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To All from PA0SE
A few additional remarks about the info in my earlier massages are
necessary.
1. I stated the field computation was based upon a radiated power
of 1kW.
That is not correct. The power fed to the antenna is 1kW.
Computer program AO gives the efficiency of the antenne as 26.4%. So the
radiated power is 264W.
I modelled the antenna as an 180m high vertical copper tube of 15mm
diameter.
AO computes the feedpoint impedance as 0.042 - j16189 ohm.
Because efficiency is 26.4% the radiation resistance is 0.264 *
0.042 = 11 milli-ohm.
To pump 1000W into the antenna the current must be SQR( 0.042 *
1000) = 6.48A.
The voltage on the antenna is then 6.48 * 16189 =
105kV.
2. Maker K6STI of AO has told me that the near field option
actually calculates the total power, so it includes the far
field.
At 2km from the antenna the phase difference between H and E is
88.8 degrees, still near the 90 degrees of the real near field. (In the far
field it would be zero degrees).
Also E/H = 872 ohms; it would be 120 * pi = 377 ohms in the far
field.
3. I will try to model a loop, as suggested by Rik, ON7YD. I will
send the E and H field figures as attachment to separate e-mail
messages.
73, Dick, PA0SE
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