In message , Wayne
writes


"Ian Jackson" wrote in message ...

In message , Wayne
writes


"Ian Jackson" wrote in message ...

In message , Wayne
writes


"Frank Turner-Smith G3VKI" wrote in message
...

On 30/10/14 14:04, Ian Jackson wrote:
In message , Frank Turner-Smith G3VKI
writes
On 30/10/14 08:47, Ian Jackson wrote:
In message , Frank Turner-Smith G3VKI
writes
In a full wave dipole the voltage at both ends will always be in


Are you sure? Think on't!

so I would expect to see a very high impedance at the feed point.

Correct.

As you point out, matching the full-wave could be difficult and very
lossy.

Double zepp?

OK, what did I miss? In a full wave dipole, at the instant the voltage
at one end is peak positive, the voltage at the other end will also be
peak positive. Similarly, at the feed point, both legs would be at
peak negative and no current would flow in the feeder, hence the high
impedance. There would be a current flowing in each leg of the dipole,
but the currents would be in anti-phase. Where have I got it wrong? Do
I need another drink?

Maybe I need a drink too. However, all dipoles/doublets have to fed
'push-pull', so when one leg goes +ve, the other leg goes -ve. The
voltage at all points along the antenna that are equidistant from the
feedpoint will be in antiphase, so if the feedpoint is in the centre,
the voltages at the ends will be in antiphase. [Or is my thinking
seriously muddled?]

# Looks like I owe you a pint. You've described the situation where a TX
# is feeding the dipole. I was trying to visualise the RX conditions, but
# it reciprocates. One of us has to be wrong, and I strongly suspect it's
# me. Time for a drink.

With drinking involved, I must throw in my 2 cents.

I'd go with Frank....for full wave assume positive peak at one end,
negative peak in the middle, and positive peak at the other end. (or
vice versa)

But, I suppose I should think about it a little more.....Laphroig
would help

# See:
# http://tinyurl.com/q8nxqep
# ten rows of images down, second from left:

# This shows the amplitude and the polarity of the voltage and current for
# a halfwave dipole. [Lots of diagrams only show the amplitude.] You will
# see that the polarities on each leg are +ve and -ve. For a fullwave,
# just imagine it continuing on for another halfwave each side.

# -- # Ian

Isn't that figure for a full wave?... lambda

# Maybe you're looking at the wrong one. I've had another look, and it's
# now 9 down, far left. It's the one with the thick black dipole, entitled
# "Halfwave Dipole Antenna (Hertz)". Ah, I've found the source, here (Fig
# 1):
# http://www.digikey.com/en/articles/t...standing-anten
# na-specifications-and-operation

# -- # Ian

OK, but I'm losing touch with what the point is.
The figures referenced both times show voltage peaks of opposite phase
at the ends of half wave dipoles, thus voltage peaks of the same phase
at the end of full wave dipoles.

I'm assuming we have agreement on that.

You assume wrongly. A centre-fed fullwave is also fed 'push-pull', ie in
antiphase - except that it's a high voltage feed instead of high
current.

Just draw a diagram similar to the halfwave, - again showing both the
voltage amplitude and polarity. The lines you draw on each side for the
amplitudes are copies of each other - not mirror images.

However, it seems to a different matter if you feed the fullwave
off-centre, a quarterwave from one end. At least on my drawing, the
voltages at the ends ARE in phase - so I guess the radiation pattern
will be different from the centre-fed.

And with that, it currently is time here in CA for Lagavulin 16.

No such luck here. FWIW, it's nearly midday, and I'm having a (rather
late) strong 'coffee-bag' coffee, with two teaspoons of maple syrup and
one teaspoon (heaped) of dark drinking chocolate powder.
--
Ian