So, you start off with a quiescent piece of wire, and launch a wave into it,
some of which will be radiated. The rest is reflected off the end, and
arrives
back at the feed point, the open end causing a reversal of the current, and
in a
short antennae, the current arriving back at the feed point is out of phase
with
the voltage and so the antenna appears reactive.

So, for those who deny that some of the power is not radiated, whence does
the reactance arise?

On 2014-10-27 09:53:36 +0000, gareth said:

So, you start off with a quiescent piece of wire, and launch a wave into it,
some of which will be radiated. The rest is reflected off the end, and arrives
back at the feed point, the open end causing a reversal of the current,
and in a
short antennae, the current arriving back at the feed point is out of
phase with
the voltage and so the antenna appears reactive.

So, for those who deny that some of the power is not radiated, whence does
the reactance arise?

I am fairly sure that it is difficult to dissipate power in a reactance.

--

Percy Picacity

gareth wrote:
So, you start off with a quiescent piece of wire, and launch a wave into it,
some of which will be radiated. The rest is reflected off the end, and
arrives
back at the feed point, the open end causing a reversal of the current, and
in a
short antennae, the current arriving back at the feed point is out of phase
with
the voltage and so the antenna appears reactive.

Yep, zero understanding of how antennas work.

Start with "a quiescent piece of wire"; a wire can never be anything
but quiescent.

Move on to "launch a wave into it"; you apply a current.

I don't have enough man years of life left to deal with all the rest of
the nonsense.

So, for those who deny that some of the power is not radiated, whence does
the reactance arise?

One does not follow the other.



--
Jim Pennino

"Brian Morrison" wrote in message
k...
On Mon, 27 Oct 2014 09:53:36 -0000
"gareth" wrote:

the open end causing a reversal of the current

There is no current flow at the end of a wire antenna, so there is
nothing to reverse. There is a voltage anti-node, the reflected wave is
reflected in phase at the open end of the wire.

Yes, and also no. In order for there to be a net no-current situation
when previously there was current heading toward the open end, the current
reflection is of opposite polarity.

"gareth" wrote in news:m2nfsl$c6b$1@dont-
email.me:

net no-current situation

The end is never connected, so there is NO current there. Not ever, at any
instant, never mind 'net'. There can be an alternating electrical charge,
hence a voltage, but I'll stop there because I'm not knowlegeable enough
about AC electricity to add much. I'm just pointing out that AC doesn't
automagically create a current through any open contact. The ONLY way a
current can flow through the end of that antenna is via contact or arc.

Imagine electrons occupying the wire. They move easily where there is a place
to go. As the charge builds, like charges repel, making it harder to move
electrons into atoms at the wire's end, so less movement, less current. My
description may be flawed now that quantum mechanics describes things
differently (and I didn't try to express it in terms of AC electrical theory
either), but the result is the same, current is not equal in all parts of the
wire, for AC. The only reason we don't think about this in a mains cable is
that the frequency is extremely low compared to its length.