"Art Unwin" wrote in message
...
On Sep 21, 7:33 pm, "christofire" wrote:
"Art Unwin" wrote in message

...
On Sep 21, 4:19 pm, "christofire" wrote:

"Art Unwin" wrote in message

- - snip - -

Chris you are being stupid as well as acting as a fool.
Acceleration of the particle only occurs while within the
electrostatic field. When it exits it has the speed of light because
it has emmerged from the intersecting two fields.and thus from the
accellerating forces. I remind you of Newtons law of ut + 1/2 ft sqd
The first expression is for the speed attained on entering the
accelerating field and the other half is for the length of the
accelerating electrostatic field. On leaving the two fields it looses
the applied accelerating force where it has arrived at a particular
speed.Maybe you should look up the workings of a CRT using Newtons
laws instead of shooting from the hip.
It was at a different time that the speed of light was measured where
it was found to equal the sppeed emerging from two intersecting fields.

* Actually, I'm well aware of the principle involved in accelerating
electrons in an electron gun as used in CRTs, klystrons, TWTs, and so on,
by
subjecting an electron cloud to a potential difference using an anode with
a
hole in it (!), but that's different from what happens in an antenna.

The acceleration of charge in an antenna results almost entirely from the
applied potential difference at its terminals. The radiated fields result
from the alternating current effectively passing through the radiation
resistance, and all the other, reactive, fields have no direct effect on
the
radiation resistance, or the component of the current that passes through
it
in phase with the voltage that is developed across it, which together, of
course, represent the radiated power. The reactive fields affect the
terminal impedance and a large imaginary part can upset the device trying
to
send power into the antenna, but that is more of a system issue. The
alternating current that passes through the radiation resistance is
composed
of charge that moves in time with each RF cycle, accelerating and
decelerating accordingly. The electrostatic field developed between the
ends of a half-wave dipole reaches its maximum value a quarter of a cycle
later than the voltage at the drive point so any effect it has on the
charge
in the antenna elements during each cycle must be reactive, and it doesn't
affect the radiation resistance or the radiated wave.

But all this can be looked up from any one of the respected books on
antennas. Krauss, and others, gives expressions for the different field
components and the theory all hangs together quite readily using Maxwell's
equations without modification.

I don't care if you think me stupid, and I'll continue to try to avoid
name
calling of individuals, although I may criticise what they write
especially
if it appears ignorant of proper science yet attempts to re-write
established theory, and therefore appears arrogant.

Chris

No it does not appear in Kraus book.

* That is incorrect, and I expect you know that to be the case. All of what
I wrote above can be traced to sources such as Kraus, Jordan and Balmain,
and Jasik; books you may never have tried to read and understand (on the
basis of your comment).


He never followed Maxwells laws
with respect to equilibrium.

* Define 'equilibrium' in trems of normal physics and cite a reference.


To do that you must think in terms of
wavelength. After all with respect to science all revolve around
boundary laws of the Universe and you blindly ignore that fact.

* How is there a boundary to the universe? It is unbounded - the universe
is known to be expanding into empty space.


Now back to radiation which applies spin to a p.

* No it doesn't - cite a reference.


Nowarticle where as
with a crt no spin is applied as it is heat that separates the resting
particle

* ... separates it from what?


Now earlier you refered to a electrostatic field that
according to what you stated did not have a border and surely you know
that just can't be unless it is in equilibrium which requires a closed
circuit.

* Fields in unbounded space tend to be unbounded. The fields around a
dipole are bounded by the surface of the dipole, which is the boundary
condition used by NEC, and usually in practice by the earth.


For this to come to fruition you go back to the boundary laws where
the arbitrary border is one that is closed i.e. in equilibrium.

* Define 'equilibrium' in terms of normal physics and cite a reference.


Now
cast your mind back to the Gauss extension where radiators and
particles lie in a closed static field. Now you should see that
equilibrium must reign for a closed static field. Now you blithely
mentioned an electrostatic field with nary a mention as to how it is
formed and how it fits into the whole picture.

* Recent discussion has been about the electric field produced by the
voltage that appears between the ends of a dipole - that's what I was
writing about, and so were you. How can you state 'with nary a mention as
to how it is formed'?


Why? Because the books do not provide an unbroken trail that fully
describes radiation
( books admit that) as I have done where everything dove tails into
the existing laws of the Universe.

* Kraus provides a trail that's as unbroken as can be comprehended by most
engineers. Physicists may wish to take it further but there's no evidence
of physicists in this newsgroup.


And then I gave you a bonus with
respect to the weak field that Einstein searched for in vain. Einstein
looked at the package presented by the Big Bag but omitted to keep
that which it was wrapped into.

* Is the universe enclosed in a Big Bag?


That was the arbitrary boundary around
which were forces or vectors that were equal and opposite when a
smallest of smallest of particles edged out towards the border. Yes it
was of a weak force but for all of that it broke the equilibrium
boundary as the forces at that point was not now equal. The breakage
was one where the opposing forces were offset to each other thus
providing a torque force that provided spin. Now we come to our own
Earth encased in a arbitrary border and outside the border we have the
same conditions of equilibrium that must equal the forces of the Big
Bang which means the outside has two vectors, gravity vector which is
straight and a vector denoting spin i.e. rotation of the Earth. On the
other side of the border you again have two vectors a straight vector
and that curly one you don't like me to call eddy current.

* Call it what you wish. Capitalising the word within a sentence, as you
did before, is usually reserved for phenomena named after the people who
discovered them - but that's an engineer thing.


Yup. Everything falls together nicely thank you, when you study
radiation from first principles instead of binding yourself to books
that readily admit to not understanding the radiation process.

* If you believe so fervently in your own version of all this then why don't
you submit it to peer review at sci.physics or sci.physics.research? It's
rather unfair to expose this only to an amateur radio newsgroup when what
you are doing is apparently re-writing physics in such a major way. Those
newsgroups would be more appropriate, considering the depth to which you are
going (i.e. well beyond amateur radio) - wouldn't you agree.


Now
this is not being arrogant when one has applied all principles

* My dictionary defines 'arrogant' as: 'having or showing an exaggerated
opinion of one's own importance , merit, ability, etc.' from the Latin
'arrogare': 'to claim as one's own'. Your re-writing of physics conforms to
this definition precisely.


but I
do think it is arrogant of you and others to asasinate the character
of "S" purely because of his english and spelling and not to help one
that wants to learn.

* He is showing no inclination of wanting to learn - he asks a question, and
then when answered responds with non-physical contradiction. His treatement
here is a direct consequence of that behaviour, like yours.


Just think about it all. I have shown how a particle moves in space in
a straight line trajectory such that the particle maintains a straight
line without the parabolic force of gravity driving it down to the
Earth. Go back to your books and show just how the electrostatic field
came about and what was the borders that it was contained in
Art
via applied spin and where gravity


* As I say, you should present your theory to sci.physics and
sci.physics.research if you have any interest in checking whether it is
correct, and not limit its exposure to this group. Do let us know when you
have posted there.

Chris

Szczepan Bia³ek wrote:
Now we should work out
the consensus on which part of the radiator radiate the radio waves.

Here's what you are missing: RF is AC. At some point
in the radiation cycle, the instantaneous radiation has
to be zero at the zero-crossing time. That is when the
magnetic field energy is essentially zero and close to
100% of the energy is contained in the electric fields
at the ends of the dipole.

You cannot have it both ways. You cannot have a radiation
peak at a net current maximum time and also have a radiation
peak at a net voltage maximum time since they are ~90 degrees
out of phase. If what you are saying were really happening,
an antenna would radiate two times the applied frequency,
but it obviously doesn't.

Or did you not realize that when the electric field is
at its maximum amplitude, the current is close to zero-
crossing all up and down the standing wave antenna? The
phase of the feedpoint current is within a couple of
degrees of the phase of the current all up and down
the antenna.

That's why the current on a standing wave antenna cannot
be used to measure the delay through a wire or a loading
coil. All such "measurements" are bogus.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

On Sep 22, 2:21*am, Szczepan Bia³ek wrote:

It apply to 1/2WL dipole. But what radiate in 0.05WL dipole? There the
"maximum radiation " is in the transmission line (1/4WL from the end).
The feed point is also in the transmissing line.

There is no radiation from a transmission line as long as the currents
in it are balanced. Attaching a transmission line to a 0.05WL center-
fed dipole does not change that balance.

The feedpoint of such a short dipole has very high capacitive
reactance, so getting much current to flow into it would be difficult
without proper impedance matching to the transmission line.

But whatever the impedance match, all of the radiation occurs from the
0.05WL dipole itself, and maximum radiation is generated at the
feedpoint, where the r-f current is greatest.

RF

On Tue, 22 Sep 2009 09:40:30 +0200, Szczepan Bia?ek
wrote:

Which has absolutely nothing to do with the failure of your swampy
metaphor - EXCEPT to demonstrate its stagnation into a cesspool by
being completely ignored by you.

It's amusing to see you wading out there tho'. ;-)

For me it is important that at the end the voltage is doubled

Yes, we can all tell what is important to you. Unfortunately it does
not equate with what is important for you. No doubt that distinction
is lost in translation and you will continue to fumble on.

I'm glad to see you shed that nonsense about hydraulics. As you
understood that topic far less than RF (which is in itself on very
shaky ground), it wouldn't have done to poison the well.

This poor understanding is quite obvious by your last comment above.

73's
Richard Clark, KB7QHC

On Tue, 22 Sep 2009 12:51:44 +0100, "christofire"
wrote:

Chris

Could you turn your newsreader's quoting mechanism back on? It is
very confusing to have to fumble with Art's babbling interleaved with
your "special" editorial marks.

73's
Richard Clark, KB7QHC

On Tue, 22 Sep 2009 10:48:43 +0200, Szczepan Bia?ek
wrote with abysmal quoting:

It will not be easy to work out here all details.
S*


Regards
Art

You guys need to get a room.

73's
Richard Clark, KB7QHC

"Richard Fry" wrote
...
On Sep 22, 2:21 am, Szczepan Bia³ek wrote:

It apply to 1/2WL dipole. But what radiate in 0.05WL dipole? There the
"maximum radiation " is in the transmission line (1/4WL from the end).
The feed point is also in the transmissing line.

There is no radiation from a transmission line as long as the currents
in it are balanced. Attaching a transmission line to a 0.05WL center-
fed dipole does not change that balance.

The feedpoint of such a short dipole has very high capacitive
reactance, so getting much current to flow into it would be difficult
without proper impedance matching to the transmission line.

The feed point is the part of the transmissing line and not radiate.

But whatever the impedance match, all of the radiation occurs from the
0.05WL dipole itself, and maximum radiation is generated at the
feedpoint, where the r-f current is greatest.

See above.

Dipole 0.05 is probably the straight. How long are the folded dipoles and
the loop antennas? Are there the short version?
S*

"Cecil Moore" wrote
...
Szczepan Bia³ek wrote:
Now we should work out the consensus on which part of the radiator
radiate the radio waves.

Here's what you are missing: RF is AC. At some point
in the radiation cycle, the instantaneous radiation has
to be zero at the zero-crossing time. That is when the
magnetic field energy is essentially zero and close to
100% of the energy is contained in the electric fields
at the ends of the dipole.

You cannot have it both ways. You cannot have a radiation
peak at a net current maximum time and also have a radiation
peak at a net voltage maximum time since they are ~90 degrees
out of phase. If what you are saying were really happening,
an antenna would radiate two times the applied frequency,
but it obviously doesn't.

We do not have the both. But we have the Luxembourg effect. Each dipole
antena radiate two times the applied frequency, The pulses from the ends
are 180 degrees apart.

Or did you not realize that when the electric field is
at its maximum amplitude, the current is close to zero-
crossing all up and down the standing wave antenna?
phase of the feedpoint current is within a couple of
degrees of the phase of the current all up and down
the antenna.

That's why the current on a standing wave antenna cannot
be used to measure the delay through a wire or a loading
coil. All such "measurements" are bogus.

S*

On Sep 22, Szczepan Bia³ek wrote:

But what radiate in 0.05WL dipole? There the
"maximum radiation" is in the transmission line (1/4WL from the end).
The feed point is also in the transmissing line.

Then later the same day he wrote:

The feed point is the part of the transmissing line and not radiate.

Pick one of the above comments, only, Szczepan.

The feed points are terminals of the antenna. On center-fed dipoles
that are 1/2WL or less in length, antenna current is highest at those
terminals.

How long are the folded dipoles and the
loop antennas? Are there the short version?

They can be any length, but some lengths have better input
characteristics and/or more useful radiation patterns than others.

RF

Szczepan Bia³ek wrote:
Each dipole
antenna radiate two times the applied frequency, ...

Sorry, 2*sin(2wt) sin(wt)

Here's the question: Is the radiated RF wave in phase
with the standing wave current or in phase with the
standing wave voltage? The radiated RF wave cannot be
in phase with both since they are 90 degrees out of
phase on the standing wave antenna.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com