That equation is valid for a homogeneous conductor. When you have a sandwich of
several layers with different conductivity you have to calculate it more
thoroughly by using the formula for each of the sections (which makes it
harder).
The point still is that a layer of (very) bad conductivity over a good
conductor will only have a small (if any) influence. Like was pointed out
earlier, it's like two resistors in parallel, the resistor with a very high
resistance will not have any major influence on the current in the resistor
with low resistance.
I do like your suggestion of making a coil and measure the resistive value of
the coil at different phases of the deterioration.
Hans - N2JFS
-----Original Message-----
From: jimlux <jimlux@earthlink.net>
To: towertalk <towertalk@contesting.com>
Sent: Wed, Dec 28, 2016 1:13 pm
Subject: Re: [TowerTalk] UV and WX deterioration of THHN insulation, and effects
On 12/28/16 7:49 AM, jimlux wrote:
> On 12/28/16 5:51 AM, Patrick Greenlee wrote:
>> Skin effect... If skin effect can force conduction into the outer limit
>> of the wire (the chemically altered part with poor conductivity) then
>> why doesn't the skin effect force conduction out into the insulation and
>> really have poor conduction? (or in bare wire out into the surrounding
>> air)
>>
>> My friend and guru (who refuses to post here) has been a ham for several
>> decades, is a retired EE, and has 35+ years antenna design experience
>> (his specialty) agrees with the concept that RF conductivity can be
>> characterized as a collection of parallel impedances, a continuum
>> actually. The depth of penetration of RF in a conductor does not have a
>> "magic" cut-off point but instead has an exponential extinction. That
>> is, the deeper into the conductor the less RF but there is no magic
>> barrier preventing RF from penetrating to any arbitrary depth, although
>> at rapidly reduced values.
>>
>
> Exactly this..
> Skin depth is a convenient way to measure the exponential fall off: it
> is the depth at which if you had a uniform slab of that thickness and
> uniform current density it would have the same resistance as an
> infinitely thick slab..
>
> That is: you can calculate the resistance by Skin depth* width * length
> * resistivity.
Oops.. resistance = length*resistivity/(skin depth * width)
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