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To All from PA0SE

A. I have modelled my aerial using Antenna Optimizer by K6STI.

This also provides the current distribution on the aerial.

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 surrounding objects.

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 resonance.

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 the surrounding.

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