LF: Estimating CFH's radiated power

[MarkusVester@aol.com]

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