What is the antenna current path or route
 

On 19 Oct, 09:04, (Richard Harrison) wrote:
Art wrote:

"And "end effect" is the confusion created at the top of the radiator."
I was carrying on the train of thought of the poster, I have made it
quite clear what my thoughts are.

When the signal arrives at the open circuit end of the antenna, current
can not continue its forward flow. It abruptly stops, no longer
producing a magnetic field.

Who says so? I suspect that in the recognition of a measurement
problem
you theorised what might have , could have etc to match your faulty
logic.


..


Energy from the magnetic field is converted to energy in the electric
field for an instant (Cecil`s famous conservation of energy). This
produces an insreased voltage at the open circuit end. This incresed
voltage has more capacitive effect, akin to the "Miller effect" caused
by the higher signal voltage on the plate of an amplifier vacuum tube
than on its grid.

All this is faulty logic manufactured to suit your intelligence.
Radiation is a function of the release of energy from capacitance and
inductance
during a cycle i.e. a parallel circuit called a "tank circuit: which
can also
be seen as a loss less system ala a pendulum.
This goes back to what I say about books. Anybody can write one. It is
up
to the reader to follow the dots of logic displayed to determine
agreement
not to swallow it and memorise it.





On a transmission line or on an antenna system, we used to call this
capacitive action the "Ferranti effect"

Best regards, Richard Harrison, KB5WZI

What is the antenna current path or route
 

Art wrote:
"Who says so?"

Many. It is commonly held opinion.

Frederick Emmons Terman writes on page 89 of his 1955 opus:

"When the load impedance is infinite, Eq. (4-14) shows that that the
coeficient of reflection will be 1 on an angle of 0. Under these
conditions the incident and reflected waves will have equal magnitudes
at the load, and the reflection will be such that the voltages of the
incident and reflected waves have the same phase. As a result, the
voltages of the two waves add mathematically so that at the load
E1=E2+EL/2.

Best regards, Richard Harrison, KB5WZI

What is the antenna current path or route
 

On 19 Oct, 11:43, (Richard Harrison) wrote:
Art wrote:

"Who says so?"

Many. It is commonly held opinion.
Right, it is opinion not factual.
There is no proof what so ever that is what happens.
So you must also consider alernatives.
Current can go out into space which doesn't seem plausible
Current can flow inside the copper wire which is plausible

So now we have a measurement problem. How can we track
one path from the other or you come up with a reason
that it can't flow thru the center of copper



Frederick Emmons Terman writes on page 89 of his 1955 opus:

"When the load impedance is infinite,

WELL,WELL,WELL!

Probably true except for one thing.... THE LOAD IMPEDANCE IS NOT
INFINITE !
See what I mean about books. It is not a case of remembering what
somebody
said it is a case of connecting the dots with good logic that one
finds
unmistakable agreement.
Which brings up another point with respect to radiation about which
many admit is not known.
If the laws of Maxwell are in agreement with the extension of Gauss
which many say
of the latter is not correct then logic states we should remove
Maxwell.
Problem is that computor programs based on Maxwell laws also confirm
Gaussian law.
So now YOU connect the dots. Maxwell is correct? Computor programs
based on Maxwell
are correct. Computor programs confirm Gausses law extensions re
adding a time varying field.
Mathematics support Gaussian verification of Gaussion law, so
where is the logic of condemming Gaussian law and yet not condemming
all other laws?

Lesson: Use your own logic before referring what you read to memory
otherwise
you are just a member of the herd without a contribution of your own.
Just follow the arse of who goes before youand ignor the smell
Art


Eq. (4-14) shows that that the
coeficient of reflection will be 1 on an angle of 0. Under these
conditions the incident and reflected waves will have equal magnitudes
at the load, and the reflection will be such that the voltages of the
incident and reflected waves have the same phase. As a result, the
voltages of the two waves add mathematically so that at the load
E1=E2+EL/2.

Meaningless jabber since it is based on "infinite impedance" which is
in error.
thus the analysis is in error for the circumstances at hand.


Best regards, Richard Harrison, KB5WZI

What is the antenna current path or route
 

Art wrote:
"THE LOAD IMPEDANCE IS NOT INFINITE!"

Relative to the impedance of a reasonable antenna. the impedance of the
atmosphere or the impedance of free-space is non-conductive.
When the signal following the conductive surface of an antenna reaches
its end of the conductive path, current reverses its direction of travel
of necessity.

Best regards, Richard Harrison, KB5WZI

What is the antenna current path or route
 

I mistyped: E1=E2+EL/2

The + sign should have been an = sign. I didn`t remove my finger from
the dhift register in time and the wrong character emerged (a + instead
of the = sign). I am sorry. It should have read:

E1=E2=EL/2 because EL is the sum of E1 and E2 which are equal.

Best regards, Richard Harrison, KB5WZI

What is the antenna current path or route
 

"art" wrote in message
oups.com...
On 19 Oct, 06:45, "Dave" wrote:
"art" wrote in message

ps.com...





On 19 Oct, 03:55, "Dave" wrote:
"art" wrote in message

groups.com...

Pseudo experts of fractional wavelength antennas.
Where does the current flow when it reaches the END of a fractional
length?
verticle antenna and why?
How does this relate to the term "end effect"?
If you have already written a book then tell us what the auther
said.
Art KB9MZ.....XG

it turns around and goes right back down the way it came.

So a electrical generater doesn't keep turning in one direction
but instead it occillates at the desired frequency.
I have never seen one do that!
And "end effect" is the confusion created at the top of the radiator

'end effect' is an effect of the capacitance seen from the end of the
antenna to ground or the other part of a dipole.

how does a generator come into this? you feed current into a wire with
an
open end, it gets to the end, reverses direction and goes back to where
it
started. no frequency was stated or implied all you asked was where
there
'current' went... that could be any kind of current including a step or
pulse or sinusoids of any frequency.- Hide quoted text -

- Show quoted text -

The current comes from a generator does it not? So how is that current
produced?
I see a generator turning in one direction only all the time where you
are suggesting that it is occillating.
Pretty hard to draw a circuit if it tracks back the way it came.
Draw a graph of an occillating current as you would see on a scope
that shows two degrees of freedom.
Does the "x" direction stop after a half cycle?

i don't need a generator to create a current. hook a battery to a piece of
wire, watch the current travel to the end of the wire then reflect back.
its easy to do the measurements and they match exactly the theory. i use
this all the time, its called a time domain reflectometer. very handy for
finding breaks or bad spots in feed lines or Beverage antennas. use one for
a while and you will become much more familiar with the true effect of
currents on wires and reflections.

What is the antenna current path or route
 

Richard Harrison wrote:
Relative to the impedance of a reasonable antenna. the impedance of the
atmosphere or the impedance of free-space is non-conductive.
When the signal following the conductive surface of an antenna reaches
its end of the conductive path, current reverses its direction of travel
of necessity.

Maybe better understood using visible EM waves.
When the medium's index of refraction changes,
reflections are the result. One can learn a
lot about EM wave reflection by looking in a
mirror. With some of the strange concepts about
EM reflected energy that exist on this newsgroup,
it is a wonder how the promoters of such concepts
can look themselves in the mirror. :-)
--
73, Cecil http://www.w5dxp.com

What is the antenna current path or route
 

On 19 Oct, 14:31, Cecil Moore wrote:
Richard Harrison wrote:
Relative to the impedance of a reasonable antenna. the impedance of the
atmosphere or the impedance of free-space is non-conductive.
When the signal following the conductive surface of an antenna reaches
its end of the conductive path, current reverses its direction of travel
of necessity.

Maybe better understood using visible EM waves.
When the medium's index of refraction changes,
reflections are the result. One can learn a
lot about EM wave reflection by looking in a
mirror. With some of the strange concepts about
EM reflected energy that exist on this newsgroup,
it is a wonder how the promoters of such concepts
can look themselves in the mirror. :-)
--
73, Cecil http://www.w5dxp.com

Probably true. I wonder if they would continue
to argue if the 1/4 wave antenna was tubular or is
there another reason that the center of a radiator
is off limits to current flow.
Still you can't blame them if all they know comes
from the ARRL books. After all it is a hobby and in
the early days many did not have the opportunity
for a college education
Best regards
Art

What is the antenna current path or route
 

On 19 Oct, 13:35, (Richard Harrison) wrote:
Art wrote:

"THE LOAD IMPEDANCE IS NOT INFINITE!"

Relative to the impedance of a reasonable antenna. the impedance of the
atmosphere or the impedance of free-space is non-conductive.
When the signal following the conductive surface of an antenna reaches
its end of the conductive path, current reverses its direction of travel
of necessity.

Best regards, Richard Harrison, KB5WZI

Keep to the discussion at hand. The route to free space was discarded
but there is another route available which you choose to ignore,
Well not you since you just quote from books without personal
voutching.
So let us say Terman chose to ignore or his secretary screwed up.
Unless ofcourse you have dotted the I's and T,s for yourself
and can personally confirm the corectness of what you have quoted.
So pray tell me, what is the impedance of current flow on the inside
of a tube
or down the center of a solid conductor assuming there is no 10 foot
fence placed at the top end of a fractional wave length antenna.
We will leave the lights on for the current to see the inside of the
tube

What is the antenna current path or route
 

On 19 Oct, 13:53, "Dave" wrote:
"art" wrote in message

oups.com...



On 19 Oct, 06:45, "Dave" wrote:
"art" wrote in message

oups.com...

On 19 Oct, 03:55, "Dave" wrote:
"art" wrote in message

groups.com...

Pseudo experts of fractional wavelength antennas.
Where does the current flow when it reaches the END of a fractional
length?
verticle antenna and why?
How does this relate to the term "end effect"?
If you have already written a book then tell us what the auther
said.
Art KB9MZ.....XG

it turns around and goes right back down the way it came.

So a electrical generater doesn't keep turning in one direction
but instead it occillates at the desired frequency.
I have never seen one do that!
And "end effect" is the confusion created at the top of the radiator

'end effect' is an effect of the capacitance seen from the end of the
antenna to ground or the other part of a dipole.

how does a generator come into this? you feed current into a wire with
an
open end, it gets to the end, reverses direction and goes back to where
it
started. no frequency was stated or implied all you asked was where
there
'current' went... that could be any kind of current including a step or
pulse or sinusoids of any frequency.- Hide quoted text -

- Show quoted text -

The current comes from a generator does it not? So how is that current
produced?
I see a generator turning in one direction only all the time where you
are suggesting that it is occillating.
Pretty hard to draw a circuit if it tracks back the way it came.
Draw a graph of an occillating current as you would see on a scope
that shows two degrees of freedom.
Does the "x" direction stop after a half cycle?

i don't need a generator to create a current. hook a battery to a piece of
wire, watch the current travel to the end of the wire then reflect back.
its easy to do the measurements and they match exactly the theory. i use
this all the time, its called a time domain reflectometer. very handy for
finding breaks or bad spots in feed lines or Beverage antennas. use one for
a while and you will become much more familiar with the true effect of
currents on wires and reflections.- Hide quoted text -

- Show quoted text -

Good idea. Where can radio hams buy a battery for an a.c. frequency of
ten metres ?