"Art Unwin" wrote

When viewing the antenna from an equilibrium point of view which is a
staple requirement of all electrical laws
one must assume that all forces/vectors equal zero (Newton )
Following this dictum physics state that foe equilibrium the charge on
the surface of a radiator does not move
linearlly there fore, there cannot be a linear force or vector to
oppose it. From this it is stated that there is no movement
in linear form else where which includes the center of the conductor/
radiator when the radfiator is one wavelength or multiple there of.
Now we have the case of a fractional wavelength radiator. In this case
one is aware that charges do move in a linear
direction as evidenced by "end effect". Therefore by following the
standard laws of physics there must be a
balancing force/vector in the opposite direction and the only place
that vector could be is in the center of the conductor
One should also be aware that a electrical curcuit for a fractional
wavelength is a series circuit and a parallelel circuit for a
fulle wavelength both of which are closed cuircuits when determining
current flow of a radiator so one can itemise the electrical circuit
in detail with respect to the components on the actual radiator to
ensure compatability.
Now according to my theory of radiation the forward current on a
radiator is opposed by closed circuit eddy current
which in combination provide a angular rotational force on any
residing particle which allows for directional levitation or
projection.
When the current of the radiator reaches the end of the radiator it
closes the circuit by entering the center of the conductor
( assuming the arrangement is not in a state of vacuum)under circular
surface current cuircuit where it is still in existance.
The internal current flow is solely resistive in nature comprising of
theseries resistance of the material used and not radiative.
Now David, if you can point to a description that differs to the above
and follows the laws of physics I would be happy to look it up and
study it , but in the final analysis one must be able to determine
the state of the conductor at it's center at all times.
David, my explanation is based on the world of physics as I know it.


Hi Art
Trying to build a picture here.
Let's say I have a 1/2 wave dipole, and I drive it such that one ampere
is flowing at the feedpoint. Let's agree to use amps RMS and volts RMS
at 14 MHz
for this example, just for clarity.
If I measure the current a short distance from the feedpoint, it's a bit
less than one amp. Correct so far?

If I tease the antenna conductor apart and measure the current flowing
on the
outside with one RF Ammeter, and the current flowing in the
center with a second ammeter, what are the two currents?

Thanks!
73
PN2222A