[TowerTalk] Balanced Line using Coax ?

[Michael Tope]

Don't feel bad, Jim. I went to bed last night thinking that it
was 1/2 the loss, but when my head hit the pillow I started
to imagine two separate generators (one driving each coax
with 1/2 the total power) which made me start to think that
the loss was the same in either case. Fortunately, when I
came back downstairs I realized that I hadn't hit send yet :):)

Anyway, I still think its the same, but perhaps are more
rigorous analysis will show otherwise. I'll leave that to
someone else as this will all be academic if I don't get
my tower application started :):)

73 de Mike, W4EF............................................

----- Original Message -----
From: "Jim Lux" <jimlux at earthlink.net>
To: "Chuck Counselman" <ccc at space.mit.edu>; <Towertalk at contesting.com>
Sent: Tuesday, July 29, 2003 10:54 AM
Subject: Re: [TowerTalk] Balanced Line using Coax ?


> Now I am really confused.. enough of the facile quick stuff.. time for the
> rigorous analysis.. stay tuned, but more facile analysis below
>
> At 12:55 PM 7/29/2003 -0400, Chuck Counselman wrote:
> >At 9:34 AM -0700 7/29/03, Jim Lux wrote:
> >>...Looking at loss as a dB/foot for the coax, you divide the power into
> >>two pieces of coax, so the absolute loss (in watts) will be half in each
> >>piece of coax, but you've got two coaxes, so the total loss is exactly
> >>the same....
> >
> >
> >No.  To deliver the _same_power_ as a single coaxial line, two coaxial
> >lines operating in "push-pull" deliver half the current at twice the
voltage.
>
> But half the power flows through each line, no?  and, half the power isn't
> half the current.
>
>
> >  At HF, virtually all of the loss is ohmic, and the power dissipated per
> > unit length of conductor is equal to I^2*R', where I is the current in
> > the conductor, I^2 is I squared, and R' is the resistance per unit
length
> > of conductor.
>
> Yup..
>
>
> >  With half the current, the power dissipated per unit length of
conductor
> > is quartered;
>
> Ok.. but, is the current halved, or 0.707'd... P/Z0 = I^2... current would
> go as the square root of the power.
>
>
> >the length of conductor is doubled; so the total power dissipated in the
> >pair of push-pull coaxial lines is one-half that dissipated in a single
> >coaxial line carrying the _same_power_.
>
>
>
>
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