On 24 jun, 16:36, "Richard Fry" wrote:
"Wimpie" wrote A broadcast transmitter (100 MHz) with ERP = 10kW produces
about 30mV/m at 3 miles and h=5 ft.

____________

Although you didn't define the height of the transmit antenna, just picking
a height of 984 feet to be on the high side, and referring to the FCC's
F(50,50) propagation curves shows that for 10 kW ERP a field of about 42
mV/m will be generated at a radius of 3 miles and an elevation of 9 meters
(29.5 feet) above the earth, assuming flat terrain.

Field strength at lower elevations for these conditions varies about in an
inverse relationship with height, so at 5 feet it would be 5/29.5, or 17% of
42 mV/m, which is about 7 mV/m.

This is quite a bit different than your number, so I was curious as to your
method.

RF

Hi Richard,

I was a little bit lazy to look into the propagation curves and do the
correction for terrain roughness. I live in the middle part of the
Netherlands where the country is really flat. Therefore I just took
the two-ray formula (and checked that the result is below free space
propagation). Normally spoken when I do not have reliable propagation
curves at hand, I use the two ray formula and add about 10..20 dB to
the loss.

The calculation example was just to show how (the procedure) someone
can calculate the received power from a non-resonant whip (that was
the original question). When using the FCC 50% curves, the output
would be even less (as your example showed).


In my previous posting I said:
"Using one of the above methods, will result in E is proportional to 1/
r^3...to 1/r^4. This matches measurements I did myself. Therefore 1/r
gives a far from realistic result (too optimistic)."

That had to be be:
"Using one of the above methods, will result in Preceived is
proportional to 1/r^3...to 1/r^4. This matches measurements I did
myself. Therefore 1/r^2 gives a far from realistic result (too
optimistic)."

Best regards,

Wim
PA3DJS.
www.tetech.nl