On 14 Oct, 08:07, "Dave" wrote:
"art" wrote in message

oups.com...





Equi is for equal, libra is the sign of Zodiac which is a pair of
scales

It means that a repetitive action must come back to its source.
AC power has a cycle as in frequency, thus its circuit is one
wavelength.

A quad antenna represents a one wave circuit where total length is
proportionat
to time .ie a picture of events during one cycle.
Now place along side of the quad a radiator of your choice where
the length of the short radiator is superimposed upon the quad.
For ease in foillowing the electrical circuit make your shorty
antenna
equal to one side of the quad

So the race begins. Both travel up for a time at the SAME velocity.
At the corner of the quad power travels onward until it gets to its
originating point to complete a cycle
The short antenna appears to have nowhere to go so what does it do?
To duplicate the quad action it must have a circuit of some sort where
the power can still travel forward during the time the power in the
quad
is still travelling to the originating point.

Complete the circuit of the short antenna showing the series of events
that occur upto the point that the events on the quad come to its
point of beginning.

Note The impedance of space is 377 ohms, the impedance of ground can
be your choice.
Art

but a wave comes back to it starting point after 1/2 cycle then reverses.
so why does your original assertion of a 1/2 wave 'gaussian' antenna still
hold? now, for a 1/4 wave monopole, where does the other 3/4's of a wave
come from?? and what happens to the power that is radiated? how does that
ever return to it's source to be in equilibrium? would a circuit that was
in equilibrium not be capable of radiating since everything has to return to
the source??- Hide quoted text -

- Show quoted text -

Oh dear I forgot to mention, there is another path for current flow
and that is at the center of the radiator. The impedance of this path
is lower than the impedance to space so naturally that is the path the
power will take.
Now on this inner path there are no electrons to dislodge as the
outer
electrons have decayed before they reached the center path so the
current
during this time is just travelling without supplying radiation until
it reaches the bottom of the radiator.
But it's counter part the quad is radiating during this time
( How about that experts!) And by the way it was mentioned somewhere
that the ground resistance was 2 ohms so one should now be able to
determine the IMPEDANCE of ground at the antenna base point
Now you take it from there....experts!
Art KB9MZ...XG

"art" wrote in message
oups.com...
On 14 Oct, 08:07, "Dave" wrote:
"art" wrote in message

oups.com...





Equi is for equal, libra is the sign of Zodiac which is a pair of
scales

It means that a repetitive action must come back to its source.
AC power has a cycle as in frequency, thus its circuit is one
wavelength.

A quad antenna represents a one wave circuit where total length is
proportionat
to time .ie a picture of events during one cycle.
Now place along side of the quad a radiator of your choice where
the length of the short radiator is superimposed upon the quad.
For ease in foillowing the electrical circuit make your shorty
antenna
equal to one side of the quad

So the race begins. Both travel up for a time at the SAME velocity.
At the corner of the quad power travels onward until it gets to its
originating point to complete a cycle
The short antenna appears to have nowhere to go so what does it do?
To duplicate the quad action it must have a circuit of some sort where
the power can still travel forward during the time the power in the
quad
is still travelling to the originating point.

Complete the circuit of the short antenna showing the series of events
that occur upto the point that the events on the quad come to its
point of beginning.

Note The impedance of space is 377 ohms, the impedance of ground can
be your choice.
Art

but a wave comes back to it starting point after 1/2 cycle then reverses.
so why does your original assertion of a 1/2 wave 'gaussian' antenna
still
hold? now, for a 1/4 wave monopole, where does the other 3/4's of a wave
come from?? and what happens to the power that is radiated? how does
that
ever return to it's source to be in equilibrium? would a circuit that
was
in equilibrium not be capable of radiating since everything has to return
to
the source??- Hide quoted text -

- Show quoted text -

Oh dear I forgot to mention, there is another path for current flow
and that is at the center of the radiator. The impedance of this path
is lower than the impedance to space so naturally that is the path the
power will take.
Now on this inner path there are no electrons to dislodge as the
outer
electrons have decayed before they reached the center path so the
current
during this time is just travelling without supplying radiation until
it reaches the bottom of the radiator.
But it's counter part the quad is radiating during this time
( How about that experts!) And by the way it was mentioned somewhere
that the ground resistance was 2 ohms so one should now be able to
determine the IMPEDANCE of ground at the antenna base point
Now you take it from there....experts!
Art KB9MZ...XG


ah, the no electron anti-skin effect return path up the middle of the
element. what about my elements that are hollow? where does the current
flow then? or where does it flow in wire cage elements?? what about coiled
elements like that 'new' short monopole, does it flow around and around
inside the wire, or take a shortcut through something else? How does this
fit into current wave reflection theory that easily explains the currents on
elements complete with easily measured and well predicted values that don't
require a return current up the middle?? too many questions left hanging
art, you gotta work on that theory a bit more to fill in all the holes. oh,
maybe thats it, try incorperating holes into your next version, maybe thats
how the current flows back along the inside, it is the holes left over when
the electrons 'decay'!

On Sun, 14 Oct 2007 16:20:21 GMT, "Dave" wrote:

what about coiled
elements like that 'new' short monopole, does it flow around and around
inside the wire

Dave,

With your accurate shorthand of validation, you are treading very
close to Arthur's loose Gaussian shoelaces and may soon find yourself
elevated by him to his Academy of Supporters of Seance when he trips.

Won't you be embarrassed when he sites you as his authority for your
cogent analaysis!

73's
Richard Clark, KB7QHC

On 14 Oct, 09:20, "Dave" wrote:
"art" wrote in message

oups.com...





On 14 Oct, 08:07, "Dave" wrote:
"art" wrote in message

groups.com...

Equi is for equal, libra is the sign of Zodiac which is a pair of
scales

It means that a repetitive action must come back to its source.
AC power has a cycle as in frequency, thus its circuit is one
wavelength.

A quad antenna represents a one wave circuit where total length is
proportionat
to time .ie a picture of events during one cycle.
Now place along side of the quad a radiator of your choice where
the length of the short radiator is superimposed upon the quad.
For ease in foillowing the electrical circuit make your shorty
antenna
equal to one side of the quad

So the race begins. Both travel up for a time at the SAME velocity.
At the corner of the quad power travels onward until it gets to its
originating point to complete a cycle
The short antenna appears to have nowhere to go so what does it do?
To duplicate the quad action it must have a circuit of some sort where
the power can still travel forward during the time the power in the
quad
is still travelling to the originating point.

Complete the circuit of the short antenna showing the series of events
that occur upto the point that the events on the quad come to its
point of beginning.

Note The impedance of space is 377 ohms, the impedance of ground can
be your choice.
Art

but a wave comes back to it starting point after 1/2 cycle then reverses.
so why does your original assertion of a 1/2 wave 'gaussian' antenna
still
hold? now, for a 1/4 wave monopole, where does the other 3/4's of a wave
come from?? and what happens to the power that is radiated? how does
that
ever return to it's source to be in equilibrium? would a circuit that
was
in equilibrium not be capable of radiating since everything has to return
to
the source??- Hide quoted text -

- Show quoted text -

Oh dear I forgot to mention, there is another path for current flow
and that is at the center of the radiator. The impedance of this path
is lower than the impedance to space so naturally that is the path the
power will take.
Now on this inner path there are no electrons to dislodge as the
outer
electrons have decayed before they reached the center path so the
current
during this time is just travelling without supplying radiation until
it reaches the bottom of the radiator.
But it's counter part the quad is radiating during this time
( How about that experts!) And by the way it was mentioned somewhere
that the ground resistance was 2 ohms so one should now be able to
determine the IMPEDANCE of ground at the antenna base point
Now you take it from there....experts!
Art KB9MZ...XG

ah, the no electron anti-skin effect return path up the middle of the
element. what about my elements that are hollow? where does the current
flow then? or where does it flow in wire cage elements?? what about coiled
elements like that 'new' short monopole, does it flow around and around
inside the wire, or take a shortcut through something else? How does this
fit into current wave reflection theory that easily explains the currents on
elements complete with easily measured and well predicted values that don't
require a return current up the middle?? too many questions left hanging
art, you gotta work on that theory a bit more to fill in all the holes. oh,
maybe thats it, try incorperating holes into your next version, maybe thats
how the current flows back along the inside, it is the holes left over when
the electrons 'decay'!- Hide quoted text -

- Show quoted text -

Well now David thge charge has arrived at the bottom of the radiator.
It is now in a quandry
it looks over at the quaud and ssees that the charge over there is
half way around!
So the charge on the small radiator finds that it can only go up
aqgain forming a loop of current.
Fortunately it is now back on track radiating just like the quad.
Ofcourse when he gets to the top
it again has only one place to go and that is down the center again
where it can't radiatiate
like the quad is doing.When the charge gets to the bottom it wsees a
path of 2 0hms (the resistor)
so it passes thru the resistor where the charge finally gets back to
the starting point at exactly the same time that the charge of the
quad got back home.
So the sequence of events is that the short antenna only radiated half
of the time neglecting the travel thru the resister of course. So the
quad radiated twice the time of the smaller radiator which amounts to
a 3 db gain.
Well some might say we knew that a quad radiated more because it is
the same as two dipoles( half wave) on top of
each other so yes for the same power expended in the same amount of
time a full wave radiator radiates twice the power that a quarter wave
does.
Art you are a genius put QED at the end of this post and also get hold
of Yuru so that he now can get to sleep at night.
Art Unwin KB9MZ.......XG (uk)

On Sun, 14 Oct 2007 10:32:53 -0700, art wrote:

So the
quad radiated twice the time of the smaller radiator which amounts to
a 3 db gain.

Hmmm, this puts Robert Vincent's antenna into a cocked hat through a
chain of your denials. Well, society needs Libras who can believe six
impossible things before breakfast. Imagine, we get to see them
posted here before brunch too!

Art wrote:
....for the same power expended in the same amount of
time a full wave radiator radiates twice the power that a
quarter wave does.
_________

So then where does the missing power applied to the 1/4-wave radiator go, if
only half of its input power is converted to EM radiation?

RF

"Richard Fry" wrote in message
...
Art wrote:
....for the same power expended in the same amount of
time a full wave radiator radiates twice the power that a
quarter wave does.
_________

So then where does the missing power applied to the 1/4-wave radiator go,
if only half of its input power is converted to EM radiation?

RF

i am convinced that art is playing dumb on purpose.... no one with any
knowledge of antennas and em fields could possible get this screwed up on
basic concepts. when he was just talking 'gaussian' and 'equilibrium' stuff
it could easily be laughed off as someone with a little bit of knowledge
trying to force the rest of the world into that single concept. but now
with decaying electrons and current down the center of the conductor and a
missing half cycle its just too far gone. i had been kind of tagging along
tweaking him into trying to explain the details of his concept, but now it
is apparent that he has none. he is just making it up and stringing us
along. you gone too far this time art, ciao and enjoy your game alone.

On 14 Oct, 15:11, "Dave" wrote:
"Richard Fry" wrote in message

...

Art wrote:
....for the same power expended in the same amount of
time a full wave radiator radiates twice the power that a
quarter wave does.
_________

So then where does the missing power applied to the 1/4-wave radiator go,
if only half of its input power is converted to EM radiation?

RF

i am convinced that art is playing dumb on purpose.... no one with any
knowledge of antennas and em fields could possible get this screwed up on
basic concepts. when he was just talking 'gaussian' and 'equilibrium' stuff
it could easily be laughed off as someone with a little bit of knowledge
trying to force the rest of the world into that single concept. but now
with decaying electrons and current down the center of the conductor and a
missing half cycle its just too far gone. i had been kind of tagging along
tweaking him into trying to explain the details of his concept, but now it
is apparent that he has none. he is just making it up and stringing us
along. you gone too far this time art, ciao and enjoy your game alone.

Awe common Dave don't go away this is fun! Now what is wrong with
current
going down the center because the impedance is less than 377 ohms?
Remember if you go up in altitude the 377 impedance will go down and
become
less that the wires internal impedance the energy will transfer to
space with
that smell of ozone. That is what happened in Quito Equador
That is what led to the design of the quad where a
lesser impedance path was provided for current flow!
Now the inner impedance of the wire must be pretty high to
create an end effect so with Isq R there has to be a lot of
heat generated. Remember Vincent saying his first antenna melted?
Now we come to rotation of a projectile.
Did you know that a projectile tumbles and comes under the influence
of the Corrioli effect as it losses energy.A shell
starts to tumble when it exceeds it's range Next time a rocket goes
into space watch for rotation. What do you think causes that?
A train speeds along on a railroad until at a certain speed
it will roll over, why do you think that happens?
Look up "hammer blow" for that one. How can
I possibly be stringing you along?
You can put everything to bed by spending
a 1/2 hour making a full wave antenna but ofcourse that
means an effort to get off the couch !
Yup
It is better to stay rested and let your mouth do
all the work at least until you start to dribble.
Art Unwin KB9MZ....XG

"Dave" wrote about Art's posts:
i am convinced that art is playing dumb on purpose.... no one with
any knowledge of antennas and em fields could possible get this
screwed up on basic concepts.

and Walt W2DU wrote:

I'm enjoying every minute of your entertainment presentations, and
I'd miss them dearly if you left the thread
___________

Art could really know better, and might be posting what he does as
some form of personal amusement.

The only problem with that is some readers of his posts may think
his posts are accurate, unless others who know they are not make
the effort to post the correct information.

RF