Gene Fuller wrote:
Cecil Moore wrote:
Once again, you have refused to answer a simple question.
If E and H are not zero, how can ExH be zero?

What is this? The RRAA version of "When did you stop beating your wife?"

No, this is the RRAA version of someone unwilling to accept
technical facts. Your name is Legion.
--
73, Cecil http://www.w5dxp.com

Jim Kelley wrote:
You fail to explain how any of that makes energy unavailable for
"radiation purposes".

This is not rocket science. If the energy is being used to
transform impedances, it is obviously unavailable to be lost
as radiation. If all the forward energy is radiated by the
antenna, there exists no impedance transformation in the
transmission line. A highly technical quote must be needed:

"You can't have your energy cake and eat it too." :-)
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Gene Fuller wrote:
The graph is complete nonsense. There is no rotation of the fields
when they undergo reflection. Any ordinary text on E&M or optics will
show you the equations and the correct sketches.

Good grief, Gene. There is a 360 degree rotation in the fields
every wavelength. The direction of rotation is associated with
the direction of travel of the wave and is displayed by EZNEC
when the current phase option is turned on. All you have to
do to see the rotation of the traveling wave is to download
http://www.w5dxp.com/rhombicT.EZ

Those simplified sketches are making you simple-minded.
Assuming you are a member of the IEEE, look up this paper:
"Rotation in electromagnetic field equations". Or Google
"Rotation and the Electromagnetic Field".

"Optics", by Hecht, is one of your ordinary texts. That is
where the material for that graph comes from. The IEEE Dictionary
says: "E and H are the electric and magnetic field vectors in
phasor notation". If that graph is nonsense to you, it is your
fault, not mine.

Hecht says "Optics", 4th edition, page 289, about standing waves:

"The composite disturbance is then:

E = Eo[sin(kx+wt) + sin(kx-wt)]

Applying the identity:

sin A + sin B = 2 sin 1/2(A+B)*cos 1/2(A-B)

yields:

E(x,t) = 2*Eo*sin(kx)*cos(wt)"

"This is the equation for a STANDING or STATIONARY WAVE, as opposed
to a traveling wave. Its profile does not move through space; it is
clearly not of the form Func(x +/- vt)."

"... a phasor rotating counterclockwise at a rate omega is equivalent
to a wave traveling to the left (decreasing x), and similarly, one
rotating clockwise corresponds to a wave traveling to the right
(increasing x)."

Hecht uses phasors to represent EM waves all through his book.
He explains the standing wave E-field based on the two traveling
waves, E1-field and E2-field, thusly:

"The resultant phasor is E1 + E2 = E ... Keeping the two [traveling
wave] phasors tip-to-tail and having E1 rotate counterclockwise as
E2 rotates (at the same rate) clockwise, generates E [total] as a
function of 't'."

[Standing wave phase] "doesn't rotate at all, and the resultant
wave it represents doesn't progress through space - its a standing
wave."

Traveling wave phase rotates. Standing wave phase doesn't.

Speaking of "... net transfer of energy, for the pure standing
wave there is none."

The forward wave and reflected wave E-field and H-field vectors
are represented by phasors just as indicated in the IEEE Dictionary.
One is rotating clockwise and the other is rotating counterclockwise.
The Poynting vector for a pure standing wave is equal to zero just
as illustrated in my graph at: http://www.w5dxp.com/EHSuper.JPG
Given those boundary conditions and solving for the angle between
the standing wave E-field and H-field yields 0 or 180 degrees.

Are you really more interested in presenting false information
and saving face than you are in valid technical facts?


Cecil,

This is truly sad. I thought you had finally begun to understand this
stuff, but you have regressed back into the same old nonsense. You are
still totally confusing phasors with field vectors. They are utterly,
totally, and absolutely unrelated.

Get help.

Call me what you like. Bye.

73,
Gene
W4SZ

Gene Fuller wrote:
Any ordinary text on E&M or optics will show
you the equations and the correct sketches.

Please stop staring at your simple-minded sketches
and stare at this diagram of an EM wave for awhile.
Maybe you are due for an epiphany.

http://www.w5dxp.com/EHWave.JPG
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

Jim Kelley wrote:

You fail to explain how any of that makes energy unavailable for
"radiation purposes".


This is not rocket science.

Or any other kind of science for that matter.

If the energy is being used to
transform impedances, it is obviously unavailable to be lost
as radiation.

Sure, assuming energy is being "used" to perform an operation in which
no work is done. You'll need to prove the first part of the sentence
in order to show the second part to be true.

ac6xg

Gene Fuller wrote:
This is truly sad. I thought you had finally begun to understand this
stuff, but you have regressed back into the same old nonsense. You are
still totally confusing phasors with field vectors. They are utterly,
totally, and absolutely unrelated.

Saying that doesn't make it true, Gene, although it may hoodwink
some of the uninitiated into believing your old wives' tales.

I just posted a more sophisticated diagram of an EM wave for
you at: http://www.w5dxp.com/EHWave.JPG

Have those simple-minded sketches that you have been staring
at made you simple-minded or are you capable of something else?
--
73, Cecil http://www.w5dxp.com

On 18 Jan, 10:52, Cecil Moore wrote:
Gene Fuller wrote:
The graph is complete nonsense. There is no rotation of the fields when
they undergo reflection. Any ordinary text on E&M or optics will show
you the equations and the correct sketches.

Good grief, Gene. There is a 360 degree rotation in the fields
every wavelength. The direction of rotation is associated with
the direction of travel of the wave and is displayed by EZNEC
when the current phase option is turned on. All you have to
do to see the rotation of the traveling wave is to downloadhttp://www.w5dxp.com/rhombicT.EZ

Those simplified sketches are making you simple-minded.
Assuming you are a member of the IEEE, look up this paper:
"Rotation in electromagnetic field equations". Or Google
"Rotation and the Electromagnetic Field".

"Optics", by Hecht, is one of your ordinary texts. That is
where the material for that graph comes from. The IEEE Dictionary
says: "E and H are the electric and magnetic field vectors in
phasor notation". If that graph is nonsense to you, it is your
fault, not mine.

Hecht says "Optics", 4th edition, page 289, about standing waves:

"The composite disturbance is then:

E = Eo[sin(kx+wt) + sin(kx-wt)]

Applying the identity:

sin A + sin B = 2 sin 1/2(A+B)*cos 1/2(A-B)

yields:

E(x,t) = 2*Eo*sin(kx)*cos(wt)"

"This is the equation for a STANDING or STATIONARY WAVE, as opposed
to a traveling wave. Its profile does not move through space; it is
clearly not of the form Func(x +/- vt)."

"... a phasor rotating counterclockwise at a rate omega is equivalent
to a wave traveling to the left (decreasing x), and similarly, one
rotating clockwise corresponds to a wave traveling to the right
(increasing x)."

Hecht uses phasors to represent EM waves all through his book.
He explains the standing wave E-field based on the two traveling
waves, E1-field and E2-field, thusly:

"The resultant phasor is E1 + E2 = E *... Keeping the two [traveling
wave] phasors tip-to-tail and having E1 rotate counterclockwise as
E2 rotates (at the same rate) clockwise, generates E [total] as a
function of 't'."

[Standing wave phase] "doesn't rotate at all, and the resultant
wave it represents doesn't progress through space - its a standing
wave."

Traveling wave phase rotates. Standing wave phase doesn't.

Speaking of "... net transfer of energy, for the pure standing
wave there is none."

The forward wave and reflected wave E-field and H-field vectors
are represented by phasors just as indicated in the IEEE Dictionary.
One is rotating clockwise and the other is rotating counterclockwise.
The Poynting vector for a pure standing wave is equal to zero just
as illustrated in my graph at:http://www.w5dxp.com/EHSuper.JPG
Given those boundary conditions and solving for the angle between
the standing wave E-field and H-field yields 0 or 180 degrees.

Are you really more interested in presenting false information
and saving face than you are in valid technical facts?
--
73, Cecil *http://www.w5dxp.com

Cecil,
Just a small question.
Do you include in your analysis of E and H waves
the effect of the field produced by a diamagnetic field produced by
current flow in the material, where the field has a tendency to be at
right angles? This is what PROF Hately was looking to do on his EH
invention without understanding the importance of the material used in
generating the fields in question.
Best regards
Art Unwin KB9MZ...xg

art wrote:
Do you include in your analysis of E and H waves
the effect of the field produced by a diamagnetic field produced by
current flow in the material, ...

Nope, I tend to ignore effects that I judge to be secondary.
Of course, I don't even pretend to exercise perfect judgment.
--
73, Cecil http://www.w5dxp.com

On 18 Jan, 12:15, Cecil Moore wrote:
art wrote:
Do you include in your analysis of E and H waves
the effect of the field produced by a diamagnetic field produced by
current flow in the material, ...

Nope, I tend to ignore effects that I judge to be secondary.
Of course, I don't even pretend to exercise perfect judgment.
--
73, Cecil *http://www.w5dxp.com

But when adding vectors as I see it there is no secondary vectors,
which in this case
says both vectors are in phase no less.I thought I would probe a bit
in this long
thread to ascertain where the controversy is. To understand that alone
in the face of thousand postings is a very difficult task indeed.Even
more so for a mechanical engineer
Very best regards
Art

On Jan 18, 12:44*am, art wrote:
On 17 Jan, 20:44, "AI4QJ" wrote:





"art" wrote in message

...
On 17 Jan, 18:25, "AI4QJ" wrote:

"art" wrote in message

....

your whole posting is just silly. First you accuse me of lieing now
you want to provoke me with stupidity. You need to upgrade to a
understanding of a tank circuit.Yes, it is used in your transmitter.
Google it and study it. Free speech is great but it also reveals your
mentality. And you are just bull ****ing in the absence of knoweledge.."

OK art, I notice a great improvement in your having only 3 spelling errors
in 5 sentences. You are a fool. If your postings could ever be elevated to
the level of mere stupidity, then it might be possible to correct your
misconceptions. However, one cannot deal with true insanity by any
rational
approach such as that. You do not drive me to use Google; you drive me to
dictionary.com so I can find the words appropriate to describe the idiocy
of
your ravings. However, it isn't worth the effort so I will leave you mired
in your intellectual wasteland, reading your postings from time to time
for
the entertainment value of comic relief. For the sake of the hobby I will
continue point out the fairy-tale nature of your postings lest the casual
reader of this newsgroup ever get the wrong impression of what the typical
ham radio operator thinks is antenna "theory".

AI4QJ

:No, I am not going to insult you,

I'm sure you have had you fill of hurling insults at me tonite. Just part of
the entertainment.

:your own actions show what manner of:
:man you are.

Pot Kettle Black

Tell me how a 1/4W tank circuit in my transmitter works, art.

AI4QJ- Hide quoted text -

- Show quoted text -

You have one in your transmitter and you don't know how it works so
you ask me.
Why? You can ask KB9..... or the MI5. If you cross post they will get
back to you.
I promise.They miss you. Maybe one of the above is actually you!- Hide quoted text -

- Show quoted text -

So you found out that I have a 1/4 wave tank circuit in my
transmitter. Only the MI5 could have known that. Hmmm...