AI4QJ wrote:
"Cecil, Can you help me understand that the end of a dipole is an open
circuit?
Nothing falls out because the current changes direction ala AC because
the end of a dipole equates with the time that requires the current
changes direction. So where is the open circuit? I also cannot see why
radiation reduces reflection if the circuit can be completed with
iron!"

OK Cecil, I'm waiting for your answer on this one! ;-)

The end of a dipole is like an open circuit transmission line.
The traveling wave hits the open-circuit and is reflected.
The forward current and reflected current are equal in magnitude
at the end point and 180 degrees out of phase. Therefore, the
current is zero. That's logical.

The forward voltage and reflected voltage are in phase. Therefore,
the voltage is at a maximum at the ends of the dipole. That's known.
That voltage can be calculated to a certain accuracy.

At the feedpoint of a dipole, the forward wave energy is about
20% higher than the reflected wave energy. That 20% difference
between forward wave and reflected wave is the amount of power
lost to losses and radiation.

A #14 horizontal wire 30 feet in the air has a characteristic
impedance of 600 ohms. That's all you need to know to perform
the voltage and current calculations for a particular power
input to the antenna.
--
73, Cecil http://www.w5dxp.com