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