Richard Harrison wrote:

Roy, W7EL wrote: "The radiation at zero elevation angle is zero for
any antenna height and ground conductivity."

It is the equal and opposite reflection which cancels the incident
wave at an equal distance from the reflection point, i.e. toward the
horizon.

Extremely low conductivity would produce radiation as in free space;
no cancellation in the horizontal plane.

No, that's not true. The radiation at zero elevation angle is zero for
any antenna height and ground conductivity, even "extremely low".

Sometimes it's necessary to look beyond simplified quotes read in books,
and read and understand the underlying math. In this case, the math can
be found as equation 1(*) on p. 717 of Kraus' _Antennas_, 2nd Ed. (I
thought this was also in the 1st Ed., but can't find it there.) This is
the equation for reflection coefficient from ground for horizontal
polarization. It's easily seen that it equals -1 at an elevation angle
(alpha) = 0 for any value of ground conductivity or permittivity. (Note
that epsilon-sub-r is the complex permittivity, as shown in eq. 4(**),
which contains the conductivity.) The total field is given in equation
3(***), which shows that its value is zero when the elevation angle is
zero and the reflection coefficient is -1.

This is, however, entirely theoretical. As the elevation angle gets
lower and lower, the reflection point becomes farther and farther from
the antenna. So a reflection at zero elevation angle takes place at an
infinite distance from the antenna. Among other problems, this requires
an observation point an infinite distance away and a perfectly flat
ground plane that's infinite in extent. While this is the standard for a
lot of theoretical analysis and in programs like NEC, MININEC, and
EZNEC, it of course can't be constructed in reality. In real life, the
Earth curves away, so if the terrain is perfectly flat, extremely low
angle radiation never strikes the ground. (Without working through the
numbers, I'd guess this to be below a small fraction of a degree for a
moderate antenna height. But it would be easy to calculate.) Again
theorectically, still using a simple reflection model but now with an
idealized model of a spherical ground, you'd get a free space pattern at
zero elevation angle and up to a fraction of a degree *regardless of the
ground conductivity*. But when slicing things this thin, you probably
need to use a better reflection model, which takes into account
dispersion and refraction. I'm not sure how that would modify the result.


(*) rho(horiz) = (er*sin(a) - sqrt(er - cos^2(a))) / (er*sin(a) +
sqrt(er - cos^2(a))) where

er = epsilon-sub-r = (complex) relative permittivity of the ground
a = alpha = elevation angle

(**) er = er' - j * sigma/(omega * e0) where

er' = epsilon-sub-r prime = (DC) dielectric constant of the ground
sigma = ground conductivity
omega = radian frequency = 2 * pi * f
e0 = permittivity of free space

(***) E(horiz) = 1 + rho(horiz) * [cos(2*Bh*sin(a) + j*sin(2*Bh*sin(a)]
where

B = beta = 2 * pi / wavelength
h = height of horizontal antenna above ground
a = alpha = elevation angle

Roy Lewallen, W7EL