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| Author |
Message |
ball
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Dipoles
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Nov 4 05:56 UTC 2006 |
What sort of velocity factor should I expect from 10 AWG
(~2.6mm) stranded (looks like about 14 strands) copper wire
at 448.45 MHz?
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| 6 responses total. |
eprom
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response 1 of 6:
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Nov 4 07:21 UTC 2006 |
1.0000 if its just a bare wire.
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ball
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response 2 of 6:
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Nov 4 15:49 UTC 2006 |
It's insulated and stranded.
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rcurl
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response 3 of 6:
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Nov 4 16:26 UTC 2006 |
The velocity factor of a bare wire is less than 1.0000, depending upon the
ratio of the diameter of the wire to the free-space wavelength. It is only
1.0000 for a hypothetical wire of zero diameter and infinite conductance.
Insulated wires have much lower velocity factors, depending upon the type
and thickness of the insulation. You can find information on this on the web,
at least in the form of the correction factor for a resonant dipole length
for different wire diameters and insulation.
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ball
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response 4 of 6:
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Nov 4 17:33 UTC 2006 |
I've been looking, but not found anything helpful.
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gull
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response 5 of 6:
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Nov 5 21:55 UTC 2006 |
You may have to use the old 'cut and try' method, with an SWR or grid
dip meter. The velocity factor will probably be greater than 70%, so
you can start there and then trim until the resonant frequency is where
you want it.
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rcurl
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response 6 of 6:
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Nov 6 01:30 UTC 2006 |
The ARRL Handbook has a graph for the correction factor for a resonant
dipole length without insulation as a function of the wire diameter to
wavelength ratio.
You could calculate this more generally with
EZNEChttp://www.dxzone.com/cgi-bin/dir/jump2.cgi?ID=447
Wire diameter and insulation properties can be included.
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