On Wed, 27 Apr 2005 20:15:03 -0400, Mike Coslo
wrote:

The main reason I ask though, is that I thought I heard here some time
ago, that a dipole would perform better over a good ground system.

Hi Mike,

You heard correctly. The ground system lowers losses which translate
to more power out. This is not method of controlling TOA, simply loss
- all angles of radiation improve.

I don't have the results published, but I have a near field study
showing interesting results for a dipole over ground, and over
shielded ground at:
http://home.comcast.net/~kb7qhc/ante...pole/index.htm

73's
Richard Clark, KB7QHC

"Richard Clark" wrote
The main reason I ask though, is that I thought I heard here some time
ago, that a dipole would perform better over a good ground system.

Hi Mike,

You heard correctly. The ground system lowers losses which translate
to more power out. This is not method of controlling TOA, simply loss
- all angles of radiation improve.
_________________

Note that the intrinsic, free space pattern/gain of a dipole does not depend
in any way on the presence or nature of a ground plane.

Reflections from the ground (and other objects) can modify the classic donut
shape of the dipole pattern and produce relative gain in some directions, at
the expense of gain in other directions. But the "power out," or absorbed
by a matched dipole will be the same in any case.

A low-resistance ground system will increase the fields radiated from an
antenna that uses the earth as an 'image' part of the complete radiation
system, such as the vertical mast radiators used in MW broadcasting. With a
perfect ground in this situation, the base current in the vertical mast is
twice what it would be for the same power applied to an equivalent wire
dipole (less a ground system) in free space -- resulting in 3 dB system
gain.

RF

Richard Fry wrote:
"With a perfect ground in this situation (MW vertical tower) the base
current in the vertical mast is twice what it would be for the same
power applied to an equivalent wire dipole (less a ground system) in
free space -- resulting in 3 dB system gain."

I`m not agreeing or disagreeing, just listing facts.

Arnold B. Bailey in "TV and Other Receiving Antennas"on p. 500 gives the
gain of a horizontal half wave wire (thin), center-fed as zero dB at its
center frequency. He should. It is his reference for all other antennas.
His authority is the famous G.H. Brown in Proc. I.R.E., Vol. 33, p. 257,
April 1945. Antenna resistance = 60 ohms.

On page 538, Bailey gives the free-apsce gain of the quarter-wave
vertical antenna. It too has a gain of zero dB = 0 dBd. His authority is
A.S. Meier & W.P. Summers in Proc. I.R.E., Vol. 37, p. 609, June 1949.
Antenna resistance 28 ohms.

Power is current squared times the resistance..

Terman says on page 886 of his 1955 edition:
"Effect of Ground on Directive Gain of Ungrounded Antennas. Consider an
antenna is far enough from ground so that the total power radiated by a
given set of antenna currents is independent of the presence or absence
of the ground. Then a ground reflection that reinforces the main lobe
will double the field strength of the main lobe, and so will increase
the directive gain of the antenna system by a factor of 4." (thet`s a
power ratio of 4)

On page 885 Terman says:
" Consequentially (due to ground reflection nulls), to obtain strong
radiation in the directions approaching the horizontal using a
horizontally polarized radiating system, it is necessary that the height
of the antenna above the earth be of the order of one wavelength or
more."

Also on page 885 Terman says:
"In the case of horizontal polarization the effect of imperfect ground
is seen to be quite small, especially at low vertical angles. With
vertical polarization the ground imperfections have greater effect; in
particular, the filling of the nulls at moderately low vertical angles
is very pronounced."

For the power to be the same in a vertical mast and a wire dipole, the I
squared R must be the same in both cases.

Best regards, Richard Harrison, KB5WZI

Mike Coslo wrote:
Here is the scenario:

Hamshack on the west side of the house.

OCF dipole between two trees running perpendicular over the house
with
the Balun directly above the shack (now *that* is handy)

Butternut vertical on the east side of the house. 12 radials so far.

Now here is what brings about the question. Over the winter months,
I
had to have my sewer line to the street replaced, which ended up
making
a huge mess out of my front yard. This means that I will probably end
up
tilling and replanting a large part of the yard.

Is there any point to laying radials in the front yard? They would
be
quite a ways (~50 feet) from the radials around the Butternut.

Emphatically NO.

The main reason I ask though, is that I thought I heard here some
time
ago, that a dipole would perform better over a good ground system.

Maybe if the "good ground" is an acre or two of sheet copper and the
dipole is a half wave above it.


- Mike KB3EIA -

w3rv

I wrote:
"With a perfect ground in this situation (MW vertical tower) the base
current in the vertical mast is twice what it would be for the same
power applied to an equivalent wire dipole (less a ground system) in
free space -- resulting in 3 dB system gain."
to which Richard Harrison responded.
____________

Excuse my inaccurate statement about that current, and thanks for catching
it.

The input resistance of a 1/4-wave monopole working against a perfect ground
plane is 36.5 ohms,* or half that of a 1/2-wave dipole in free space. A
given input power then results in 1.414X more current in the monopole than
the dipole. Hence the monopole radiates 1.414X the field of the dipole.
And, as shown in Kraus 3rd edition, Table 6-2, a 1/4-wave monopole against a
perfect ground has 3 dB more gain than a 1/2-wave dipole in free space.

Increasing the field from the 1/2-wave, free space dipole by 1.414X in the
above example would require doubling its input power (3dB), which would
result in the same 1.414X increase in its current -- to the same value as
seen in the monopole with 1/2 that power.

* per Kraus 3rd Edition, p 567

RF

Richard Harrison wrote:

Mike, KB3EIA wrote:
"The main reason I ask, is that I thought I heard here some time ago,
that a dipole would perform better over a good ground system."

It may not.

At a distant point, the received signal is probably composed of two
parts that started their journey as an incident ray and a ray which was
the incident ray`s reflection off on a tangent from the surface of the
earth.

If by good fortune these two rays happened to arrive at the distant
receiving point in-phase they would present a stronger signal than the
direct ray alone, and certainly a stronger signal than a combination of
two out-of-phase signals.

Unfortunately, the incident wave`s reflection is always out-of-phase
with the incident wave which produces it at the reflection point. A
perfect reflector would ensure the reflection was equal in magnitude as
well as out-of-phase to the incident ray.

Unless you get a difference in path length between incident and
reflected rays to invert the phase of one of the rays as compared with
the other, they will tend to cancel. You might be better off without the
reflected ray.

The ground connection in a vertical antenna system is entirely
different. Half the antenna system is the antenna`s image in the earth.
The connection to the earth or to a capacitive coupling to the earth
(elevated radials or ground-plane) carries the r-f current to the earth
side of the system. Any resistance in your gtound system directly adds
to loss in the system.With the usual vertical antenna system, radials
are essential.for efficiency..

This was a long-winded way to say you don`t need radials with a
horizontal dipole for r-f efficiengy. You do need a ground connection
for electrical safety and lightning protection. Radials work well for
these too.

I'm beginning to think that what makes an antenna "good" is the time at
which a person uses it!

- Mike KB3EIA -

This is unfortunately an example of arriving at the right result by
using the wrong (actually, incomplete) method.

The field from a conductor is proportional not only to the current
flowing in it, but also the length of the conductor. So if you put the
same current into a dipole and monopole, and assuming they have the same
current distribution, the field from the dipole will be twice the field
from the monopole.

So let's start again. The input R of a monopole over an infinite perfect
ground is 1/2 the resistance of a free space dipole, so for a given
power input the monopole current is 1.414 times the dipole current, as
you said. Ok so far. But because the dipole is twice as long and with
the same current distribution (and oriented in such a way that the
fields from the two halves add in phase), the field from the dipole is
2/1.414 = 1.414 times the field from the monopole. However, each ray
from the monopole is reflected from ground, resulting in two rays adding
in phase at a distant point. This doubles the field from the monopole,
so it's now 2/1.414 = 1.414 times the field from the free space dipole.
This is the same result, but with the two additional important factors
of radiator length and ground reflection included.

A good check of the final result is to note that, neglecting loss, the
average field intensity from *any* antenna in free space has to be 3 dB
less than the average field intensity from *any* antenna over an
infinite ground plane, if the same power is applied to each. The reason
is simply that the supplied power is spread over half the volume when
the ground plane is present.

Roy Lewallen, W7EL

Richard Fry wrote:
. . .
The input resistance of a 1/4-wave monopole working against a perfect
ground plane is 36.5 ohms,* or half that of a 1/2-wave dipole in free
space. A given input power then results in 1.414X more current in the
monopole than the dipole. Hence the monopole radiates 1.414X the field
of the dipole. And, as shown in Kraus 3rd edition, Table 6-2, a 1/4-wave
monopole against a perfect ground has 3 dB more gain than a 1/2-wave
dipole in free space.

Increasing the field from the 1/2-wave, free space dipole by 1.414X in
the above example would require doubling its input power (3dB), which
would result in the same 1.414X increase in its current -- to the same
value as seen in the monopole with 1/2 that power.

* per Kraus 3rd Edition, p 567

RF

Hmmm, I was working on the "phases of the moon", you might have something...
grin

Regards,
John

Richard Clark wrote:

On Wed, 27 Apr 2005 20:15:03 -0400, Mike Coslo
wrote:


The main reason I ask though, is that I thought I heard here some time
ago, that a dipole would perform better over a good ground system.


Hi Mike,

You heard correctly. The ground system lowers losses which translate
to more power out. This is not method of controlling TOA, simply loss
- all angles of radiation improve.

I don't have the results published, but I have a near field study
showing interesting results for a dipole over ground, and over
shielded ground at:

Interesting indeed!

- Mike KB3EIA -

You heard correctly. The ground system lowers losses which
translate
to more power out. This is not method of controlling TOA, simply
loss
- all angles of radiation improve.

======================================

Some buried wires under a horizontal dipole, at a height of 1/4 or 1/2
wavelengths, will, in theory, reduce losses. Some old wife, once upon
a time, must have read something about it in a book without bothering
about the magnitude of the effect.

It's not enough to be detectable. So don't listen to your voices or
waste time digging up your back yard and getting back ache.

There's far too much of old wives reading things in books and
ill-written radio magazines, getting the wrong ideas, and then
plagiarising them. Which innocent people hear about 3rd or 4th hand
or n'th hand.

The fabled SWR meter is another example.
----
Reg.