On 27 Mar 2007 16:39:52 -0700, "art" wrote:

On 27 Mar, 15:12, (Richard Harrison) wrote:
Art wrote:

"I have his (Kraus`) second edition (Antennas) and I find no mention of
radiation from the beginning where current is applied onward."

I think I have that edition too. If you review the chapter on "Point
Sources" you`ll find: power patterns, a power theorem and its
application to isotropic sources, rediation intensity, source with
hemispheric power pattern, unidirectional cosine power pattern, etc.,
etc..

The new, now available 3rd ed. of "Antennas" by Kraus, Marhefka, and a
host of others is greatly expanded and improved. It is worth the
investment.

Being uncertain of what Art really wants, doesn`t stop me from advising
him to start by having a look at the famous Sommerfeld formula on page
804 of Terman`s 1955 opus.
It predicts 1 kilowatt will produce 186 mv per m at a distance of 1 mile
from a short vertical transmitting antenna given a certain ground
conductivity and other conditions.

Best regards, Richard Harrison, KB5WZI

Richard, I have lots of books but as yet have not found the answer
even tho many have posted none appear to really have an answer other
than to throw stones. You apparently have found the answer! Could you
quote from the books that you are refering to the angle of radiation
relative to the radiator, thats it ? If you can't understand that then
relay to me the angle of a radiation front relative to a radiator, I'm
sure some other people are interested in what you found. Even better,
let me know the TOA of a dipole in free space and how much it varies
to that of the same dipole over a perfect ground. Use a computor
program if you like, anything that sheds light on the matter . The
books say that a horizontal "v" antenna should be tipped for max gain,
doesn't that raise your interest about the reasoning and mathematics
behind this? Jimmie D asked me to state this angle but I have only a
expensive computor program that doesn't give the math with the answer.
Please read off the angle and the specifics so we all can move on, I
don't want a 160 thread postings some thrust upon Walt
Art

The take-off angle of a dipole in free space? The angle with respect to what?

Walt, W2DU

On 31 Mar, 14:36, Walter Maxwell wrote:
On 27 Mar 2007 16:39:52 -0700, "art" wrote:





On 27 Mar, 15:12, (Richard Harrison) wrote:
Art wrote:

"I have his (Kraus`) second edition (Antennas) and I find no mention of
radiation from the beginning where current is applied onward."

I think I have that edition too. If you review the chapter on "Point
Sources" you`ll find: power patterns, a power theorem and its
application to isotropic sources, rediation intensity, source with
hemispheric power pattern, unidirectional cosine power pattern, etc.,
etc..

The new, now available 3rd ed. of "Antennas" by Kraus, Marhefka, and a
host of others is greatly expanded and improved. It is worth the
investment.

Being uncertain of what Art really wants, doesn`t stop me from advising
him to start by having a look at the famous Sommerfeld formula on page
804 of Terman`s 1955 opus.
It predicts 1 kilowatt will produce 186 mv per m at a distance of 1 mile
from a short vertical transmitting antenna given a certain ground
conductivity and other conditions.

Best regards, Richard Harrison, KB5WZI

Richard, I have lots of books but as yet have not found the answer
even tho many have posted none appear to really have an answer other
than to throw stones. You apparently have found the answer! Could you
quote from the books that you are refering to the angle of radiation
relative to the radiator, thats it ? If you can't understand that then
relay to me the angle of a radiation front relative to a radiator, I'm
sure some other people are interested in what you found. Even better,
let me know the TOA of a dipole in free space and how much it varies
to that of the same dipole over a perfect ground. Use a computor
program if you like, anything that sheds light on the matter . The
books say that a horizontal "v" antenna should be tipped for max gain,
doesn't that raise your interest about the reasoning and mathematics
behind this? Jimmie D asked me to state this angle but I have only a
expensive computor program that doesn't give the math with the answer.
Please read off the angle and the specifics so we all can move on, I
don't want a 160 thread postings some thrust upon Walt
Art

The take-off angle of a dipole in free space? The angle with respect to what?

Walt, W2DU- Hide quoted text -

- Show quoted text -

Always zero Walt since it takes two dipoles to tango.

Art wrote:
"Even better, let me know the TOA of a dipole in free space and how much
it varies to that of the same dipole over perfect ground."

If possible, look at Terman`s1955 opus page 882:
"Effect of Ground on the Directional Pattern of Ungrounded Antennas -
Image Antennas."

In free space, no reflecting surface is nearby to distort the pattern of
a radiator.

Terman gives the familiar radiation of a 1/2-wave dipole in space in
Fig. 23-4(a) on page 867.

Terman says on page 883:
"For purposes of calculation, it is convenient to consider that the
reflected wave is generated, not by reflection, but rather by an "image"
antenna located below the surface of the ground."

The summation of the direct and reflected waves from a horoizontal wire
above the earth often has serious consequences as Terman notes on page
885:
"Consequently, to obtain strong radiation in directions approaching the
horizontal using a horizontally polarized radiating system, it is
necessary that the height of the antenna above the earth be in the order
of one wavelength or more."

Terman shows the vertical radiation patterns for a horizontal wire at
various heights above the earth in Fig. 23-21 on page 884. Note that
half-wave elevation concentrates most energy into a good elevation angle
for sky wave reflection at some frequencies and distances.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

[snip]

Terman says on page 883:
"For purposes of calculation, it is convenient to consider


Richard,

Great quote!

If people listened to the words of the master, there would be far fewer
arguments on RRAA.

73,
Gene
W4SZ

Art wrote:
"Even better, let me know the TOA of a dipole in free space and how much
it varies to that of the same dipole over perfect ground."

If possible, look at Terman`s1955 opus page 882:
"Effect of Ground on the Directional Pattern of Ungrounded Antennas -
Image Antennas."

In free space, no reflecting surface is nearby to distort the pattern of
a radiator.

Terman gives the familiar radiation of a 1/2-wave dipole in space in
Fig. 23-4(a) on page 867.

Terman says on page 883:
"For purposes of calculation, it is convenient to consider that the
reflected wave is generated, not by reflection, but rather by an "image"
antenna located below the surface of the ground."

The summation of the direct and reflected waves from a horoizontal wire
above the earth often has serious consequences as Terman notes on page
885:
"Consequently, to obtain strong radiation in directions approaching the
horizontal using a horizontally polarized radiating system, it is
necessary that the height of the antenna above the earth be in the order
of one wavelength or more."

Terman shows the vertical radiation patterns for a horizontal wire at
various heights above the earth in Fig. 23-21 on page 884. Note that
half-wave elevation concentrates most energy into a good elevation angle
for sky wave reflection at some frequencies and distances.

Best regards, Richard Harrison, KB5WZI

On 1 Apr, 09:34, (Richard Harrison) wrote:
Art wrote:

"Even better, let me know the TOA of a dipole in free space and how much
it varies to that of the same dipole over perfect ground."

If possible, look at Terman`s1955 opus page 882:
"Effect of Ground on the Directional Pattern of Ungrounded Antennas -
Image Antennas."

In free space, no reflecting surface is nearby to distort the pattern of
a radiator.

Terman gives the familiar radiation of a 1/2-wave dipole in space in
Fig. 23-4(a) on page 867.

Terman says on page 883:
"For purposes of calculation, it is convenient to consider that the
reflected wave is generated, not by reflection, but rather by an "image"
antenna located below the surface of the ground."

The summation of the direct and reflected waves from a horoizontal wire
above the earth often has serious consequences as Terman notes on page
885:
"Consequently, to obtain strong radiation in directions approaching the
horizontal using a horizontally polarized radiating system, it is
necessary that the height of the antenna above the earth be in the order
of one wavelength or more."

Terman shows the vertical radiation patterns for a horizontal wire at
various heights above the earth in Fig. 23-21 on page 884. Note that
half-wave elevation concentrates most energy into a good elevation angle
for sky wave reflection at some frequencies and distances.

Best regards, Richard Harrison, KB5WZI

Very good, all accepted, I am now in a position to ask the real
question of "why" where the starting point is acceptable to all.
..
I model a dipole in free space to obtain its impedance, this dipole
was resonant at right angles to the ground and resting on the ground.
I also modelled another dipole over the ground and tipped it over
about 5% of its WL but letting the length grow until the dipole was
resonant over ground ( ground means perfect ground as per Mininec) I
then placed this tipped dipole in free space and noted its impedance.
In all cases the impedance in free space was the same though not
resonant! In all cases over ground the dipoles were resonant. Why were
both dipoles over ground resonant? Why did the program model the
dipole at a tipped angle for resonance when given total freedom
instead of modeling a dipole at right angles to ground?
Regards
Art

On 31 Mar, 14:36, Walter Maxwell wrote:
On 27 Mar 2007 16:39:52 -0700, "art" wrote:





On 27 Mar, 15:12, (Richard Harrison) wrote:
Art wrote:

"I have his (Kraus`) second edition (Antennas) and I find no mention of
radiation from the beginning where current is applied onward."

I think I have that edition too. If you review the chapter on "Point
Sources" you`ll find: power patterns, a power theorem and its
application to isotropic sources, rediation intensity, source with
hemispheric power pattern, unidirectional cosine power pattern, etc.,
etc..

The new, now available 3rd ed. of "Antennas" by Kraus, Marhefka, and a
host of others is greatly expanded and improved. It is worth the
investment.

Being uncertain of what Art really wants, doesn`t stop me from advising
him to start by having a look at the famous Sommerfeld formula on page
804 of Terman`s 1955 opus.
It predicts 1 kilowatt will produce 186 mv per m at a distance of 1 mile
from a short vertical transmitting antenna given a certain ground
conductivity and other conditions.

Best regards, Richard Harrison, KB5WZI

Richard, I have lots of books but as yet have not found the answer
even tho many have posted none appear to really have an answer other
than to throw stones. You apparently have found the answer! Could you
quote from the books that you are refering to the angle of radiation
relative to the radiator, thats it ? If you can't understand that then
relay to me the angle of a radiation front relative to a radiator, I'm
sure some other people are interested in what you found. Even better,
let me know the TOA of a dipole in free space and how much it varies
to that of the same dipole over a perfect ground. Use a computor
program if you like, anything that sheds light on the matter . The
books say that a horizontal "v" antenna should be tipped for max gain,
doesn't that raise your interest about the reasoning and mathematics
behind this? Jimmie D asked me to state this angle but I have only a
expensive computor program that doesn't give the math with the answer.
Please read off the angle and the specifics so we all can move on, I
don't want a 160 thread postings some thrust upon Walt
Art

The take-off angle of a dipole in free space? The angle with respect to what?

Walt, W2DU- Hide quoted text -

- Show quoted text -

Perhaps Walter you are really serious with the above question!!!!!!
You have two reference planes, the radiator and the plane of maximum
gain(ie. for vertical polarization) Therefor 90 degrees minus the
angle between both planes equal "take off angle".
Regards
Art

art wrote:
On 31 Mar, 14:36, Walter Maxwell wrote:
On 27 Mar 2007 16:39:52 -0700, "art" wrote:





On 27 Mar, 15:12, (Richard Harrison) wrote:
Art wrote:

"I have his (Kraus`) second edition (Antennas) and I find no mention of
radiation from the beginning where current is applied onward."

I think I have that edition too. If you review the chapter on "Point
Sources" you`ll find: power patterns, a power theorem and its
application to isotropic sources, rediation intensity, source with
hemispheric power pattern, unidirectional cosine power pattern, etc.,
etc..

The new, now available 3rd ed. of "Antennas" by Kraus, Marhefka, and a
host of others is greatly expanded and improved. It is worth the
investment.

Being uncertain of what Art really wants, doesn`t stop me from advising
him to start by having a look at the famous Sommerfeld formula on page
804 of Terman`s 1955 opus.
It predicts 1 kilowatt will produce 186 mv per m at a distance of 1 mile
from a short vertical transmitting antenna given a certain ground
conductivity and other conditions.

Best regards, Richard Harrison, KB5WZI

Richard, I have lots of books but as yet have not found the answer
even tho many have posted none appear to really have an answer other
than to throw stones. You apparently have found the answer! Could you
quote from the books that you are refering to the angle of radiation
relative to the radiator, thats it ? If you can't understand that then
relay to me the angle of a radiation front relative to a radiator, I'm
sure some other people are interested in what you found. Even better,
let me know the TOA of a dipole in free space and how much it varies
to that of the same dipole over a perfect ground. Use a computor
program if you like, anything that sheds light on the matter . The
books say that a horizontal "v" antenna should be tipped for max gain,
doesn't that raise your interest about the reasoning and mathematics
behind this? Jimmie D asked me to state this angle but I have only a
expensive computor program that doesn't give the math with the answer.
Please read off the angle and the specifics so we all can move on, I
don't want a 160 thread postings some thrust upon Walt
Art

The take-off angle of a dipole in free space? The angle with respect to what?

Walt, W2DU- Hide quoted text -

- Show quoted text -

Perhaps Walter you are really serious with the above question!!!!!!
You have two reference planes, the radiator and the plane of maximum
gain(ie. for vertical polarization) Therefor 90 degrees minus the
angle between both planes equal "take off angle".
Regards
Art

Of course he is serious.

There are no planes in free space for a dipole.

The dipole defines a line, not a plane.

Since the pattern of a dipole in free space is a circle viewed from
the end, and a cardioid from the side, there is no plane there either.

You are plain confused about planes.


--
Jim Pennino

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