Part 2 Is it possible to ask questions here?
 

It has been stated on this antenna newsgroup that with
short antennas the current goes up the radiator and then turns back
and goes down. If this is so then it must be radiating all the time,
yes?
If a radiator is radiating all the time then the efficiency is the
same
as a full leght antenna. Yes?
This does not conform with reality Right?
So is it possible that the circuit (current)
returns along the path down the center of the radiator which is
bordered by decaying electrons which thus would prevent
radiation?
IF it was possible then radiation figures accepted by hams
would coincide with respect to short antennas. Yes?.
Then why do all the "experts" reject the notion of the circuit
continueing down the center of the radiator?
What exacly preventing such a circuit becoming a reality?
Why does current go up the radiator in the first place
knowing it has nowhere to go?
Just a silly question for the self perceived experts
Best regards and waiting in unabaited attention to responses
by the experts, as I can't find it in the books which tell all
that is known.
Your friend and eager listener
Art KB9MZ.....XG

Part 2 Is it possible to ask questions here?
 

"art" wrote in message
ups.com...
It has been stated on this antenna newsgroup that with
short antennas the current goes up the radiator and then turns back
and goes down. If this is so then it must be radiating all the time,
yes?
If a radiator is radiating all the time then the efficiency is the
same
as a full leght antenna. Yes?
This does not conform with reality Right?
So is it possible that the circuit (current)
returns along the path down the center of the radiator which is
bordered by decaying electrons which thus would prevent
radiation?
IF it was possible then radiation figures accepted by hams
would coincide with respect to short antennas. Yes?.
Then why do all the "experts" reject the notion of the circuit
continueing down the center of the radiator?
What exacly preventing such a circuit becoming a reality?
Why does current go up the radiator in the first place
knowing it has nowhere to go?
Just a silly question for the self perceived experts
Best regards and waiting in unabaited attention to responses
by the experts, as I can't find it in the books which tell all
that is known.
Your friend and eager listener
Art KB9MZ.....XG

Art

You should know better (and I should know better for responding). :-}

The signal going down the centre would cancel the
signal going up the outside and nothing would be radiated except heat due to
the electrical impedence and resistance of the antenna conductor.

What keeps the two paths separated? A simple experiment with a length of
solid copper rod and a similar length of copper water pipe will demonstrate
that what you suggest doesn't in fact happen. Both will display similar
radiation characteristics and no sign of reverse current flows down the
centre of the rod or the inside bore of the tube. The only difference in
characteristics will be caused by a difference in the outer dimensions of
the copper rod and tube. At high enough frequencies, the tube will act as a
waveguide, but that's a completely different matter altogether.

Why does a light bulb glow? How do the photons know what direction to travel
in?

If electrons are raised to a higher energy level (usually referred to as
moved to occupy a higher orbital shell) by the input of energy, after a
short period, they will return to their original energy state, emitting a
photon to carry away the excess energy. The energy of the photon being
directly in proportion to the energy input in the first place.

This has been verified repeatedly in published laboratory experiments. The
photons are
emitted at around 300,000 Km/sec at right angles from the surface of the
conductor. They don't need to 'know' which way to go or to be sucked out by
some mystic force. Neither do we need degenerate or decaying electrons to
direct the flow. An electron is an electron and nothing else. All electrons
are inherently the same. When they form part of an atom, they can absorb or
emit photons to balance the energy in the atom. The photon is not part of
the electron, it is just the manifestation of temporarily stored, excess
energy, being emitted to restore the atom back to its lowest energy state.

Mike G0ULI

Part 2 Is it possible to ask questions here?
 

On 10 Nov, 19:17, "Mike Kaliski" wrote:
"art" wrote in message

ups.com...



It has been stated on this antenna newsgroup that with
short antennas the current goes up the radiator and then turns back
and goes down. If this is so then it must be radiating all the time,
yes?
If a radiator is radiating all the time then the efficiency is the
same
as a full leght antenna. Yes?
This does not conform with reality Right?
So is it possible that the circuit (current)
returns along the path down the center of the radiator which is
bordered by decaying electrons which thus would prevent
radiation?
IF it was possible then radiation figures accepted by hams
would coincide with respect to short antennas. Yes?.
Then why do all the "experts" reject the notion of the circuit
continueing down the center of the radiator?
What exacly preventing such a circuit becoming a reality?
Why does current go up the radiator in the first place
knowing it has nowhere to go?
Just a silly question for the self perceived experts
Best regards and waiting in unabaited attention to responses
by the experts, as I can't find it in the books which tell all
that is known.
Your friend and eager listener
Art KB9MZ.....XG

Art

You should know better (and I should know better for responding). :-}

The signal going down the centre would cancel the
signal going up the outside and nothing would be radiated except heat due to
the electrical impedence and resistance of the antenna conductor.

What keeps the two paths separated? A simple experiment with a length of
solid copper rod and a similar length of copper water pipe will demonstrate
that what you suggest doesn't in fact happen. Both will display similar
radiation characteristics and no sign of reverse current flows down the
centre of the rod or the inside bore of the tube. The only difference in
characteristics will be caused by a difference in the outer dimensions of
the copper rod and tube. At high enough frequencies, the tube will act as a
waveguide, but that's a completely different matter altogether.

Why does a light bulb glow? How do the photons know what direction to travel
in?

If electrons are raised to a higher energy level (usually referred to as
moved to occupy a higher orbital shell) by the input of energy, after a
short period, they will return to their original energy state, emitting a
photon to carry away the excess energy. The energy of the photon being
directly in proportion to the energy input in the first place.

This has been verified repeatedly in published laboratory experiments. The
photons are
emitted at around 300,000 Km/sec at right angles from the surface of the
conductor. They don't need to 'know' which way to go or to be sucked out by
some mystic force. Neither do we need degenerate or decaying electrons to
direct the flow. An electron is an electron and nothing else. All electrons
are inherently the same. When they form part of an atom, they can absorb or
emit photons to balance the energy in the atom. The photon is not part of
the electron, it is just the manifestation of temporarily stored, excess
energy, being emitted to restore the atom back to its lowest energy state.

Mike G0ULI- Hide quoted text -

- Show quoted text -

Mike I want to wait until some of the others present there thoughts
but one thing I have to comment on. The term is an electron is an
electron
and nothing else! I won't argue on the nomenclature of an electron
but must point out that an electrons has different properties!
You can have an electron that is bound to an atom via it's orbit
and that word bound cannot be underestimated. You can also have free
electrons
that attach themselves within to what we understand as matter.You can
also have
static particles where there are several states of decay right down
to a particle with chemical atributes but no electric atributes.
Now everybody brings photons into this picture that I will not comment
on
but as far as electrons one must state the electrons status before
debating
property changes that some suggest are foisted upon them.Lets wait to
see
what the others present though you didn't refer to the fact that if
the
circuit is always on the surface with respect to time then the
radiation
should be the same as a full size radiator. But that can wait.
Somebody may yet refer to a description from a book !
Best regards
Art

Part 2 Is it possible to ask questions here?
 

On 10 Nov, 20:25, "Stefan Wolfe" wrote:
"art" wrote in message

ups.com...

It has been stated on this antenna newsgroup that with
short antennas the current goes up the radiator and then turns back
and goes down. If this is so then it must be radiating all the time,
yes?

Yes, as long as the sinusoidal EMF is applied to the antenna. This is a
forcing function. Let is assume a sinusoidal carrier wave is generated at
the source.

If a radiator is radiating all the time then the efficiency is the
same
as a full leght antenna. Yes?

No. The efficiency will be based on the ratio of radiation resistance to
total resistance. Shorter antennas tend to require coils that increase ohmic
resistive losses. Ground image plane losses can be huge on short mobile
80/160m mobile antennas. At lower frequencies, image plane losses can
greatly exceed radiation resistance thus lower the efficiency greatly. Yet,
the antenna is radiating 100% of the time, just not as efficiently.

This does not conform with reality Right?

Yes, it does, per above. I guess with my elementary explanation, it is not
helpful to go on the other questions since known science is contradicting
your assumption (that some people, not necessarily you-yourself believe)
antenna radiation efficiency is somehow related to the amount of time that
current is flowing in the short antenna. I am not sure if you are intending
to advocate this model or oppose it as I am not clear as to whose side you
are on in this rather curiously controversial discussion of something that I
thought was simpler. However, I am very open to expanding my horizons to new
ideas since even today, antenna theory is to me a 'black art' (meaning that
it is not fully mathematically understood by any one person that I am aware
of also there are plenty of antenna companies making money based on
empirical designs).

I am trying to conceptualize the design of your unique antenna model that
you say is based on a gaussian extension to maxwell's equations. I have read
your archives and I would like to try to understand your positions more
specifically. You can throw the math at me. Can you point me to exactly
which Gaussian extension formula I must apply to maxwell? Yes, I am familiar
with the Gaussian area integrals of E*d(A) and how to solve them...I do have
an EE degree. This area integral is actually a part of maxwell's equations
and I do not know what extension you are referring to. I understand a Dr.
Davis proved your work; can you point me to the calculations he did? That is
where I think I might learn a lot. BTW I have never used antenna modelling
programs as I do not find the analyisis of repeated antenna segments
particularly interesting. However, I may have to try it out to understand
your stuff. Know of a good program I could download? It's called EZNEC or
something like that? Thanks Art.

I am attacking present day theory on the basis of what my reseach has
revealed over the past decade. I placed it in front of my peers
ie this newsgroup for examination. It was not examined in the normal
scientific way that I am accustomed to as an engineer.

If it was, it would quickly determine if the basis of my assault on
present
dogma were correct or in error. If you have existing progams you will
know
that sometimes it runs away for UNKNOWN reasons so YOU have to
determine
what is right and what is wrong. My research points to the inclusion
of
the sino soidal properties exist at every segment point and that is
known
by all parties concerned. It was shown to be correct at some segment
points but not all! But this asumption was kept in the absence of
known
alternatives. If you refuse to review mty work and its associated
mathematics then you are assuming that all is known even tho your
assumption
prove to be in error.I accepted that there were errors made and still
are
and have now found where the problem is. At the same time the solution
I
found resolves questions that scientists have puzzled about for over
100 years.
So if my peers will not review my work then I have to place parts that
are subject to contention which is going to be a verry long task and
perhaps irratable to many but I have no other options if all consider
everything is known. Now let me make just one point. I reffered to
the helix angle being determed empirically. Well I can show how
Maxwells
determines that angle using his laws. You also can determine using any
program that the angle is the summation of all vectors involed in
radiation.
The vector is not aligned with the radiator axis so with any program
move away
from planar forms to find that angle which produces maximum gain of
a desired polarity. How many of you used that method to avoid
relying
on a empirical method? Was it already known as with all the other
facts.
How many of you knew that an array that is non planar could exceed
the
attributes of a yagi? Or is that fallacious becuase all is known and
the books
cannot be wrong in any way when you have to acknoweledge that the
present
assumptions are known to create errors. I know, ignore the facts that
you know about and put your head in the sand and I you don't care
whether
the review by Dr Davis showed I was correct. To hell with the
mathematics
we know all is known because we have books that say so.
I am getting weary of all this but I cannot let it drop since I hold
to a
regimen that all engineers follow.
Go ahead Richard you can attack me now and not the subject as you
always do.
Art Unwin KB9MZ....xg

Part 2 Is it possible to ask questions here?
 

"Stefan Wolfe" wrote in message
...

Art, I'm not sure if you read my entire post.

and he wouldn't understand it even if he did.


Where can I find a copy of your work? I have gone through the archives but
no luck.

you won't. he hasn't published anything other than handwaving distortions
and misconceptions based on some weird distortion of gauss's law and his
concept of 'equilibrium', which he can't define either. you will be happier
when you add him and anyone who responds to him to your kill file.

Part 2 Is it possible to ask questions here?
 

"art" wrote
It has been stated on this antenna newsgroup that with
short antennas the current goes up the radiator and then turns back
and goes down. If this is so then it must be radiating all the time,
yes? If a radiator is radiating all the time then the efficiency is the
same as a full leght antenna. Yes? This does not conform with
reality Right?

Wrong, as regards your "reality." Using the classic definition of
efficiency, an antenna of ANY length (including a point source) will radiate
nearly 100% of the power it accepts from the r-f source driving it.

The radiation patterns of those antennas will vary. Some will radiate more
relative field in some directions and less in some directions than others
will. But, disregarding dielectric and conductor I^2R losses, ALL antennas
radiate ALL of the power they accept from their driving source (ie, their
efficiencies are equal).

So is it possible that the circuit (current) returns along
the path down the center of the radiator...

No, it's not possible. No matter the direction of flow along a solid
conductor, alternating current tends to travel on/near its outer surface.
This is due to the greater number of enclosed lines of magnetic flux
generated by current flowing at/near its center, which increases the
inductive reactance of the conductor in those areas. The result is a
redistribution of the current to the parts of the conductor cross-section
having the least reactance, ie, on and near its outer surface. Read
Terman's RADIO ENGINEERS' HANDBOOK, 1943 edition, pp 30-31 for more on this
(or many other sources).

IF it was possible then radiation figures accepted by hams
would coincide with respect to short antennas. Yes?.
Then why do all the "experts" reject the notion of the circuit
continueing down the center of the radiator?

Because it doesn't do that.

RF

Part 2 Is it possible to ask questions here?
 

On Nov 11, 2:35 pm, "Stefan Wolfe" wrote:


Hi Stefan
If you search "John E Davis" you will find the link. He
made the post's regarding the math on March 10-11-14 and 15 the
heading is re gaussian statics law

Derek

Part 2 Is it possible to ask questions here?
 

A lot of this 'discussion' depends on how you define 'efficiency'.
A 'point source' can be very efficient, in it's self. It can also be
very inefficient when compared to another type 'source'.
It's true that any antenna can radiate all of the signal getting to
it. The 'catch' is just how much 'signal' is getting to it and how/
where is it being radiated. If it's going to where you want it, and
if a usable amount of 'signal' gets there, then it's efficient for
that particular situation. If not... then it isn't very efficient, is
it?
- 'Doc

(With the 'proper' mind-set, you can apply the above to anything, not
just antennas.)

Part 2 Is it possible to ask questions here?
 

If it's going to where you want it, and if a usable amount of
'signal' gets there, then it's efficient for that particular situation.
If not... then it isn't very efficient, is it?
- 'Doc
____________

In a pure sense, the radiator itself is.

It just may not be as useful in that application as an antenna of
another configuration that provides the system result being sought.

RF

Part 2 Is it possible to ask questions here?
 

On 11 Nov, 06:15, wrote:
A lot of this 'discussion' depends on how you define 'efficiency'.
A 'point source' can be very efficient, in it's self. It can also be
very inefficient when compared to another type 'source'.
It's true that any antenna can radiate all of the signal getting to
it. The 'catch' is just how much 'signal' is getting to it and how/
where is it being radiated. If it's going to where you want it, and
if a usable amount of 'signal' gets there, then it's efficient for
that particular situation. If not... then it isn't very efficient, is
it?
- 'Doc

(With the 'proper' mind-set, you can apply the above to anything, not
just antennas.)

I like that last comment regarding mind set. Just look how people are
not viewing the subject without predisposition. No onw is willing to
deal only what has been proffered to the exclusion of every thing
else.
Everybody will use a text gained from somewhere to side line true
examination.
Stephan,. you
wanted out I took you at your word. I don't know how many times
This discussion will end the same as always, I don't understand what
you are saying
To heck with mathematics. Iknow what I know is correct.sSme will
change the content of what I state . And as always shown in history
ridicule is turned to when all other efforts fail.
But nobody will question the fact that all computor programs support
my
addition to Gaussian law to those of Maxweell. True, other scientists
concluded
that radiation is created via a time varience. No body has found
correllation to prove it
With a legitamate addition to a known law by Gauss I have given a
method where as
the hows of radiation is revealed that is consistent with Maxwells
laws.
The mathematics have been given that support it but they have been
swept aside
Existing programs support it but it is left to the user to determine
whether
"garbage in is garbage out" or to only accept what the program
supplies with
the appearance with known reality and junk the rest.
And make no mistake about it, when programmers placed an assumed
condition
to a known law they did it with deliberation.
When it supplied error they covered it up by changing the program to
concurr
with traditional thought. This is no different to when NASA ignored
what engineers told them about O rings and science was pushed aside.
Mathematical laws were broken and all that deal with these programs
are part and parcel of this mathematical fraud.
Best regards to all
Art Unwin....KB9MZ...xg

Part 2 Is it possible to ask questions here?
 

On 11 Nov, 05:19, "Richard Fry" wrote:
"art" wrote

It has been stated on this antenna newsgroup that with
short antennas the current goes up the radiator and then turns back
and goes down. If this is so then it must be radiating all the time,
yes? If a radiator is radiating all the time then the efficiency is the
same as a full leght antenna. Yes? This does not conform with
reality Right?

Wrong, as regards your "reality." Using the classic definition of
efficiency, an antenna of ANY length (including a point source) will radiate
nearly 100% of the power it accepts from the r-f source driving it.

The radiation patterns of those antennas will vary. Some will radiate more
relative field in some directions and less in some directions than others
will. But, disregarding dielectric and conductor I^2R losses, ALL antennas
radiate ALL of the power they accept from their driving source (ie, their
efficiencies are equal).

So is it possible that the circuit (current) returns along
the path down the center of the radiator...

No, it's not possible. No matter the direction of flow along a solid
conductor, alternating current tends to travel on/near its outer surface.
This is due to the greater number of enclosed lines of magnetic flux
generated by current flowing at/near its center, which increases the
inductive reactance of the conductor in those areas. The result is a
redistribution of the current to the parts of the conductor cross-section
having the least reactance, ie, on and near its outer surface. Read
Terman's RADIO ENGINEERS' HANDBOOK, 1943 edition, pp 30-31 for more on this
(or many other sources).

IF it was possible then radiation figures accepted by hams
would coincide with respect to short antennas. Yes?.
Then why do all the "experts" reject the notion of the circuit
continueing down the center of the radiator?

Because it doesn't do that.

RF

To say that an AC current will not flow in copper unless it has clear
axis
access to the copper surface is balderdash. Cover the copper with an
insulator
with any thickness that you desire for safety incase you are in error
and then drill
into the center of the copper. Without a safe guard you will die!
What provides resistance on the outside als skin depth can by the
reverse contain current flow to the inside. You like many use the word
"tends"
with respect to external current flow. The word "tends" does not
make the current passage an undeniable fact. Yet you have hung
your hat on that premise. I repeat...balderdash
Art

Part 2 Is it possible to ask questions here?
 

On 11 Nov, 07:05, "Richard Fry" wrote:
If it's going to where you want it, and if a usable amount of
'signal' gets there, then it's efficient for that particular situation.
If not... then it isn't very efficient, is it?
- 'Doc

____________

In a pure sense, the radiator itself is.

It just may not be as useful in that application as an antenna of
another configuration that provides the system result being sought.

RF

There you go again, "may" does not affirm fact.
Art

Part 2 Is it possible to ask questions here?
 

I repeat...balderdash
Art


Artsy,
Finally you summarized your "writings" and your repeating of it.
Balderdash trophy of the year goes to Art da ex G man.

Have you considered fishing or other activities?

bada BUm

Part 2 Is it possible to ask questions here?
 

"art"
What provides resistance on the outside als skin depth can
by the reverse contain current flow to the inside.

This is another of your beliefs that not supported either by theory
or practice.

The word "tends" does not make the current passage
an undeniable fact.

I wrote "tends" because there is no discrete boundary near the outer surface
of a conductor where ALL of the alternating current flowing near its surface
is confined. But almost all of that current flows within several "skin
depths."

The 1.8 MHz skin depth in a round, copper conductor is about 0.06 mm, which
means that a tubular conductor with a wall thickness at least 3 times that
can be used in place of a solid conductor of the same outer diameter, with
no practical change in performance at that frequency.

RF

Part 2 Is it possible to ask questions here?
 

"art"
There you go again, "may" does not affirm fact.
_________

OK, then.

A 1/2-wave dipole absolutely HAS more directivity than an isotropic radiator
(and so does every other practical antenna).

But when any/all of them accept the same amount of power from an r-f source,
then they ALL will radiate the same total amount of power.

So they are all equally efficient, by the classic definition of total power
in vs. total power out.

Antenna directivity/gain is not a measure of efficiency.

RF

Part 2 Is it possible to ask questions here?
 

Richard Fry wrote:
"Read Terman`s RADIO ENGINEERS` HANDBOOK, 1943 edition, pp 30-31 for
more on this (or many other sources)."

Amen. Terman doesn`t say different things in different places. He is
consistent. In Terman`s 1955 edition of "Electronic and Radio
Engineering" he writes on page 21:
"It is to be noted that some of this (magnetic) flux exists within the
conductor and therefore links with, i.e., encircles, current near the
center of the conductor while not linking current flowing near the
surface. The result is that inductance of the central part of the
conductor is greater than the part of the conductor nesr the surface;
this is because of the greater number of flux linkages existing in the
central region.

That`s why we have skin effect and why hollow pipes carry HF current as
effectively as solid rods with the same external surface area in most
cases. The pipe`s interior doesn`t carry current unless its diameter is
at least 1/2 wavelength (its cutoff as a waveguide).

Best regards, Richard Harrison, KB5WZI

Part 2 Is it possible to ask questions here?
 

Art wrote:
"But nobody will question the fact that all computer programs support my
addition to Gaussian law to those of maxwell."

That`s an I dare you.

Roy may tell us if EZNEC needs Art`s embellishment for accuracy.

Art did not answer my question of Nov 8, 10:27am in the "An instructive
trick" thread. It was: "why would we use the time constant without the
angular frequency?"

On page 904 0f the 3rd edition of Kraus` "Antennas" is found:
"The availability of computers in the 1960s provided antenna designers
with an alternative. They could develop software to simulate the
performance of antennas. In general, these techniques either numerically
solve Maxwell`s equations by descretizing the problem using integral
techniques, such as Moment Methods (MoM) as discussed in Sec. 14-11, or
differential techniques, such as finite elements or finite
difference-time domain."

Best regards, Richard Harrison, KB5WZI

Part 2 Is it possible to ask questions here?
 

Richard Harrison wrote:
Richard Fry wrote:
"Read Terman`s RADIO ENGINEERS` HANDBOOK, 1943 edition, pp 30-31 for
more on this (or many other sources)."

Amen. Terman doesn`t say different things in different places. He is
consistent. In Terman`s 1955 edition of "Electronic and Radio
Engineering" he writes on page 21: "It is to be noted that some of this
(magnetic) flux exists within the conductor and therefore links with,
i.e., encircles, current near the center of the conductor while not
linking current flowing near the surface. The result is that inductance
of the central part of the conductor is greater than the part of the
conductor nesr the surface; this is because of the greater number of
flux linkages existing in the central region.


What Terman says is true, for the particular example that he chooses.
But it may leave an incorrect impression that the conductor needs to be
completely encircled by flux linkages.

In fact the skin effect will develop on the surface of any conducting
material of any shape, wherever there is RF current flowing.

Here is a link to a detailed mathematical proof, from 'Transmission
Lines for Communications' by C W Davidson (Macmillan Press, 1978, ISBN 0
333 32738 1): http://www.ifwtech.co.uk/g3sek/misc/skin.htm

Davidson's analysis starts with the most general assumption possible:
that RF current is flowing over any small patch of a conductor's
surface. No assumption is required about the reason for the RF current
to be present, only that it is. Likewise no assumption is required about
the cross-section of the conductor, only that it has an exposed surface
(and by implication, that there are no constraints due to a small radius
or insufficient depth). Davidson then derives all the usual equations
for the skin effect. The only drawback of this derivation is that it is
highly mathematical, and difficult to put into words; but it's still
physically correct.

To repeat, I am not saying that Terman's explanation is incorrect; only
that the skin effect is a far more general phenomenon than his
particular examples imply.

This is important because, by taking the existence of the skin effect as
a guaranteed starting-point, the explanations for the behaviour of
coaxial cables, 'bazooka' baluns, 'shielded' loops and many other
devices will all fall neatly into place.


--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Part 2 Is it possible to ask questions here?
 

"Richard Harrison" wrote in message
...
Art wrote:
"But nobody will question the fact that all computer programs support my
addition to Gaussian law to those of maxwell."

That`s an I dare you.

Gauss's law IS one of Maxwell's equations. In fact both Ramo Whinnery and
Van Duzer's "Fields and Waves in Communications Electronics" (pg 237 in the
1st edition) and Jackson's "Classical Electrodynamics" (compare pg 2 and 33
in the 2nd edition). So every time art makes that assertion he is just
showing his ignorance of the facts.

Part 2 Is it possible to ask questions here?
 

Dave wrote:
"Gauss`s law IS one of Maxwell`s equations."

Yes. I`ve suggested Kraus to Art but he seems not to have pored through
Kraus yet. On page 395 of the 3rd edition of Antennas is a table of
Maxwell`s equations in integral form. One column is from Ampere, another
from Faraday, and the last two are from Gauss.

Best regards, Richard Harrison, KB5WZI

Part 2 Is it possible to ask questions here?
 

On 11 Nov, 14:18, Ian White GM3SEK wrote:
Richard Harrison wrote:
Richard Fry wrote:
"Read Terman`s RADIO ENGINEERS` HANDBOOK, 1943 edition, pp 30-31 for
more on this (or many other sources)."

Amen. Terman doesn`t say different things in different places. He is
consistent. In Terman`s 1955 edition of "Electronic and Radio
Engineering" he writes on page 21: "It is to be noted that some of this
(magnetic) flux exists within the conductor and therefore links with,
i.e., encircles, current near the center of the conductor while not
linking current flowing near the surface. The result is that inductance
of the central part of the conductor is greater than the part of the
conductor nesr the surface; this is because of the greater number of
flux linkages existing in the central region.

What Terman says is true, for the particular example that he chooses.
But it may leave an incorrect impression that the conductor needs to be
completely encircled by flux linkages.

In fact the skin effect will develop on the surface of any conducting
material of any shape, wherever there is RF current flowing.

Here is a link to a detailed mathematical proof, from 'Transmission
Lines for Communications' by C W Davidson (Macmillan Press, 1978, ISBN 0
333 32738 1):http://www.ifwtech.co.uk/g3sek/misc/skin.htm

Davidson's analysis starts with the most general assumption possible:
that RF current is flowing over any small patch of a conductor's
surface. No assumption is required about the reason for the RF current
to be present, only that it is. Likewise no assumption is required about
the cross-section of the conductor, only that it has an exposed surface
(and by implication, that there are no constraints due to a small radius
or insufficient depth). Davidson then derives all the usual equations
for the skin effect. The only drawback of this derivation is that it is
highly mathematical, and difficult to put into words; but it's still
physically correct.

To repeat, I am not saying that Terman's explanation is incorrect; only
that the skin effect is a far more general phenomenon than his
particular examples imply.

This is important because, by taking the existence of the skin effect as
a guaranteed starting-point, the explanations for the behaviour of
coaxial cables,

Ian, I have no disagreement to your reply above other than you are
being to king in your response
I personaly would have put more emphasis on what you stated with
respect
to RF traveling along a path that has no external surface .With
emphasising
where many have about RF travel without which one CANNOT understand
coaxial cables or braid The inside of braid on a coax CAN and DOES
carry
RF current but it does NOT radiate, because it does NOT have an
exposed surface
other than a dielectric interface. The outside surface can and DOES
radiate if a
RF current flows on the outside of the braid. I would also add that
copper/braid
itself does not turn into a dielectric or contain a diode thus it
also WILL
also pass a RF current at its centre but of course does NOT radiate.
This very fact was refuted by popular vote on this newsgroup where
poll
standings always overule science. So yes, without true understandings
errors
are sure to congregate and eventually will create a "fact".
Art KB9MZ...xg





'bazooka' baluns, 'shielded' loops and many other
devices will all fall neatly into place.

--

73 fromIanGM3SEK 'In Practice' columnist for RadCom (RSGB)http://www.ifwtech.co.uk/g3sek

Part 2 Is it possible to ask questions here?
 

"art" wrote
I would also add that copper/braid itself does not turn into a dielectric
or contain a diode thus it also WILL also pass a RF current at its
centre...
____________

art... so by your post you reject the theory and experience of physical
science?

RF

Part 2 Is it possible to ask questions here?
 

"art" wrote
I would also add that copper/braid itself does not turn into
a dielectric or contain a diode thus it also WILL also
pass a RF current at its centre but of course does NOT radiate.
_____________

art, you really need to buy and read Terman's RADIO ENGINEERS' HANDBOOK or
similar source, instead of relying on your intuition. Terman provides the
following equation for the r-f attenuation of air-insulated, copper coaxial
transmission line:

a = 0.00362 SQRT(f)*(1+ D/d) / D*log(D/d) dB per 1,000 feet

where f = frequency in MHz, D = inner diameter of outer conductor, d =
outer diameter of inner conductor.

Note that the attenuation is the same whether the inner conductor is solid
or tubular. This is the result of "skin effect," which for r-f frequencies
1.8 MHz and higher confines the r-f current on the inner conductor from its
outer surface to a depth of less than 0.18 mm.

RF

Part 2 Is it possible to ask questions here?
 

Richard Fry wrote:
"art" wrote

I would also add that copper/braid itself does not turn into
a dielectric or contain a diode thus it also WILL also
pass a RF current at its centre but of course does NOT radiate.

_____________

art, you really need to buy and read Terman's RADIO ENGINEERS' HANDBOOK
or similar source, instead of relying on your intuition. Terman
provides the following equation for the r-f attenuation of
air-insulated, copper coaxial transmission line:

a = 0.00362 SQRT(f)*(1+ D/d) / D*log(D/d) dB per 1,000 feet

where f = frequency in MHz, D = inner diameter of outer conductor, d =
outer diameter of inner conductor.

Note that the attenuation is the same whether the inner conductor is
solid or tubular. This is the result of "skin effect," which for r-f
frequencies 1.8 MHz and higher confines the r-f current on the inner
conductor from its outer surface to a depth of less than 0.18 mm.


One should be aware that this formula applies only to "large" coaxial
transmission lines, where the skin depth is a small fraction of the
conductor thickness.

It's not like the current is confined in a uniform band of the skin
depth, and zero elsewhere. The skin depth is a convenient mathematical
fiction.. it's the depth at which the current density is 1/e, so you can
calculate things like voltage drop by assuming a uniform current density
in a layer that thick, instead of actually integrating it.

On a smallish round conductor, where the circumference isn't many, many
skin depths, there's a broken assumption in the skin depth formula of an
infinite flat plane. Actually solving for the true AC resistance (or
current distribution) involves elliptic integrals which only have
infinite series solutions.


Which is why there are nifty tables and empirical formulas for AC
resistance of round conductors (solid and tubular) that get you
arbitrarily close. See, e.g., NBS Circular 75 or Grover or Reference
Data for Radio Engineers.


Lest you think I am nit picking here.. take a piece of venerable RG-8
style coax, with the AWG13 inner conductor (0.072" diameter, 1.83 mm).
The skin depth at 1.8 MHz (per the above post) is 0.18mm, so the wire is
10 skin depths across, so it's probably a reasonable assumption.

However, let's take something a bit smaller, like RG-8X or RG-58 type
coaxes, which have a inner conductor on the order of 0.9mm. Now, you're
talking only 4-5 skin depths, and the assumption of an infinite plane
probably doesn't hold.


So.. Terman's equation probably holds for coax where the inner conductor
is 20 skin depths, and, as posted, it would make no difference whether
it's a tube (with wall thickness5 skin depth) and a solid conductor.

RF

Part 2 Is it possible to ask questions here?
 

Jim Lux wrote:
. . .
Lest you think I am nit picking here.. take a piece of venerable RG-8
style coax, with the AWG13 inner conductor (0.072" diameter, 1.83 mm).
The skin depth at 1.8 MHz (per the above post) is 0.18mm, so the wire is
10 skin depths across, so it's probably a reasonable assumption.

However, let's take something a bit smaller, like RG-8X or RG-58 type
coaxes, which have a inner conductor on the order of 0.9mm. Now, you're
talking only 4-5 skin depths, and the assumption of an infinite plane
probably doesn't hold.

That would be nit picking unless very high accuracy is required.

As Jim said, the current density actually decays from the surface in an
exponential manner. The skin depth is the depth at which it's dropped to
1/e its density at the surface. If a conductor is infinitely thick, the
total loss is exactly the same as if the current density was uniform to
the skin depth and zero below. So this approximation is widely used when
it can be assumed that the conductor is at least several skin depths
thick. A rigorous calculation for a round wire really requires a
computer, since it involves evaluating complex Bessel functions, and I
believe that closed form equations for many other wire shapes don't
exist at all.

But there are two levels of approximation you can make with the
assumption that the current is all flowing in a uniform layer. If you
calculate the cross sectional area of the ring of current, you come up
with (from simple geometry)

Area = pi * delta * (OD - delta)

where OD is the outer diameter of the wire, and delta is the skin depth.
The material's bulk resistivity is divided by this area to find the
wire's resistance per unit length.

If the diameter is much greater than the skin depth (OD delta), an
even simpler approximation can be and is often made:

Area ~ pi * delta * OD

I assume this is the infinite diameter assumption Jim mentions.

If you use this infinite diameter assumption, the error in the
calculated resistivity of a copper wire 0.9 mm diameter at 1.8 MHz is
5.4% (compared to a rigorous calculation). This error isn't a big deal
for most purposes. But by simply using the first rather than the second
equation for area, the error drops to less than 0.1%. You're still using
the approximation that the current is flowing in a uniform layer one
skin depth thick, so the entire calculation can easily be done on a
pocket calculator in a minute or so.

Roy Lewallen, W7EL

Part 2 Is it possible to ask questions here?
 

Jim Lux wrote:
Richard Fry wrote:
[...]
Note that the attenuation is the same whether the inner conductor is
solid or tubular. This is the result of "skin effect," which for r-f
frequencies 1.8 MHz and higher confines the r-f current on the inner
conductor from its outer surface to a depth of less than 0.18 mm.


One should be aware that this formula applies only to "large" coaxial
transmission lines, where the skin depth is a small fraction of the
conductor thickness.

It's not like the current is confined in a uniform band of the skin
depth, and zero elsewhere. The skin depth is a convenient mathematical
fiction.. it's the depth at which the current density is 1/e, so you
can calculate things like voltage drop by assuming a uniform current
density in a layer that thick, instead of actually integrating it.

On a smallish round conductor, where the circumference isn't many, many
skin depths, there's a broken assumption in the skin depth formula of
an infinite flat plane. Actually solving for the true AC resistance (or
current distribution) involves elliptic integrals which only have
infinite series solutions.


Which is why there are nifty tables and empirical formulas for AC
resistance of round conductors (solid and tubular) that get you
arbitrarily close. See, e.g., NBS Circular 75 or Grover or Reference
Data for Radio Engineers.


Lest you think I am nit picking here.. take a piece of venerable RG-8
style coax, with the AWG13 inner conductor (0.072" diameter, 1.83 mm).
The skin depth at 1.8 MHz (per the above post) is 0.18mm, so the wire
is 10 skin depths across, so it's probably a reasonable assumption.


No, that wasn't nit picking; those are all fair points.

The underlying point is that engineering is ultimately about numbers. We
all like to think in words and mental images if we can, but in marginal
cases these simple slogans and cartoons won't work.

On the other hand, the marginal cases don't invalidate the point that
the skin effect *will* be present. If there isn't enough conductor depth
to allow the skin effect to develop unhindered, it only affects our
estimates of the AC/RF resistance.

If the available depth of conductor is too small, the inside boundary
will push the current density profile outward towards the surface. For a
round conductor, we can think of it as 'current crowding' along the
centreline. A closely related case is copper-plated steel, where the
magnetic nature of the steel increases its AC/RF resistance by a further
factor of sqrt(mu), which squeezes a much higher fraction of the total
current into the thin layer of copper.

However, let's take something a bit smaller, like RG-8X or RG-58 type
coaxes, which have a inner conductor on the order of 0.9mm. Now,
you're talking only 4-5 skin depths, and the assumption of an infinite
plane probably doesn't hold.

We can see a little further into this without the need for detailed
math. The radius of the conductor is 2.5 skin depths (again using
0.18mm) so the current density at this depth would normally be 1/e^2.5
or about 1/12 of its surface value. That suggests that the perturbation
in RF resistance due to insufficient depth is only taking place at
around the 10% level. In the context of *estimating* the RF resistance
to help us decide whether to buy a drum of cable, that wouldn't be a
serious error.

However, it warns of a very serious error if the centre conductor was
made of copper-plated steel instead of solid copper.


So.. Terman's equation probably holds for coax where the inner
conductor is 20 skin depths,

Sorry, Jim, you lost me: why such a large number as 20?

At 2.5 skin depths, the current density is 10% of the surface value;
at 5 skin depths, 1%. If at least 5 skin depths are available, we can
be confident in the accuracy of the standard, uncorrected equation for
most purposes.

A more serious effect of insufficient conductor depth may be in
estimating the effectiveness of shielding. The residual fields at the
opposite side of an extremely thin shield can be very significant if
we're looking for attenuations of 40dB, 60dB or more.



--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Part 2 Is it possible to ask questions here?
 

Ian White GM3SEK wrote:

So.. Terman's equation probably holds for coax where the inner
conductor is 20 skin depths,


Sorry, Jim, you lost me: why such a large number as 20?

At 2.5 skin depths, the current density is 10% of the surface value; at
5 skin depths, 1%. If at least 5 skin depths are available, we can be
confident in the accuracy of the standard, uncorrected equation for most
purposes.


But it's round... (unless Terman rolled that into his constants)

Consider if you peeled that 2.5 skin depth layer and made it flat. It
would look like a pyramid, not a rectangular bar.

Of course, if you assume that the cross sectional area is an annulus
(the pi*( r^2-(r-skindepth)^2) style calculation) this partially gets
taken into account.

The other factor is that in a wire that is comparable to skin depth in
radius, the current on the far side of the wire also contributes to
squeezing the current towards the near side surface. (and that's why
the actual math gets hairy.. you can't use a simple exponential
approximation for the current density)

At 20 skin depths, the difference is negligble.

As a practical matter, if you have an application that actually cares
about this level of detail, you probably have the resources to deal with
the exact calculations, so the simple "thin layer on the surface" is
close enough.

For what it's worth, this kind of thing is why the loss in coax doesn't
follow a nice k1*f + k2*sqrt(f) characteristic at low frequencies. The
second term is essentially assuming that the skin depth in the
conductors is "small" compared to conductor size.



A more serious effect of insufficient conductor depth may be in
estimating the effectiveness of shielding. The residual fields at the
opposite side of an extremely thin shield can be very significant if
we're looking for attenuations of 40dB, 60dB or more.

Especially at low frequencies... (shielding mains frequency
interference, or PWM switcher noise, for instance)

Part 2 Is it possible to ask questions here?
 

Jim Lux wrote:
Ian White GM3SEK wrote:

So.. Terman's equation probably holds for coax where the inner
conductor is 20 skin depths,
Sorry, Jim, you lost me: why such a large number as 20?
At 2.5 skin depths, the current density is 10% of the surface
value; at 5 skin depths, 1%. If at least 5 skin depths are
available, we can be confident in the accuracy of the standard,
uncorrected equation for most purposes.


But it's round... (unless Terman rolled that into his constants)

I don't believe so.

Consider if you peeled that 2.5 skin depth layer and made it flat. It
would look like a pyramid, not a rectangular bar.

I see your point, just wouldn't have imagined that the circular geometry
would make such a very large difference compared with a flat surface.


Of course, if you assume that the cross sectional area is an annulus
(the pi*( r^2-(r-skindepth)^2) style calculation) this partially gets
taken into account.

The other factor is that in a wire that is comparable to skin depth in
radius, the current on the far side of the wire also contributes to
squeezing the current towards the near side surface. (and that's why
the actual math gets hairy.. you can't use a simple exponential
approximation for the current density)

Agreed. I do have the more detailed equations involving Bessel
functions, but no time to compute them just now.

At 20 skin depths, the difference is negligble.

No doubt; but it's the journey towards "negligible" that concerns us
more than the destination :-)


--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek