After reading this thread it seems there is significant confusion between E and
H fields created locally from electric current and the E and H aspects of a
passing RF photon stream. It is easy to see how this can occur. That said the
tools for working with them are entirely different.

E and H fields from capacitors, inductors, and transformers are predominantly
local effects. Photons are not involved, they are created by electric current
and are explained by Maxwell equations. They can be created separately, that is
a capacitor is pretty much a pure E field and an inductor an H field, albeit
none of these are perfect. Photons are not involved in the operation of a

On the other hand EM waves and photons which are RF waves and visible light and
beyond are entirely different. Although they are referred to as
electro-magnetic, they are in fact massless particles with a well defined energy
that is a function of frequency. These particles, traveling at the speed of
light, create what is referred to as RF including RF fields. The E and M
components cannot be separated. Antennas are devices which convert RF currents
into photons for transmission and to convert photons into RF currents on
reception, all this at the frequencies we are mostly interested in.

These are two very different physical constructs. Now at RF frequencies -
Faraday shields are local electrical devices, they have nothing to do with
photons and antennas. (Although they may be effective at blocking their
passage.) Electric currents in equipment create both E and H fields. These
fields are not associated with photons and can be separated. Faraday shields are
a way to block the E portions of those local fields while allowing the passage
of the H.

Dan

Roy Lewallen wrote:
*Sigh*

Richard Harrison wrote:

Roy Lewallen, W7EL wrote: "A short while ago, I explained why your
Faraday cage doesn`t separate the E and H fields as you claim." n I
am misunderstood. I never used the tem Faraday cage. I understand the
Faraday cage to be a completely shielded enclosure which could be a
metal automobiole body, a steel rebar reinforced concrete structure
or a screened room. These all tend to completely block both the
E-field and the H-field components of an electromagnetic wave.


Sorry, I meant "Faraday screen", which is the term you used, and I used
in my posting explaining its operation.

If you block either the E or H field, you also block the other. You
can't independently block one or the other.

. . .


Why would one pay a lot of extra money for a transformer which
eliminated capacitive coupling if it didn`t work, especially in
Havana, Cuba?


Because in order to "work" it doesn't need to "eliminate capacitive
coupling". All it needs to do is locally reduce the E/H field ratio,
which is what it does.

Roy Lewallen, W7EL

Dan Sawyeror wrote:
"They can be created separately, that is a capacitor is pretty much a
pure E field and an inductor an H field."

Agreed.

Whenever a current grows or shrinks (with acceleration) it produces both
an E-field and an H-field. I didn`t clam a permanent divorce, only an
instantaneous separation.

Best regards, Richard Harrison, KB5WZI

"dansawyeror" wrote -
After reading this thread it seems there is significant confusion
between E and
H fields created locally from electric current and the E and H
aspects of a
passing RF photon stream.
========================================

When the geometric dimensions of the structures are short compared
with a wavelength, the amps and volts are associated with the NEAR
FIELD and Ohm's Law applies.

The radiation field is so weak in comparison it can be forgotten
about. KISS.

Why does everybody HAVE to unnecessarily complicate matters in order
to understand what really goes on?

What have photons to do with winning a contest? ;o)
----
Reg.

Reg Edwards wrote:
What have photons to do with winning a contest? ;o)

Just try winning a contest without them. :-)
--
73, Cecil, http://www.qsl.net/w5dxp

Reg, G4FGQ wrote:
"What do photons have to do with winning a contest?"

I don`t know. But, B.Whitfield Griffith, Jr. has some observations in
"Rado-Eledtronic Transmission Fundamentals" that may be a useful check
on your computations. I expect he checked, rechecked, then checked again
before publication. He put a transmitter on an elemental antenna but it
would work the same in reverse. From page 325:

"An Elemental Antenna

Since the length of an antenna is commonly expressed in electrical
degrees and since this allows the convenient use of trigonometric tables
in computing the ratios of currents in various parts of an antenna, let
us choose as our elemental antenna a piece of wire which is 1-degree in
length and in which the current is constant from one end to the other.We
shall first assume that this elemental antenna is located far out in
space, so that its field is not disturbed by reflections from the
surface of the earth or from any other object. This, of course, is a
most improbable set of conditions, but we can certainly imagine that we
have a situation such as this and compute from the field equations its
electromagnetic result.

These computations will show, first of all, that the maximum field
intensity will be produced in the directions which are at right angles
to the direction of current flow. This is a reasonable result, since the
magnetic field which is produced by the current surrounds the wire in
concentric rings and thus gives rise to a radiation field which moves
outward at right angles to the wire. As a matter of fact, the field
intensity, measured according to our standard procedure at a distance of
1 mile in any direction from the radiating element, will be found to be
proportional to the sine of the angle between the direction of the
current flow and the direction in which the measurement is taken. If we
represent the field intensity at 1 mile in any direction by the length
of a vector starting at the center of the element and extending in that
direction, the tips of the vectors will mark out the radiation pattern
of the antenna element. A cross section of the entire radiation pattern
of this element is shown in Fig. 39-2; the entire pattern would be
obtained by rotating the figure about the axis of the antenna element.

But this pattern tells us only the relative signal strength in various
directions; it is a normalized pattern, with the intensity in the
direction of maximum radiation being considered simply as unity. We need
much more information than this; we must know the relationship between
the current and the actual value of the field it produces. Further
computation from the field equations gives this relalationship; we find
that a current of 1 amp flowing in the antnna element will produce a
field intensity of 0.3253 mv/m at a distance of 1 mile iin the direction
of maximum radiation. We have said nothing about the frequency, and we
do not need to; as long as the wavelength is shorter than 1 mile, so
that there are no serious induction-ffield effects to upset our
calculations, this figure will be correct at any frequencyfor which the
length of the element is 1/360 wavelength.

The field intensity of the elemental antenna is directly proportional to
the current. Therefore, if the current in the element is 15 amp, the
field intensity will be 15 X 0.3253, or 4.8795, mv/m at 1 mile.
Similarly, the field intensity is directly proportional to the length of
the element; an element which is 2-degrees in length, carrying a current
of 1 amp, will thus produce a maximum field intensity of 0.6506 mv/m at
1 mile.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

I don`t know. But, B.Whitfield Griffith, Jr. has some observations in
"Rado-Eledtronic Transmission Fundamentals" that may be a useful check
on your computations. I expect he checked, rechecked, then checked again
before publication. He put a transmitter on an elemental antenna but it
would work the same in reverse. From page 325:

Yes, it should.

. . .
. . . We need
much more information than this; we must know the relationship between
the current and the actual value of the field it produces. Further
computation from the field equations gives this relalationship; we find
that a current of 1 amp flowing in the antnna element will produce a
field intensity of 0.3253 mv/m at a distance of 1 mile iin the direction
of maximum radiation. . .

The field intensity of the elemental antenna is directly proportional to
the current. Therefore, if the current in the element is 15 amp, the
field intensity will be 15 X 0.3253, or 4.8795, mv/m at 1 mile.
Similarly, the field intensity is directly proportional to the length of
the element; an element which is 2-degrees in length, carrying a current
of 1 amp, will thus produce a maximum field intensity of 0.6506 mv/m at
1 mile.

A short dipole antenna with 1 amp of current at the center has an
average of 0.5 amp of current along the whole length. So it should be
obvious from the above analysis that the field from a dipole is half the
field from the elemental antenna with uniform current which the author
is discussing. Or, instead of just taking the average, you can integrate
I * delta L to get the same result.

Roy Lewallen, W7EL

Richard Harrison wrote:
Reg, G4FGQ wrote:
"What do photons have to do with winning a contest?"

I don`t know. But, B.Whitfield Griffith, Jr. has some observations in
"Rado-Eledtronic Transmission Fundamentals" that may be a useful check
on your computations. I expect he checked, rechecked, then checked again
before publication. He put a transmitter on an elemental antenna but it
would work the same in reverse. From page 325:

"An Elemental Antenna

Since the length of an antenna is commonly expressed in electrical
degrees and since this allows the convenient use of trigonometric tables
in computing the ratios of currents in various parts of an antenna, let
us choose as our elemental antenna a piece of wire which is 1-degree in
length and in which the current is constant from one end to the other.We
shall first assume that this elemental antenna is located far out in
space, so that its field is not disturbed by reflections from the
surface of the earth or from any other object. This, of course, is a
most improbable set of conditions, but we can certainly imagine that we
have a situation such as this and compute from the field equations its
electromagnetic result.

These computations will show, first of all, that the maximum field
intensity will be produced in the directions which are at right angles
to the direction of current flow. This is a reasonable result, since the
magnetic field which is produced by the current surrounds the wire in
concentric rings and thus gives rise to a radiation field which moves
outward at right angles to the wire. As a matter of fact, the field
intensity, measured according to our standard procedure at a distance of
1 mile in any direction from the radiating element, will be found to be
proportional to the sine of the angle between the direction of the
current flow and the direction in which the measurement is taken. If we
represent the field intensity at 1 mile in any direction by the length
of a vector starting at the center of the element and extending in that
direction, the tips of the vectors will mark out the radiation pattern
of the antenna element. A cross section of the entire radiation pattern
of this element is shown in Fig. 39-2; the entire pattern would be
obtained by rotating the figure about the axis of the antenna element.

But this pattern tells us only the relative signal strength in various
directions; it is a normalized pattern, with the intensity in the
direction of maximum radiation being considered simply as unity. We need
much more information than this; we must know the relationship between
the current and the actual value of the field it produces. Further
computation from the field equations gives this relalationship; we find
that a current of 1 amp flowing in the antnna element will produce a
field intensity of 0.3253 mv/m at a distance of 1 mile iin the direction
of maximum radiation. We have said nothing about the frequency, and we
do not need to; as long as the wavelength is shorter than 1 mile, so
that there are no serious induction-ffield effects to upset our
calculations, this figure will be correct at any frequencyfor which the
length of the element is 1/360 wavelength.

The field intensity of the elemental antenna is directly proportional to
the current. Therefore, if the current in the element is 15 amp, the
field intensity will be 15 X 0.3253, or 4.8795, mv/m at 1 mile.
Similarly, the field intensity is directly proportional to the length of
the element; an element which is 2-degrees in length, carrying a current
of 1 amp, will thus produce a maximum field intensity of 0.6506 mv/m at
1 mile.

Best regards, Richard Harrison, KB5WZI



How did you get .3253 mv/m at 1 miles?

Dave

How did you quote damned near 75 lines of text to ask a 9 word question?

Jim


"EE123" wrote in message
oups.com...

Best regards, Richard Harrison, KB5WZI



How did you get .3253 mv/m at 1 miles?

Dave

EEE 123 wrote:
"How do you get .3253 mv/m at 1 mile?"

The formula to calculate the field strength produced by an dlementaarry
doublet is difficult for me to reproduce with my keyboard. It is found
on page 770 of Terman`s 1943 "Radio Sngineers` Handbook. In my prior
posting, Griffith hhas done the work for us.

Best regards, Richard Harrison, KB5WZI