art wrote:
What is the ratio of magnetic electrons emitted from a radiator
compared with
with the number of electrons emmitted due to current flow?
What is the combination ratio required of both types of electrons to
form a radiation field?
Regards
Art


You mean the radiation from an antenna driven by a radio transmitter?
It doesn't emit electrons, but it does emit photons at that radio
frequency.

Go find a book on electromagnetism and fields, but be prepared for
calculus level math in that book. I took such a class 30 years ago, and
got a "C", and remember even less now. A high school physics book might
be enough depending on your needs.

robert casey wrote:
art wrote:
What is the ratio of magnetic electrons emitted from a radiator
compared with
with the number of electrons emmitted due to current flow?
What is the combination ratio required of both types of electrons to
form a radiation field?
Regards
Art


You mean the radiation from an antenna driven by a radio transmitter?
It doesn't emit electrons, but it does emit photons at that radio
frequency.

Go find a book on electromagnetism and fields, but be prepared for
calculus level math in that book. I took such a class 30 years ago, and
got a "C", and remember even less now. A high school physics book might
be enough depending on your needs.

The current (May/June) issue of QEX contains the article:

Electromagnetic Radiation: A Brief Tutorial

It contains equations but no calculus that I noticed.

"magnetic electrons emitted from a radiator" isn't mentioned, but
that isn't a surprise to most people.

--
Jim Pennino

Remove .spam.sux to reply.

On 10 May, 11:45, wrote:
robert casey wrote:
art wrote:
What is the ratio of magnetic electrons emitted from a radiator
compared with
with the number of electrons emmitted due to current flow?
What is the combination ratio required of both types of electrons to
form a radiation field?
Regards
Art

You mean the radiation from an antenna driven by a radio transmitter?
It doesn't emit electrons, but it does emit photons at that radio
frequency.
Go find a book on electromagnetism and fields, but be prepared for
calculus level math in that book. I took such a class 30 years ago, and
got a "C", and remember even less now. A high school physics book might
be enough depending on your needs.

The current (May/June) issue of QEX contains the article:

Electromagnetic Radiation: A Brief Tutorial

It contains equations but no calculus that I noticed.

"magnetic electrons emitted from a radiator" isn't mentioned, but
that isn't a surprise to most people.

--
Jim Pennino

Remove .spam.sux to reply.- Hide quoted text -

- Show quoted text -

Jim, you have no legitamacy in the subject of radiation, your past
posts prove that. On top of that you do not talk for most people as
you intimate.
You haven't yet capitulated on the static subject or negated the truth
of the mathematics and examples supplied. Just stating consistently
that you can't this or you can't that just doesn't elevate your
stature.

On 10 May, 12:06, art wrote:
On 10 May, 11:45, wrote:





robert casey wrote:
art wrote:
What is the ratio of magnetic electrons emitted from a radiator
compared with
with the number of electrons emmitted due to current flow?
What is the combination ratio required of both types of electrons to
form a radiation field?
Regards
Art

You mean the radiation from an antenna driven by a radio transmitter?
It doesn't emit electrons, but it does emit photons at that radio
frequency.
Go find a book on electromagnetism and fields, but be prepared for
calculus level math in that book. I took such a class 30 years ago, and
got a "C", and remember even less now. A high school physics book might
be enough depending on your needs.

The current (May/June) issue of QEX contains the article:

Electromagnetic Radiation: A Brief Tutorial

It contains equations but no calculus that I noticed.

"magnetic electrons emitted from a radiator" isn't mentioned, but
that isn't a surprise to most people.

--
Jim Pennino

Remove .spam.sux to reply.- Hide quoted text -

- Show quoted text -

Jim, you have no legitamacy in the subject of radiation, your past
posts prove that. On top of that you do not talk for most people as
you intimate.
You haven't yet capitulated on the static subject or negated the truth
of the mathematics and examples supplied. Just stating consistently
that you can't this or you can't that just doesn't elevate your
stature.- Hide quoted text -

- Show quoted text -

Some have stated that 377 ohms is a ratio. I don't understand that
assertion
since I understood that a ratio is not confined to a specific unit and
in fact
does not have units. I am aware that the impedance of a particular
atmosphere is 377 ohms but that is certainly not a ratio. Some say one
must have a knoweledge of calculus to understand radiation. Another
declares he read a book on radiation that did not use calculus which
is just as well if one becomes careless with terms such as a ratio.
But no matter, this newsgroup is a living example of the use of free
speech where amateurs can take on the guise of professionalism despite
their lowly education level.

art wrote:
I am aware that the impedance of a particular
atmosphere is 377 ohms but that is certainly not a ratio.

From: http://whatis.techtarget.com/definit...845268,00.html

Mathematically, the Zo of free space is equal to the square root of the
ratio of the permeability of free space (µo) in henrys per meter (H/m)
to the permittivity of free space (o) in farads per meter (F/m):

Zo = (µo/o)1/2

= [(1.257 x 10-6 H/m)/(8.85 x 10-12 F/m)]1/2

= 377 ohms (approximately)

The exact value of the Zo of free space is 120 pi ohms, where pi is the
ratio of the circumference of a circle to its diameter.
--
73, Cecil, w5dxp.com

On 10 May, 12:58, Cecil Moore wrote:
art wrote:
I am aware that the impedance of a particular
atmosphere is 377 ohms but that is certainly not a ratio.

From:http://whatis.techtarget.com/definit...845268,00.html

Mathematically, the Zo of free space is equal to the square root of the
ratio of the permeability of free space (µo) in henrys per meter (H/m)
to the permittivity of free space (o) in farads per meter (F/m):

Zo = (µo/o)1/2

= [(1.257 x 10-6 H/m)/(8.85 x 10-12 F/m)]1/2

= 377 ohms (approximately)

The exact value of the Zo of free space is 120 pi ohms, where pi is the
ratio of the circumference of a circle to its diameter.
--
73, Cecil, w5dxp.com

O.k. Cecil I will for the moment embrace that a ratio must have a unit
of measurement which puts me in line with all the amateurs of this
newsgroup (see I am flexible). I will also change from particles now,
to electrons and now to protons as requested ( see I am flexible
again)
For my interest, what is the unit that must be used for the front to
back ratio of a directive antenna?
Regards
Art

art wrote:
For my interest, what is the unit that must be used for the front to
back ratio of a directive antenna?

Power ratios are commonly stated in dB.
--
73, Cecil http://www.w5dxp.com

Art wrote:
"For my interest, what is the unit that must be used for front to back
ratio of a directive antenna?"

I must be an idiot for venturing an answer, but ratios can be just
numbers, but numbers have origins. If radiated power in one direction is
twice that in another (reference), we can say it has a directive gain of
two or we can say it has a 3 dB gain. Front to back ratios have the same
origins and units. For legitimacy, Terman says on page 871 of his 1955
opus:
"The directive gain can be expressed either as a power ratio, or in
terms of the equivalent number of decibels.

Best regards, Richard Harrison, KB5WZI

"Cecil Moore" wrote in message
t...
art wrote:
I am aware that the impedance of a particular
atmosphere is 377 ohms but that is certainly not a ratio.

From: http://whatis.techtarget.com/definit...845268,00.html

Mathematically, the Zo of free space is equal to the square root of the
ratio of the permeability of free space (µo) in henrys per meter (H/m) to
the permittivity of free space (o) in farads per meter (F/m):

Zo = (µo/o)1/2

= [(1.257 x 10-6 H/m)/(8.85 x 10-12 F/m)]1/2

= 377 ohms (approximately)

The exact value of the Zo of free space is 120 pi ohms, where pi is the
ratio of the circumference of a circle to its diameter.
--
73, Cecil, w5dxp.com

Also the ratio of E/H. [(V/m)/(A/m)] = [ohms].

Frank

Frank's wrote:
Also the ratio of E/H. [(V/m)/(A/m)] = [ohms].

Is that a cause or an effect? :-)
--
73, Cecil http://www.w5dxp.com