An easier way to measure the EM properties of a tree at HF would be to wind
a big coil around the tree, and hook it up to your antenna analyzer and
measure the change in Q, compared to the same measurement in free space.
You could also just wind a single turn, resonate it with a capacitor, and
look at the Q that way.
This technique is used to measure water content in various things (wood,
soil, etc.), since the loss is highly dependent on the water content. Once
you've calibrated it against some known standards (typically by taking the
sample and gradually drying it), it's fast, easy, and repeatable.
----- Original Message -----
From: "Guy Olinger, K2AV" <olinger@bellsouth.net>
To: "Michael Tope" <W4EF@dellroy.com>; "Tom Rauch" <W8JI@contesting.com>;
"RICHARD BOYD" <ke3q@msn.com>; "towertalk reflector"
<towertalk@contesting.com>; "Jim Lux" <jimlux@earthlink.net>
Sent: Wednesday, September 15, 2004 6:06 PM
Subject: Re: [TowerTalk] verticals in woods vs. in a field
> I'm not sure the dielectric properties of such a poor conductor will
> mean much. The dielectric of something touching a conductor is another
> matter, but these are not "close" and the field around conductors
> diminishes very quickly.
Close, in antenna near field terms, would be within a wavelength or so
(depending on the size of the antenna, etc.), so for a 40m or 80m, that's a
fair distance. Whether a lossy tree would have significant loss
contribution at 80 meters away is another story. It might happen to be 377
ohms, and look like free space.
>
> I went outside and took the sharp picks of my Fluke multimeter and
> stuck them through the bark into the wet of several different kinds of
> live trees. An oak measured over 1 meg across two feet vertical on the
> trunk. One of my maples measured over 100K and interestingly did a
> very slow charge like an electrolytic capacitor, measuring over 500K
> after five minutes.
>
What you're seeing is the electrolytic decomposition of the tree with a DC
bias. (also called "electrode polarization")
AC resistance and DC resistance will tend to be different.
I think that the most conductive layer of the tree is going to be the
cambium, right under the bark, with the heartwood not being such a great
conductor (doesn't most of the water transport occur in the cambium?). A
two point measurement will read quite high, even for a sheet of a fairly
good conductor. Consider a sheet of space cloth with 377 ohms/square. If
you had a 1 cm wide electrode on each end, and measured over 60 cm distance
would read 22K. You've got electrodes that are a lot smaller than 1cm wide
(maybe a couple mm?). That same sheet resistance with 2mm wide electrodes
would be 100K.
> All you need is a multimeter with sharp enough prods to dig into the
> wet layer of the tree.
>
> If there isn't something in the tree that will allow a fair amount of
> current, there won't be enough current to induce loss by the
> dielectric properties. 50K or more ohms per foot not exactly conducive
> to current.
But a sheet that's 2 feet across and 2 feet long will be only 300-500 ohms.
>
> 73, Guy.
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