n8de@thepoint.net wrote:
> Instead of bringing the ladderline directly down the side of the
> tower, 'twist' it as you do, thereby changing the distributed
> capacitance of the two wires to the tower and 'equalizing' any
> capacitive problems.
> 3 or 4 turns should suffice for 50'.
>
> One foot sounds good to me, too.
>
A foot sounds convenient..
But, one can actually calculate this pretty easily.
It's mostly a magnetic field problem, rather than a capacitance problem,
but the equations are the same. The field from a single wire drops off
as 1/r.
So, you have two wires, carrying opposite currents/voltages.
The field at some point is then 1/r1  1/r2, where r1 is the distance
from the first wire and r2 is the distance from the second wire.
On the center line, symmetric to the two wires, there's no net field..
the worst case is when you are to the side, directly lined up..
If the two wires are spaced by D, then the distance to one wire is, say,
r and the other is r+D..
the field is 1/r  1/(r+D).
It's nice to think of these things as "how many wire diameters or wire
spacings away"... so you normalize by D and get
1/(R*(R+1)) where R is the distance in "cable widths"
Let's say our cable is an inch wide.. and we're a foot away.. that makes
R = 12
The field is 1/(12*13) or about 0.006.
relative to what? The field of one wire, at the other wire, is "1" (i.e.
r=D, so R=1)
That is, if the wires were carrying a current of 1 amp, it would induce
a current of 6mA in the tower.
Turning this into a dB kind of thing, we take 20log10(0.006) and we get
about 40 dB..
Just for grins, for R = 5, you get 29dB, for R = 2, 15dB...
So, if your 1" balanced line is spaced just 2 inches away, and oriented
edge on, you're still looking at a sub 10% kind of effect.
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