Richard I need more explanation than that because power is somewhat
irrelevant.
Radiation in my mind (and I must be now worst off than I really thougt) is
accelleration and decelleration of current
which is certainly not power which in my mind is Isquared R. Thus if voltage
is increased current decreases
and therefore radiation decreases.( see effects of very close coupling of a
parallel circuit)
I really do not know where you are coming from, it seems so glib! This is
not meant in a demeaning way, I just
don't want to mix apples with oranges at this point in time. .
Current through a member travels at a constant speed but with cyclic
variations
in radiation. Travel in a circle is also at constant speed but with a
CONSTANT radiation per unit length.
Where exactly is the error in my logic? I suspect it has to do with relative
phase angles but I need it explained
in every day language
regards
Art

"Richard Clark" wrote in message
news
On Mon, 16 Aug 2004 20:31:03 GMT, "
wrote:

If I am incorrect I need to understand why

Hi Art,

Because "Extra" in the budget does not balance.

If you get more power OUT because it is going in a circle, then you
FIRST have to put more power IN to get into that same circle. There
is no extra left.

73's
Richard Clark, KB7QHC

On Mon, 16 Aug 2004 23:40:12 GMT, "
wrote:

Richard I need more explanation than that because power is somewhat
irrelevant.

You then broach the remainder of your post in terms of acceleration -
which requires power, and deceleration which begats power.

There is no room for "extra" in the power budget.

Radiation in my mind (and I must be now worst off than I really thougt) is
accelleration and decelleration of current

I recognize this as a commonplace expression in this group (not unique
to you by any means). It is one of the most ill-conceived statements
ever to come down the pike and your conundrum (as for others) in
trying to retrofit it into a theory is part and parcel to its poor
analogy.

I need it explained
in every day language

Hi Art,

The plain, every day language of budgets is there. If you cannot
balance that simple ledger, then you have scant chance of
understanding the larger enigma.

73's
Richard Clark, KB7QHC

wrote:
I suspect it has to do with relative phase angles but I need it explained
in every day language

Art, a single RF traveling-wave wire has a natural Z0 in the
ballpark of 600 ohms which sets the V/I (E/H field) ratio.
The feedpoint current and the current at any point on a
standing-wave antenna is the phasor sum of the forward current
and the reflected current. Vf/If and Vr/Ir remain in the
ballpark of 600 ohms but Vf+Vr and If+Ir vary up and down
the wire. The current is high at a 1/2WL dipole feedpoint
because If and Ir are in phase at that point. The current is
low at a 1WL dipole feedpoint because If and Ir are out of
phase at that point. But If is approximately the same for
both antennas and Ir is approximately the same for both antennas.

That's why the superposed If+Ir at the bottom of a loading
coil is not usually equal to the superposed If+Ir at the
top of a loading coil. Each of the two currents indeed has
close to a constant magnitude through the coil but the phases
of If and Ir are changing in opposite directions. Thus their
sums are different except for the special case where the
relative phase difference is equal at both ends of the coil.

But for understanding RF radiation, If is fairly constant
and Ir is fairly constant, and each of those currents cause
radiation. Thinking about a terminated Rhombic Vs an unterminated
Rhombic will separate the two processes out in your mind.
--
73, Cecil http://www.qsl.net/w5dxp



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