On 15 Jun 2006 11:51:24 -0700, "
wrote:

so I can use the 2:1 SWR bandwidth as
per the description he

http://lists.contesting.com/_amps/2006-01/msg00179.html

Once you know the Q, you need to know the antenna's radiation
resistance using some method, so that you can figure out the loss
resistance.

Hi Dan,

Conceptually close, but certainly not to the degree of accuracy you
might expect. 2:1 SWR points are not half power, as would be the
classic determinant for Q bandwidth. As your link provides, it is 88%
power points.

73's
Richard Clark, KB7QHC

Yes, I guess alternatively you could calculate the SWR at the
half-power points from

(E/Eo)^2=(1-((1-SWR)/(1+SWR))^2)

With E/Eo = 1/sqrt(2)

I get that this is about SWR = 5.8, and I understand that the MFJ-259B
is less and less accurate as the load moves away from 50 ohms
resistive, so it might be better to use the 88% power, 2:1 SWR
bandwidth points.

- - - - - -

I've thought of an additional question regarding this method. The
ground has an effect on antenna feedpoint impedance. This effect
contains both modification to the radiation resistance and the loss
resistance components of the feedpoint resistance.

Is the effect on the radiation resistance different for lossy earth and
for perfect earth? Obviously the loss increases with proximity to
ground, but can the effect of ground reflections on radiation
resistance be accurately modeled by putting the antenna over perfect
ground?

73,
Dan
N3OX

On 15 Jun 2006 12:37:20 -0700, "
wrote:

I've thought of an additional question regarding this method. The
ground has an effect on antenna feedpoint impedance.

Hi Dan,

To say the least.

This effect
contains both modification to the radiation resistance and the loss
resistance components of the feedpoint resistance.

Well, it doesn't change radiation resistance, that is a property of
the height of the radiator (as in length, not how high above ground).
As for feedpoint resistance, the loss of ground is cast into the
determination of feedpoint Z (which contains the sum of Rr, Ohmic loss
of the structure, and dielectric loss of the soil).

Is the effect on the radiation resistance different for lossy earth and
for perfect earth?

To say the least.

Obviously the loss increases with proximity to
ground, but can the effect of ground reflections on radiation
resistance be accurately modeled by putting the antenna over perfect
ground?

Ground reflections are post-hoc determinations of what happens to the
available power AFTER it has escaped ground loss and left the
radiator. Ground serves to influence the radiation pattern, but only
ground at a distance. This is basically ray-tracing and trigonometry
between the antenna as a point, and the far field earth. Ground also
serves to influence the radiation pattern insofar as how much power is
divided between heating the ground and radiating out to that ground at
a distance which then bounces off and away.

Perfect ground has no loss, and is perfectly reflective. Poor ground
(very poor ground in fact) has little loss, but little reflectivity.
Better ground has more loss, but higher reflectivity. Ground is a
two-edged sword and the degrees of variation don't swing very far for
the majority of Hams.

If you happen to be one of those minority Hams, you possibly have your
vertical planted into very low loss sand dune overlooking the ocean.
If you happen to reside at the other end of the bell curve, that
minority probably lives in a mountain valley swamp.

All situations bear upon the drivepoint Z, and the launch
characteristics, but bear in mind they are separate characteristics
that have different major variables.

73's
Richard Clark, KB7QHC