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.

"Wayne" wrote in
news:7L%Gh.506$Ih.268@trnddc02:


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

You probably mean 1 less the magnitude of the reflection coefficient
squared, which I would write as 1-|Gamma|^2 or 1-rho^2. Your implication
is that "reflected power" is (necessarily) lost, that is wrong.

If you do not have a transmission line, why are you trying to use
transmission line concepts to solve the problem? The power delivered to
the antenna is the real part of V^2/Z where V is the feedpoint voltage
and Z is the feedpoint impedance.

resistance of 1 ohm in the dipole conductors the efficiency I
calculate is about 98.6% (72/73).

The number you guess could be reasonable, and if it is truly the
equivalent R at the feedpoint (rather than R/unit length * length), it
does impact the antenna efficiency in the way you calculate.

Owen

Owen Duffy wrote in news:Xns98EB57182E4B6nonenowhere@
61.9.191.5:

the antenna is the real part of V^2/Z where V is the feedpoint voltage

the antenna is the real part of V^2/Z where V is the feedpoint RMS voltage

Wayne wrote:

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.

There is no relationship between SWR and efficiency.

Roy Lewallen, W7EL

"Roy Lewallen" wrote in message
...
Wayne wrote:

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.

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.

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, 07:54, "Yuri Blanarovich" wrote:
"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- Hide quoted text -

- Show quoted text -

I don't know which patent you are refering to but I will examine your
particular statement.
Length alone is not an indication of reflective or atractive atributes
especially when viewed in combination of nearby elements. When
elements are in close association it is the cumulative effect that
determines whether it attracts or repels action from the driven
element and where the increased diametrical action of two intertwined
parasitic elements come to the fore. Knowing that english is not your
first language I must point out to you that the term reflector and
director are names applied to the function of a parasitic and not
necessarily a
physical description of length.
Art