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Old April 5th 10, 10:30 PM posted to rec.radio.amateur.antenna
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First recorded activity by RadioBanter: Dec 2006
Posts: 1,169
Default velocity factor, balanced line

Bob wrote in
:

....
It mentioned the voltage maximum problems, the current maximum
problems, and then said, "Is there a better option?" And I don't
understand the few sentences that follow that query. In other words, I
don't understand the solution -- i.e. "line lengths around 135
degrees longer than voltage maximum" :-)


The location of voltage maxima depends on the load on the line. If you
were to plot the impedance at various lengths of line, it is highest (and
purely resistive) when fed at a voltage maximum. As the line is lengthed,
that impedance becomes capacitive, and lower, eventually becoming lowest
(purely resistive again) at the current maximum (90° longer than the
point of voltage maximum). Increasing the length further, impedance
becomes inductive and increases eventually becoming highest at the next
voltage maximum. At a point of about 135° longer than the voltage
maximum, the impedance presented to the T match is in the region where it
is most efficient. Alternatively, you could state this as 45° shorter
than a voltage maximum.

This is not your odd eighth wave (from a resonant load) rule, because
that also encourages the capacitive region where losses are higher.

Owen
 
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