Return-Path: Received: (qmail 24490 invoked from network); 21 Nov 2004 22:45:48 -0000 Received: from unknown (HELO ptb-spamcore01.plus.net) (192.168.71.1) by ptb-mailstore01.plus.net with SMTP; 21 Nov 2004 22:45:48 -0000 Received: from mailnull by ptb-spamcore01.plus.net with spamcore-l-b (Exim 4.32; FreeBSD) id 1CW0kc-0007La-63 for dave@picks.force9.co.uk; Sun, 21 Nov 2004 23:04:12 +0000 Received: from [192.168.67.2] (helo=ptb-mxcore02.plus.net) by ptb-spamcore01.plus.net with esmtp (Exim 4.32; FreeBSD) id 1CW0kc-0007LX-3N for dave@picks.force9.co.uk; Sun, 21 Nov 2004 23:04:10 +0000 Received: from post.thorcom.com ([193.82.116.20]) by ptb-mxcore02.plus.net with esmtp (Exim) id 1CW0Sn-000CKf-Cl for dave@picks.force9.co.uk; Sun, 21 Nov 2004 22:45:45 +0000 Received: from majordom by post.thorcom.com with local (Exim 4.14) id 1CW0Rt-0004AA-Su for rs_out_1@blacksheep.org; Sun, 21 Nov 2004 22:44:49 +0000 Received: from [193.82.116.30] (helo=relay.thorcom.net) by post.thorcom.com with esmtp (Exim 4.14) id 1CW0Rt-0004A1-Cf for rsgb_lf_group@blacksheep.org; Sun, 21 Nov 2004 22:44:49 +0000 Received: from imo-m22.mx.aol.com ([64.12.137.3]) by relay.thorcom.net with esmtp (Exim 4.41) id 1CW0Rp-0001Yl-6j for rsgb_lf_group@blacksheep.org; Sun, 21 Nov 2004 22:44:49 +0000 Received: from MarkusVester@aol.com by imo-m22.mx.aol.com (mail_out_v37_r3.8.) id l.90.514dcae7 (4254) for ; Sun, 21 Nov 2004 17:44:29 -0500 (EST) From: MarkusVester@aol.com Message-ID: <90.514dcae7.2ed2744d@aol.com> Date: Sun, 21 Nov 2004 17:44:29 EST To: rsgb_lf_group@blacksheep.org MIME-Version: 1.0 X-Mailer: 8.0 for Windows sub 6104 X-SPF-Result: relay.thorcom.net: domain of aol.com designates 64.12.137.3 as permitted sender X-Spam-Score: 0.7 (/) X-Spam-Report: autolearn=no,HTML_MESSAGE=0.001,HTML_TAG_BALANCE_HTML=0.411,NO_REAL_NAME=0.285 Subject: LF: Estimating CFH's radiated power Content-Type: text/html; charset=windows-1252 X-Spam-Checker-Version: SpamAssassin 2.63 (2004-01-11) on post.thorcom.com X-Spam-Status: No, hits=0.7 required=5.0 tests=HTML_MESSAGE, HTML_TAG_BALANCE_HTML,NO_REAL_NAME autolearn=no version=2.63 X-SA-Exim-Scanned: Yes Sender: owner-rsgb_lf_group@blacksheep.org Precedence: bulk Reply-To: rsgb_lf_group@blacksheep.org X-Listname: rsgb_lf_group X-SA-Exim-Rcpt-To: rs_out_1@blacksheep.org X-SA-Exim-Scanned: No; SAEximRunCond expanded to false X-Spam-Filtered: by PlusNet SpamCORE (v3.00) Content-transfer-encoding: 8bit Dear John and group,

ok, I'll try...

One kilowatt EMRP across 700 km of lossless flat earth would give
109.6 - 20 log (697) dBµV/m = 52.6 dBµV/m.
The CCIR 1963 graph for seawater (4 S/m) says 45 dBµV/m at 150kHz, so we have lost ~8dB, mostly to the spherical diffraction. As a first approximation, we could take that and then simply add the extra ground loss for 200 km. Extracting this from the 1mS/m curve gives another 8dB, but I'd use 6dB as a realistic guess for 1.6mS/m conductivity. Thus you'd get 39 dBµV/m from a kW, give or take a couple of dB.

Your measurement of 235 µV/m = 47.4 dBµV/m would thus indicate an EMRP of
8.4 dBkW = 7 kW,
more or less in line with Alan's suggestion that it could be somewhat less than their 20kW maximum rating. Through ~ 4500 km of free space, CFH could theoretically produce
109.6 + 8.4 - 73 dBµV/m = 45 dBµV/m
in Europe. At 20:00 today the plot beneath my grabber showed around 4 dBµV/m, which means 41dB path loss for reflections and D-layer absorption.

An amateur at comparable distance using full 1W ERP (ie. 0.55W EMRP) would be 41 dB weaker than 7kW CFH. To copy him here in a bandwidth of 10.5mHz (QRSS 60), the SNR of CFH in my plot would have to climb from currently ~40dB to at least ~50dB.

73 and best wishes

Markus, DF6NM

In einer eMail vom 21.11.2004 21:12:14 Westeuropäische Normalzeit schreibt w1tag@w1tag.com:

Markus,

Curiosity took over, and I attempted a mid-afternoon measurement of the CFH
carrier's field strength. I get 235 uV/M at 697 kM. A quick check with a
great circle plot shows about 70% of that path to be over sea water. Ground
conductivity for the rest of the path is awful (blame the last ice age), no
better than 1/2500 of sea water.

If anyone can factor in the earth's curvature and calculate an unattenuated
field from this, be my guest.

John Andrews