On 16 Sep 2005 18:12:59 -0700,
wrote:
He did IMPLY that all conditions were equal in the case of 1 watt and
the case of 25 watts. Given that, all thats left is the old 1/r^2
which means that 1 watt gives 1/5 the distance of 25 watts.



There are some other practical considerations, mostly minimum
sensitivity of the receiver, and it's noise floor, and power losses in
antennas, which vary non-linearly with power.


--
Jim Richardson http://www.eskimo.com/~warlock
Chaos, panic, & disorder - my work here is done