Hi Yuri,
Well, Cecil's example is a good explanation for why there is
a current standing wave on an resonant antenna like a dipole.
But, how do you get from that observation to the conclusion
that the "delay" through the inductor is the same as the "delay"
through the section of antenna which the inductor replaces?
Unless the "delay" of the inductor is the same as the "delay"
of the section of antenna it replaces, then the current tapers
won't be the same. That is what is at issue, and what has
been at issue in this debate. Now when you are talking
about inductor, what is its "delay"? If you have current
going into the inductor which doesn't come out right away,
then you have a concentration of excess charge building
up in the inductor - No?. This concentration of charge will
give rise to an electric field between the excess charge
and the rest of the antenna's enviroment which causes
charge to move elsewhere in response. But this is nothing
more than displacement current. So the "delay" through the
inductor (and hence the taper) will depend on the distributed
capacitance between the inductor and the rest of the environment
(imagine a bunch of little tiny distributed capacitors between
segments of the inductor and the other half of the dipole), but
this is what all the proponents of Kirchoff's law have been
saying all along. You see, the circuit view of things has to be
consistent with the distributed view of things.
Okay now me head doesn't hurt as much as it did this
morning. Your turn.
73 de Mike, W4EF........................................................
----- Original Message -----
From: <K3BU@aol.com>
>
>
> We are not dealing with closed circuit, but distributed and an antenna
with
> standing waves.
> Here are some explanations for the phenomena:
>
> W5DXP:
> The key to understanding is to realize that the net current is the
> phasor sum of the forward current and reflected current (on a standing-
> wave antenna). Assume a 10 degree phase delay through the coil on the
> frequency of operation. Ifwd-in and Iref-out are on the same side of
> the coil. Ifwd-out and Iref-out are on the other side of the coil.
>
> Ifwd-in--> coil Ifwd-out-->
> -----------------------////////////-------------------------
> <--Iref-out <--Iref-in
>
> Assume that |Ifwd-in| = |Ifwd-out| which satisfies Kirchhoff
>
> Assume that |Iref-in| = |Iref-out| which satisfies Kirchhoff
>
> Ifwd-in + Iref-out = net current on left side of the coil
>
> Ifwd-out + Iref-in = net current on right side of the coil
>
> Ifwd-out lags Ifwd-in by 10 degrees
>
> Iref-out lags Iref-in by 10 degrees (Iref-in leads Iref-out)
>
> Now let's assume that Ifwd-in and Iref-out are in phase. So current
> on the left side of the coil equals Ifwd-in at zero degrees plus
> Iref-out at zero degrees which is a current maximum point.
>
> Ask yourself: Can we have a current maximum point on both sides of
> the coil? I trust that answer is obvious.
>
> Ifwd-out lags Ifwd-in by 10 degrees. Iref-in leads Iref-out by 10 degrees.
> So current on the right side of the coil equals Ifwd-out at -10 degrees
> plus Iref-in at +10 degrees, NOT a current maximum point.
>
> Therefore, in this example, net current on the left side of the coil
> cannot possibly be equal to net current on the right side of the coil.
> --
> 73, Cecil http://www.qsl.net/w5dxp
>
>
> and summarized by W4JLE:
> If we feed an antenna at the current point, the current decreases as the
> voltage increases along the antenna element from feed point to end..
>
> That being said, a coil replacing a segment of an antenna (in order to
> physically shorten it) will exhibit the same properties (relating to
> currents) as the segment it replaced.
> _______________________________________________
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