"Szczepan Bialek" wrote in message
...

"Cecil Moore" wrote
...
Szczepan Bia³ek wrote:
Could you tell us if the "almost double voltage" is the measured of
theoretically predicted?

Since a voltage reference point is difficult to achieve
at the ends of a dipole, we rely on the conservation of
energy principle. Since the current is zero at the ends
of a dipole, all the energy must be contained in the
electric field. With that knowledge, the voltage can
be estimated.

Such is easy to understand by taking voltage measurements
on a 1/4WL open-circuit stub. If one uses resistance wire
for the stub, one can simulate radiation loss in a dipole.

The following is a transmission line simulation of a 1/4WL
monopole designed to run on the free demo version of EZNEC
available from www.eznec.com

http://www.w5dxp.com/stub_dip.EZ

The user defined resistivity of the wire is what causes
the 35 ohm feedpoint resistance akin to a 1/4WL monopole.
The 10 megohm load allows us to look at the voltage
across the "open" end of the stub. It is 1033 volts for
a 100 watt input. We can assume the forward voltage and
the reflected voltage at the end of the stub are equal
at 516 volts.

The acoustic analogy predict it: "Between the nodes are places where the
amplitude is doubled. So the places
with doubled amplitude are standing. Pressure pulse travel.
In antennas is electron gas. The first place where the doubled amplitude
(amplitude means voltage or electron density) appear is end of the
radials.
The next is halve wave apart from the end. Such places radiate strong
electric waves. They are the source of radiation."

no, they are called electro-magnetic waves for a REASON! it takes BOTH
fields to make up a propagating wave. So it is not the ends that radiate,
it is the whole length where there are both electric and magnetic fields
generated in smooth transitions of the sine wave, not in pulses.