Mike W4EFI wrote:
I get that at any point in the far field there is RF current in the ground
due to the space wave from the transmitter reflecting obliquely off
ground.
If the earth had perfect conductivity then an EM wave radiated into space
from a vertical monopole would have no tilt where it encountered the earth
boundary, and there would be no horizontal voltage component of it to
produce r-f currents at, and just below the surface of the earth beyond ~1/2
wavelength from the radiator.
It is the lossy conductive nature of the earth for r-f energy that produces
those far-field ground currents, not because radiation from the transmit
antenna is reflecting obliquely off the surface of the earth.
I was, however, under the impression that at very low angles the field
from this reflection cancelled direct radiation for earth of finite
conductivity. I think you can see this in the Distance=20 mile E-field
plot you provided for the 1.83 MHz radiator. In that plot there is an
inflection in the E-field versus height very close to the horizon. If you
ignore that inflection and extrapolate a smooth curve from the E-field at
higher elevation angles down to zero elevation angle you can see that the
field would go to zero. <snip>
An h-pol, direct path E-field tends to be cancelled at/near zero elevation
by its reflection from a flat ground plane -- but not so for a v-pol, direct
path E-field. The ground plane reflection of a v-pol E-field can add up to
3.01.. dB field intensity to the direct wave.
Note that the fields in these NEC calculations were made for one degree
elevation steps starting at zero degrees. The plotted field for zero-degree
elevation at 20 miles is a CALCULATED field. So that field value cannot be
ignored, and replaced by a presumed extrapolation of zero field at an
elevation angle of zero degrees..
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