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G0FTD > BEACON 09.07.03 01:42l 49 Lines 1683 Bytes #999 (0) @ WW
BID : CD1284G0FTD
Read: DF6DBF DB0FHN GUEST
Subj: Re: 10mW BCN is heard on 14.318,4mc
Path: DB0FHN<DB0RGB<OK0PPL<RZ6HXA<ER3KAZ<DB0RES<ON0AR<IK1ZNW<GB7CRV<GB7CIP<
GB7SXE
Sent: 030708/1915Z @:GB7SXE.#38.GBR.EU #:8529 [Hastings] FBB7.00i $:CD1284G0FTD
From: G0FTD@GB7SXE.#38.GBR.EU
To : BEACON@WW
PA1SDB sent:-
> Until now the 10mW QRP signal has been heard by 4 listeners. The max. DX
> distance is 1063 km. The most close distance is 288km.
Congratulations to Peter for a worthy experiment.
For my personal study I tried to see what this meant.
Firstly 10mW @ 14Mhz and at a distance of 1063Km means a free space path loss
of 116dB. Thus assuming NO reflection loss via the F-Layer would mean the
beacon would be recieved at a strength of -106dbm.
Now there is usually an figure thrown about of 10db attenuation per hop, but
this is VERY variable :-)
Now for one hop F Layer signals, this means a take off angle of 30 degrees
for 1000km and for 288km being about 65 degrees takes off.
It is interesting that typical antennas would have have the main lobe in
such a range.
I conclude that:
1) Each time the signal was heard only 1 hop was involved.
2) That assuming 10db per hop (=10db in this case) the signal received
was at a level typically of -116 or less. This would equate on a modern
S-meter as about S3.
3) That 10mW would enable AT best upto 1000km communication, no more due to
signal attenuation on the second hop bring the signal down below the
noise floor of a *typical* modern day terrestial receiving apparatus.
4) If 10mW is enough for *Typical* 1 hop 1000km range, then to get from
northern europe to New Zealand would take about 17 hops (assuming 30 degree
take off) would be 10mW multiplied by 17 which equals 170mW BUT BUT,
do we *really* know what each F - Layer hop attenuation is ?
Ok, I've started the ball rolling, does anyone want to take it further ?
- Andy -
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