Mime-Version: 1.0 (Apple Message framework v1078)
From: John Rabson <john.rabson@numeo.fr>
In-Reply-To: <90CFBF56-FB16-42F3-B830-E7D695D2B184@hurontel.on.ca>
Date: Thu, 27 May 2010 19:00:11 +0200
Message-Id: <2153A39D-8114-42E9-89CF-45325F8684A2@numeo.fr>
References: <AANLkTinG-0mJJ3CDqFwxeLXfOJXFFfs42gjLYz8yX3rW@mail.gmail.com><38A51B74B884D74083D7950AD0DD85E82A1C00@File-Server-HST.hst.e-technik.tu-darmstadt.de><AANLkTilyd93InqPvp5qUVkonthTgtMaIVqVfH3zyen2B@mail.gmail.com><Pine.LNX.4.64.1005252313060.26690@paranoid.lipkowski.org><AANLkTimudswbmeqd9Ej52egRUOJ4RHIf5VtYiQ7_7d8T@mail.gmail.com><BF4A524700075746A6467658DFC7102C5106464EEA@ICTS-S-EXC2-CA.luna.kuleuven.be><AANLkTimXcAWlXyxfo4gAThfv3maPkFNyKK6HYMNRxhk3@mail.gmail.com><BF4A524700075746A6467658DFC7102C5106464EEB@ICTS-S-EXC2-CA.luna.kuleuven.be><BF4A524700075746A6467658DFC7102C5106464EEC@ICTS-S-EXC2-CA.luna.kuleuven.be> <AANLkTik6aEnxQbPCvn16LZS-hVb9L3uvwwNFBMgrVRg_@mail.gmail.com>, <38A51B74B884D74083D7950AD0DD85E82A1C0A@File-Server-HST.hst.e-technik.tu-darmstadt.de> <1016715129.3429947.1274945594131.JavaMail.fmail@mwmweb033> <90CFBF56-FB16-42F3-B830-E7D695D2B184@hurontel.on.ca>
To: rsgb_lf_group@blacksheep.org
Subject: Re: LF: Earth electrodes -  depth or surface area?
Content-Type: text/plain; charset=us-ascii
Content-Transfer-Encoding: quoted-printable
Sender: owner-rsgb_lf_group@blacksheep.org
Precedence: bulk
Reply-To: rsgb_lf_group@blacksheep.org


JB, Roger, LF,

It isn't  quite that simple. The standard analysis of a long thin ground=
  rod  (as per David Gibson's article) is to  estimate  the radius of the=
 buried hemisphere  which would give the same capacitance. Knowing the res=
istivity of the ground and  the estimated capacitance you can work out the=
 equivalent resistance. I don't have the analysis to hand but I seem to re=
call it models the rod as an oblate spheroid (as long as the length/diamet=
er ratio is not too great which it isn't in the caving situation). =20

I seem to recall that the effective radius of a thin rod is about one fift=
h of its length. To get the benefits of multiple ground rods you need to=
 space them by at least twice their  length.

73
John F5VLF

On 27 May 2010, at 15:13CEST, John Bruce McCreath wrote:

>=20
> Hello LFers,
>=20
> While laying in bed this morning pondering the inside of my eyelids I go=
t to thinking about the electrodes used
> for "through the earth" communications.  What is more important, the dep=
th of the electrode or its surface area?
> A typical 3/4 inch diameter by 4 foot long ground rod has a surface area=
 of 113 square inches, while a metal
> plate 1 foot square has a surface area of 288 square inches.  If depth=
 is the key, then obviously the rods have
> the advantage, but if it's surface area, the plates win hands down.  Eve=
n in stoney soil, it's relatively easy to
> make a slit-like hole into which could be slipped a sheet of galvanized=
 metal with a lead attached.  To make
> a good connection to the surrounding soil, pour some "kitty litter" into=
 the slit and moisten it with water so as
> to improve the contact between the plate and the surrounding soil.  Am=
 I onto something here or have I
> overlooked some important detail?  =20
>=20
> 73, J.B., VE3EAR
>=20
> LowFER Beacon "EAR"
> 188.830 kHz. QRSS30
>=20
>=20
>=20
>=20
>=20
>=20
>=20