I have tried various combinations of multiple dipoles fed from a common
point, and they all worked. At resonance, the feedpoint impedance will
actually be closer to 50 Ohms than for a single dipole.

Note from Cecil's numbers that the other dipole add reactance. You will have
to compensate for that by adjusting the lengths of the dipoles. Start with
the lower frequency dipole first, but since 10 & 15 are rather close, you
might have to go back and forth a few times.

The main thing is to get separation between the ends. Having the two dipoles
at right angles would be optimum, but 15 degrees, or so, will work.

If you use Cecil's numbers to calculate the feedpoint impedances, don't
forget to convert everything to a parallel equivalence first.

Tam/WB2TT
"Cecil Moore" wrote in message
...
Cecil Moore wrote:
On 21.4 MHz, the 10m dipole will have a feedpoint impedance of 200+j450

On 28.4 MHz, the 15m dipole will have a feedpoint impedance of 29-j384

Well, I obviously got these impedances reversed. Should have been:

On 28.4 MHz, the 15m dipole will have a feedpoint impedance of 200+j450

On 21.4 MHz, the 10m dipole will have a feedpoint impedance of 29-j384

Thanks, Jerry, for catching my mistake. But the paralleling advice stands.
--
73, Cecil http://www.qsl.net/w5dxp



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