Reg Edwards wrote:
"Ken Bessler" wrote in message
news:VEmag.22577$4H.10017@dukeread03...

Lets assume a single hop 40m signal from 400 miles away. What
elevation angle does it arrive at? Both stations are using inverted

V's

at nominal height. There are no large bodies of water in between.

Daytime and/or nighttime.

--
73's de Ken KG0WX

===================================

The type of antenna or its radiation pattern has nothing whatever to
do with the path taken by the radio wave through the ionosphere. The
take-off angle and its name, generated by EZNEC, can be very
misleading.

It is hard to look at a radiation pattern, conclude that the take-off
angle is the only angle of radiation, and then blame it on EZNEC! Most
of the antennas that I have modeled seem to have radiation in lots of
directions. 8^) Otherwise you are correct.



The radio path is simply a matter of trigonometry involving only the
groundpath distance between transmitter and receiver and the height of
the reflecting layer.

The height of the reflecting layer changes between day and night. And
there may be more than one layer present in daylight. The layer
actually used depends on frequency.

If the Tx and Rx stations are far apart, the trigonometry becomes a
little bit complicated because of the curvature of the Earth's
surface. But for groundpath distances up to 500 miles a flat earth can
be assumed. Get a sheet of paper and a pencil and sketch the triangle
to be solved. The average height of the F-Layer in darkness is about
200 miles. In daylight it is about 300 miles.

To do the actual calculations download program SKYTRIG from website
below in a few seconds and run immediately. SKYTRIG is near the bottom
of the list on the "Download Progs From Here" page. Just left-click
on it.


- 73 de Mike KB3EIA -