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

"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?

The difference in ground conductivity at different frequencies. On the AM
band, all else being equal, a lower frequency will cover much more than a
high one.

"Bob Haberkost" wrote in message
news:LIBee.16270$c86.2254@trndny09...


David - I take exception to this assertion. Sure, low-end ground wave is
better...MUCH better...than the high end, but it's more like a factor of
5,
maybe 10...not 50. In fact, referenencing the FCC graphs available as
PDFs, the
560kHz graph shows that, for the 20mmhos conductivity curve, for example,
the
curve intersects 10mV/m at 9.6km, whereas for the 1550kHz graph, this
point is
at 7.6km....hardly even a factor of 2, and when squared (for coverage
area) only
a factor of 1.5 or so. So, sure, a 1kW station may have decent coverage
at
550kHz of perhaps 1000 sq.km, but a 50kW operation at 1600 kHz is still
going to
have much, much more.
--

I actually ran maps of a 5 kw below 600 and a 50 kw above 1500. Both located
in areas of the same conductivity, within a few miles of each other. The low
frequency station covers slightly less than the 50 kw right above 1500.

A good practical example is the 5 kw coverage of WMT and the 50 kw of KXEL,
a few miles apart in identical high conductivity areas in Iowa. one on 600.
The other on 1540. A run on the coverage of each reveals 600 with slightly
better coverage than 1540. Or, take the 550 real coverage of KFYR in Bismark
or the daytime 5 KW of KNAX in Yankton, and compare with WHO on 1040. Both
cover greater areas for comparable signal levels. In fact, KWMT, a 5 kw on
540 in Ft. Dodge, IA, used to advertise in the 50's as having the largest
land coverage area in the USA, and the ads were certified by one of the
major consulting engineering firms of the day.

On Fri, 06 May 2005 04:12:20 GMT, "David Eduardo"
wrote:

Bigger waves contain more energy.

"David" wrote in message
...
On Fri, 06 May 2005 04:12:20 GMT, "David Eduardo"
wrote:

Bigger waves contain more energy.


That is pretty funny.

"Bob Haberkost" wrote in message
news:LIBee.16270$c86.2254@trndny09...

David - I take exception to this assertion. Sure, low-end ground wave is
better...MUCH better...than the high end, but it's more like a factor of
5,
maybe 10...not 50. In fact, referenencing the FCC graphs available as
PDFs, the
560kHz graph shows that, for the 20mmhos conductivity curve, for example,
the
curve intersects 10mV/m at 9.6km, whereas for the 1550kHz graph, this
point is
at 7.6km....hardly even a factor of 2, and when squared (for coverage
area) only
a factor of 1.5 or so. So, sure, a 1kW station may have decent coverage
at
550kHz of perhaps 1000 sq.km, but a 50kW operation at 1600 kHz is still
going to
have much, much more.
--

I did a little more research on this, and the key issue is, of course,
conductivity. I am guilty of using the old adage of 1 kw at the bottom is
equal to 50 kw at the top... which is true on average conductivity paths to
a great extent.

In fact, I've seen different numbers like 50kw at 1600 would equal 280w at
540. I had some sites run in CommStudy and found it depends largely on
ground conductivity and what contour you want to reach.
We used a theoretical site at sea where all signals propagate over 5,000 ms
and a site in Newfoundland in Canada where it was 1ms.

At sea, it took 1,010 watts of 1600 to equal 1000 of 540 if one wanted to
get to the 25mv contour. It took 1,070 to match the 5mvs, and 2,100 watts to
match the 0.5s.

On nasty land, it took 21kw to match the 25 and 5mv contours (they weren't
the same but just eyeballing them, they were a 99% fit. It took 42+ kw to
get them to match at 0.5mv.

So, while milage may vary, the fact is that there can be as much as a 40;1
ratio on contour coverage just in these examples.

On Fri, 06 May 2005 15:38:41 GMT, "David Eduardo"
wrote:


"David" wrote in message
.. .
On Fri, 06 May 2005 04:12:20 GMT, "David Eduardo"
wrote:

Bigger waves contain more energy.


That is pretty funny.


Relationship between wavelength and idstance to the ionosphere?
Wavefront coherence?

"David Eduardo" wrote in message
m...
I did a little more research on this, and the key issue is, of course,
conductivity. I am guilty of using the old adage of 1 kw at the bottom is
equal to 50 kw at the top... which is true on average conductivity paths
to
a great extent.


Another way of explaining the frequency dependence is to consider the
phenomenon of skin effect in conductors. The higher frequencies do not
penetrate a conductor (or the earth in the case of an AM band ground wave)
as deeply as do the lower frequencies. Try to visualize the signal being
conducted via a narrow conductor vs. a thicker one, which is essentially the
case.

"David" wrote in message
...
| On Fri, 06 May 2005 15:38:41 GMT, "David Eduardo"
| wrote:
|
|
| "David" wrote in message
| .. .
| On Fri, 06 May 2005 04:12:20 GMT, "David Eduardo"
| wrote:

| Bigger waves contain more energy.

| That is pretty funny.

| Relationship between wavelength and idstance to the ionosphere?
| Wavefront coherence?

No, it's all related to conductivity, as David Eduardo suggests. The graphs
show that the 5000mmho sea-water conductivity is very nearly the same as the
ideal inverse-km line, but slightly more divergent at the higher frequency. The
less conductivity, per the graph, the more distinctly superior a watt at 540 is
over the same power at 1600. As conductivity is a resistance measurement, the
issue is that a path with less conductivity results in more losses in the
radiated wave, attenuating it more at the higher frequencies. The greater
attenuation, however, probably relates to the laws of physics in the same way as
electromagnetis spectra work at the higher frequencies....at high UHF, for
example, it's the size of the molecules in the air that attentuate the higher
frequencies more than the lower. At MW frequencies, this effect is exacted by
the size of the earth (which also is the reason why groundwaves go
over-the-horizon, as it's the earth which diffracts the wavefront).
--
-----------------------------------------------------------------------------
Men born to freedom are naturally alert to repel invasion of their liberty by
evil-minded rulers. The greatest dangers to liberty lurk in insidious
encroachment by men of zeal, well-meaning but without understanding." -- Justice
Brandeis
-----------------------------------------------------------------------------
For direct replies, take out the contents between the hyphens. -Really!-

In article ,
David Eduardo wrote:

"David" wrote in message
.. .
On Fri, 06 May 2005 04:12:20 GMT, "David Eduardo"
wrote:

Bigger waves contain more energy.


That is pretty funny.




Surf's up!

--
a d y k e s @ p a n i x . c o m

Don't blame me. I voted for Gore.