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

Owen Duffy wrote:
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.

What is the hourly rental charge on a helicopter? :-)
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Owen Duffy wrote:

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.


What is the hourly rental charge on a helicopter? :-)

that's what balloons and blimps are for, or model airplanes..

Owen wrote:
"Is that true?"

Only for limited conditions. Owen is correct that enough samples must be
taken to catch all variations in signal strength, in elevation and
azimuth if total radiated power is to be determined.

A simple vertical antenna rests on the earth and has a null at its tip.
It is far from an isotropic radiator.

The sampling I described is done on AM broadcast antennas. Only the wave
which travels along the earth is usually of any interest. The minimum
distance from the broadcast antenna for any field strength measurements
is usually one mile to be sure the far field is being measured.

The FCC`s ground wave intensity charts assume that if you have ground
conductivity such as sea water, a certain power and antenna efficiency
deliver 100 millivolts per meter at a distance of one mile from the
antenna. Field intensity in millivolts declines linearly with distance
so that at 10 miles you might have 10 mv/m if you have 100 mv/m at one
mile.Power is proportional to the square of the voltage.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

The FCC`s ground wave intensity charts assume that if you have ground
conductivity such as sea water, a certain power and antenna efficiency
deliver 100 millivolts per meter at a distance of one mile from the
antenna. Field intensity in millivolts declines linearly with distance
so that at 10 miles you might have 10 mv/m if you have 100 mv/m at one
mile. . . .

The attenuation of the ground wave is the same as the free space
attenuation?

Roy Lewallen, W7EL

Roy Lewallen, W7EL wrote:
"The attenuation of the ground wave is the same as the free space
attenuation?"

No, it is greater. The attenuation of the ground wave is more rapid
because the soil has resistance and imposes a loss on the passing wave.

In free space, there is no ground loss and the signal loss is from the
spreading of the signal which reduces its density. It declines 6 dB (its
power is quartered) every time the distance from the transmitter
doubles, that is its volts per meter or signal strength is cut in half.

Ground waves decrease more rapidly with distance from the transmitter
due to imperfect conductivity in the earth`s surface. Loss goes up with
the operating frequency and with higher resistivity in the soil.

The FCC`s ground-wave field intensity charts cover limited frequency
ranges and each curve is for a specific soil conductivity.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
Roy Lewallen, W7EL wrote:
"The attenuation of the ground wave is the same as the free space
attenuation?"

No, it is greater. The attenuation of the ground wave is more rapid
because the soil has resistance and imposes a loss on the passing wave.
. . .

I should hope so. That's why your statement that the field strength
decreases as the inverse of the distance in the context of AM broadcast
measurements at ground level was puzzling.

Roy Lewallen, W7EL

On Mon, 23 Apr 2007 17:30:20 -0700, Roy Lewallen
wrote:

Richard Harrison wrote:
Roy Lewallen, W7EL wrote:
"The attenuation of the ground wave is the same as the free space
attenuation?"

No, it is greater. The attenuation of the ground wave is more rapid
because the soil has resistance and imposes a loss on the passing wave.
. . .

I should hope so. That's why your statement that the field strength
decreases as the inverse of the distance in the context of AM broadcast
measurements at ground level was puzzling.

Richard had been reciting decade-for-decade declines of FS faithfully
from the FCC charts -over seawater- which he had specifically called
out.

"Faithfully" except for an inadvertent substitution of decade miles
for decade kilometers distance.

73's
Richard Clark, KB7QHC