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Old April 28th 07, 06:54 PM posted to rec.radio.amateur.antenna
Cecil Moore[_2_] Cecil Moore[_2_] is offline
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First recorded activity by RadioBanter: Mar 2007
Posts: 3,521
Default Current in 1/2WL Dipoles

Jimmie D wrote:
"Cecil Moore" wrote in message
...
The point I was trying to make on the "Rotational Speed"
thread got lost in semantics so here goes again.

Given - a 75m dipole modeled by EZNEC with 90 segments
in each leg of the dipole. 90 segments was chosen
to correspond to the number of degrees from the
feedpoint. Illustrating 1/2 of the dipole with
0 ohm loads at the following segments:

FP--------23--------46--------68--------90

EZNEC reports the following results:

Feedpoint current: 1 amp at 0 deg

Current at seg 23: 0.9281 amp at -1.06 deg

Current at seg 46: 0.7154 amp at -1.78 deg

Current at seg 68: 0.4049 amp at -2.31 deg

Current at seg 90: 0.0122 amp at -2.76 deg

Since the segment numbers correspond to the
number of degrees, it's obvious that the
segment numbers correspond to the expected
phase shift in the traveling waves.

Question: Assuming the current reported by
EZNEC is a ~cosine function, how does one use
that current to determine the traveling wave
phase shift in the wire?


What is the current phase relative to. The feedpoint , the voltage in that
segment? As I have read this thread there seems to be some confusion on
this point.


I believe that EZNEC references net current to the
source current of 1A at 0 degrees. I consider it
to be a *snapshot* of conditions when the source
current is 1A at 0 degrees. This is what I have
been assuming in all of my postings. I believe
that is also what Kraus does in his graphs.
--
73, Cecil http://www.w5dxp.com