On Sun, 23 May 2004 17:18:52 GMT, "Henry Kolesnik"
wrote:

OK, if we shine a flashlight at a mirror the light bounces back and what
ever is caught by the reflector will be reflected. Take away the reflector
and the reflection just keeps going. If someone can tell me what the hot
filament does perhaps I can understand what happens in the finals, or
whatever.
tnx
Hank WD5JFR

Hi Hank,

Well I see in correspondence following this as "replies," that they
are far from satisfactory. All part of the arcana that precedes the
convolutions of math you would have had to endure.

The abandonment of this lead is simply a matter of Cecil's lack of
experience in the metaphor of light.

To answer your question above. Removing the reflector is unnecessary
as it is part of the initial condition and has nothing to do with it
serving as the correlative to a tuner that you want to remove from the
argument (which is a perfectly acceptable imposition of conditions).

Your question also goes to the heart of the matter. We begin with a
hot filament which emits radiation (and yes, no reflector is required
so we will skip that as one of your conditions). The amount of
radiation is directly correlated to the amount of heat. This will
simplify matters, but in the end it will yield a failure of metaphors
(which always occur if you cannot bridge the logic).

The radiation strikes a reflection (immaterial whether complete or
partial) and that portion which returns, impinges upon the filament,
the source. The filament absorbs the power, which in turn raises its
temperature (everyday experience proves the heat of such radiation).
This will, in turn, cause a higher radiation (given the quid-pro-quo
of heat and radiation). In a sense, this means the reflected power is
re-radiated. The confirmation of this is that if you achieved full
reflection, you then define total insulation of the radiation (no heat
escapes) and temperature rises accordingly, and this may lead to
catastrophic failure of the filament (a very bright illumination if
you could see it, and consequent fusing current - electric kilns use
this principle but tolerate the current by under rating the source).
You can imagine the correlative to transmitter failure for the same
conditions.

The failure of the metaphor? RF is not heat (common light is) and the
return of power to be rendered into heat does not result in a higher
RF output.

I will anticipate the sophomore's comments that RF reflections do not
become heat, in and of itself:

The returned power (either through wave mechanics or lumped circuitry)
must result in either a higher potential across the source, or a
higher current through it. Elevated potentials yield the everyday
experience of an arc (heat). Elevated currents yield the everyday
experience of current density through the same element (much like the
filament of our metaphor - heat). One failure mode comes with the
peak power snap, followed by the muttering of "Oh ****!" Or it comes
through the more progressive thermal runaway, followed by the
muttering of "what's that funny smell?"

73's
Richard Clark, KB7QHC