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Old May 23rd 06, 04:52 PM posted to rec.radio.amateur.antenna
 
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Default sgc dipole?

Tom,

I expect there's not supposed to be any feedline between the antenna
and the tuner in this installation, so I think that it probably is to
moderate the impedance for a ham band antenna cut for the lowest band.


It is witchcraft, but if you're selling antenna tuners to people who
don't know anything about the impedance issues with an all band wire,
it's better to give them a rule of thumb to avoid cooking the tuner,
especially when they don't read the bit in the manual where it's a
"100W" tuner on SSB and they fire up on RTTY.

BKR, where did you get that explanation?

I think you should measure the RF on the outside of your coax when it's
an odd multiple of a quarter wave long including the loading effect of
the case of the radio and the ground half of your dipole is made 5%
longer than the other half.

Listen to Tom. He knows common-mode current on the feedline!

Dan
N3OX

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Old May 24th 06, 01:03 AM posted to rec.radio.amateur.antenna
Ed
 
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Default sgc dipole?




If I have a different length on the case of the tuner compared to the
length on the hot terminal, how does that moderate impedance?



Assuming you're speaking of a resonant dipole or verticle, it would
seem to me that by doing that you are moving the feedpoint away from the 50
ohm (70 ohm?) feedpoint at the center.... and probably towards a higher
impedance feedpoint. Not sure what effect that would have on those
autotuners, but perhaps they work better with a bit higher impedance than
50 ohms?




Ed K7AAT
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Old May 24th 06, 04:58 PM posted to rec.radio.amateur.antenna
 
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Default sgc dipole?

I was thinking along the same lines as Jon, LA4RT when I posted my
comment... using an 80m dipole on 40m, maybe you want to move the
feedpoint away from the current node.

How about this. If you make one leg 5% longer than the other: The
length of leg 1 is L, the length of leg 2 is 1.05L. What do you need
to have an integer number of half wavelengths on each side of the
feedpoint? (This is the only time you can get a mid element current
node, right?)

L=m*lambda/2
1.05L=n*lambda/2

= 1.05(m*lambda/2)=n*lambda/2

The lowest integers that satisfy this: n/m = 105/100 = 21/20.

So until you hit 21 halfwaves on one side and 20 halfwaves on the
other, I think you don't have a current node at the feedpoint.

Is this desirable? I think it probably is, but I am not a tuner
expert.

73,
Dan



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