In article ,
"Brenda Ann" wrote:

"Telamon" wrote in message
..
.
In article ,
"David Eduardo" wrote:

"Michael Lawson" wrote in message
news:16b7f$427a2948$d8c4c9e6 That big
test is happening now in Cincy, where some of AA's
content is on one of two clear channel stations
in the area (WCKY). For comparision, Rush, Savage
and the gang are not on the other clear channel station,
WLW, but are on 550 WKRC, which has significantly
less reach than WCKY.

550 has better daytime coverage and overall coverage of the local market
than 1530. As a rule of thumb, 1 kw on 540 is equal to 50 kw on 1600 at
the
same location, so 550 and 1530 are going to get about equal local
coverage... with 550 winning in some areas.

Snip

What is the basis of this power/frequency rule?


Brain fade.. I accidentally posted this response in the wrong thread (then
inadvertantly clicked too many times and sent a blank answer here)

I can't vouch for the exact math, but it has to do with ground conductivy
and I squared R losses versus frequency. Lower frequencies have better
ground conductivity (hence ground wave) than higher ones. Much more of the
signal is sky wave at the high end of the dial. Many 50KW stations on the
high end of the dial don't even cover their service areas at night (When I
lived in west Portland, OR, in the 70's, I used to get nearly as much signal
on 1520 from KOMA in OKC as I did from (what was then KYXI) on the same
frequency a few miles away in Clackamas.

I would never have expected that large a difference in a loss result for
a vertically polarized signal except at the transmitter antenna. At the
transmitter antenna I would expect that the ground radial systems
employed would reduce the differences there between 500 KHz and 1500 KHz.

--
Telamon
Ventura, California