** Reply to note from [EMAIL PROTECTED] Mon, 26 Mar 2007 00:40:32 EDT How about this... This information is obvious once you consider the problem. This explanation was what I was given for line of sight frequencies. DRAW a large circle (4" diameter) on a piece of paper (this is the earth). Draw a line extending from the center along a radius about 2-1/2" long (extends out of the circle). Label the center of the circle "A", and the point on the other end of this line "B". Label the line from "A" to "B" with the letter "H". >From the outer end of the line (top of antenna), draw a line that touches the >circle (tangent point), label this point "C", and label the line from "B" to "C" with the letter "D". Complete the triangle by drawing another line from the center "A" to the tangent "C" (forms a right angle), and label this line "R". Okay, the earth radius is about 4000 mi. convert to feet, this length assign to "R" (radius). Add to that the height above average terrain (HAAT) of this antenna system (same units), assign this length "H". This is a right triangle so it follow Pythagorean thereom. H*H = R*R + D*D Solve for D. "D" is the distance from the top of the tower to the horizon. To receive beyond this distance, one must be elevated. Of course, local terrain, obstacles and environmental conditions will restrict this range, but it is a good "rule of thumb". 73
Regards, Jim Scott -------------------------- Energy and persistence alter all things. ~Benjamin Franklin~
