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 phase,

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.
--
;-)
..
73 de Frank Turner-Smith G3VKI - mine's a pint.
..
http://turner-smith.co.uk
..
Ubuntu 12.04
Thunderbirds are go.

"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 phase,

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


Wayne
W5GIE/6

"Wayne" wrote in :

Laphroig would help


Oban. That comes at you like a strong onshore wind. Good stuff.

About fullwave dipoles, I read something that said just stay with halfwave
for easier matching and an easier time getting it high enough. The writer had
a callsign and everything. Seriously, I think he's right because those
things will make up for any 'advantage' you might gain with a fullwave,
apparently.

"Lostgallifreyan" wrote in message
. ..

"Wayne" wrote in :

Laphroig would help


# Oban. That comes at you like a strong onshore wind. Good stuff.

# About fullwave dipoles, I read something that said just stay with halfwave
# for easier matching and an easier time getting it high enough. The writer
had
# a callsign and everything. Seriously, I think he's right because those
# things will make up for any 'advantage' you might gain with a fullwave,
# apparently.

IIRC one might choose a full wave over half wave if the pattern lobes are
more suitable.

I might look at that later today.

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 phase,

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

"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
phase,

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

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
phase,

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

On 30/10/14 20:44, Ian Jackson wrote:
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
phase,

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

The original proposal in this thread was that long antennas performed
better than short ones. If that was true you'd get a good 600MHz UHF TV
picture using a 132ft end fed longwire. I've not tried it, but it
doesn't seem very likely.

--
;-)
..
73 de Frank Turner-Smith G3VKI - mine's a pint.
..
http://turner-smith.co.uk
..
Ubuntu 12.04
Thunderbirds are go.

In message , Frank Turner-Smith G3VKI
writes
On 30/10/14 20:44, Ian Jackson wrote:
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
phase,

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

The original proposal in this thread was that long antennas performed
better than short ones. If that was true you'd get a good 600MHz UHF TV
picture using a 132ft end fed longwire. I've not tried it, but it
doesn't seem very likely.

A 132' endfed will have one hell of a gain on 600MHz - but it will be
almost straight off the ends.
--
Ian

"Ian Jackson" wrote in message
...
The original proposal in this thread was that long antennas performed
better than short ones. If that was true you'd get a good 600MHz UHF TV
picture using a 132ft end fed longwire. I've not tried it, but it doesn't
seem very likely.

A 132' endfed will have one hell of a gain on 600MHz - but it will be
almost straight off the ends.
--
Ian
I know the gain will be off the end of the wire, but still wonder if an
antenna that long (in wavelengths) will actually work or will it be too
long and the gain does not meet the expectations or if programs like NEC
will predict it or fall apart.



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