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

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?

On 5 Mar, 09:04, 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



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 -

If you adopt my expansion of gauss it can be calculated.
If you look at the example in Chapter 21 of the Rutgers book on fields
on the net maybe, but maybe, you will think a bit different but I
expect you to fight off the idea of change
Art
Art

"art" wrote in message
oups.com...
On 5 Mar, 09:04, 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



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 -

If you adopt my expansion of gauss it can be calculated.
If you look at the example in Chapter 21 of the Rutgers book on fields
on the net maybe, but maybe, you will think a bit different but I
expect you to fight off the idea of change
Art

Art--
Thanks for your previous reponse. I don't know if the comment above is
directed to me or Roy. But I can tell you that I am not the person who is
going to understand an expansion of gauss
Wayne

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.