To All from PA0SE
A. I have modelled my aerial using Antenna Optimizer by
This also provides the current distribution on the
The coil is modelled as a lumped inductance.
As expected the current above and below the coil is
shown to be the same.
Now the books have it that a coil with a length of L metres
radiates like a straight wire of the same length. To find the radiation
from my 58cm high coil I looked at the current distribution on the first 58cm
of aerial wire immediately above the coil, where the current that enters at
the bottom end is the same as in the coil . AO indicates that the
current at the top end of that piece of wire is 1.55% lower than at the bottom
end. So that is current lost by radiation.
I measured a difference of 10% between currents at the top and
bottom of my coil. So most of the current lost must be due to capacitance to
B. I also modelled Steve's 12m high vertical. The
program assumes that the aerial is over perfect earth. The
real earth may be some distance below the surface so the actual aerial may be
longer than 12 m. (Remember Jim's aerial on a hill at Puckeridge that radiated
better than expected from its physical length?)
The program shows that a coil of 12.87mH will be needed for
For the difference in current between 2.2A at the bottom and
1.8A at the top of the coil to be entirely caused by radiation the coil
must be 1.85m high. Looking at the picture at Steve's website
this is not the case. So some current must be escaping through capacitance to
C. I entirely agree with those who state
that a difference in current between the two ends of a coil can only be caused
by radiation and capacitance to surrounding objects.
Any other effects, like loss in the coil, distributed capacitance
etc. remain internal to the coil and cannot cause a difference between current
flowing in and out.
73, Dick, PA0SE