----- Original Message -----
From: "Al Williams" <email@example.com>
To: "Towertalk" <firstname.lastname@example.org>
Sent: Saturday, December 04, 2004 9:43 AM
Subject: Re: [TowerTalk] elevated short vertical dipole or
> ----- Original Message -----
> >. Did you ever contemplate where the difference was coming from?
> > Ground losses?
> I have difficulty in understanding or accepting the notion of "ground
> losses--...wasted power heating the earth, etc..."
> I have no problem with using a dc, ac, or rf resistance meter to measure
> between two earth points, but radio waves striking the earth, I dunno!
RF propagating through a lossy medium heats it up. Think of 2.5 GHz radio
waves hitting that cup of coffee in the microwave.
> What follows is what is puzzling; maybe someone can help answer them.
> 1. Radio waves from a horizontal antenna striking the ground are
> reflected (exactly what is going on by reflection?) and add signal
> strength to the skyward wave. However radio waves from a 1/4 wave
> vertical striking the ground are absorbed (all? partially ? and as a
> result do not provide the ~2 db gain of a dipole and the ~ 4 db gain
> from ground "reflection".
Not quite as you've described it. EM waves (whether light or RF) striking
the ground are partially reflected and partially transmitted into the soil
depending on the relative index of refraction and the polarization. A good
example is looking into a body of water (river or lake). At some angles,
the light reflects completely, at others the light penetrates. And, it
depends on the polarization (why polarized sunglasses reduce glare). The
reflection process changes both the amplitude and phase of the reflected
The horizontally polarized waves reflect fairly well, regardless of
incidence angle. Vertically polarized waves only reflect at angles sharper
than a certain value, otherwise they penetrate (and are absorbed). So, a
horizontal dipole (which has the same gain at all elevation angles, at least
broadside) picks up gain in some elevations where the reflected and direct
wave are in phase. A significant case is where the dipole is 1/2 wavelength
above the reflector. The reflected wave is exactly in phase with the ray
heading straight away from the reflector (towards the zenith in the case of
ground), so this antenna has most of its gain straight up. 1/4 wavelength
high actually produces a null straightoverhead (the reflected wave is 180
out of phase with the direct wave).
For vertically polarized antennas the situation is a bit different. First,
they already have a null straight up. Second, the reflection is phased
differently, and the "apparent source" isn't as obvious in the vertical
> 2. The horizontal dipole radiates because each half of the dipole
> alternately charges and discharges oppositely and equally. But what
> goes on when the feedpoint is moved off center so that metal to charge
> on each side is unequal?
Any antenna radiates because charge is accelerating and decelerating. A
dipole is a special case which is symmetrical. An asymmetrical antenna
still moves charge around, so it still radiates. To a first order, the
effect is an impedance transformation. The basic pattern (given that the
antenna is the same length and you assume lossless) will remain pretty much
> 3. The 1/4 vertical alternately charges the vertical portion against the
> ground (or ground wires, salt water). Therefore is the problem not that
> the ground is lossy but rather that the vertical part of the antenna
> cannot be charged as well it is over metal wires or salt water? Has
> anyone measured the power going into the vertical part when over ground
> vs when over wires or salt water?
Lots of people have done this measurement. It's how programs like NEC are
validated. And, the ground IS lossy. In a vertical monopole there are two
sources of loss: the ground return (the current flowing in and out of the
antenna) and the eddy and dielectric losses from having the field penetrate
the earth. You could put up a radial field above the ground and practically
eliminate the ground return losses (think about an inverted V dipole turned
on it's side...) , but you'd still have the losses due to the electric and
magnetic fields interacting with a lossy substance in the near field.
> 4. Is the "ground loss" frequently referred to related or confined to
> the Brewster angle consideration?
No... The pseudo-brewster angle is a far field pheonomenon.
> 5. Again, what is actually going on by ionosphere and ground
> reflection--is it similar to metal interception radio waves and becoming
> a generator?
Ionospheric propagation is more of a refraction as the ray passes through an
area with changing index of refraction, so it bends (just like light in a
lens). The index is different for different angles of incidence and
polarization, and, is dispersive (different with small changes in frequency)
and nonuniform to boot. The ionosphere isn't a very good conductor, but
there's a lot of it. More than reflecting from metal, it's more like
reflecting from water, or, even better, like a mirage, where you see an
image of the sky on the road in front of you that looks like water.
However, from a "large scale" ray tracing standpoint, you can consider it as
a reflection from a plane at the "virtual height" of the ionosphere.
Ground reflection is similar.. it's the propagation of an EM wave across an
interface with different indices of refraction, which makes for a big
reflection coefficient. If you like, you can think of the path being like a
transmission line with a really bad VSWR at the air/ground interface.
See: http://www.mscomputer.com for "Self Supporting Towers", "Wireless Weather
Stations", and lot's more. Call Toll Free, 1-800-333-9041 with any questions
and ask for Sherman, W2FLA.
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