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1. Re: LF: RU6LWZ report (score: 1): Author: "Ed Lesnichy" <[email protected]>; Date: Thu, 10 Oct 2002 16:39:05 +0300; Hello Alan! I think that: Rrad = 1600*(H effective/lamb)^2= 1600*(80...100/2200)^2=2...3 Ohm P = I^2*Rrad=2.5^2*(2...3)=12.5...18.75w ERP but the aerial is close located to a metal mast... but...but.; /rsgb_lf_group-archives/html/rsgb_lf_group/2002-10/msg00292.html (8,539 bytes)
2. LF: RU6LWZ report (score: 1): Author: "Alan Melia" <[email protected]>; Date: Thu, 10 Oct 2002 13:00:01 +0100; Hi all, a quick crude calucation from Ed's drawing suggests that the Rrad of the aerial he was using could well have been in the order of 6 to 7 ohms. This would result in an ERP of about 35 to 45 wa; /rsgb_lf_group-archives/html/rsgb_lf_group/2002-10/msg00294.html (8,152 bytes)

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