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Re: [TowerTalk] TT SHUTDOWN

To: JC Smith <>,TowerTalk <>
Subject: Re: [TowerTalk] TT SHUTDOWN
From: Red <>
Date: Wed, 12 Jul 2006 14:04:24 -0500
List-post: <>
Sri, gang;  the Scot, Maxwell, and the Englishman, Heaviside, decree 
that RF will still concentrate on the outside of a conductor.  The 
distribution of current in depth is an exponential function, but the 
effective skin thickness, in copper, is 0.0026" at 1 MHz, 0.000826" at 
10 MHz, and 0.00026" at 100 MHz.  Reference _Radio-Electronic 
Transmission Fundamentals,_ B. Whitfield, Jr.  It is a semi-technical 
book, long out of print, that offers good explanations for the 
non-engineer qualified ham.  It is worth searching for on the used book 
market.  Meanwhile, copy those numbers and paste them where you can find 
them.  When calculating the resistance of a copper conductor, don't use 
the cross sectional area, use the circumference times the effective skin 
depth to calculate the effective area of conduction.

Another advantage of strap, or widely spaced wires, is reduced 
inductance per unit length, compared with a cylinder of similar 

1994 Radio Amateurs' Handbook, page 2-20, providing  the following 

For a flat strap:

L = 0.00508b[ln (2b/(w+h) + 0.5 + 0.2236 (w+h)/b]

    L = inductance in microhenrys
    b = length in inches
    w = width in inches
    h = thickness in inches.

For a round wire:

L = 0.0002b[(ln(2b/a)) - 0.75]

where L = inductance in microhenrys
    a = wire radius in mm
    b = wire length in mm
    ln is the natural logarithm

Sorry about the change from inches to mm.  I'm quoting ARRL.  I believe 
the following is correct with the dimensions in inches:

L = 0.00508b[(ln(2b/a)) - 0.75]

Inductance of parallel wires spaced apart approximates the inductance of 
parallel inductors.  Their paralleled resistance is also lower than that 
of a single wire at HF frequency if the total circumference or surface 
area is greater, assuming the material is the same.

73 de WOØW

JC Smith wrote:

>  What if
>you slit the copper tube before using it, would that be enough to allow both
>the inside and outside to conduct high frequency current?

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