"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.