I should elaborate a little.

The average gain is the ratio of the total power in all directions at a
great distance (beyond the point where the surface wave has decayed to a
negligible value) to the power into the antenna from all the sources.
(There's a factor of two also involved when using a ground plane with
NEC but not with EZNEC.) So the average gain is the efficiency if you
consider ground reflection and the decay of the surface wave to be part
of the loss.

Roy Lewallen, W7EL

Roy Lewallen wrote:
Frank wrote:

Correct Owen. NEC shows 97.3% for free space, and 100 %,
as expected, with perfect conductors. Certainly the loss does
include absorption of the reflected rays. As mentioned before, in
previous threads, it is very tedious to determine what percentage
of the "Loss" is due to ground wave radiation. One of these
days I will write the code necessary to compute the actual
TRP including ground wave.

That capability is already built into NEC, as the average gain calculation.

Roy Lewallen, W7EL

On 5 Mar, 20:41, Roy Lewallen wrote:
I should elaborate a little.

The average gain is the ratio of the total power in all directions at a
great distance (beyond the point where the surface wave has decayed to a
negligible value) to the power into the antenna from all the sources.
(There's a factor of two also involved when using a ground plane with
NEC but not with EZNEC.) So the average gain is the efficiency if you
consider ground reflection and the decay of the surface wave to be part
of the loss.

Roy Lewallen, W7EL

Lets have another look at this.

Roy inferred that the radiation field volume is the total useful
output

He then goes on to say that average gain what ever that means relative
to the final radiation field is the efficiency.
He also adds a condition relative to the definition of efficiency that
this is only true IF you count ground reflection and the decay of the
surface wave to be part of the loss


Hmmm I don't think anybody would deny that surface wave represents a
loss
relative to usefull work though some might say it contributes to
current flow, but why single out ground reflection as a loss since
that can be useful?
So Roy is classifying efficiency as something he considers usefull
and ground reflection is not usefull. He also throws average gain into
the equation without providing a definition of average
gain ( like gain is an advance over something he doesn't want to
state)
Jimminy cricket
I agree that you need to provide more elaboration

Why is it you can't say the useful result of what you provided is the
radiation volume where efficiency is useful output over input times
100?
Why does one have to place conditions on :

efficiency = useful output/ actual input x 100 ?

Seems like efficiency in radiation is not the same as efficiencies
in other sciences. Possibly a definition supplied by a related
commitee
solely for their own interpretation even though it is not in
accordance with other diciplines. Also possibly based on the number of
books on a particular shelf.And then the following week they placed
conditions to clarify what efficiency includes and does not include
such as certain portions of radiation, possibly with a different color
to the norm
phew




Roy Lewallen wrote:
Frank wrote:

Correct Owen. NEC shows 97.3% for free space, and 100 %,
as expected, with perfect conductors. Certainly the loss does
include absorption of the reflected rays. As mentioned before, in
previous threads, it is very tedious to determine what percentage
of the "Loss" is due to ground wave radiation. One of these
days I will write the code necessary to compute the actual
TRP including ground wave.

That capability is already built into NEC, as the average gain calculation.

Roy Lewallen, W7EL- Hide quoted text -

- Show quoted text -

"Roy Lewallen" wrote in message
...
I should elaborate a little.

The average gain is the ratio of the total power in all directions at a
great distance (beyond the point where the surface wave has decayed to a
negligible value) to the power into the antenna from all the sources.
(There's a factor of two also involved when using a ground plane with NEC
but not with EZNEC.) So the average gain is the efficiency if you consider
ground reflection and the decay of the surface wave to be part of the
loss.

Roy Lewallen, W7EL

I was using average gain for my calculation of efficiency; i.e.
XNDA = 1001, or 1002. I have also been considering
the factor of "2" in the results. To be accurate, and to
determine the radiation resistance of a structure, you do
need to include the surface wave. The only way I
can think of doing this is to sum "E X H" close enough
to the radiating structure so as to include all its
elements. At the moment I am using Excel
to compute the Poynting vector.
Even then, there is some question as
to how much ground absorption effects the results
between antenna and the hemispherical radius
of computation. I have noticed some weird results
if you get too close to the ends of a buried radial system.

Regards,

Frank (VE6CB)

"art" wrote in message
oups.com...
On 5 Mar, 20:41, Roy Lewallen wrote:
I should elaborate a little.

The average gain is the ratio of the total power in all directions at a
great distance (beyond the point where the surface wave has decayed to a
negligible value) to the power into the antenna from all the sources.
(There's a factor of two also involved when using a ground plane with
NEC but not with EZNEC.) So the average gain is the efficiency if you
consider ground reflection and the decay of the surface wave to be part
of the loss.

Roy Lewallen, W7EL

Lets have another look at this.

Roy inferred that the radiation field volume is the total useful
output

He then goes on to say that average gain what ever that means relative
to the final radiation field is the efficiency.
He also adds a condition relative to the definition of efficiency that
this is only true IF you count ground reflection and the decay of the
surface wave to be part of the loss


Hmmm I don't think anybody would deny that surface wave represents a
loss
relative to usefull work though some might say it contributes to
current flow, but why single out ground reflection as a loss since
that can be useful?
So Roy is classifying efficiency as something he considers usefull
and ground reflection is not usefull. He also throws average gain into
the equation without providing a definition of average
gain ( like gain is an advance over something he doesn't want to
state)
Jimminy cricket
I agree that you need to provide more elaboration

Why is it you can't say the useful result of what you provided is the
radiation volume where efficiency is useful output over input times
100?
Why does one have to place conditions on :

efficiency = useful output/ actual input x 100 ?

Seems like efficiency in radiation is not the same as efficiencies
in other sciences. Possibly a definition supplied by a related
commitee
solely for their own interpretation even though it is not in
accordance with other diciplines. Also possibly based on the number of
books on a particular shelf.And then the following week they placed
conditions to clarify what efficiency includes and does not include
such as certain portions of radiation, possibly with a different color
to the norm
phew

"Ground reflection loss" is probably a more precise term.

Frank

Frank wrote:
On 5 Mar, 20:41, Roy Lewallen wrote:
I should elaborate a little.

The average gain is the ratio of the total power in all directions at a
great distance (beyond the point where the surface wave has decayed to a
negligible value) to the power into the antenna from all the sources.
(There's a factor of two also involved when using a ground plane with
NEC but not with EZNEC.) So the average gain is the efficiency if you
consider ground reflection and the decay of the surface wave to be part
of the loss.

"Ground reflection loss" is probably a more precise term.

Yes, that is what I meant and what I should have said, instead of just
"ground reflection". Thanks for the correction.

Roy Lewallen, W7EL

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

Jimmie D reported Roy Lewallen to write:
"There is no direct way to measure the total power being radiated other
than sampling the field at many points in all directions and
integrating."

That sounds right to me. An approximation is sometimes made by taking 36
samples of field strength in volts per meter at 10-fegrees of azimuth
intervals at the same distance from the central antenna system. Each of
these sample values is squared and the sum of these squared samples is
divided by 36, the number of samples, to get their average. The square
poot of this quotient is then the average field strength at that
distance from the antenna. A true average signal strength should be the
same as the value an isotropic antenna would radiate at a given
distance.

Knowing the field strength, one could calculate the watts per square
meter of the envelope of radiation at a given distance and total the
watts per square meter of all the squares to get the total power being
radiated.

Since 1960, I`ve used the Bird wattmeter satisfactorily to get the total
power being delivered by the transmitter and radiated by the antenna. It
should be the same if the transmission line and antenna have low losses.
It is simply the difference between the forward power indication and the
reverse power indication. Many lines and antennas have very high
efficiencies.

Best regards, Richard Harrison, KB5WZI

(Richard Harrison) wrote in news:10571-
:

Jimmie D reported Roy Lewallen to write:
"There is no direct way to measure the total power being radiated other
than sampling the field at many points in all directions and
integrating."

That sounds right to me. An approximation is sometimes made by taking
36
samples of field strength in volts per meter at 10-fegrees of azimuth
intervals at the same distance from the central antenna system. Each of
these sample values is squared and the sum of these squared samples is
divided by 36, the number of samples, to get their average. The square
poot of this quotient is then the average field strength at that
distance from the antenna. A true average signal strength should be the
same as the value an isotropic antenna would radiate at a given
distance.

Is that true?

Firstly you seem to assume that your 36 samples around the azimut circle
adequately fulfill Roy's "sampling the field at many points in all
directions", surely he mean't all elevation angles as well as all azimuth
angles.

Secondly, your suggestion that the average field strength (presumably for
a 100% efficient antenna) at zero elevation is the same as for an
isotropic antenna at the same distance seems to preclude the antenna
having directivity in the elevation dimension.

Owen