"Szczepan Bialek" wrote in message
...

Here you are the simple:
http://www.ac-grenoble.fr/yre/agency...e/rap-nuc2.htm

Are the dipole elements symmetrical?
It seems that one of them is a counterpoise.
S*
Wiki says that a rectenna is "a dipole antenna with a diode connected across
the dipole elements".
A dipole is a balanced, symmetrical aerial and does not have a counterpoise.
I can't be responsible for you being unable to understand what
"symmetrical", "balanced" and "counterpoise" mean. I can suggest / recommend
that you acquire a reference book on aerials from the ARRL or RSGB. You'd
find them immensely helpful.
Regards, Ian.

Szczepan Bialek wrote:

"Ian" napisa³ w wiadomo¶ci
...
"Szczepan Bialek" wrote in message
.. .

See:
http://www.google.com/patents?id=j3h...434678&f=false

S*
Ah - we're back to the rectenna which is defined in Wiki thus "A simple
rectenna element consists of a dipole antenna with a diode connected
across the dipole elements".

Here you are the simple:
http://www.ac-grenoble.fr/yre/agency...e/rap-nuc2.htm

Are the dipole elements symmetrical?

Yes

It seems that one of them is a counterpoise.

Why? I see symmetrical elements.

What you can also see here is that this is another example of what
I wrote earlier: there is a circle in the schematic starting at the
diode where you can go around with DC current and end at the same
point.
This circle is where the DC current flows, not from the elements into
the air.

"Rob" napisa³ w wiadomo¶ci
...
Szczepan Bialek wrote:
I don't care what people have written in the 19th century. Please
stop bringing that up. I am only interested in how things are
explained today.

Everything was discovered in XIX (for the radio):

But then later it was found that the first discoveries were not
entirely correctly described.

EM was "correctly" described by Heaviside and Pointing in XIX century before
the Hertz experiment.

"The electrical waves produced by the oscillations at A traveled along
the
wires and were reflected at the far ends. Lodge knew that the longer
spark
at B3 was due to what he called the "recoil impulse" or "recoil kick" at
the
end of the wires where the waves were reflected.[4] At spark gap B3 both
the
incident wave and the reflected wave had their maximum values and were in
phase. This produced a voltage twice as large as the voltage at spark gap
A.
From: http://www.antiquewireless.org/otb/lodge1102.htm

Is it still true?

A reflected wave along a nonterminated transmission line will result in
doubled voltage at the open end.

Not always. The "nonterminated transmission line" may be the Lodge's wire or
Your antenna (short dipole).
In your antenna the electrons are not reflected and do not destroy your
transmitter. They JUMP OFF Periodically = pressure waves.
Do not write that I claim it. It is the explanation by Faraday, Lorenz,
Tesla and Dirac. The all is in the each textbooks. But in different chapters
(lessons).

If yes, than you must admit that the leakage must be stronger at "recoil
kick" when the voltage is doubled.
"So there is the unsymmetrical flow of electrons." Do you agree?

But here you are talking complete hogwash again. The effect you describe
above has nothing to do with leakage or unsymmetrical flow of electrons.

In each textbooks is the Richardson equation. The electron field emission is
voltage and temperature dependent.

Do you understand the Pointing explanation that nothing if flowing in the
conductor?

For me: " But then later it was found that the first (Pointing) discoveries
were not
entirely correctly described."
S*

"Rob" napisa³ w wiadomo¶ci
...
Szczepan Bialek wrote:


Here you are the simple:
http://www.ac-grenoble.fr/yre/agency...e/rap-nuc2.htm

Are the dipole elements symmetrical?

Yes

Mechanically.
Also electrically?

It seems that one of them is a counterpoise.

Why? I see symmetrical elements.

I see that the voltages are not symmetrical when the diode shines.
I bet that is the such orientation vs a oven that is the symmetry and no
shine.

What you can also see here is that this is another example of what
I wrote earlier: there is a circle in the schematic starting at the
diode where you can go around with DC current and end at the same
point.
This circle is where the DC current flows, not from the elements into
the air.

From Maxwell's time all circuits are closed. So sometimes the current must
flow in the air. Of course sometimes is enough if only part flows in the
air. It is the leakage.
At HF the leakage is rather large.
S*

Szczepan Bialek wrote:
From Maxwell's time all circuits are closed. So sometimes the current must
flow in the air. Of course sometimes is enough if only part flows in the
air. It is the leakage.
At HF the leakage is rather large.

But today we know that this is not true.

Szczepan Bialek wrote:
In your antenna the electrons are not reflected and do not destroy your
transmitter. They JUMP OFF Periodically = pressure waves.

Not in my antenna.
Not in your antenna either, because you have no antenna.

Do not write that I claim it. It is the explanation by Faraday, Lorenz,
Tesla and Dirac. The all is in the each textbooks. But in different chapters
(lessons).

But not in textbooks written today. Because today we know that no electrons
are jumping off antennas.

(maybe this evening they will, thunderstorms announced. but not because
of transmissions)

On Friday, July 27, 2012 7:39:28 AM UTC-5, Szczepan Bialek wrote:
From Maxwell's time all circuits are closed. So sometimes the
current must flow in the air.

Maxwell did not believe in current flowing "in the air". That's why he used the concept of "displacement current". He was ignorant of photon energy flow being the equivalent of current flow. Nowadays, we know that when it appears that coherent current is flowing in the air or indeed through any dielectric (including free space), it is actually photons that are doing the flowing, i.e. photons are accomplishing the energy transfer *as if* current were flowing through charge carriers. All EM field/wave phenomena involve quantized photons.

For those who still believe in displacement current, it is easy to prove that, under certain conditions, the displacement current would have to violate the speed of light limit in order to accomplish the energy transfer in the measured time.

Do you understand the Pointing explanation that nothing if flowing in
the conductor?

Quoting "Fields and Waves ...", by Ramo & Whinnery: "A perfect conductor is usually understood to be a material in which there is no electric field at any frequency. Maxwell's equations ensure that there is then also no time-varying magnetic field in the perfect conductor."

In the non-perfect real world, something is flowing in the skin effect depth of a conductor which is the cause of the power losses in the conductor.
--
73, Cecil, w5dxp.com

"Rob" napisa³ w wiadomo¶ci
...
Szczepan Bialek wrote:
In your antenna the electrons are not reflected and do not destroy your
transmitter. They JUMP OFF Periodically = pressure waves.

Not in my antenna.
Not in your antenna either, because you have no antenna.

Also not in Heaviside-Pointing because there no electrons.

Do not write that I claim it. It is the explanation by Faraday, Lorenz,
Tesla and Dirac. The all is in the each textbooks. But in different
chapters
(lessons).

But not in textbooks written today. Because today we know that no
electrons
are jumping off antennas.

The whole XX century was the century of intensive egzamination/explanation
of the field electron emmission.

If not in your antenna than you should be able to explain what the electrons
do in your antenna if they are not reflected (VSWR = 1).
Do you try?

(maybe this evening they will, thunderstorms announced. but not because
of transmissions)

Your antenna goes into receiving.
But I do not know if the electrons are injected into your antenna.
Are they?

S*

Szczepan Bialek wrote:
(maybe this evening they will, thunderstorms announced. but not because
of transmissions)

Your antenna goes into receiving.
But I do not know if the electrons are injected into your antenna.
Are they?

I hope not! I don't like electrons injected in my antenna.
It is very costly because of all the damaged equipment.

"W5DXP" napisal w wiadomosci
...
On Friday, July 27, 2012 7:39:28 AM UTC-5, Szczepan Bialek wrote:
From Maxwell's time all circuits are closed. So sometimes the
current must flow in the air.

Maxwell did not believe in current flowing "in the air". That's why he used
the concept of "displacement current". He was ignorant of photon energy
flow being the equivalent of current flow. Nowadays, we know that when it
appears that coherent current is flowing in the air or indeed through any
dielectric (including free space), it is actually photons that are doing
the flowing, i.e. photons are accomplishing the energy transfer *as if*
current were flowing through charge carriers. All EM field/wave phenomena
involve quantized photons.

For those who still believe in displacement current, it is easy to prove
that, under certain conditions, the displacement current would have to
violate the speed of light limit in order to accomplish the energy transfer
in the measured time.

Do you understand the Pointing explanation that nothing if flowing in
the conductor?

Quoting "Fields and Waves ...", by Ramo & Whinnery: "A perfect conductor is
usually understood to be a material in which there is no electric field at
any frequency. Maxwell's equations ensure that there is then also no
time-varying magnetic field in the perfect conductor."

In the non-perfect real world, something is flowing in the skin effect
depth of a conductor which is the cause of the power losses in the
conductor.

Why you do not like the electrons?
For what you need the next " the equivalent of current flow".
S*