On 2011-05-16, at 9:09 PM, Stephen Airy wrote: > Hi all... > > Does anyone know how to calculate the effective radiated power (or whatever > is the correct term) for an AM directional station along a particular heading? > Using 1170 KCBQ as an example, I know: > the actual transmitter power (50kW), > the RMS field at 1 km (2,499.50 mV/m - assuming it's operating > non-directional - found this in the FCC database) > the field at 1km in a particular direction (3,750 mV/m - > estimating/interpolating between the 185° 3559.01mV/m and 190° 3935.08mV/m - > actual heading is 187.3°) > >
ERP = xmtr power x (field at 1km /RMS field at 1 km)^^2 In your example. 50 kW x (3750/2499.5)^^2 = 50 kW x 2.251 = 112.5 kW ERP 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]
