Thanks Deanne. Inquiring minds want to know (maybe just me, hi), how can you figure ERP from power (25 Kw) and the numbers from the plot info? That'd be good info to know (i.e. I know that my local WNDE 1260 is stronger at night and want to know what the ERP is versus the actual 5 Kw, transmitted).
73, Dave ------------------------------ Message: 2 Date: Wed, 29 Sep 2010 13:31:38 -0600 From: Deane McIntyre <[email protected]> To: Mailing list for the International Radio Club of America <[email protected]> Subject: Re: [IRCA] 1130 - interference reported by CKWX Message-ID: <[email protected]> Content-Type: text/plain; charset=us-ascii On 2010-09-29, at 1:20 PM, HASCALL, DAVID CIV DFAS wrote: > I do not know how to figure out effective radiated power but I could see > where 25Kw., directly NE of the station, would sound like 50 Kw., or > better. > > http://www.fcc.gov/ftp/Bureaus/MB/Databases/AM_DA_patterns/1355584-11043 > 2.pdf > You are correct - effective radiated power works out to 81.6 kW at an azimuth of 45 degrees (and 51.2 kW straight north) 73, Deane McIntyre VE6BPO ********************** _______________________________________________ IRCA mailing list [email protected] http://montreal.kotalampi.com/mailman/listinfo/irca Opinions expressed in messages on this mailing list are those of the original contributors and do not necessarily reflect the opinion of the IRCA, its editors, publishing staff, or officers For more information: http://www.ircaonline.org To Post a message: [email protected]
