Roy Lewallen wrote:
A ground plane is a poor model of how currents flow along a car body.

If the car body was 1/2WL in the air, would the antenna
be more efficient?
--
73, Cecil http://www.qsl.net/w5dxp

Dan Richardson wrote:
Roy is quite correct in stating that a vehicle's body behaves as one
side of a dipole. A lopsided dipole to be sure, but one half the
antenna just the same.

Seems the truth might lie somewhere in between. If the ground
plane of a vertical antenna is near the ground, there are
losses. If the ground plane of a vertical antenna is located
1/2WL above ground, the losses are a lot less. I'll bet that
if the vehicle were located 1/2WL in the air, the efficiency
would increase.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
Dan Richardson wrote:

Roy is quite correct in stating that a vehicle's body behaves as one
side of a dipole. A lopsided dipole to be sure, but one half the
antenna just the same.


Seems the truth might lie somewhere in between. If the ground
plane of a vertical antenna is near the ground, there are
losses. If the ground plane of a vertical antenna is located
1/2WL above ground, the losses are a lot less. I'll bet that
if the vehicle were located 1/2WL in the air, the efficiency
would increase.

Kind of tough though going under power lines, bridges and overpasses :-)

Amos Keag wrote:

Cecil Moore wrote:
I'll bet that
if the vehicle were located 1/2WL in the air, the efficiency
would increase.

Kind of tough though going under power lines, bridges and overpasses :-)

What if the vehicle is a helicopter? :-)
--
73, Cecil http://www.qsl.net/w5dxp

On Thu, 02 Mar 2006 17:46:19 -0500, Amos Keag
wrote:

Quite interesting reading. Have you received peer comments?

Naw.

However, those nec models are available at my web site and you can run
you own analysis if you like.

Danny




email: k6mheatarrldotnet
http://www.k6mhe.com/

Danny K6MHE wrote:
"Boy Richard you sure missed on that one!"

A broadcast tower over a perfect ground system is the source of radiated
energy even though its image in the ground system produces a pattern
which behaves as if there were a dipole, the lower half of which is
buried.

The earth is not radiating. It is conducting. The tower above the earth
is the source of radiation.

Every ground radial in the broadcast system (usually all 120 of them),
has a twin running in the opposite direction. All radials are tied
together at the base of the tower. So the current in the radials all
starts out in the same phase and stays roughly in the same phase as it
progresses outward. It declines in magnitude away from the feedpoint.
That`s the reason ground radials don`t need to be unlimited in length.
You don`t need radials after the current plays out. As current travels
in opposite directions in the groind radials. the fields they prodoce
add to zero.

The two halves of a dipole are fed with opposite polarities at their
feedpoint. this puts the two halves running in opposite directions
in-phase. Their fields thus reinforce.

A 1/4-wave ground plane in free space has the same power gain as a
center-fed 1/2-wave dipole.

The matched power radiated by either ground plane or dipole is the same,
but the resistance at the feedpoint of the ground plane is only 50% that
of the dipole. Radiation resistance is defined as the resistance at the
high current point of the antenna unless otherwise specified. Radiated
power is (I) squared times the radiation resistance.

Danny did not specify where he thought I erred in my previous posting. I
said that a whip mounted on a vehicle is not exactly like a dipole. I
meant that the whip did most of the radiating because it carried a
concentration of current in the same direction while in the car body the
current is dispersed in various directions, some of which canncel in
their effects.

I still insist that is the case.

Best regards, Richard Harrison, KB5WZI

ORIGINAL MESSAGE:

Dan Richardson wrote:

That may have some validity in the VHF and higher ranges, but on HF -
particularly on 80 meters - a car body's size is a small fraction of a
wavelength (as is the whip portion). Consequently the vehicle body
acts like the one half of a dipole antenna.


~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Q. How can a car body which is a "small fraction" of a wavelength act
like one half of a dipole?

A. It can't.

Q. Well, what does it do then?

A. It acts like a short piece of wire leading from the bottom of the
whip to the actual ground plane, namely the earth itself.

Q. Does that help any?

A. Probably a little, but remember the piece of wire (the car body) is
only a few feet long. Not very much on 80 meters.

Q. Thanks, I get it now.

A. You're welcome.

73, Bill W6WRT

Bill Turner wrote:
ORIGINAL MESSAGE:

Dan Richardson wrote:


That may have some validity in the VHF and higher ranges, but on HF -
particularly on 80 meters - a car body's size is a small fraction of a
wavelength (as is the whip portion). Consequently the vehicle body
acts like the one half of a dipole antenna.



~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Q. How can a car body which is a "small fraction" of a wavelength act
like one half of a dipole?

A. It can't.

Q. Well, what does it do then?

A. It acts like a short piece of wire leading from the bottom of the
whip to the actual ground plane, namely the earth itself.

Q. Does that help any?

A. Probably a little, but remember the piece of wire (the car body) is
only a few feet long. Not very much on 80 meters.

Q. Thanks, I get it now.

A. You're welcome.

73, Bill W6WRT
Actually it is acting as one half of a dipole. It is just a non-resonant
half of a dipole. Remember "di" means two.

Dave WD9BDZ

I'm afraid people are getting too hung up by trying to squeeze
everything into various pigeon holes like "dipole" and "ground". You'll
have to think beyond those narrow and poorly defined and understood
categories and look at the basics of antenna operation in order to
understand what's happening.

The field radiated from a conductor is determined by two things: the
amount of current on it, and the length of the path the current takes.
Theorists have known this for well over a century. The most
sophisticated antenna analysis programs break the current paths into
very short pieces ("segments"), calculate the current on each piece, and
then calculate the resulting field from the product of the current and
the segment length. Fields from various parts of the conductors can
cancel or reinforce to any degree. (Mathematically, they add vectorally.)

If you don't or can't believe this, you needn't bother continuing.

For those still reading, let's imagine a 16 foot vertical wire with a
tiny 3.5 MHz signal generator at the center. This is known in textbooks
as a "dipole", but how things behave aren't dictated by what we call
them, so feel free to insist it's a "seagull", "pizza", "xfppftm", or
whatever makes you comfortable. The signal generator has two terminals,
and any generator must have equal currents in and out of its two
terminals. If you don't or can't believe that, brush up on Kirchoff's
current law. If that doesn't do it, there's no need to continue further.

Let's suppose the generator is producing one amp RMS of RF current. If,
say, 0.2 amp is flowing upward out of the top terminal at a given
instant, 0.2 amp is flowing upward into the bottom terminal at the same
instant. By inspection, one amp RMS is flowing upward in the vertical
wire immediately above and below the generator. By a number of
techniques, we can show that the current decreases nearly linearly from
the center to the ends. That is, four feet from the center, either above
or below the source, the current is 1/2 amp. At the antenna tips, the
current is zero, which we should expect: there's nowhere for it to go.

It should be obvious that the wire above the source is radiating the
same field strength as the wire below the source -- for each little
piece of the wire above the source there's a piece below the source
carrying exactly the same current. And as it turns out, the fields from
all parts of both wires add completely in phase directly broadside to
the wire, and only partially in phase in other directions. So at least
directly broadside, we can say that the contribution from each wire is
equal and proportional to its total field strength.

Ok, now let's make one of the wires "ground" and the other a "whip",
because we like to do that, right? Let's call the top wire a "whip", and
bottom load it. We add an inductor (very small, physically, to avoid
adding another dimension to this analysis) between the signal
generator's top terminal and its connection with the upper wire. We can
make the inductor the proper value to make the upper wire/inductor self
resonant if it were grounded, or we can make the inductor about twice as
large to make the whole dipole resonant. It doesn't matter. Now let's
see what happened to the radiation from the "whip" and "ground" wires.

There's no change whatsoever! The currents are exactly the same as they
were before, on both wires. They still taper from the center to the tips
as before. They both radiate equally. All we've done is change the
impedance seen by the generator. If you don't believe this, perhaps you
can explain why they won't.

Next, let's replace the lower wire with a cylinder like a tank, say 10
feet in diameter but still 8 feet high. What happens then? Surely it
must now be "ground", and "ground" doesn't radiate, does it?

Well, it does radiate. The one amp flowing into the bottom generator
terminal spreads out radially over the top of the cylinder. Although the
current density decreases as we move out from the center, the total
current also decreases. If only the cylinder top was present and the
rest of the cylinder missing, the current would drop to nearly zero at
the edge. But because of the presence of the rest of the cylinder, the
current at the edge drops to about half the value at the center. The
half which remains flows down the cylinder sides. This would result in
the field from the cylinder being about half the field from the "whip"
if the current decreased to zero at the bottom of the cylinder as it
does at the top of the whip. But the current along the sides of the
cylinder doesn't drop to zero at the bottom of the walls because it can
flow onto the bottom of the cylinder. The average current on the whip is
0.5 amp, and on the cylinder (from a model) about 0.35 amp, so the
cylinder's field is about 3 dB less than that of the "whip". Not quite
what most of envision when we think of a "ground".

If we top load the whip with a 10 foot diameter top hat, its average
current increases to about 0.9 amp. But its presence also reduces the
amount of current drop from the center to the edge of the cylinder top
due to mutual coupling. The end result is larger current along the
cylinder sides and very nearly the same field strength ratio between the
"whip" and cylinder.

So far this analysis has taken place in free space. What happens if we
put the cylinder bottom just above the ground, say six inches? Now,
surely, the cylinder is "ground"! But the current still flows down the
sides and radiates just like the old original vertical lower wire did.
And putting the bottom close to ground increases the current along the
sides! The coupling between the cylinder bottom and ground acts somewhat
like a top hat does to a whip, and increases the average current.
Instead of 0.35 amp, it increases to about 0.42. Now the cylinder's
field is only about 1.5 dB less than that of the "whip".

I hope this has encouraged at least a few people to think a little
before declaring every conductor to be either an "antenna" or a "ground
plane" and assuming that by doing so they'll somehow cause it to behave
in some predetermined and only vaguely understood fashion.

Roy Lewallen, W7EL

Bill Turner wrote:
ORIGINAL MESSAGE:

Dan Richardson wrote:

That may have some validity in the VHF and higher ranges, but on HF -
particularly on 80 meters - a car body's size is a small fraction of a
wavelength (as is the whip portion). Consequently the vehicle body
acts like the one half of a dipole antenna.


~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Q. How can a car body which is a "small fraction" of a wavelength act
like one half of a dipole?

A. It can't.

Q. Well, what does it do then?

A. It acts like a short piece of wire leading from the bottom of the
whip to the actual ground plane, namely the earth itself.

Q. Does that help any?

A. Probably a little, but remember the piece of wire (the car body) is
only a few feet long. Not very much on 80 meters.

Q. Thanks, I get it now.

A. You're welcome.

73, Bill W6WRT

David G. Nagel wrote:

Actually it is acting as one half of a dipole. It is just a
non-resonant half of a dipole. Remember "di" means two.

Dave WD9BDZ

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~

In a strict sense you are correct, but in the context here where one
half of the dipole is an eight-foot whip and the other half is four
feet of car body, we don't have much of an 80 meter antenna without the
coupling from car body to earth ground.

Bill, W6WRT