On Tue, 12 May 2009 18:00:09 +0200, Szczepan Bia?ek
wrote:

The topics is polarisation. This word has the two meaning. The both have
wrong explanations in texbooks. The one I have verified with the comb.
The second "wave polarisation" is explained with transverse waves. No
transverse waves. If receiver (resonator) must be parallel to emmiter you
can explain it in many ways. But to verify it the comb is not enough. So I
need help.


That is fair. Let's start with some serious misunderstandings with a
few questions to test them.

First, let us return to that link you offered with the Hertzian Loop
with its spark gap. Let us say that this loop is 1 meter of wire
(about the actual size anyway). Let us say there is a current
detector at each end of this loop. Let us say we have closed a switch
that applies voltage to the loop, and the first meter has indicated
current flow. This is our time reference point. Now the questions:

1. For the electron that went through the first current detector, how
long does it take for that SAME electron to get to the second
detector?

2. How long does it take for the second detector to indicate there is
current flow?

Hint: the answer for 1. is very, very different for the answer for 2.

Now, let us say that before that SAME electron gets to the second
current detector, that path is broken open (maybe 1 pico second before
the SAME electron arrival). The SAME electron sees an open circuit.
What is the amount of energy required for the electron to break out of
the metal conductor, and into the air?

73's
Richard Clark, KB7QHC

"Richard Clark" wrote
...
On Tue, 12 May 2009 18:00:09 +0200, Szczepan Bia?ek
wrote:

The topics is polarisation. This word has the two meaning. The both have
wrong explanations in texbooks. The one I have verified with the comb.
The second "wave polarisation" is explained with transverse waves. No
transverse waves. If receiver (resonator) must be parallel to emmiter you
can explain it in many ways. But to verify it the comb is not enough. So I
need help.


That is fair. Let's start with some serious misunderstandings with a
few questions to test them.

First, let us return to that link you offered with the Hertzian Loop
with its spark gap. Let us say that this loop is 1 meter of wire
(about the actual size anyway). Let us say there is a current
detector at each end of this loop. Let us say we have closed a switch
that applies voltage to the loop, and the first meter has indicated
current flow. This is our time reference point. Now the questions:

1. For the electron that went through the first current detector, how
long does it take for that SAME electron to get to the second
detector?

2. How long does it take for the second detector to indicate there is
current flow?

Hint: the answer for 1. is very, very different for the answer for 2.

Now, let us say that before that SAME electron gets to the second
current detector, that path is broken open (maybe 1 pico second before
the SAME electron arrival). The SAME electron sees an open circuit.
What is the amount of energy required for the electron to break out of
the metal conductor, and into the air?

You went to details. Early you wrote: "An antenna radiates in ALL directions
from EVERYPOINT of
the antenna. "
Textbooks say that EM transversial waves are emitted by current (the sparks
in Hertz apparatus - not from the ends).
I say that from the ends (as electric waves similar to acoustics).
The directional pattern must be different.
The directional patterns of loudspeakers and Herts dipoles are very
similar.
So I try to find evidences.
Now I do not know if you prefer EM or electric waves.
S*

On Tue, 12 May 2009 21:29:31 +0200, Szczepan Bia?ek
wrote:

You went to details. Early you wrote: "An antenna radiates in ALL directions
from EVERYPOINT of
the antenna. "

They do.

Textbooks say that EM transversial waves are emitted by current (the sparks
in Hertz apparatus - not from the ends).

That is not the same thing as radiation. Repeating poor quotations
does not make it better.

I say that from the ends (as electric waves similar to acoustics).

Compressional waves or longitudinal waves? In solid or air or in
liquid? The answers to these questions lead to very, very different
behavior. As I say, these simplicities you use are nonsense.

The directional pattern must be different.

The directional pattern is a combination of EVERYPOINT radiating in
ALL directions. The differences in their position contribute to an
unique pattern. This is the whole basis of the "method of moments"
application of modeling radiation emitters.

The directional patterns of loudspeakers and Herts dipoles are very
similar.

The are more differences than similarities.

So I try to find evidences.
Now I do not know if you prefer EM or electric waves.

That shouldn't keep you from answering the simple physics of:
Let's start with some serious misunderstandings with a
few questions to test them.

First, let us return to that link you offered with the Hertzian Loop
with its spark gap. Let us say that this loop is 1 meter of wire
(about the actual size anyway). Let us say there is a current
detector at each end of this loop. Let us say we have closed a switch
that applies voltage to the loop, and the first meter has indicated
current flow. This is our time reference point. Now the questions:

1. For the electron that went through the first current detector, how
long does it take for that SAME electron to get to the second
detector?

2. How long does it take for the second detector to indicate there is
current flow?

Hint: the answer for 1. is very, very different for the answer for 2.

Now, let us say that before that SAME electron gets to the second
current detector, that path is broken open (maybe 1 pico second before
the SAME electron arrival). The SAME electron sees an open circuit.
What is the amount of energy required for the electron to break out of
the metal conductor, and into the air?

73's
Richard Clark, KB7QHC

We will take this by parts:

On Tue, 12 May 2009 18:00:09 +0200, Szczepan Bia?ek
wrote:

The topics is polarisation. This word has the two meaning. The both have
wrong explanations in texbooks.

Blaming references is a very strong indicator of you having the
problem, not the textbooks.

The one I have verified with the comb.

I've read that, and it proves nothing that is part of the topic (I
shall explain below).

The second "wave polarisation" is explained with transverse waves. No
transverse waves.

If "no," then what "yes?"

Actually you have mixed up two different characteristics. Polarity
and polarization are NOT the same thing. With RF radiation, the wave
is constantly changing polarity (that is why the source of RF is
called alternating current), but within the "line of sight" of the
antenna, the polarization for a dipole is defined by its angle to the
earth as viewed by the observer.

If you see an horizontal dipole, it produces alternating polarities of
waves with horizontal polarization. If you see a vertical dipole, it
produces alternating polarities of waves with vertical polarization.

RF energy is ALWAYS changing polarity.

If receiver (resonator) must be parallel to emmiter you
can explain it in many ways. But to verify it the comb is not enough. So I
need help.

The comb was useless in proving anything about RF, antennas, or
polarization. What it demonstrates is induced polarity in a
dielectric. Interesting, but entirely unrelated to the topic.

73's
Richard Clark, KB7QHC

"Richard Clark" wrote
...
We will take this by parts:

On Tue, 12 May 2009 18:00:09 +0200, Szczepan Bia?ek
wrote:

The topics is polarisation. This word has the two meaning. The both have
wrong explanations in texbooks.

Blaming references is a very strong indicator of you having the
problem, not the textbooks.

Is in old books the same as in todays?
Sometimes they are changed.

The one I have verified with the comb.

I've read that, and it proves nothing that is part of the topic (I
shall explain below).

The second "wave polarisation" is explained with transverse waves. No
transverse waves.

If "no," then what "yes?"

You wrote: "Compressional waves or longitudinal waves? In solid or air or
in
liquid? The answers to these questions lead to very, very different
behavior. As I say, these simplicities you use are nonsense"

In EM are many simplifications. Electric waves are like the acoustic. Of
course they kick the electrons in antennas not a membranes.

Actually you have mixed up two different characteristics. Polarity
and polarization are NOT the same thing. With RF radiation, the wave
is constantly changing polarity

In transvers waves something changes the direction of rotation. The source
makes rotating oscillations.

(that is why the source of RF is called alternating current), but within
the "line of sight" of the
antenna, the polarization for a dipole is defined by its angle to the
earth as viewed by the observer.)

You decribe the electric waves.

If you see an horizontal dipole, it produces alternating polarities of
waves with horizontal polarization. If you see a vertical dipole, it
produces alternating polarities of waves with vertical polarization.

!00% agreement with me. RF waves are electrical.

RF energy is ALWAYS changing polarity.

If receiver (resonator) must be parallel to emmiter you
can explain it in many ways. But to verify it the comb is not enough. So I
need help.

The comb was useless in proving anything about RF, antennas, or
polarization. What it demonstrates is induced polarity in a
dielectric. Interesting, but entirely unrelated to the topic.

The Acoustic analogy will be also interesting.
S*

Richard Clark wrote:
If you see an horizontal dipole, it produces alternating polarities of
waves with horizontal polarization.

That is broadside to the antenna. Directly off the
ends of the dipole, the polarization is vertical.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

On Wed, 13 May 2009 09:01:35 +0200, Szczepan Bia?ek
wrote:

Blaming references is a very strong indicator of you having the
problem, not the textbooks.

Is in old books the same as in todays?
Sometimes they are changed.

Engineering? Never.

You wrote: "Compressional waves or longitudinal waves? In solid or air or
in
liquid? The answers to these questions lead to very, very different
behavior. As I say, these simplicities you use are nonsense"

In EM are many simplifications. Electric waves are like the acoustic.

They are not. You are repeating your nonsense again.

Of
course they kick the electrons in antennas not a membranes.

This is more nonsense.

Actually you have mixed up two different characteristics. Polarity
and polarization are NOT the same thing. With RF radiation, the wave
is constantly changing polarity

In transvers waves something changes the direction of rotation. The source
makes rotating oscillations.

You are showing very little capacity to understand alternating
current, RF, and radiation.

(that is why the source of RF is called alternating current), but within
the "line of sight" of the
antenna, the polarization for a dipole is defined by its angle to the
earth as viewed by the observer.)

You decribe the electric waves.

I am describing radiation.

The Acoustic analogy will be also interesting.
S*

No, it would not. Radiation is wholly unlike acoustics and very few
people actually understand acoustics. Having the experience of
"hearing" does not qualify you as being proficient in its discussion.
Human sensation is vastly more illusion than science.

This is quite evident in your postings where you rely on crude
simplifications and incorrect metaphors. The proof of their failure
in your being to understand the topic of wave polarization is you
cannot answer:

First, let us return to that link you offered with the Hertzian Loop
with its spark gap. Let us say that this loop is 1 meter of wire
(about the actual size anyway). Let us say there is a current
detector at each end of this loop. Let us say we have closed a switch
that applies voltage to the loop, and the first meter has indicated
current flow. This is our time reference point. Now the questions:

1. For the electron that went through the first current detector, how
long does it take for that SAME electron to get to the second
detector?

2. How long does it take for the second detector to indicate there is
current flow?

Hint: the answer for 1. is very, very different for the answer for 2.

Now, let us say that before that SAME electron gets to the second
current detector, that path is broken open (maybe 1 pico second before
the SAME electron arrival). The SAME electron sees an open circuit.
What is the amount of energy required for the electron to break out of
the metal conductor, and into the air?

These are very fundamental concepts that, if you cannot respond to
them, reveal your range of involvement is extremely thin and you
probably have no real interest in the topic.

Why are you posting here?

73's
Richard Clark, KB7QHC

"Richard Clark" wrote
...
On Wed, 13 May 2009 09:01:35 +0200, Szczepan Bia?ek
wrote:

Blaming references is a very strong indicator of you having the
problem, not the textbooks.

Is in old books the same as in todays?
Sometimes they are changed.

Engineering? Never.

Now we have the two different physics. One for students and the second in
engineering.
Textbooks are for students. In engineering are Handbooks and specifications.

You wrote: "Compressional waves or longitudinal waves? In solid or air or
in
liquid? The answers to these questions lead to very, very different
behavior. As I say, these simplicities you use are nonsense"

In EM are many simplifications. Electric waves are like the acoustic.

They are not. You are repeating your nonsense again.

Of
course they kick the electrons in antennas not a membranes.

This is more nonsense.

Actually you have mixed up two different characteristics. Polarity
and polarization are NOT the same thing. With RF radiation, the wave
is constantly changing polarity

In transvers waves something changes the direction of rotation. The source
makes rotating oscillations.

You are showing very little capacity to understand alternating
current, RF, and radiation.

(that is why the source of RF is called alternating current), but within
the "line of sight" of the
antenna, the polarization for a dipole is defined by its angle to the
earth as viewed by the observer.)

You decribe the electric waves.

I am describing radiation.

In engineering sense. I agree with you.

The Acoustic analogy will be also interesting.
S*

No, it would not. Radiation is wholly unlike acoustics and very few
people actually understand acoustics. Having the experience of
"hearing" does not qualify you as being proficient in its discussion.
Human sensation is vastly more illusion than science.

This is quite evident in your postings where you rely on crude
simplifications and incorrect metaphors. The proof of their failure
in your being to understand the topic of wave polarization is you
cannot answer:

First, let us return to that link you offered with the Hertzian Loop
with its spark gap. Let us say that this loop is 1 meter of wire
(about the actual size anyway). Let us say there is a current
detector at each end of this loop. Let us say we have closed a switch
that applies voltage to the loop, and the first meter has indicated
current flow. This is our time reference point. Now the questions:

1. For the electron that went through the first current detector, how
long does it take for that SAME electron to get to the second
detector?

2. How long does it take for the second detector to indicate there is
current flow?

Hint: the answer for 1. is very, very different for the answer for 2.

Now, let us say that before that SAME electron gets to the second
current detector, that path is broken open (maybe 1 pico second before
the SAME electron arrival). The SAME electron sees an open circuit.
What is the amount of energy required for the electron to break out of
the metal conductor, and into the air?

These are very fundamental concepts that, if you cannot respond to
them, reveal your range of involvement is extremely thin and you
probably have no real interest in the topic.

Why are you posting here?

To lern something from engineering people. I was sure that people in this
group must know such phenomenon. It was Brian Howie. He wrote: "You can get
ionospheric mixing of radio waves. e.g Luxembourg Effect; so
doubling is possible."

Today I found this:
http://durenberger.com/resources/doc...EFFECT0235.pdf
So I have known what I want to know.
Of course my explanation of the phenomenon is as I described here.
It is not easy to understand me. I am sure that Wim does. He wrote "If you
would like to know more about EM-fields related to antennas and electronics,
just start with classical EM theory. This is a solid
tool, existing over 100 years and is used by many people with succes
to predict behaviour of circuits and antennas. If this will change of
today, I will close my business activities next monday"

Most engineering people very quickly forget the "classical EM theory" as was
tought in schools. In the "classical EM theory" no electrons. Engineering
use his own physics with electrons.
S*

On Wed, 13 May 2009 21:41:55 +0200, Szczepan Bia?ek
wrote:

Blaming references is a very strong indicator of you having the
problem, not the textbooks.

Is in old books the same as in todays?
Sometimes they are changed.

Engineering? Never.

Now we have the two different physics. One for students and the second in
engineering.

The students are different, not the Physics. There is also college
physics and University physics. The sense here is that you are adept
at calculus (University) or you are not (college). Your explanations
of physics are not even High School (Gymnasium) level.

Why are you posting here?

To lern something from engineering people. I was sure that people in this
group must know such phenomenon. It was Brian Howie. He wrote: "You can get
ionospheric mixing of radio waves. e.g Luxembourg Effect; so
doubling is possible."

This is useless. It is like you asking "Will it rain?", and someone
telling you the story of Noah and the flood. Would you build an Ark
based on this answer to your question?

Today I found this:
http://durenberger.com/resources/doc...EFFECT0235.pdf

Again, a very poor source. You are merely wandering along the shelves
and picking up random information.

So I have known what I want to know.
Of course my explanation of the phenomenon is as I described here.
It is not easy to understand me. I am sure that Wim does. He wrote "If you
would like to know more about EM-fields related to antennas and electronics,
just start with classical EM theory. This is a solid
tool, existing over 100 years and is used by many people with succes
to predict behaviour of circuits and antennas. If this will change of
today, I will close my business activities next monday"

What was the point of quoting this when you have done nothing from his
advice?

Most engineering people very quickly forget the "classical EM theory" as was
tought in schools. In the "classical EM theory" no electrons. Engineering
use his own physics with electrons.
S*

This worn out phrase is repeated so much that I don't think you are
really interested in anything more than trolling.

If you actually found a solution in the Luxembourg effect you could
answer:

First, let us return to that link you offered with the Hertzian Loop
with its spark gap. Let us say that this loop is 1 meter of wire
(about the actual size anyway). Let us say there is a current
detector at each end of this loop. Let us say we have closed a switch
that applies voltage to the loop, and the first meter has indicated
current flow. This is our time reference point. Now the questions:

1. For the electron that went through the first current detector, how
long does it take for that SAME electron to get to the second
detector?

2. How long does it take for the second detector to indicate there is
current flow?

Hint: the answer for 1. is very, very different for the answer for 2.

Now, let us say that before that SAME electron gets to the second
current detector, that path is broken open (maybe 1 pico second before
the SAME electron arrival). The SAME electron sees an open circuit.
What is the amount of energy required for the electron to break out of
the metal conductor, and into the air?

73's
Richard Clark, KB7QHC

"Richard Clark" wrote
...
On Wed, 13 May 2009 21:41:55 +0200, Szczepan Bia?ek
wrote:

Blaming references is a very strong indicator of you having the
problem, not the textbooks.

Is in old books the same as in todays?
Sometimes they are changed.

Engineering? Never.

Now we have the two different physics. One for students and the second in
engineering.

The students are different, not the Physics. There is also college
physics and University physics. The sense here is that you are adept
at calculus (University) or you are not (college). Your explanations
of physics are not even High School (Gymnasium) level.

On all levels is: "James Clerk Maxwell's mathematical theory of 1873 had
predicted that electromagnetic disturbances should propagate through space
at the speed of light and should exhibit the wave-like characteristics of
light propagation."

Next on all levels is that the light waves are transversal because they can
be polarised.

Maxwell described it in his Treatise in words and in equations (are on-line)


Why are you posting here?

To lern something from engineering people. I was sure that people in this
group must know such phenomenon. It was Brian Howie. He wrote: "You can
get
ionospheric mixing of radio waves. e.g Luxembourg Effect; so
doubling is possible."

This is useless. It is like you asking "Will it rain?", and someone
telling you the story of Noah and the flood. Would you build an Ark
based on this answer to your question?

I have never heard about the Luxembourg Effect. I predict it and ask proper
people (if is "frequency doubling).

Today I found this:
http://durenberger.com/resources/doc...EFFECT0235.pdf

Again, a very poor source. You are merely wandering along the shelves
and picking up random information.

The only I need.

So I have known what I want to know.
Of course my explanation of the phenomenon is as I described here.
It is not easy to understand me. I am sure that Wim does. He wrote "If you
would like to know more about EM-fields related to antennas and
electronics,
just start with classical EM theory. This is a solid
tool, existing over 100 years and is used by many people with succes
to predict behaviour of circuits and antennas. If this will change of
today, I will close my business activities next monday"

What was the point of quoting this when you have done nothing from his
advice?

I know the classical EM theory.

Most engineering people very quickly forget the "classical EM theory" as
was
tought in schools. In the "classical EM theory" no electrons. Engineering
use his own physics with electrons.
S*

This worn out phrase is repeated so much that I don't think you are
really interested in anything more than trolling.


If you actually found a solution in the Luxembourg effect you could
answer:

First, let us return to that link you offered with the Hertzian Loop
with its spark gap. Let us say that this loop is 1 meter of wire
(about the actual size anyway). Let us say there is a current
detector at each end of this loop. Let us say we have closed a switch
that applies voltage to the loop,

....voltage by additionl wire ( in what point of the ring?) or by space?
S*