Reg, G4FGQ wrote:
"What do photons have to do with winning a contest?"

I don`t know. But, B.Whitfield Griffith, Jr. has some observations in
"Rado-Eledtronic Transmission Fundamentals" that may be a useful check
on your computations. I expect he checked, rechecked, then checked again
before publication. He put a transmitter on an elemental antenna but it
would work the same in reverse. From page 325:

"An Elemental Antenna

Since the length of an antenna is commonly expressed in electrical
degrees and since this allows the convenient use of trigonometric tables
in computing the ratios of currents in various parts of an antenna, let
us choose as our elemental antenna a piece of wire which is 1-degree in
length and in which the current is constant from one end to the other.We
shall first assume that this elemental antenna is located far out in
space, so that its field is not disturbed by reflections from the
surface of the earth or from any other object. This, of course, is a
most improbable set of conditions, but we can certainly imagine that we
have a situation such as this and compute from the field equations its
electromagnetic result.

These computations will show, first of all, that the maximum field
intensity will be produced in the directions which are at right angles
to the direction of current flow. This is a reasonable result, since the
magnetic field which is produced by the current surrounds the wire in
concentric rings and thus gives rise to a radiation field which moves
outward at right angles to the wire. As a matter of fact, the field
intensity, measured according to our standard procedure at a distance of
1 mile in any direction from the radiating element, will be found to be
proportional to the sine of the angle between the direction of the
current flow and the direction in which the measurement is taken. If we
represent the field intensity at 1 mile in any direction by the length
of a vector starting at the center of the element and extending in that
direction, the tips of the vectors will mark out the radiation pattern
of the antenna element. A cross section of the entire radiation pattern
of this element is shown in Fig. 39-2; the entire pattern would be
obtained by rotating the figure about the axis of the antenna element.

But this pattern tells us only the relative signal strength in various
directions; it is a normalized pattern, with the intensity in the
direction of maximum radiation being considered simply as unity. We need
much more information than this; we must know the relationship between
the current and the actual value of the field it produces. Further
computation from the field equations gives this relalationship; we find
that a current of 1 amp flowing in the antnna element will produce a
field intensity of 0.3253 mv/m at a distance of 1 mile iin the direction
of maximum radiation. We have said nothing about the frequency, and we
do not need to; as long as the wavelength is shorter than 1 mile, so
that there are no serious induction-ffield effects to upset our
calculations, this figure will be correct at any frequencyfor which the
length of the element is 1/360 wavelength.

The field intensity of the elemental antenna is directly proportional to
the current. Therefore, if the current in the element is 15 amp, the
field intensity will be 15 X 0.3253, or 4.8795, mv/m at 1 mile.
Similarly, the field intensity is directly proportional to the length of
the element; an element which is 2-degrees in length, carrying a current
of 1 amp, will thus produce a maximum field intensity of 0.6506 mv/m at
1 mile.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

I don`t know. But, B.Whitfield Griffith, Jr. has some observations in
"Rado-Eledtronic Transmission Fundamentals" that may be a useful check
on your computations. I expect he checked, rechecked, then checked again
before publication. He put a transmitter on an elemental antenna but it
would work the same in reverse. From page 325:

Yes, it should.

. . .
. . . We need
much more information than this; we must know the relationship between
the current and the actual value of the field it produces. Further
computation from the field equations gives this relalationship; we find
that a current of 1 amp flowing in the antnna element will produce a
field intensity of 0.3253 mv/m at a distance of 1 mile iin the direction
of maximum radiation. . .

The field intensity of the elemental antenna is directly proportional to
the current. Therefore, if the current in the element is 15 amp, the
field intensity will be 15 X 0.3253, or 4.8795, mv/m at 1 mile.
Similarly, the field intensity is directly proportional to the length of
the element; an element which is 2-degrees in length, carrying a current
of 1 amp, will thus produce a maximum field intensity of 0.6506 mv/m at
1 mile.

A short dipole antenna with 1 amp of current at the center has an
average of 0.5 amp of current along the whole length. So it should be
obvious from the above analysis that the field from a dipole is half the
field from the elemental antenna with uniform current which the author
is discussing. Or, instead of just taking the average, you can integrate
I * delta L to get the same result.

Roy Lewallen, W7EL

Richard Harrison wrote:
Reg, G4FGQ wrote:
"What do photons have to do with winning a contest?"

I don`t know. But, B.Whitfield Griffith, Jr. has some observations in
"Rado-Eledtronic Transmission Fundamentals" that may be a useful check
on your computations. I expect he checked, rechecked, then checked again
before publication. He put a transmitter on an elemental antenna but it
would work the same in reverse. From page 325:

"An Elemental Antenna

Since the length of an antenna is commonly expressed in electrical
degrees and since this allows the convenient use of trigonometric tables
in computing the ratios of currents in various parts of an antenna, let
us choose as our elemental antenna a piece of wire which is 1-degree in
length and in which the current is constant from one end to the other.We
shall first assume that this elemental antenna is located far out in
space, so that its field is not disturbed by reflections from the
surface of the earth or from any other object. This, of course, is a
most improbable set of conditions, but we can certainly imagine that we
have a situation such as this and compute from the field equations its
electromagnetic result.

These computations will show, first of all, that the maximum field
intensity will be produced in the directions which are at right angles
to the direction of current flow. This is a reasonable result, since the
magnetic field which is produced by the current surrounds the wire in
concentric rings and thus gives rise to a radiation field which moves
outward at right angles to the wire. As a matter of fact, the field
intensity, measured according to our standard procedure at a distance of
1 mile in any direction from the radiating element, will be found to be
proportional to the sine of the angle between the direction of the
current flow and the direction in which the measurement is taken. If we
represent the field intensity at 1 mile in any direction by the length
of a vector starting at the center of the element and extending in that
direction, the tips of the vectors will mark out the radiation pattern
of the antenna element. A cross section of the entire radiation pattern
of this element is shown in Fig. 39-2; the entire pattern would be
obtained by rotating the figure about the axis of the antenna element.

But this pattern tells us only the relative signal strength in various
directions; it is a normalized pattern, with the intensity in the
direction of maximum radiation being considered simply as unity. We need
much more information than this; we must know the relationship between
the current and the actual value of the field it produces. Further
computation from the field equations gives this relalationship; we find
that a current of 1 amp flowing in the antnna element will produce a
field intensity of 0.3253 mv/m at a distance of 1 mile iin the direction
of maximum radiation. We have said nothing about the frequency, and we
do not need to; as long as the wavelength is shorter than 1 mile, so
that there are no serious induction-ffield effects to upset our
calculations, this figure will be correct at any frequencyfor which the
length of the element is 1/360 wavelength.

The field intensity of the elemental antenna is directly proportional to
the current. Therefore, if the current in the element is 15 amp, the
field intensity will be 15 X 0.3253, or 4.8795, mv/m at 1 mile.
Similarly, the field intensity is directly proportional to the length of
the element; an element which is 2-degrees in length, carrying a current
of 1 amp, will thus produce a maximum field intensity of 0.6506 mv/m at
1 mile.

Best regards, Richard Harrison, KB5WZI



How did you get .3253 mv/m at 1 miles?

Dave

How did you quote damned near 75 lines of text to ask a 9 word question?

Jim


"EE123" wrote in message
oups.com...

Best regards, Richard Harrison, KB5WZI



How did you get .3253 mv/m at 1 miles?

Dave

EEE 123 wrote:
"How do you get .3253 mv/m at 1 mile?"

The formula to calculate the field strength produced by an dlementaarry
doublet is difficult for me to reproduce with my keyboard. It is found
on page 770 of Terman`s 1943 "Radio Sngineers` Handbook. In my prior
posting, Griffith hhas done the work for us.

Best regards, Richard Harrison, KB5WZI

Reg, G4FGQ wrote:
"Just a number please."

Given 1 volt per m as the field strength, and a 1-m antenna parallel to
the electric vector of the wave, the open-circuit voltage at the end of
the wire is 1 volt. The best you can get across the receiver input is
0.5 volt when there is a conjugate match between the receiver and the
antennna.

Most of the explanations are irrelevant when the field strength is
specified at the antenna.

Terman preached scientific gospel. He had proof to back what he said. In
Terman`s 1943 "Radio Engineers` Handbook" he wrote:

"The strength of a radio wave is expressed in terms of the voltage
stress produced in space by the electric field of the wave, and is
usually expressed in either millivolts or microvolts stress per meter.
The stress expressed this way is exactly the same voltage that the
magnetic flux of the wave induces in a conductor one meter long when rhe
wave sweeps across the conductor with the velocity of light." This is
found on page 770.

There is no qualification or equivocation. Terman also posts the same
statement with no change in substance on page 2 of his 1955 edition of
"Electronic and Radio Engineerinng". It means what it plainly says.

Best regards, Richard Harrison, KB5WZI

Richard (Harrison)

I'm sorry Richard, you have not anwered my question. I will repeat it.

"What is the voltage measured between the bottom end and ground of a
1metre high vertical antenna when the field strength is 1 volt per
metre?"

Assume a perfect ground and antenna height is much less than
1/4-wavelength.

A measured voltage is always relative to or is between TWO specified
points.
----
Reg, G4FGQ

Reg Edwards wrote:
. . .
A measured voltage is always relative to or is between TWO specified
points.

And when measuring in the presence of a time-varying H field, those two
specified points have to be physically very close together.

Roy Lewallen, W7EL

Richard Harrison wrote:
Reg, G4FGQ wrote:
"Just a number please."

Given 1 volt per m as the field strength, and a 1-m antenna parallel to
the electric vector of the wave, the open-circuit voltage at the end of
the wire is 1 volt.

Relative to what? The other terminal has to be extremely close to the
end of the wire in order for the voltage to be single valued.

The best you can get across the receiver input is
0.5 volt when there is a conjugate match between the receiver and the
antennna.

Sorry, that's not just a little wrong, it's wrong by orders of
magnitude. For example, a 1 meter long 10 mm diameter dipole, terminated
in the complex conjugate of its self impedance (load Z = 0.8855 + j6030
ohms), in a 1 V/m field, has about 1667 volts across the load. Hardly a
half volt!

. . .

Roy Lewallen, W7EL

Roy Lewallen wrote:
. . .
Sorry, that's not just a little wrong, it's wrong by orders of
magnitude. For example, a 1 meter long 10 mm diameter dipole, terminated
in the complex conjugate of its self impedance (load Z = 0.8855 + j6030
ohms), in a 1 V/m field, has about 1667 volts across the load. Hardly a
half volt!

For the casual reader, I should emphasize that this analysis, like all
the others on this thread, assumes zero loss. The conditions I described
could never be achieved in real life because of unavoidable loss. But
it's important to first understand how a lossless system behaves before
we add the complicating factor of loss.

Roy Lewallen, W7EL