Message-ID: <001701c0dbb9$d035e6c0$1a8c01d4@g4jnt>
From: "Andrew Talbot" <G4JNT@thersgb.net>
To: rsgb_lf_group@blacksheep.org
Subject: Re: LF: Measurement of antenna current
Date: Sun, 13 May 2001 15:34:26 +0100
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You've made me think about it a lot more now.  The AC/DC resistance argument
is certainly valid.   It does indeed mean that any capacitive field will be
captured with less loss than letting it end up in lossy ground,  or it would
if the sheet covered the area captured by the E field due to the WHOLE
antenna.

However, just concentrating on the loading coil, which has a huge H field
and very low E component is a different matter.   What is the loss due to
the H travelling through ground of poor conductivity ? - not too much I
would estimate

More thought required..............

Andy  G4JNT


>I thought it means that the current density over the thickness of the foil
>is almost uniform with result that AC and DC resistance are almost equal.
>But even with the thin foil that resistance will be so much smaller than
the
>resistance via the real earth that I expect  almost all capacitive current
>from the coil is captured by the foil.
>Am I wrong?
>
>73, Dick, PA0SE
>
>
>
>