Re: Tesla Coil RF Transmitter

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: Paul Nicholson <paul@xxxxxxxxxxxxxxxxxxx>

Dan (DUllfig) wrote:

> Suppose you have a flexible beam ...

> ...before you start saying that my analogy is faulty,

Your analogy is fine, Dan.   As you say, if you excite
the system (beam or earth resonance) at or near a resonant
frequency, then the system will respond extra well and a
'global' resonance is stimulated.

> the amplitude of the motion will be nearly identical on both
> halves,

Yes,

> even though nothing is physically shaking the other end.

Well, not nothing.  The beam is carrying a shear wave, enough
leaking through the mid-point clamp to ensure that the two
halves are coupled to produce the 'global' resonance of the
beam.

We use 'Q factor' to indicate the extent to which a resonator
responds extra well at a resonant frequency, compared to its
response away from a resonance.   The higher the Q factor, the
less excitation is needed to achieve a given level of response
from the system.

A resonant system with a high Q factor will respond well to
even a small excitation.   Why?  Because Q measures the ability
of the system to store energy.  High Q means good energy storage,
so that the successive little pushes of a weak excitation are
steadily accumulated as resonant energy, until a point is reached
where each little bit of excitation is just enough to cover the
losses incurred since the last bit of excitation.   At this point
the stored energy is roughly Q times higher than the per-cycle
excitation energy.  So you can think of Q as a kind of
magnification factor.

> Tesla did not even contemplate world power transmission, until
> he measured standing waves produced by lightning strikes.

I strongly doubt that Tesla actually measured an earth resonance.
A lot of things have been claimed by and about Tesla - things
which turn out to be exaggerations, nonsense, misunderstandings,
or just plain impossible.  Tesla himself was quite big on claims
but often short on detail and evidence.  Add to that the inevitable
hype which surrounds such a cult figure, and the result is you
have to be very wary of accepting things at face value, unless you
want to be on a wild goose chase.

Unfortunately, the Schumann resonances don't have a very high
Q factor.   They are also very difficult to couple to - meaning it's
hard to devise an antenna which will insert or remove energy from
the 'storage tank' of the Schumann resonance.   This is because
the wavelengths are so long, and the 'cavity' is so large, compared
with realistically sized antennas.   And yes - an antenna is needed,
there's no magic way to excite the Schumann cavity field by 'pushing
current into the earth' or some such vague term.  You must design
an antenna which will couple to the Schumann field reasonable well.
A high Schumann Q would help, allowing weaker coupling to be used,
but I'm afraid it looks like the combination of low Q and difficult
coupling kills off this particular variation of Tesla's power
distribution schemes.

> Do lightning strikes actually produce standing waves or not ?!

Yes, they do.  There's about 100 lightning pulses a second feeding
energy into the Schumann resonances.  Their overall effect is to
produce some excitation of all the Schumann resonances, although the
effect of each individual lightning stroke is not discernable.

(Imagine a hundred fleas spread along your beam, pronking at random.
 Unless you train them to pronk in unison in time to the resonance
 of the beam, the random excitation will produce just a small (not
 zero) response.)

The resonant field strength from all that lightning is of the order
of 1mV/metre, and this is what we measure with ELF receivers. The
lightning makes a useful test signal and allows the characteristics
(frequency and Q) of the Schumann resonances to be measured.

> let me remind you that at an atomic level we never really "grab"
> anything, the atoms of your hand never come in contact with the
> atoms of the beam :)

Yes, indeed.   What we think of as mechanical contacts ultimately
reduce to electromagnetic interaction between the molecules of
the moving parts.

Stork wrote:

> F = qE        Lorentz Foece Law.  Defines an E field really.

> dW = F.l      dl is distance moved related tp work differential,dw.

> dW = qE.dl    E.dl is a scalar product.  E has only a single
>               component in the radial direction to q whereas dl
>               has a component in the radial direction also, but
>               may have components in other directions
>               perpendicular to the radial direction.  E has no
>               effect on components perpendicular to r (dot
>               product reduces to zero)so this reduces to:

> dW = qEdr

> W = q INTG Edr  Integrating gives the total work done by E in the
>                 radial direction (longitudinal direction in our
>                 experiment) on q.

Fine so far.   You've established
a) the force is along the line of the E-field.
b) no work is done unless the target object is moved by the force.
c) when it does move, the work done on the object is q INTG Edr,
   which is q times the E potential through which the object has
   fallen, or in other words, q times the voltage drop.

The next steps in your calculation are:
a) Show that, in addition to the 0.5mV^2 kinetic energy of the
   object, you also now have some energy stored in the H field of
   the moving charge.  The total energy required to accelerate
   the object is the sum of the object's 'mechanical' energy that
   you've just calculated, plus the 'field' energy of the object.

b) Show how the gain in energy by the object and its field, is
   matched by an equal loss of energy from the incident field.
   If you work it out from first principles (the Maxwell/Lorentz
   set of equations), you arrive at a function ExH showing in detail
   how energy flows from the initial static field to both the object
   and its H field.

For very light, slowly moving charges, such as electrons in
conductors, the mechanical 0.5mV^2 energy is very, very small
compared with the energy stored in the H field of the moving charge.
At frequencies below umpteen terahertz we can ignore the electron's
kinetic energy altogether and take the total energy of motion of
the object to be given just by the energy of its H field.

For any kind of macroscopic object, it is the other way around.
Most of the input energy goes into the 0.5mV^2.   But in both
cases, the flow of energy into the object and its field is 'plotted'
by ExH.

There's a similar thing for that other conserved quantity - momentum.
You get DxB for the momentum density.

Let me remind those that are worried about the meaning and 'reality
status' of things like 'displacement current', 'EMF', E, H, etc, that
these are all just names for terms and vectors in mathematical
equations.   The basis of EM theory is that if you rattle a charge,
other charges nearby will respond.   EM theory is just a mathematical
edifice which describes how those charges respond.   Only the final
answers of the theory (eg the charged object will do such and such)
have any real meaning.  All the intermediate quantities that we talk
about all the time are just mathematical terms in equations and it's
a bit silly to ask if, say, 'displacement current' is real.

Instead, the question to ask is 'Do the EM equations still work if
you leave out the displacement current term?'

The answer in this case is 'No'.  You no longer get a consistent set
of equations (they cannot represent Ampere's law, for eg) and the
whole EM edifice falls apart.  You don't get any answers at all.

So, if you want to work out the effect one charge has on another,
you have to include the term that we call 'displacement current'
if you want the right answers.  That's the only sense in which
we can claim that any component part of the overall mathematical
scheme of EM theory is 'real'.
--
Paul Nicholson
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