Cec, there you go again, quoting the bibles!

If the F2 is present then it must be daylight.

And if its daylight then the E-Layer must also be present.

For a groundpath distance of 400 miles the most likely reflector is
the E-Layer. In daylight the E-Layer is an excellent, stable
reflector. Furthermore, the radio path distance is considerably
shorter than via the higher F1 or F2-Layers.

The height of the E-Layer is about 70 miles.

Assuming a flat Earth, the elevation angle is -

ArcTan(70/200) = 19 degrees.

It depends to some extent on the E-Layer critical frequency and the
MUF. If the MUF is low enough and the transmit frequency is high
enough, e.g., 14 or 21 MHz, then the wave may pass right through the
E-Layer and be reflected most likely by the F1-Layer at a height of
roughly 300 miles.

Signals received via the F-Layers, if received at all, will be weaker
than via the E-Layer, if only because the path length is greater.

The elevation angle via the F1-Layer will be about -

ArcTan(300/200) = 56 degrees.

At night the E and F2 layers disappear. There remains only the
nighttime F at a height of about 200 miles to give an elevation angle
of 45 degrees.

That is, of course, if the nighttime-F critical frequency and MUF
allows propagation. There are such things as skip distances.

( Roy, no, I havn't made arrangements with high power broadcasters at
known distances to make tests to prove the foregoing predictions. And
yes, I know it's refraction and not reflection.)
----
Reg, G4FGQ.
========================================

"Cecil Moore" wrote in message
. net...
Ken Bessler wrote:
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.

That chart is in the ARRL Antenna Book. Assuming F2 layer
reflection, the arriving angle is reported to be in excess
of 50 degrees.
--
73, Cecil http://www.qsl.net/w5dxp