Wayne wrote:
"Roy Lewallen" wrote in message

There is no relationship between SWR and efficiency.

Roy Lewallen, W7EL

Agreed. For example, SWR into a dummy load. The equation given was just to
show the approximation I was trying to make for the dipole case.


I don't seem to be communicating well, but that's not unusual. It's a
shortcoming I have. So I'll try again.

There is not even an approximate relationship between SWR and
efficiency, so what you calculated was not an approximation of the
efficiency. It was simply a number which has no relationship whatsoever
to the efficiency. You could have used any equation you might dream up,
and the result would be as meaningless as the result you got.

Roy Lewallen, W7EL

On 5 Mar, 14:14, "Wayne" wrote:
Wayne wrote:
When the subject of antenna efficiency comes up, it often involves a
discussion of ground losses on verticals. What about, for example, a
dipole? Could one calculate "power out/power in" by measuring the VSWR
and declaring that everything not reflected was transmitted? It would
seem more accurate to actually measure power out and power in, but that
introduces inaccuracies by having to calibrate the setup. Thoughts?
"Roy Lewallen" wrote in message

...

There's no direct way to measure the total power being radiated other than
sampling the field at many points in all directions and integrating.
"Reflected" power is not power that isn't transmitted. You can find the
power being applied to the antenna by subtracting the "reverse" or
"reflected" power from the "forward" power, but that tells you nothing
about what fraction is radiated and what fraction lost as heat.

Roy Lewallen, W7EL

Thanks for the reply. My dipole example is intended to avoid transmission
line issues by not having one, and the elements are assumed to be reasonably
low loss. If I do some quick back-of-the-envelope calculations, for a VSWR
of 1.3:1, I get an efficiency of about 98.3% (using the equation
1-gamma^2). Assuming a resistance of 1 ohm in the dipole conductors the
efficiency I calculate is about 98.6% (72/73).

Are there any other loss issues missing in this example.

I would say you are close enough to say you are correct. Because you
chose a dipole which is in a state of equilibrium and thus particles
projected from the dipole cannot collide with other particles from
other parasitic radiators. Aren't you not basically refering to the
foundations of Poyntings vector which like Gauss is refering to an
item in equilibrium when subjected to a time variable of zero value ?
If the item is not in a state of equilibrium collision
of particles may well occur without a radiation field reaction thus
one cannot calculate the resultant field since energy transfer due to
particle collision prevents the return of particles to the mother
radiator.

"art" wrote:

I would say you are close enough to say you are correct. Because you
chose a dipole which is in a state of equilibrium and thus particles
projected from the dipole cannot collide with other particles from
other parasitic radiators. Aren't you not basically refering to the
foundations of Poyntings vector which like Gauss is refering to an
item in equilibrium when subjected to a time variable of zero value ?
If the item is not in a state of equilibrium collision
of particles may well occur without a radiation field reaction thus
one cannot calculate the resultant field since energy transfer due to
particle collision prevents the return of particles to the mother
radiator.

I've been reading Art's posts for some time now, mostly for the
entertainment value. Some of the recent posts were starting to make
sense, and this was causing me some concern: that my bafflegab
filter in the computor (sic) had gone out. Well, the above quote was
a wakeup call... I don't have a clue what he's saying. Can someone
translate, please.

Mike W5CHR

"Mike Lucas" wrote in message
...

"art" wrote:

I would say you are close enough to say you are correct. Because you
chose a dipole which is in a state of equilibrium and thus particles
projected from the dipole cannot collide with other particles from
other parasitic radiators. Aren't you not basically refering to the
foundations of Poyntings vector which like Gauss is refering to an
item in equilibrium when subjected to a time variable of zero value ?
If the item is not in a state of equilibrium collision
of particles may well occur without a radiation field reaction thus
one cannot calculate the resultant field since energy transfer due to
particle collision prevents the return of particles to the mother
radiator.

I've been reading Art's posts for some time now, mostly for the
entertainment value. Some of the recent posts were starting to make
sense, and this was causing me some concern: that my bafflegab
filter in the computor (sic) had gone out. Well, the above quote was
a wakeup call... I don't have a clue what he's saying. Can someone
translate, please.

Mike W5CHR


Maybe if you put this in a word blender and spun long enough, maybe you
could get one sentence that would make some sense. :-)
I gave up way back when I could not understand what is the POLARITY in
antennas, which end up? So I guess humanity has to evolve for few more
centuries to catch up with "antenna wizard" and understand and appreciate
his piosneering work.
Judging by some other posts on other more earthly subjects, looks like there
are some missing gears in the gear box. What you expect from inventor that
has a patent on reflector beeing director and vice-voica. Looks like that
one will not ever be copyright violated. So take it with grain of salt and
enjoy the mumbo-jumbo-entoitenmeint. :-)

73 bada BUm

On 29 Apr, 04:48, "Mike Lucas" wrote:
"art" wrote:

I would say you are close enough to say you are correct. Because you
chose a dipole which is in a state of equilibrium and thus particles
projected from the dipole cannot collide with other particles from
other parasitic radiators. Aren't you not basically refering to the
foundations of Poyntings vector which like Gauss is refering to an
item in equilibrium when subjected to a time variable of zero value ?
If the item is not in a state of equilibrium collision
of particles may well occur without a radiation field reaction thus
one cannot calculate the resultant field since energy transfer due to
particle collision prevents the return of particles to the mother
radiator.

I've been reading Art's posts for some time now, mostly for the
entertainment value. Some of the recent posts were starting to make
sense, and this was causing me some concern: that my bafflegab
filter in the computor (sic) had gone out. Well, the above quote was
a wakeup call... I don't have a clue what he's saying. Can someone
translate, please.

Mike W5CHR

Mike, you must first understand that that two unlike particles in
combination from the same radiator creates radiation. Lock that into
your mind. Other collisions or combinations do not creat radiation so
their energy has to be ascertained so they do not finish up on the
radiation side of the equation. If the radiating article is in
equilibrium there is no other radiator in competition in the same
space to create radiation thus the single radiator is free to emit
particles in isolation where errent collisions or combinations can not
occur. You must also note that all particles emitted from a single
radiator do not all finish up on the radiation side of the equation
since some return to the mother element with the same kinetic impact
that was imparted on them in the first place and thus these must be
accounted for in any equation. If one is to ascertain the final
arrangement of any energy transfer from a black box it is desirable
not to introduce energy exchange phenomina by introducing rival or
cumulative exchanges which cannot be accounted for. All equations that
do not allow for the unit of time are purely mathematical exercises of
theory thus one must be absolutely sure that the numbers add up on
both sides of the equation to justify the addition of that very
important symbol called equal. That is why the subject of ther terms
of "time" and "equilibrium" is important to transform an equation from
a fantasy form to one of reality in both Poyntings vector and the
Gaussian equation.
Art

On Sun, 29 Apr 2007 06:48:18 -0500, "Mike Lucas"
wrote:


"art" wrote:

I would say you are close enough to say you are correct. Because you
chose a dipole which is in a state of equilibrium and thus particles
projected from the dipole cannot collide with other particles from
other parasitic radiators. Aren't you not basically refering to the
foundations of Poyntings vector which like Gauss is refering to an
item in equilibrium when subjected to a time variable of zero value ?
If the item is not in a state of equilibrium collision
of particles may well occur without a radiation field reaction thus
one cannot calculate the resultant field since energy transfer due to
particle collision prevents the return of particles to the mother
radiator.

I've been reading Art's posts for some time now, mostly for the
entertainment value. Some of the recent posts were starting to make
sense, and this was causing me some concern: that my bafflegab
filter in the computor (sic) had gone out. Well, the above quote was
a wakeup call... I don't have a clue what he's saying. Can someone
translate, please.

Equi Librium is a drug, that is too big to swallow, so you have to
crush it into particles on one mother of a radiator. If you try to
swallow it whole, you will get one massive field reaction (a buzz) in
a time variable of zero value (like really fast) which may induce
projectile vomiting.

Maybe.

If not, then we will have to Gauss again.

73's
Richard Clark, KB7QHC

On Apr 28, 8:07 pm, art wrote:
On 5 Mar, 14:14, "Wayne" wrote:





Wayne wrote:
When the subject of antenna efficiency comes up, it often involves a
discussion of ground losses on verticals. What about, for example, a
dipole? Could one calculate "power out/power in" by measuring the VSWR
and declaring that everything not reflected was transmitted? It would
seem more accurate to actually measure power out and power in, but that
introduces inaccuracies by having to calibrate the setup. Thoughts?
"Roy Lewallen" wrote in message

...

There's no direct way to measure the total power being radiated other than
sampling the field at many points in all directions and integrating.
"Reflected" power is not power that isn't transmitted. You can find the
power being applied to the antenna by subtracting the "reverse" or
"reflected" power from the "forward" power, but that tells you nothing
about what fraction is radiated and what fraction lost as heat.

Roy Lewallen, W7EL

Thanks for the reply. My dipole example is intended to avoid transmission
line issues by not having one, and the elements are assumed to be reasonably
low loss. If I do some quick back-of-the-envelope calculations, for a VSWR
of 1.3:1, I get an efficiency of about 98.3% (using the equation
1-gamma^2). Assuming a resistance of 1 ohm in the dipole conductors the
efficiency I calculate is about 98.6% (72/73).

Are there any other loss issues missing in this example.

I would say you are close enough to say you are correct. Because you
chose a dipole which is in a state of equilibrium and thus particles
projected from the dipole cannot collide with other particles from
other parasitic radiators. Aren't you not basically refering to the
foundations of Poyntings vector which like Gauss is refering to an
item in equilibrium when subjected to a time variable of zero value ?
If the item is not in a state of equilibrium collision
of particles may well occur without a radiation field reaction thus
one cannot calculate the resultant field since energy transfer due to
particle collision prevents the return of particles to the mother
radiator.- Hide quoted text -

- Show quoted text -

what is a particle?

On 5 Mar, 10:04, Roy Lewallen wrote:
There's no direct way to measure the total power being radiated other
than sampling the field at many points in all directions and
integrating. "Reflected" power is not power that isn't transmitted. You
can find the power being applied to the antenna by subtracting the
"reverse" or "reflected" power from the "forward" power, but that tells
you nothing about what fraction is radiated and what fraction lost as heat.

Roy Lewallen, W7EL



I believe this to be untrue.
If an array is in equilibrium you have skin depth that give you a
resistance figure as well as d.c. resistance . You also know power
input thus all input and output power is therefore known. Pray tell
what energy cannot be accounted for?
Remember radiation does not begin to occur until the arbitary border
is punctured thus at that time it can be considered as output.
Movement of flux cannot begin until the clock starts
or time begins So now you have a beginning. An enclosed arbitary
border that containes energy and a end section that represents
radiation. You could also use the potential momentum theorem to
determine the exit proportions of electric and magnetic particles to
generate a electromagnetic field as well determining what particles
return to the radiator to serve in the formation of skin depth or
radiation resistance which serves to augument time changing current as
well as accounting for decay. All this requires a smattering of
understanding with respect to Einstein law of relativity which most
will probably tend to dismiss as hog wash especially those who view
the subject of static particles as being useless.
The answer Roy gave is only applicable when an array has parasitics
that either deflect or attract energy with respect to polarity( Walter
note the use of the word "polarity" with respect to antennas)
Ofcourse some will drop back to point out Pointings Vector so this
could be interesting.
If you want to disagree start off with a resonant dipole to ensure
applicability of ones auguments.

Art Unwin



Wayne wrote:
When the subject of antenna efficiency comes up, it often involves a
discussion of ground losses on verticals. What about, for example, a
dipole? Could one calculate "power out/power in" by measuring the VSWR and
declaring that everything not reflected was transmitted? It would seem more
accurate to actually measure power out and power in, but that introduces
inaccuracies by having to calibrate the setup. Thoughts?- Hide quoted text -

- Show quoted text -

"art" wrote in message
oups.com...
On 5 Mar, 10:04, Roy Lewallen wrote:
There's no direct way to measure the total power being radiated other
than sampling the field at many points in all directions and
integrating. "Reflected" power is not power that isn't transmitted. You
can find the power being applied to the antenna by subtracting the
"reverse" or "reflected" power from the "forward" power, but that tells
you nothing about what fraction is radiated and what fraction lost as
heat.

Roy Lewallen, W7EL



I believe this to be untrue.
If an array is in equilibrium you have skin depth that give you a
resistance figure as well as d.c. resistance . You also know power
input thus all input and output power is therefore known. Pray tell
what energy cannot be accounted for?
Remember radiation does not begin to occur until the arbitary border
is punctured thus at that time it can be considered as output.
Movement of flux cannot begin until the clock starts
or time begins So now you have a beginning. An enclosed arbitary
border that containes energy and a end section that represents
radiation. You could also use the potential momentum theorem to
determine the exit proportions of electric and magnetic particles to
generate a electromagnetic field as well determining what particles
return to the radiator to serve in the formation of skin depth or
radiation resistance which serves to augument time changing current as
well as accounting for decay. All this requires a smattering of
understanding with respect to Einstein law of relativity which most
will probably tend to dismiss as hog wash especially those who view
the subject of static particles as being useless.
The answer Roy gave is only applicable when an array has parasitics
that either deflect or attract energy with respect to polarity( Walter
note the use of the word "polarity" with respect to antennas)
Ofcourse some will drop back to point out Pointings Vector so this
could be interesting.
If you want to disagree start off with a resonant dipole to ensure
applicability of ones auguments.

Art Unwin



Art
Roy said "There's no direct way to measure" , what you are describing isnt a
direct way.

Jimmie

On 22 Apr, 18:18, "Jimmie D" wrote:
"art" wrote in message

oups.com...





On 5 Mar, 10:04, Roy Lewallen wrote:
There's no direct way to measure the total power being radiated other
than sampling the field at many points in all directions and
integrating. "Reflected" power is not power that isn't transmitted. You
can find the power being applied to the antenna by subtracting the
"reverse" or "reflected" power from the "forward" power, but that tells
you nothing about what fraction is radiated and what fraction lost as
heat.

Roy Lewallen, W7EL

I believe this to be untrue.
If an array is in equilibrium you have skin depth that give you a
resistance figure as well as d.c. resistance . You also know power
input thus all input and output power is therefore known. Pray tell
what energy cannot be accounted for?
Remember radiation does not begin to occur until the arbitary border
is punctured thus at that time it can be considered as output.
Movement of flux cannot begin until the clock starts
or time begins So now you have a beginning. An enclosed arbitary
border that containes energy and a end section that represents
radiation. You could also use the potential momentum theorem to
determine the exit proportions of electric and magnetic particles to
generate a electromagnetic field as well determining what particles
return to the radiator to serve in the formation of skin depth or
radiation resistance which serves to augument time changing current as
well as accounting for decay. All this requires a smattering of
understanding with respect to Einstein law of relativity which most
will probably tend to dismiss as hog wash especially those who view
the subject of static particles as being useless.
The answer Roy gave is only applicable when an array has parasitics
that either deflect or attract energy with respect to polarity( Walter
note the use of the word "polarity" with respect to antennas)
Ofcourse some will drop back to point out Pointings Vector so this
could be interesting.
If you want to disagree start off with a resonant dipole to ensure
applicability of ones auguments.

Art Unwin

Art
Roy said "There's no direct way to measure" , what you are describing isnt a
direct way.

Jimmie- Hide quoted text -

- Show quoted text -

I stand corrected. You cannot wet your finger and stick it up into the
air expecting that one could obtain a direct measure of radiation.
Most observant of you Jimmie
Art