If short antennae radiate all the power that is fed to them, then why would
anyone use long antennae, because the first part of such an antenna, the
short
part, would radiate all the power, and then there'd be nothing left for the
extra bit, making up the rest of the long antenna, to do?

The answer is, of course, because it is more difficult to feed a short
antenna
because of its reactance.

So, whence does this reactance arise?

Simple.

It is the power that has NOT been all radiated by the short antenna arriving
back at
the feed point with an awkward phase relationship with the incident power.

What happens to that power that has not ALL been radiated when it arrives
back
at the feed point?

Simple.

It passes back into the matching network, which, together with the short
bit, form
the resonant artefact, where much of it disappears as heat in the matching
network
before being fed back to the short antenna to start all over again.

Now, Stephen Thomas Cole, that well-respected font of all technical
knowledge over
in uk.radio.amateur is saying that all you Yanks are a bunch of dopes if you
do not understand
the above, so take it up with him over there.

In rec.radio.amateur.antenna gareth wrote:
If short antennae radiate all the power that is fed to them, then why would
anyone use long antennae, because the first part of such an antenna, the
short
part, would radiate all the power, and then there'd be nothing left for the
extra bit, making up the rest of the long antenna, to do?

An antenna radiates as a whole, not as parts.

The answer is, of course, because it is more difficult to feed a short
antenna
because of its reactance.

Nope, the reactance is fairly easily canceled.

The answer is because the radiation resistance is measured in milliohms
and a matching network to match 50 Ohms to milliohms has huge resistive
losses.

So, whence does this reactance arise?

As your base assumtion is nonsense, this is rather irrelevant.

Simple.

It is the power that has NOT been all radiated by the short antenna arriving
back at
the feed point with an awkward phase relationship with the incident power.

What happens to that power that has not ALL been radiated when it arrives
back
at the feed point?

Simple.

It passes back into the matching network, which, together with the short
bit, form
the resonant artefact, where much of it disappears as heat in the matching
network
before being fed back to the short antenna to start all over again.

Now, Stephen Thomas Cole, that well-respected font of all technical
knowledge over
in uk.radio.amateur is saying that all you Yanks are a bunch of dopes if you
do not understand
the above, so take it up with him over there.

Babbling nonsense based on yet another false assumption.



--
Jim Pennino

wrote in message
...

The answer is because the radiation resistance is measured in milliohms
and a matching network to match 50 Ohms to milliohms has huge resistive
losses.

Afraid you've just shot yourself in the foot, there, Old Chap, because the
reason that the apparent radiation resistance is so low is because so little
is radiated!

On 04/11/14 18:57, gareth wrote:
wrote in message
...

The answer is because the radiation resistance is measured in milliohms
and a matching network to match 50 Ohms to milliohms has huge resistive
losses.

Afraid you've just shot yourself in the foot, there, Old Chap, because the
reason that the apparent radiation resistance is so low is because so little
is radiated!


I despair, and I don't think I'm alone in doing so.

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

gareth wrote:
wrote in message
...

The answer is because the radiation resistance is measured in milliohms
and a matching network to match 50 Ohms to milliohms has huge resistive
losses.

Afraid you've just shot yourself in the foot, there, Old Chap, because the
reason that the apparent radiation resistance is so low is because so little
is radiated!

Cart, horse, reversed.

And it is not the "apparent radiation resistance " it is the real, calculable,
and measurably radiation resistance, you gas bag.



--
Jim Pennino

wrote in message
...
gareth wrote:
wrote in message
...

The answer is because the radiation resistance is measured in milliohms
and a matching network to match 50 Ohms to milliohms has huge resistive
losses.

Afraid you've just shot yourself in the foot, there, Old Chap, because
the
reason that the apparent radiation resistance is so low is because so
little
is radiated!

And it is not the "apparent radiation resistance " it is the real,
calculable,
and measurably radiation resistance, you gas bag.

Why do you have this compulsion to shout out insults in the manner of a
5-year-old?

There are two major ways in which the power is dissipated. One is radiation,
and the
other is the i2r losses in the metal.

It is easier for us to model things as though they were resistances, even if
they were not.
(By the same token is the BJT modelled as a combination of resistances,
capacitances
and current generators)

So, the power that is dissipated as radiation is modelled as though it is a
resisitance, although
it is not a resistance, but a mechanism by which power is dissipated.

In terms of the resistance model, that so-called radiation resistance
behaves as though it
is a resistance in series with the resistance of the wire, and it matters
not what current you
manage to force into the antenna, as the antenna shortens, and the apparent
radiation
resistance decreases, the i2r losses start to dominate, and therefore the
short antenna
is a poor radiator in not radiating all the power fed to it.

gareth wrote:
wrote in message
...
gareth wrote:
wrote in message
...

The answer is because the radiation resistance is measured in milliohms
and a matching network to match 50 Ohms to milliohms has huge resistive
losses.

Afraid you've just shot yourself in the foot, there, Old Chap, because
the
reason that the apparent radiation resistance is so low is because so
little
is radiated!

And it is not the "apparent radiation resistance " it is the real,
calculable,
and measurably radiation resistance, you gas bag.

Why do you have this compulsion to shout out insults in the manner of a
5-year-old?

Why do you have this compulsion to post utter nonsense, gas bag?

There are two major ways in which the power is dissipated. One is radiation,
and the
other is the i2r losses in the metal.

They are not the "two major ways", they are the only two ways.

It is easier for us to model things as though they were resistances, even if
they were not.

No, it is not; it makes no difference.

(By the same token is the BJT modelled as a combination of resistances,
capacitances
and current generators)

Gas bag babble.

So, the power that is dissipated as radiation is modelled as though it is a
resisitance, although
it is not a resistance, but a mechanism by which power is dissipated.

Nope, the radiation resistance is the result of the model. Once again
you have the cart and horse reversed.

In terms of the resistance model, that so-called radiation resistance
behaves as though it

There is no "resistance model", gas bag.

is a resistance in series with the resistance of the wire, and it matters
not what current you
manage to force into the antenna, as the antenna shortens, and the apparent
radiation
resistance decreases, the i2r losses start to dominate, and therefore the
short antenna
is a poor radiator in not radiating all the power fed to it.

Yet more gas bag babble and poor logic based on a false premise.




--
Jim Pennino

wrote in message
...
And it is not the "apparent radiation resistance " it is the real,
calculable,
and measurably radiation resistance, you gas bag.

Lo! And behold! When you calculate it, one of the terms is the ratio
between antenna length and wavelength

QED

gareth wrote:
wrote in message
...
And it is not the "apparent radiation resistance " it is the real,
calculable,
and measurably radiation resistance, you gas bag.

Lo! And behold! When you calculate it, one of the terms is the ratio
between antenna length and wavelength

Therefore it is not "apparent radiation resistance" as you said, it is
real and calculable as I said, gas bag.



--
Jim Pennino

On 11/4/2014 12:48 PM, gareth wrote:
If short antennae radiate all the power that is fed to them, then why would
anyone use long antennae, because the first part of such an antenna, the
short
part, would radiate all the power, and then there'd be nothing left for the
extra bit, making up the rest of the long antenna, to do?

The answer is, of course, because it is more difficult to feed a short
antenna
because of its reactance.

So, whence does this reactance arise?

Simple.

It is the power that has NOT been all radiated by the short antenna arriving
back at
the feed point with an awkward phase relationship with the incident power.

What happens to that power that has not ALL been radiated when it arrives
back
at the feed point?

Simple.

It passes back into the matching network, which, together with the short
bit, form
the resonant artefact, where much of it disappears as heat in the matching
network
before being fed back to the short antenna to start all over again.

Now, Stephen Thomas Cole, that well-respected font of all technical
knowledge over
in uk.radio.amateur is saying that all you Yanks are a bunch of dopes if you
do not understand
the above, so take it up with him over there.

It is so amazing how a simple post of nonsense will make all the
gullible members of this group dance like puppets on a string.

--

Rick