Practical_Antenna_Handbook_0071639586

36 p a r t I I : F u n d a m e n t a l s
is that the earth has a radio radius equal to about 4 3 (K = 1.33) of its physical radius, as
discussed previously.
Distance d is found from the expression
d = 2r o
h
(2.26)
where d = distance to radio horizon in statute miles
r o = radius of earth in statute miles
h = antenna height in feet
Accounting for all constant factors, the expression reduces to
d
= 1.414 h
(2.27)
all factors being the same as defined previously.
Example 2.2 A radio tower has a UHF radio antenna that is mounted 150 ft above the
surface of the earth. Calculate the radio horizon (in statute miles) for this system.
Solution
r
d = 1.414 (h) 1/2
= (1.414)(150
ft) 1/2
= (1.414)(12.25)
= 17.32 mi
For other units of measurement:
d (nmi) = 1.23 h
(2.28)
and
d ( km)
= 1.30 h
(2.29)
Surface Waves
The surface wave travels in direct contact with the earth’s surface, and it suffers a severe
frequency-dependent attenuation from absorption by the ground.
The surface wave extends to considerable heights above the ground level, although
its intensity drops off rapidly at the upper end. The surface wave is subject to the same
attenuation factors as the space wave but, in addition, it suffers ground losses. These
losses are caused by ohmic (resistive) losses in the conductive earth and by the dielectric
properties of the earth. In short, the signal heats up the ground. Horizontally polarized
waves are not often used for surface-wave communications because the earth
tends to short-circuit the E-field component. (A perfectly conducting plane has no voltage
between any two points on it; hence, no E-field can exist on the plane.) For verti-