On Dec 22, 11:13*am, "Frank" wrote:
In this example the vertical half wave dipole, with the base 30 ft above
an average ground, on 147.3 MHz, shows a field strength at ground
level of: 0.418 uV/m from 30 W into the antenna.

And, obviously, at 50 km.
________________

Here is another method (Longley-Rice) for calculating the field
intensity produced at the receive site by your model. But the NEC
approach is less accurate than L-R for long path lengths (due to earth
curvature), and for specific terrain contours.

In your model the path loss calculated using L-R is about 68.8 dB more
than the free space loss. The peak, free space field produced by a
1/2-wave, linear dipole radiating 30 watts over a 50 km path is about
770 uV/m. This voltage reduction of 68.8 dB is a field multiplier of
about 0.00036, so the 770 uV/m field is reduced to about 0.28 uV/m --
a bit less than your NEC model predicts. Agreement probably would be
better over shorter paths (as long as no specific terrain profile
needed to be applied), and worse for longer paths.

In the L-R example I set the path over the middle of Lake Michigan in
order to get a smooth earth contour, such as used in NEC models.

This all just illustrates that analyses made using NEC and any other
method need to consider the limits inherent in their algorithms with
respect to the physical reality being analyzed.

http://i62.photobucket.com/albums/h8...strialPath.gif

RF