Owen Duffy wrote:
http://www.vk1od.net/lost/W551Example.htm

Is the shape of the curve (the cyclic variation over each electrical
half wave diminishing away from the load, and the general shape of the
curve a surprise? The effects plotted here might not be explained by
the ARRL charts.

The graph is unclear. What does it mean that 6% loss occurs at
100 neters? Is that 6% loss per meter at the source? There's 4%
loss at 50 meters. Does that mean the average loss per meter is
4%? Where is the 4% loss in the meter closest to the load plotted?
--
73, Cecil http://www.qsl.net/w5dxp

On Wed, 09 Nov 2005 14:46:07 GMT, Cecil Moore wrote:

Owen Duffy wrote:
http://www.vk1od.net/lost/W551Example.htm

Is the shape of the curve (the cyclic variation over each electrical
half wave diminishing away from the load, and the general shape of the
curve a surprise? The effects plotted here might not be explained by
the ARRL charts.

The graph is unclear. What does it mean that 6% loss occurs at
100 neters? Is that 6% loss per meter at the source? There's 4%
loss at 50 meters. Does that mean the average loss per meter is
4%? Where is the 4% loss in the meter closest to the load plotted?

The loss scale is in dB, it is the loss in dB at position x metres
from the load.

If you examine the graph, you will find that the slope of the loss vs
position line is as high as about -22dB/100m at the load, it has a
minimum slope of close to 0dB/100m, and you can see that at large x,
the slope approaches the matched line loss of -1dB/100m.

(You find the -22dB/100m by using a ruler to scale off the slope.
-22dB/100m is -0.22dB/m, or 10**-0.022 which is 0.9506, which
corresponds to a loss of almost 5% in that one metre of line nearest
the load. These aren't mental gymnastics!)

You could calculate an average loss per meter figure, but I don't know
what you could you use it for? The fact that this line is not straight
(as some people seem to assume) means that working with average
numbers is inherently inaccurate.

Owen
--

Owen Duffy wrote:
The loss scale is in dB, it is the loss in dB at position x metres
from the load.

Aha, I see that noted at the bottom now that I scroll down.
It didn't make sense to me if the loss scale was in percent.
So what do I get if I integrate the area under the curve?

Incidentally, I was engaging in fuzzy republican thinking when
I came up with eleven current maximum points in 100 meters of
feedline. Of course, there are twice that, i.e. 22 current maximum
points which can be counted on your graph.

You could calculate an average loss per meter figure, but I don't know
what you could you use it for? The fact that this line is not straight
(as some people seem to assume) means that working with average
numbers is inherently inaccurate.

Owen, most hams are not rocket scientists like you :-) Quite
often, a rule-of-thumb average beats total ignorance. All
measurements contain errors and are inherently inaccurate.
Some of us live with that reality. Some of us rant and rave
about it. Next time you are on your motorcycle, note that
your speedometer is "inherently inaccurate" as is your gas
gauge as are your reflexes. If you are not inherently inaccurate
throwing darts in the local pub, you are a very unusual homo sapien.

I, for one, am satisfied with average losses, presumably averaged
over one half wavelength. The way I came up with that 25:1 limit
on my open-wire SWR is that 600/25 equals 24 ohms and that is an
acceptable impedance to my IC-256PRO's autotuner. Noting that the
losses in 100 ft. of open-wire line running at an SWR of 25:1 are
acceptable was an afterthought.
--
73, Cecil http://www.qsl.net/w5dxp