necessity requires incredibly high
> circulating currents so that any loss at all destroys the efficiency.

I must be wrong then.  I was under the impression that the currents were
far lower than those in the 60Hz power grid as a whole.  Obviously we have
to look at current density in some small volume of earth or ionosphere,
just as we'd look at the amperes in one piece of power line.  We'd better
not add up all the power lines west of the rockies and decide that the
current in the AC power grid is astronomical.

> As for commercial power lines, the operating voltage is a compromise
> between corona loss at high voltages and lower currents and the

But if those total corona/resistive losses at present are in the
megawatts, then a Tesla system with megawatt losses wouldn't be ridulous
in comparison.  The world system isn't like a power line, it's like a
worldwide power grid, so we need to compare it to at least the USA power
grid.


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William J. Beaty                            SCIENCE HOBBYIST website
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