On Sun, 08 Apr 2007 23:56:37 -0700, Roy Lewallen
wrote:

Indeed it is. Can you point me to a reference where I can get a more
detailed explanation of this circumferential current and its cause?

Hi Roy,

Brown, Lewis and Epstein.

73's
Richard Clark, KB7QHC

Richard,

What coupling does this imply? Is it direct as is contact or is it a
field coupling as capacitive or inductive? The question is: what would
insulation or corrosion due to buried radials?

- Dan

Richard Clark wrote:
On Sun, 08 Apr 2007 20:20:36 -0700, dansawyeror
wrote:

To focus on the question is: Why is the ground resistance so high?

Hi Dan,

Focus on Why? Why? This is something that you have absolutely no
control over without a huge investment in new dirt several 10s of
meter deep out to at least as far as the antenna is tall.

You want less resistance? then dig a hole and fill it with quartz
sand. Build a dune to drive the loss down further.

A suitable substitute is to lay out a ground field (radial wires). The
more the better. Here is the Why? you should be asking (Why more
wires?).

The path through the earth is shortened between adjacent ground wires.
Less path, less loss. The earth current travels not IN toward the
center as the radials do. The earth current travels ACROSS or
circumferentially towards the radial wires. This is due to the phase
lag between the induced earth current and the radial current. The
greater the distance between radials, the more path loss from the
average distance between the radials, to the radials.

As this is very difficult to treat in words alone, it is undoubtedly
confusing in the description above.

73's
Richard Clark, KB7QHC

On Mon, 09 Apr 2007 08:21:36 -0700, dansawyeror
wrote:

What coupling does this imply? Is it direct as is contact or is it a
field coupling as capacitive or inductive?

Hi Dan,

Yes to all three.

The question is: what would
insulation or corrosion due to buried radials?

Not much, practically; unless the wire is extremely thin, and the
currents are large. Loss is in the earth. You can, of course, build
a very crummy radial system if you try hard. For instance, using
expensive piano wire in place of cheap house wiring.

73's
Richard Clark, KB7QHC

To focus on the question is: Why is the ground resistance so high? It
is not important at this stage to determine its precise value. The
point is it is high enough to cause a return of 0 degrees. This puts
the 'system' at over 50 degrees. Even if the antenna were 6 to 8 Ohms
the ground loss would be at least 42 to 44 Ohms.

Simple answer- add more radials. Especially if the impedance is varying
noticeably depending on ground moisture, you need more radials.

Another possibility is that if there is some other conductor in the near
field (tower, house, other wires, etc), the impedance might not be what
you expect.

Tor
N4OGW

On Sun, 08 Apr 2007 20:20:36 -0700, dansawyeror wrote:

Owen,

I did not want to concentrate on that; however it is a fair question.
The frequency is 3970 kc and the methodology is based on use of an hp
network analyzer. The analyzer has both polar and Cartesian displays.
The outputs were cross checked.f

[snip]

To focus on the question is: Why is the ground resistance so high? It is
not important at this stage to determine its precise value. The point is
it is high enough to cause a return of 0 degrees. This puts the 'system'
at over 50 degrees. Even if the antenna were 6 to 8 Ohms the ground loss
would be at least 42 to 44 Ohms.

How is the antenna loaded and what is the Q of the loading coil? Coil
losses are almost certainly 5 ohms and could easily be as high as 35 ohms.

bart

Bart,

Thanks for the thought. Modeling predicts it is much lower then that;
however I don't want to overlook anything. The Q looks very good on the
network analyzer, however I have never run the math. The coil is about
3.3 inches in diameter. I will let you know.

Dan

Bart Rowlett wrote:
On Sun, 08 Apr 2007 20:20:36 -0700, dansawyeror wrote:

Owen,

I did not want to concentrate on that; however it is a fair question.
The frequency is 3970 kc and the methodology is based on use of an hp
network analyzer. The analyzer has both polar and Cartesian displays.
The outputs were cross checked.f

[snip]

To focus on the question is: Why is the ground resistance so high? It is
not important at this stage to determine its precise value. The point is
it is high enough to cause a return of 0 degrees. This puts the 'system'
at over 50 degrees. Even if the antenna were 6 to 8 Ohms the ground loss
would be at least 42 to 44 Ohms.

How is the antenna loaded and what is the Q of the loading coil? Coil
losses are almost certainly 5 ohms and could easily be as high as 35 ohms.

bart