Standing waves
Szczepan Białek
pressure pulse. When the two waves travel in opposite directions they
interfere. See: http://en.wikipedia.org/wiki/Kundt's_tube
"The sound generator is turned on and the piston is adjusted until the sound
from the tube suddenly gets much louder. This indicates that the tube is at
resonance, which means its length is a multiple of the wavelength of the
sound wave. At this point the sound waves in the tube are in the form of
standing waves, and the amplitude of vibrations of air are zero at equally
spaced intervals along the tube, called the nodes."
Between the nodes are places where the amplitude is doubled. So the places
with doubled amplitude are standing. Pressure pulse travel.
In antennas is electron gas. The first place where the doubled amplitude
(amplitude means voltage or electron density) appear is end of the radials.
The next is halve wave apart from the end. Such places radiate strong
electric waves. They are the source of radiation.
Of course the next source is weaker because some part of energy is radiated.
But such is stronger than the trawled source (normal voltage pulse)
If antenna has only one source it is omnidirectional. If two or more is
directional because the waves from different sources interfere.
The halve wave dipole has the two sources. The next two appear than a
dipole is longer than the wave length.
So that what R. Clark wrote is obvious: "[* What is this proportional and
proportionate mean? For a dipole of
0.05 WL to a dipole of 0.5WL, the far field change for that 10:1
variation is negligible. However, for a dipole of 0.5WL to a dipole
of 1.25WL, the far field change for that 2.5:1 (a smaller proportion)
variation is very noticeable.]
The above is the antenna with the Helmholtz' pressure wave.
Could anybody describe the antenna with the Heavisde's TEM waves?
In a few words. Do not send mi to library.
S*
Richard Clark
wrote:
>If antenna has only one source it is omnidirectional. If two or more is
>directional because the waves from different sources interfere.
>
>The halve wave dipole has the two sources. The next two appear than a
>dipole is longer than the wave length.
>So that what R. Clark wrote is obvious: "[* What is this proportional and
>proportionate mean? For a dipole of
>0.05 WL to a dipole of 0.5WL, the far field change for that 10:1
>variation is negligible. However, for a dipole of 0.5WL to a dipole
>of 1.25WL, the far field change for that 2.5:1 (a smaller proportion)
>variation is very noticeable.]
In the space of two sentences you contradict yourself. You don't get
it, do you?
73's
Richard Clark, KB7QHC
K7ITM
> Frankly speaking no standing waves. Waves always travel. In air travel the
> pressure pulse. When the two waves travel in opposite directions they
> interfere. See:http://en.wikipedia.org/wiki/Kundt's_tube
> "The sound generator is turned on and the piston is adjusted until the sound
> from the tube suddenly gets much louder. This indicates that the tube is at
> resonance, which means its length is a multiple of the wavelength of the
> sound wave. At this point the sound waves in the tube are in the form of
> standing waves, and the amplitude of vibrations of air are zero at equally
> spaced intervals along the tube, called the nodes."
>
> Between the nodes are places where the amplitude is doubled. So the places
> with doubled amplitude are standing. Pressure pulse travel.
> In antennas is electron gas. The first place where the doubled amplitude
> (amplitude means voltage or electron density) appear is end of the radials.
> The next is halve wave apart from the end. Such places radiate strong
> electric waves. They are the source of radiation.
The problem I immediately see here is that you're probably interested
in electromagnetic radiation, not just the electric field. The
result, as I see it, is that the rest of your discussion is based on a
completely false premise. But see below.
...
> The above is the antenna with the Helmholtz' pressure wave.
> Could anybody describe the antenna with the Heavisde's TEM waves?
> In a few words. Do not send mi to library.
> S*
Others have tried to describe radiation from linear antennas in
reasonably simple terms. One of the best I know is Joseph Boyer's
pair of articles from May and June, I think it was, 1978 "Ham Radio"
magazine: "The Antenna-Transmission Line Analog." It's a non-
mathematical work; it will leave you with answers with not a lot to
back them up, but they do match what we observe, as far as I
understand it. I have these as a PDF, along with a fairly important
section from a book referenced by the articles.
Cheers,
Tom
Richard Fry
> If antenna has only one source it is omnidirectional. If two or more is
> directional because the waves from different sources interfere.
More study is needed on your part, S*. Consult the textbooks of the
authors that already have been listed, and quoted here.
Your unproven, personal opinions are not sufficient to support some of
the statements you post-- which accounts for the "negative" comments
responding to them.
Every single real-world antenna in existence has more than one source
along its length that contributes to its radiation pattern, and
therefore has some directionality.
Your study and accurate understanding of the works of the authors
mentioned will prove this.
RF
Szczepan Białek
"Richard Clark" <kb7...@comcast.net>wrote
news:c9s4b5l6rarmdou19...@4ax.com...
"If antenna has only one source" means the monopole antena (the second
source is "graved")
"The halve wave dipole has the two sources" means "dipole has visible the
both ends".
O.K now?
S*
Szczepan Białek
"K7ITM" <k7...@msn.com> wrote
news:d4c3bb73-431a-495a...@d9g2000prh.googlegroups.com...
On Sep 17, 10:15 am, Szczepan Bia�ek <sz.bia...@wp.pl> wrote:
>...
>> The above is the antenna with the Helmholtz' pressure wave.
> Could anybody describe the antenna with the Heavisde's TEM waves?
> In a few words. Do not send mi to library.
> S*
>Others have tried to describe radiation from linear antennas in
reasonably simple terms. One of the best I know is Joseph Boyer's
pair of articles from May and June, I think it was, 1978 "Ham Radio"
magazine: "The Antenna-Transmission Line Analog." It's a non-
mathematical work; it will leave you with answers with not a lot to
back them up, but they do match what we observe, as far as I
understand it. I have these as a PDF, along with a fairly important
section from a book referenced by the articles.
You send me to library. Try use his own words. Christofire do it (in
"Spherical radiation patern"):
">> Szczepan Bia�ek wrote:
>>> So I will start "Mr. Bialek's lecture series" as a new topic.
>>> The first will be on a "standing waves". A will try to explain the
>>> paradox: "Hence there is something of a paradox that where two currents
>>> reside (the metal elements are continuous and conductive) it is said no
>>> current flows (R. Clark).
>>
>> No NET current flows. The forward current associated with the
>> forward EM wave is moving forward at the speed of light in
>> the medium. The reverse current associated with the reverse
>> EM wave is moving backwards at the speed of light in the
>> medium. When the forward current amplitude equals the reverse
>> current amplitude and is 180 degrees out of phase, the net
>> current flow is zero. 1/4WL away, those equal currents are in
>> phase and add. If the net current is measured at the maximum
>> (anti-node) point, it can be divided by two to obtain an estimate
>> of the magnitude of the forward current at the node point where
>> the net current is zero.
>>
>> The forward current at the ends of a 1/2WL copper dipole is
>> approximately 95% of the forward current at the feedpoint.
>> The reverse current at the feedpoint is approximately 95%
>> of the reverse current at the ends of the dipole. It's a
>> STANDING WAVE ANTENNA! That ~10% decrease between forward
>> current and reverse current at the feedpoint is caused by
>> radiation "loss" and I^2*R loss in the antenna.
>>
>> The SWR on the antenna element is approximately 20:1 at the
>> feedpoint. It is a STANDING WAVE ANTENNA! Most of the net
>> current on the antenna is in the standing wave and cannot
>> be used to measure the delay through the wire or through a
>> loading coil because the current phase changes by only about
>> 2-3 degrees in 90 degrees of wire. The standing wave contains
>> about 80% of the steady-state energy on the antenna. Only about
>> 20% of the steady-state energy/sec is lost to radiation and
>> ohmic losses.
>>
>> One can prove such for oneself. Set up a low 10m 1/2WL dipole
>> with current probes at the 30 deg (x) and 60 deg (y) points.
>> One will not be able to detect more than about a degree of
>> phase shift in that 30 degrees of standing wave antenna.
>>
>> 8ft 2ft 2ft 2ft
>> ------------------------fp-------x-------y--------
>>
>> Why would anyone expect that if the 30 degrees of antenna
>> is replaced by a loading coil that the delay through the
>> loading coil could be measured using standing wave current
>> phase on a standing wave antenna?
Chris "
S*
Szczepan Białek
"Richard Fry" <rf...@adams.net>
news:51205169-be41-45cc...@q14g2000vbi.googlegroups.com...
On Sep 17, 12:15 pm, Szczepan Bia�ek <sz.bia...@wp.pl> wrote:
>> If antenna has only one source it is omnidirectional. If two or more is
> directional because the waves from different sources interfere.
>More study is needed on your part, S*. Consult the textbooks of the
authors that already have been listed, and quoted here.
>Your unproven, personal opinions are not sufficient to support some of
the statements you post-- which accounts for the "negative" comments
responding to them.
I wrote: "Do not send me to librery".
>Every single real-world antenna in existence has more than one source
along its length that contributes to its radiation pattern, and
therefore has some directionality.
I wrote" "Pressure pulse travel" > I means that in monopole antena is one
strong source on the and and traveling source "along its length" .
>Your study and accurate understanding of the works of the authors
mentioned will prove this.
Tell me who is right: Helmholtz or Heaviside?
S*
Richard Fry
> I wrote" "Pressure pulse travel" I means that in monopole
> antena is one strong source on the and and traveling
> source "along its length" .
I assume from what you posted before that you meant to write
"in a monopole antenna there is one strong source on the END..."
Could you please post the reason(s) you think so?
Note that only the change in current and charge, over time,
produces EM radiation. At the top of a monopole, and at the ends
of a dipole the net current is almost zero -- so those locations
generate very little of the total EM radiation from these antennas.
RF
Cecil Moore
> "The halve wave dipole has the two sources" means "dipole has visible
> the both ends".
A 1/2WL dipole is a standing wave antenna.
Are you saying that the forward energy is one source
and the energy reflected from the ends is a second
source?
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com
Cecil Moore
> "The Antenna-Transmission Line Analog."
It's a good analog. For instance a 1/4WL open-wire
open-circuit stub made with resistance wire with
a resistivity of 0.0000021 ohms-m has a feedpoint
impedance of 35 ohms according to EZNEC.
The current distribution is a close approximation
to a 1/4WL monopole. The additional resistance
over copper approximates the energy lost to
radiation in a 1/4WL monopole.
Cecil Moore
> At the top of a monopole, and at the ends
> of a dipole the net current is almost zero -- so those locations
> generate very little of the total EM radiation from these antennas.
Since the forward current and reflected current are equal
in magnitude and opposite in phase at the ends, they act
like transmission line currents and the magnetic fields
cancel at the ends. They are in phase at the feedpoint -
hence the maximum radiation at that point.
Richard Clark
wrote:
>> In the space of two sentences you contradict yourself. You don't get
>> it, do you?
>
>"If antenna has only one source"
Thank you for confirming that in spite of quoting me, you just don't
get it.
Szczepan Białek
"Cecil Moore" <nos...@w5dxp.com>
news:sZJsm.42132$JG1....@newsfe24.iad...
> Szczepan Bia�ek wrote:
>> "The halve wave dipole has the two sources" means "dipole has visible
>> the both ends".
>
> A 1/2WL dipole is a standing wave antenna.
> Are you saying that the forward energy is one source
> and the energy reflected from the ends is a second
> source?
Each dipole has visible the two radials (sometimes end parts of the
transmitting line). Each radials has the one strong source at the end like
the Kundt's tube. See: http://en.wikipedia.org/wiki/Kundt's_tube
A dipole has the two radials so it has at least the two strong sources.
Only the two if it is shorter then WL. If it is longer than WL then appear
the next sources on the both radials at each 1/2WL from the end.
I will be absent till Monday evening.
S*
Szczepan Białek
"Richard Clark" <kb7...@comcast.net> wrote
news:c877b5hq7pebddl6k...@4ax.com...
You go into details.
In the Gas Analogy the monopole antena is exactly like the Kundt's tube.
See: http://en.wikipedia.org/wiki/Kundt's_tube
A dipole has the two Kundt's tubes.
S*
Richard Clark
wrote:
>so those locations
>generate very little of the total EM radiation from these antennas.
Hi Richard,
The entire radiator radiates, not just portions of it. The phase,
time, distance relationships along the length contribute to a myriad
of characteristics, but they are not separable from the complete
contribution.
Richard Clark
wrote:
>You go into details.
Yes, I do go into the details.
>In the Gas Analogy the monopole antena is exactly like the Kundt's tube.
Analogy is a false arguement. In the car-seen-at-a-distance analogy,
this proves that only midgets or pygmies drive cars because we are too
big to fit into such small things seen in the distance.
Cecil Moore
> The entire radiator radiates, not just portions of it. The phase,
> time, distance relationships along the length contribute to a myriad
> of characteristics, but they are not separable from the complete
> contribution.
Using the method of moments, each segment contributes
radiation proportional to the net current in the segment.
Szczepan Białek
"Richard Clark" <kb7...@comcast.net> wrote
news:frd7b5l5fa8vp973i...@4ax.com...
> On Fri, 18 Sep 2009 18:33:00 +0200, Szczepan Bia?ek <sz.b...@wp.pl>
> wrote:
>
>>You go into details.
>
> Yes, I do go into the details.
>
>>In the Gas Analogy the monopole antena is exactly like the Kundt's tube.
>
> Analogy is a false arguement.
Heaviside did the Hydraulic Analogy. All is exactly the same like in the
fluids mechanics.
Next the electrons were discovered. Automatically Heaviside is a history and
the Gas Analogy is in power.
> In the car-seen-at-a-distance analogy,
> this proves that only midgets or pygmies drive cars because we are too
> big to fit into such small things seen in the distance.
But you, radio people, are very close to waves and should be easy for you to
work out the answer for the Question:
Which Analogy is right?
Richard Clark
wrote:
>All is exactly the same like in the fluids mechanics.
"Exactly" makes it very, very easy to show how an analogy fails:
Describe the laminar flow in terms of
the Reynolds number for
the interface between RF and a Biconical Antenna
and
the interface between RF and a thin wire Antenna.
If you do not understand
1. the terms of fluid mechanics and/or
2. cannot complete this request, then
your analogy has failed.
I won't wait for that obvious failure. This is several steps above
your pay-grade. So, you should really attempt to work on first
principles rather than rummaging in the attic for impressive artifacts
of science. The musty chestnuts you find would poison a dog.
Dave
"Szczepan Bialek" <sz.b...@wp.pl> wrote in message
news:h903ub$opk$1...@node1.news.atman.pl...
neither analogy is 'right'. they are useful in limited circumstances to
demonstrate some basic pressure wave physics to young students. but neither
one properly reproduces electromagnetic waves.
tom
> On Fri, 18 Sep 2009 19:13:17 +0200, Szczepan Bia?ek <sz.b...@wp.pl>
> wrote:
>
>> All is exactly the same like in the fluids mechanics.
>
> "Exactly" makes it very, very easy to show how an analogy fails:
> Describe the laminar flow in terms of
> the Reynolds number for
Hmm, I think I'd start with a very long K1FO yagi, say 50 elements.
Maybe even extend one to 100 elements to getting very fine details.
Then we look at the longitoodordinal current along the horizontal
element by element. I'll have to work on it a while though.
What are you thinking?
> 73's
> Richard Clark, KB7QHC
tom
K0TAR
tom
>
> You send me to library. Try use his own words. Christofire do it (in
> "Spherical radiation patern"):
God forbid that you should actually do some research! What a terrible
thought!
tom
K0TAR
Richard Clark
I am thinking that Stefan by lacking a demonstration of this
employment of his own chosen metaphor displays a vacuum in two subject
areas. As it stands, he stumbles through the nuances of RF. Instead,
he is trying to extrapolate them through a second subject, where, of
course, he tumbles over the nuances of fluidics.
The best we can expect is for him to haul a book to the nearest Xerox
and lean on the copy button to produce a snow job. The deepest
impression he will get of that intellectual experience is a paper cut.
tom
>>
>> What are you thinking?
>
> I am thinking that Stefan by lacking a demonstration of this
> employment of his own chosen metaphor displays a vacuum in two subject
> areas. As it stands, he stumbles through the nuances of RF. Instead,
> he is trying to extrapolate them through a second subject, where, of
> course, he tumbles over the nuances of fluidics.
>
> The best we can expect is for him to haul a book to the nearest Xerox
> and lean on the copy button to produce a snow job. The deepest
> impression he will get of that intellectual experience is a paper cut.
>
> 73's
> Richard Clark, KB7QHC
Very perspective.
tom
K0TAR
Szczepan Białek
"Richard Clark" <kb7...@comcast.net> wrote
news:00g7b5hnd4530cn0r...@4ax.com...
>>
> I won't wait for that obvious failure. This is several steps above
> your pay-grade. So, you should really attempt to work on first
> principles rather than rummaging in the attic for impressive artifacts
> of science.
And what should do Richard Harrison who wrote: "At the open circuited ends
of a resonant antenna there is almost double
the forward voltage but zero total current due to cancellation of the
dorward and reflected currents at the open circuit. At the open circuit
in the wire, all the energy in the wave is transferred to the electric
field. "
S*
Szczepan Białek
"Dave" <no...@nowhere.com>wrote
news:OJPsm.2659$Jd7...@nwrddc02.gnilink.net...
>>>
>> fluids mechanics.
>> Next the electrons were discovered. Automatically Heaviside is a history
>> and the Gas Analogy is in power.
>>
>> to work out the answer for the Question:
>> Which Analogy is right?
>
> demonstrate some basic pressure wave physics to young students. but
> neither one properly reproduces electromagnetic waves.
"electromagnetic waves" are paper waves. Radio waves are real waves. Now we
must not know what the waves are like. Now we should estabilish from which
part of the radiator radiate the radio waves.
Do you agree with Richard Harisson:
Szczepan Białek
"Cecil Moore" <nos...@w5dxp.com> wrote
news:UeKsm.42134$JG1....@newsfe24.iad...
EM means elecro- magnetic. Radiation can start from any of them, See what
Richard Harrison wrote: "At the open circuited ends of a resonant antenna
Szczepan Białek
"tom" <news...@taring.org>
news:4ab41e80$0$42842$8046...@newsreader.iphouse.net...
> Szczepan Bia�ek wrote:
>
> God forbid that you should actually do some research! What a terrible
> thought!
All necessary resarch are done by radio people. You all know how antennas
work. Monopole and dipole means the electric pole because no magnetic poles.
You only do not realize that EM waves can start from the ELECRIC field. The
electric field is radiated from the ends where is high voltage and no
current.
S*
Richard Fry
> You only do not realize that EM waves can start from the ELECRIC field. The
> electric field is radiated from the ends where is high voltage and no
> current.
No, Szczepan, it is you that does not realize that voltage, alone,
cannot produce an electromagnetic field.
Only the change in current and charge flowing along a conductor, over
time, produces far-field EM radiation. That radiation includes both
the magnetic and electric fields, at right angles to each other and to
the direction of travel.
It is untrue that one part of a conductor or antenna radiates the
magnetic field, and another part radiates the electric field, no
matter the claims of the proponents of the E-H antenna (which have not
been demonstrated).
The fact that the ends of a dipole, and the top of a monopole have
very little net current flowing means that those locations cannot
contribute very much to the EM radiation from those antennas.
You really should form your opinions from research in modern textbooks
on antennas, rather than using Wikipedia and inapplicable analogies to
sound waves. At a minimum you could recognize the quotes from them on
this subject that already have been posted here.
RF
christofire
"Szczepan Bia�ek" <sz.b...@wp.pl> wrote in message
news:h97vr4$rp$1...@node1.news.atman.pl...
What Richard wrote is correct, if written in a slightly provocative manner
(deliberately?). But he wasn't stating that the electric field 'At the
open-circuited ends of a resonant antenna' passes energy into a radiated
radio wave. The energy that makes it that far (i.e. isn't radiated on
account of current in the element) is stored temporarily in an
'electrostatic' field which is one of several 'reactive' or 'induction'
field components that surround a dipole antenna and decay with distance much
faster than the radiation field components (i.e. those that make up a radio
wave). As I've noted before, the term 'electrostatic' should not be
interpreted literally as an unchanging field - it is used to differentiate
between the reactive components and the radiation components of electric
field - if this offends you, just call it a 'reactive' component of electric
field. This stored energy is passed back into the antenna during the
following RF quarter cycle.
And guess what ... one of the reactive field components is longitudinal!
... but it isn't part of a radio wave - both parts of a radio wave, the
magnetic field and the attendant electric field, are directed transverse to
the direction of propagation, but now I'm repeating myself from a week or
more ago. Power cannot be abstracted from the reactive fields, including
the longitudinal one; they affect the imaginary part of the terminal
impedance of the antenna.
Of course, I expect you will contradict all this but I still recommend that
you read a proper account of the fields around a dipole rather than making
up your own version. Since you appear to have a phobia of libraries, you
could buy a second-hand copy of Kraus, Antennas for only $15 online:
http://www.abebooks.com/servlet/SearchResults?an=Kraus&sts=t&tn=Antennas&x=55&y=10
and there are many, many other sources.
Failing that, you could always search the web for a bootleg copy, or one of
the MIT Radiation Laboratory series of books. I don't condone bootlegging
but someone in another newsgroup recently gave a link to a collection of
illegal copies and, in the hope of ending these ridiculous arguments, I'll
pass on what he wrote:
http://cer.ucsd.edu/~james/notes/MIT%20OpenCourseWare/MIT%20Radiation%20Lab/
Chris
christofire
See my post earlier in this thread. You are misinterpreting what Richard
Harrison wrote to suit your own, incorrect, made-up version of how antennas
work.
Please think on this, Szczepan Bia�ek: the likelihood that your personal
version of the physics is correct is vanishingly small - when it conflicts
with the version everyone else (except perhaps Art Unwin) appears to
understand from their education, which is derived from the basis for
antennas that have been in use for more than 100 years.
Chris
Cecil Moore
>
> "Cecil Moore" <nos...@w5dxp.com> wrote
>> in magnitude and opposite in phase at the ends, they act
>> like transmission line currents and the magnetic fields
>> cancel at the ends. They are in phase at the feedpoint -
>> hence the maximum radiation at that point.
>
> "At the open circuited ends of a resonant
> antenna there is almost double
> the forward voltage but zero total current due to cancellation of the
> in the wire, all the energy in the wave is transferred to the electric
> field."
What I wrote agrees with what Richard H. wrote and vice versa.
Richard Clark
wrote:
>And what should do Richard Harrison who wrote
Which has absolutely nothing to do with the failure of your swampy
metaphor - EXCEPT to demonstrate its stagnation into a cesspool by
being completely ignored by you.
It's amusing to see you wading out there tho'. ;-)
Cecil Moore
> The electric field is radiated from the ends where is
> high voltage and no current.
An individual photon is emitted by an individual
decelerating free electron complete with a
self-contained electric and magnetic field.
It is impossible for the electric field of
a single particular photon to originate 1/4WL
away from the origin of the magnetic field.
Art Unwin
>
> > Szczepan Bia³ek wrote:
>
> > God forbid that you should actually do some research! What a terrible
> > thought!
>
> All necessary resarch are done by radio people. You all know how antennas
> work. Monopole and dipole means the electric pole because no magnetic poles.
> You only do not realize that EM waves can start from the ELECRIC field. The
> electric field is radiated from the ends where is high voltage and no
> current.
> S*
S*
These guys are not helping you! What they are doing is using you for
cannon fodder.
Try looking at things my way. You know that when a time varying
current is applied to a radiator that it also supports a reacting
current with spin, known as as an Eddy current
You also know that the current applied produces a electrical field and
a magnetic field that interchange energy between each other in the
form of a tank circuit.
Now look at the sequence of actions.We do know that the Eddy current
produces a lifting force and a spin force and we also know that there
is a electro static field surrounding the radiator. First we must
recognise that particles encapsulate the whole radiator but can be
individually lifted from the radiator with spin applied a short
distance. At this point it enters the electrostatic field around the
radiator where at the same time the generated magnetic field is
intersecting the electrostatic field. The moment that the lifted
particle enters the electro static field mix it is subjected to a
accelarating force exactly the same way as a electron in a CRT is
impacted upon. If you refer to the actions within a electron tube you
will note that the electrostatic field offsets the direction of the
accellerated particle into an exiting parabolic direction. The
combined fields will only accelerate the particle while it is within
the electrostatic field proper, after which it has a straight line
projection with spin. The time that it is within the electrostatic
field is the total accelerating time ie Newtons law 1/2 ft sqd.
The acceleration imparted to the particle happens to be the speed of
light which implies that this particle is able to emit light. If you
have difficulty then read up on the CRT.
As a point of interest the Eddy current itself must be balanced by an
equal and opposite force per Newton and if we look at it in boundary
terms we see that the opposing force is the combination of Gravity and
the rotation of the Earth. What this shows is that the particle has
spin and accelleration where the vector associated with Gravity is now
neutralized such that it retains its straight line action with spin
as it traverses the boundaries of the Earth Another thing of
importance is that Newtons laws are based on the condition of mass
where the particle becomes an excellent fit as opposed to a field or a
wave.
Back to the radiator itself. If it is a full wave length then it is a
closed circuit of the tank circuit form. If a radiator is less than a
wave length then yes, charges will form at the end of a radiator but
is hampered from further movement by the opposing impedance of the
environment. The charges will still leak but with out a spinning
action it will remain in the near field. Ofcourse changing the
environment will give an instantaneus charge in a spark form because
as an open circuit it always searches for the closed circuit function.
Regards
Art
christofire
"Art Unwin" <arthu...@gmail.com> wrote in message
news:2540a899-797b-432d...@p15g2000vbl.googlegroups.com...
That's hilarious!
We're not using Mr. Bialek as 'cannon fodder' - he is presenting himself as
cannon fodder voluntarily and he has the power to stop this happening. The
same applies to you. But the concept of him looking at things your way is
akin to a nun on a clowns head.
Eddy currents are not named after someone called Eddy and the speed of light
is a speed so it cannot be an amount of acceleration. Any takers for the
other faux pas?
Chris
Richard Clark
<chris...@btinternet.com> wrote:
>
>"Art Unwin" <arthu...@gmail.com> wrote in message
>the Eddy current itself must be balanced by an
>equal and opposite force
>Eddy currents are not named after someone called Eddy and the speed of light
>is a speed so it cannot be an amount of acceleration. Any takers for the
>other faux pas?
I can imagine an Uncle Eddy, but for equal and opposite - Auntie Eddy?
If an eddy circulates in one direction (maintaining the hydrological
metaphor of Stefan's) Auntie Eddy must run opposite. So now the Art
of Antenna Bris finds itself in a backwash.
Richard Fry
> S*
> These guys are not helping you! What they are doing is using you for
Interesting how Art Unwin has tried to hijack this thread into yet
another pulpit of his for those willing to believe his bizarre, and
unprovable concepts about antenna theory and performance.
RF
Art Unwin
> "Art Unwin" <arthurun...@gmail.com> wrote in message
Chris you are being stupid as well as acting as a fool.
Acceleration of the particle only occurs while within the
electrostatic field. When it exits it has the speed of light because
it has emmerged from the intersecting two fields.and thus from the
accellerating forces. I remind you of Newtons law of ut + 1/2 ft sqd
The first expression is for the speed attained on entering the
accelerating field and the other half is for the length of the
accelerating electrostatic field. On leaving the two fields it looses
the applied accelerating force where it has arrived at a particular
speed.Maybe you should look up the workings of a CRT using Newtons
laws instead of shooting from the hip.
It was at a different time that the speed of light was measured where
it was found to equal the sppeed emerging from two intersecting fields.
Dave
and yes, richard's statements are true, but a bit too restrictive, it
doesn't HAVE to be resonant. Voltage doubles and current=0 at the end of
any wire fed with a time varying current, it doesn't even have to be a sine
wave... note the effect of sending square waves from a time domain
reflectometer down an open circuited wire.
christofire
On Sep 21, 4:19 pm, "christofire" <christof...@btinternet.com> wrote:
> "Art Unwin" <arthurun...@gmail.com> wrote in message
- - snip - -
Chris you are being stupid as well as acting as a fool.
Acceleration of the particle only occurs while within the
electrostatic field. When it exits it has the speed of light because
it has emmerged from the intersecting two fields.and thus from the
accellerating forces. I remind you of Newtons law of ut + 1/2 ft sqd
The first expression is for the speed attained on entering the
accelerating field and the other half is for the length of the
accelerating electrostatic field. On leaving the two fields it looses
the applied accelerating force where it has arrived at a particular
speed.Maybe you should look up the workings of a CRT using Newtons
laws instead of shooting from the hip.
It was at a different time that the speed of light was measured where
it was found to equal the sppeed emerging from two intersecting fields.
* Actually, I'm well aware of the principle involved in accelerating
electrons in an electron gun as used in CRTs, klystrons, TWTs, and so on, by
subjecting an electron cloud to a potential difference using an anode with a
hole in it (!), but that's different from what happens in an antenna.
The acceleration of charge in an antenna results almost entirely from the
applied potential difference at its terminals. The radiated fields result
from the alternating current effectively passing through the radiation
resistance, and all the other, reactive, fields have no direct effect on the
radiation resistance, or the component of the current that passes through it
in phase with the voltage that is developed across it, which together, of
course, represent the radiated power. The reactive fields affect the
terminal impedance and a large imaginary part can upset the device trying to
send power into the antenna, but that is more of a system issue. The
alternating current that passes through the radiation resistance is composed
of charge that moves in time with each RF cycle, accelerating and
decelerating accordingly. The electrostatic field developed between the
ends of a half-wave dipole reaches its maximum value a quarter of a cycle
later than the voltage at the drive point so any effect it has on the charge
in the antenna elements during each cycle must be reactive, and it doesn't
affect the radiation resistance or the radiated wave.
But all this can be looked up from any one of the respected books on
antennas. Kraus, and others, gives expressions for the different field
components and the theory all hangs together quite readily using Maxwell's
equations without modification.
I don't care if you think me stupid, and I'll continue to try to avoid name
calling of individuals, although I may criticise what they write especially
if it appears ignorant of proper science yet attempts to re-write
established theory, and therefore appears arrogant.
Chris
Art Unwin
> components and the theory all hangs together quite readily using Maxwell's
> equations without modification.
>
> I don't care if you think me stupid, and I'll continue to try to avoid name
> calling of individuals, although I may criticise what they write especially
> if it appears ignorant of proper science yet attempts to re-write
> established theory, and therefore appears arrogant.
>
> Chris
No it does not appear in Kraus book. He never followed Maxwells laws
with respect to equilibrium. To do that you must think in terms of
wavelength. After all with respect to science all revolve around
boundary laws of the Universe and you blindly ignore that fact.
Now back to radiation which applies spin to a p. Nowarticle where as
with a crt no spin is applied as it is heat that separates the resting
particle Now earlier you refered to a electrostatic field that
according to what you stated did not have a border and surely you know
that just can't be unless it is in equilibrium which requires a closed
circuit.
For this to come to fruition you go back to the boundary laws where
the arbitrary border is one that is closed i.e. in equilibrium. Now
cast your mind back to the Gauss extension where radiators and
particles lie in a closed static field. Now you should see that
equilibrium must reign for a closed static field. Now you blithely
mentioned an electrostatic field with nary a mention as to how it is
formed and how it fits into the whole picture.
Why? Because the books do not provide an unbroken trail that fully
describes radiation
( books admit that) as I have done where everything dove tails into
the existing laws of the Universe. And then I gave you a bonus with
respect to the weak field that Einstein searched for in vain. Einstein
looked at the package presented by the Big Bag but omitted to keep
that which it was wrapped into. That was the arbitrary boundary around
which were forces or vectors that were equal and opposite when a
smallest of smallest of particles edged out towards the border. Yes it
was of a weak force but for all of that it broke the equilibrium
boundary as the forces at that point was not now equal. The breakage
was one where the opposing forces were offset to each other thus
providing a torque force that provided spin. Now we come to our own
Earth encased in a arbitrary border and outside the border we have the
same conditions of equilibrium that must equal the forces of the Big
Bang which means the outside has two vectors, gravity vector which is
straight and a vector denoting spin i.e. rotation of the Earth. On the
other side of the border you again have two vectors a straight vector
and that curly one you don't like me to call eddy current.
Yup. Everything falls together nicely thank you, when you study
radiation from first principles instead of binding yourself to books
that readily admit to not understanding the radiation process. Now
this is not being arrogant when one has applied all principles but I
do think it is arrogant of you and others to asasinate the character
of "S" purely because of his english and spelling and not to help one
that wants to learn.
Just think about it all. I have shown how a particle moves in space in
a straight line trajectory such that the particle maintains a straight
line without the parabolic force of gravity driving it down to the
Earth. Go back to your books and show just how the electrostatic field
came about and what was the borders that it was contained in
Art
via applied spin and where gravity
tom
>
> Chris you are being stupid as well as acting as a fool.
> Acceleration of the particle only occurs while within the
> electrostatic field. When it exits it has the speed of light because
> it has emmerged from the intersecting two fields.and thus from the
Nice job on the name calling, dingbat.
So, which charged particle are you proposing is being accelerated here?
There are none that have no mass, which would mean that no particle
capable of being accelerated by an electrostatic field could exit at
"the speed of light because it has emmerged from the intersecting two
fields".
Have you now discovered a new particle which has charge and no mass?
tom
K0TAR
Art Unwin
> Art Unwin wrote:
>
> > Chris you are being stupid as well as acting as a fool.
> > Acceleration of the particle only occurs while within the
> > electrostatic field. When it exits it has the speed of light because
> > it has emmerged from the intersecting two fields.and thus from the
>
> Nice job on the name calling, dingbat.
You do it all the time TOM , no reasons, just because!
>
> So, which charged particle are you proposing is being accelerated here?
> There are none that have no mass, which would mean that no particle
> capable of being accelerated by an electrostatic field could exit at
> "the speed of light because it has emmerged from the intersecting two
> fields".
You are talking garbage Tom in a attempt to sound educated.
When you introduce Newtons laws you are standing on the existence of
mass.
It was a particle with mass that broke the equilibrium of the Sun
which starts the whole cycle.
This mass is debris from the burning and release of energy from the
sun which ebbs and flows according to the 11 year sun spot cycle. It
is acknowledged that particles from the Universe are in the billions
per cubic meter in the Earths system and all these particles have
mass. The idea that neutrinos, a particle from the Sun had no mass was
dispelled years ago, so the Laws of Newton remains in place. This very
same action is shown with respect to a CRT where the particle which in
that case does not have spin. You will see that the particle is
accelerated within the borders of the electrostatic field which is
also intersected by a mgnetic field such that its path is parabolic
within the confines of the electrostatic field which becomes a
straight line travel without acelleration as it leaves the confines of
the electrostatic field. It is all quite simple Tom but you continue
to implant erronius data into the description as a basis to supply
insults.
>
> Have you now discovered a new particle which has charge and no mass?
There you go again, all particles have mass and Newtons laws are still
in place
Stop playing the fool
>
> tom
> K0TAR
tom
>> Have you now discovered a new particle which has charge and no mass?
>
> There you go again, all particles have mass and Newtons laws are still
> in place
> Stop playing the fool
So explain how your particle which has mass can attain light speed.
tom
K0TAR
derek unwin
derek unwin
Read before you post. Plank!
Art Unwin
You could go to college and learn all this but here goes.
Planck did a treatise on heat radiation where he stated that heat
radiation
is enabled by two independent functions. First was by conduction and
the second by a ray.
He went on to enlarge on the notion of a ray when is studies moved on
to radio and light radiation. He often consulted with Einstein and was
well aware of his efforts to find the weak force and where he gave up
on that and started a new science where he came up with mc sqd,. With
the abandoned search of Einstein Planck settled hard on the idea that
light was a radiated beam or wave of sorts and it was natural of him
to see radiation of heat in a similar manner i.e a wave or ray in a
similar way that light was thought of.
In fact his work moved later to the nature of light and radio
radiation where the theme of a ray or wave was retained. If Einstein
had included the wrappings of the Big Bang in his studies particles
would have retained the high ground for both him and Planck.
as well as this group together with science which has been misled to
this very day.
Thus Planck started on the wrong track with respect to heat radiation
and all of his other studies. Now he may have known what the speed of
light was but he would see it as a ray of light the same way as a ray
of heat. It was later found independently that a particle at the exit
from the combined field intersection was the speed of light which now
can be seen as the cause of light in connection with particle speed
i.e an effect and not a cause.
Since we note that wavelengths can denote a change of color the
connection of wavelength, spin ,deccelleration or what ever, directly
relates to the emmission of light.
So to answer your question the cause of this particular speed is the
interaction of two intesecting fields on a particle of mass. This same
particle is the same particle that emits light which cannot exceed the
speed attained in the combined fields since there is no
applied accellerating force applied to that same particle ,only
retardation. And this is the same particle that sits on a diamagnetic
material that is removed from a heated anode without spin but with an
increase in potential energy. Without spin this same particle
deccellerates before impacting on a luminescent glass surface of a CRT.
tom
>
> Read before you post. Plank!
Plank?
You don't even know the standard word which means "I'm blocking you
forever because your intelligence is less than dirt".
tom
K0TAR
tom
Bafflegab.
> of heat. It was later found independently that a particle at the exit
> from the combined field intersection was the speed of light which now
> can be seen as the cause of light in connection with particle speed
> i.e an effect and not a cause.
Give references. This is the key to your argument, in case you can't
figure that out. It also breaks all current laws of physics.
> Since we note that wavelengths can denote a change of color the
> connection of wavelength, spin ,deccelleration or what ever, directly
> relates to the emmission of light.
> So to answer your question the cause of this particular speed is the
> interaction of two intesecting fields on a particle of mass. This same
> particle is the same particle that emits light which cannot exceed the
> speed attained in the combined fields since there is no
> applied accellerating force applied to that same particle ,only
> retardation. And this is the same particle that sits on a diamagnetic
> material that is removed from a heated anode without spin but with an
> increase in potential energy. Without spin this same particle
> deccellerates before impacting on a luminescent glass surface of a CRT.
So you admit you can't explain how to get a particle with mass to light
speed.
Fraud.
tom
K0TAR
Art Unwin
All that time I took to educate you was all for nothing
On top of that you accused me of breaking the laws of physics which
can only be the laws as you under stand them which is different to the
rest of the World. I also note that you have insulted another person
of the group calling him dirt.
Really you need to read a book of the World as we know it where the
magnetic fields of the Earth become inert to change and the forces of
the standard model reverse to the point of beginnings. When this
reverse begins the dead will rise and it will be heaven
as we then grow younger. Ofcourse those who decided to stay on the
side of the Red sea
will not rejoin the others that stayed on the right path which suggest
that the Middle East will be destroyed as the majority of the tribe
descendants lives there. Ofcourse if you have lived a life where you
insulted all you certainly will not be "one of the meek" that gets
blessed. Yup, read a different book next time and check out the laws
that we all have violated.
> Fraud.
>
> tom
> K0TAR
jaroslav lipka
------------ Current laws of physics ---------------
Wowl he got something right for a change
Jaro
Szczepan Białek
"Cecil Moore" <nos...@w5dxp.com> news:ImQtm.49222$ec2....@newsfe13.iad...
> Szczepan Bia�ek wrote:
>>
>> "Cecil Moore" <nos...@w5dxp.com> wrote
>>> Since the forward current and reflected current are equal
>>> in magnitude and opposite in phase at the ends, they act
>>> like transmission line currents and the magnetic fields
>>> cancel at the ends. They are in phase at the feedpoint -
>>> hence the maximum radiation at that point.
It apply to 1/2WL dipole. But what radiate in 0.05WL dipole? There the
"maximum radiation " is in the transmission line (1/4WL from the end). The
feed point is also in the transmissing line.
"R. Clark wrote : "[* What is this proportional and
proportionate mean? For a dipole of
0.05 WL to a dipole of 0.5WL, the far field change for that 10:1
variation is negligible. However, for a dipole of 0.5WL to a dipole
of 1.25WL, the far field change for that 2.5:1 (a smaller proportion)
variation is very noticeable.]"
>>
>> See what Richard Harrison wrote:
>> "At the open circuited ends of a resonant antenna there is almost double
>> the forward voltage but zero total current due to cancellation of the
>> forward and reflected currents at the open circuit. At the open circuit
>> in the wire, all the energy in the wave is transferred to the electric
>> field."
>
> What I wrote agrees with what Richard H. wrote and vice versa.
Rather no. Richard do not write about a feed point.
S*
Szczepan Białek
"christofire" <chris...@btinternet.com> wrote
news:JJSdndnC9PUWUyrX...@bt.com...
>
> "Szczepan Bia�ek" <sz.b...@wp.pl> wrote in message
> news:h97vr4$rp$1...@node1.news.atman.pl...
>>
>>
>> And what should do Richard Harrison who wrote: "At the open circuited
>> ends of a resonant antenna there is almost double
>> the forward voltage but zero total current due to cancellation of the
>> dorward and reflected currents at the open circuit. At the open circuit
>> in the wire, all the energy in the wave is transferred to the electric
>> field. "
>> S*
>
>
> What Richard wrote is correct, if written in a slightly provocative manner
> (deliberately?). But he wasn't stating that the electric field 'At the
> open-circuited ends of a resonant antenna' passes energy into a radiated
> radio wave.
Look at this: ""At the open circuited ends of a resonant antenna there is
almost double the forward voltage".
"Almost double voltage" is like the pressure in the Kundt's tube.
> The energy that makes it that far (i.e. isn't radiated on account of
> current in the element) is stored temporarily in an 'electrostatic' field
> which is one of several 'reactive' or 'induction' field components that
> surround a dipole antenna and decay with distance much faster than the
> radiation field components (i.e. those that make up a radio wave). As
> I've noted before, the term 'electrostatic' should not be interpreted
> literally as an unchanging field - it is used to differentiate between the
> reactive components and the radiation components of electric field - if
> this offends you, just call it a 'reactive' component of electric field.
> This stored energy is passed back into the antenna during the following RF
> quarter cycle.
>
> And guess what ... one of the reactive field components is longitudinal!
>
> ... but it isn't part of a radio wave - both parts of a radio wave, the
> magnetic field and the attendant electric field, are directed transverse
> to the direction of propagation, but now I'm repeating myself from a week
> or more ago. Power cannot be abstracted from the reactive fields,
> including the longitudinal one; they affect the imaginary part of the
> terminal impedance of the antenna.
>
> Of course, I expect you will contradict all this but I still recommend
> that you read a proper account of the fields around a dipole rather than
> making up your own version.
It is not my version. The electrons were discovered 100 years ago.
> Since you appear to have a phobia of libraries, you could buy a
> second-hand copy of Kraus, Antennas for only $15 online:
> http://www.abebooks.com/servlet/SearchResults?an=Kraus&sts=t&tn=Antennas&x=55&y=10
> and there are many, many other sources.
I am sure that there is all about electrons. It is not easy to read with
understanding. Take a glance once more and try to find compressible
electrons.
>
> Failing that, you could always search the web for a bootleg copy, or one
> of the MIT Radiation Laboratory series of books. I don't condone
> bootlegging but someone in another newsgroup recently gave a link to a
> collection of illegal copies and, in the hope of ending these ridiculous
> arguments,
All arguments are from this Group posts.
> I'll pass on what he wrote:
> http://cer.ucsd.edu/~james/notes/MIT%20OpenCourseWare/MIT%20Radiation%20Lab/
S*
Szczepan Białek
U�ytkownik "Richard Clark" <kb7...@comcast.net> napisa� w wiadomo�ci
news:c6lfb5t0fqiqdu9vs...@4ax.com...
For me it is important that at the end the voltage is doubled: "At the open
circuited ends
of a resonant antenna there is almost double the forward voltage".
I am colecting such arguments from radio people posts.
S*
Szczepan Białek
"Richard Fry" <rf...@adams.net> wrote
news:5a0a5d03-ac80-4bcf...@g1g2000vbr.googlegroups.com...
On Sep 21, 12:23 pm, Szczepan Bia�ek <sz.bia...@wp.pl> wrote:
>> You only do not realize that EM waves can start from the ELECRIC field.
>> The
> electric field is radiated from the ends where is high voltage and no
> current.
>No, Szczepan, it is you that does not realize that voltage, alone,
cannot produce an
Let us assume that electromagnetic field is a proposition by Maxwell.
The electric field is more realistic.
>Only the change in current and charge flowing along a conductor, over
time, produces far-field EM radiation. That radiation includes both
the magnetic and electric fields, at right angles to each other and to
the direction of travel.
>It is untrue that one part of a conductor or antenna radiates the
magnetic field, and another part radiates the electric field, no
matter the claims of the proponents of the E-H antenna (which have not
been demonstrated).
But it is experimentally proved. Stationary charge - electric field, Moving
charge - magnetic field.
Probably the both fields are the same. Only instruments are different.
>The fact that the ends of a dipole, and the top of a monopole have
very little net current flowing means that those locations cannot
contribute very much to the EM radiation from those antennas.
But there are the doubled voltage. Very strong pulses must appear in space.
>You really should form your opinions from research in modern textbooks
on antennas, rather than using Wikipedia and inapplicable analogies to
sound waves. At a minimum you could recognize the quotes from them on
this subject that already have been posted here.
Up to now the acoustic analogy is fully applicable.
S*
RF
Szczepan Białek
"Dave" <no...@nowhere.com>
news:LETtm.2963$Jd7....@nwrddc02.gnilink.net...
>
> "Szczepan Bialek" <sz.b...@wp.pl> wrote in message
> news:h980ba$tt$1...@node1.news.atman.pl...
>>>>>
>>>>>>tube.
>>>>>
>>>> Heaviside did the Hydraulic Analogy. All is exactly the same like in
>>>> the fluids mechanics.
>>>> Next the electrons were discovered. Automatically Heaviside is a
>>>> history and the Gas Analogy is in power.
>>>>
>>>> But you, radio people, are very close to waves and should be easy for
>>>> you to work out the answer for the Question:
>>>> Which Analogy is right?
>>>
>>> neither analogy is 'right'. they are useful in limited circumstances to
>>> demonstrate some basic pressure wave physics to young students. but
>>> neither one properly reproduces electromagnetic waves.
>>
>> "electromagnetic waves" are paper waves. Radio waves are real waves. Now
>> we must not know what the waves are like. Now we should estabilish from
>> which part of the radiator radiate the radio waves.
>> Do you agree with Richard Harisson:
>> "At the open circuited ends of a resonant antenna there is almost double
>> the forward voltage but zero total current due to cancellation of the
>> dorward and reflected currents at the open circuit. At the open circuit
>> in the wire, all the energy in the wave is transferred to the electric
>> field."
>> S*
>>
>>
> we know what they are like, you just have to understand the mathematics.
I understand. Without that it was be impossible to know that Maxwell
proposed the displacement current to save the incompressible electric fluid.
In Maxwell times AC current was known. To pass the incompressible fluid
through a capacitor the displacement current is necessary. I prefere the
compressible electrons. They compress in the plates and nothing flow between
them. The polarization is not the macro flow.
> and yes, richard's statements are true, but a bit too restrictive, it
> doesn't HAVE to be resonant. Voltage doubles and current=0 at the end of
> any wire fed with a time varying current, it doesn't even have to be a
> sine wave... note the effect of sending square waves from a time domain
> reflectometer down an open circuited wire.
Yes. But antennas are in resonance.
S*
>
Szczepan Białek
"Art Unwin" <arthu...@gmail.com> wrote
news:2540a899-797b-432d...@p15g2000vbl.googlegroups.com...
On Sep 21, 12:23ďż˝ pm, Szczepan BiaĹ ek <sz.bia...@wp.pl> wrote:
> ďż˝ "tom"
> <news4...@taring.org>news:4ab41e80$0$42842$8046...@newsreader.iphouse.net...
>
> > Szczepan BiaÂłek wrote:
>
> > God forbid that you should actually do some research! ďż˝ What a terrible
> > thought!
>
>> All necessary resarch are done by radio people. You all know how antennas
> work. Monopole and dipole means the electric pole because no magnetic
> poles.
> You only do not realize that EM waves can start from the ELECRIC field.
> The
> electric field is radiated from the ends where is high voltage and no
> current.
> S*
S*
>These guys are not helping you! What they are doing is using you for
cannon fodder.
Day after day they state the new for me aguments that the acoustic analogy
is the winner.
>Try looking at things my way. You know that when a time varying
current is applied to a radiator that it also supports a reacting
current with spin, known as as an Eddy current
You also know that the current applied produces a electrical field and
a magnetic field that interchange energy between each other in the
form of a tank circuit.
You go into details. Now is not time for that. Now we should work out the
consensus on which part of the radiator radiate the radio waves.
It will not be easy to work out here all details.
S*
Regards
Art
Richard Fry
> Yes. But antennas are in resonance.
That is rarely true.
A naturally resonant antenna has zero reactance at its feedpoint. But
that is not a requirement for it to produce EM radiation efficiently
from all of the r-f current flowing on it.
The monopoles used by MW broadcast stations vary from 60 to 225
electrical degrees in height. Only a few of those heights are
naturally resonant.
Yet they all radiate nearly 100% of the applied power, because any
antenna reactance is offset by an impedance matching network at the
feedpoint.
RF
Cecil Moore
> For me it is important that at the end the voltage is doubled: "At the
> open circuited ends
> of a resonant antenna there is almost double the forward voltage".
The same thing happens with an open-circuit stub yet it
radiates a negligible amount. The only place where an
antenna radiates more than a negligible amount is where
the forward and reverse currents are unbalanced.
At the ends of a 1/2WL dipole, the forward and reverse
currents are perfectly balanced, i.e. they are 100%
transmission line currents which are known not to
radiate.
At the 1/2WL dipole feedpoint, the forward and reverse
currents are most unbalanced, i.e. since they are in
phase at the feedpoint, they can be considered to be
100% antenna currents.
This is just one more example of the illogic of
forcing forward and reflected waves into a mashed
potatoes theory of energy. The underlying laws of
physics are lost in the process.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com
Richard Fry
> Up to now the acoustic analogy is fully applicable.
Not if one understands the physics of radiation.
> But it is experimentally proved. Stationary charge - electric field,
> Moving charge - magnetic field.
Untrue, and I challenge you to cite any credible experimental data
that you think proves your belief.
Far-field EM radiation is produced only by the current flow on the
antenna, and that radiation contains BOTH the electric and the
magnetic fields.
You may have missed the accurate description posted by Chris, and
pasted below.
RF
christofire
* That is incorrect, and I expect you know that to be the case. All of what
I wrote above can be traced to sources such as Kraus, Jordan and Balmain,
and Jasik; books you may never have tried to read and understand (on the
basis of your comment).
He never followed Maxwells laws
with respect to equilibrium.
* Define 'equilibrium' in trems of normal physics and cite a reference.
To do that you must think in terms of
wavelength. After all with respect to science all revolve around
boundary laws of the Universe and you blindly ignore that fact.
* How is there a boundary to the universe? It is unbounded - the universe
is known to be expanding into empty space.
Now back to radiation which applies spin to a p.
* No it doesn't - cite a reference.
Nowarticle where as
with a crt no spin is applied as it is heat that separates the resting
particle
* ... separates it from what?
Now earlier you refered to a electrostatic field that
according to what you stated did not have a border and surely you know
that just can't be unless it is in equilibrium which requires a closed
circuit.
* Fields in unbounded space tend to be unbounded. The fields around a
dipole are bounded by the surface of the dipole, which is the boundary
condition used by NEC, and usually in practice by the earth.
For this to come to fruition you go back to the boundary laws where
the arbitrary border is one that is closed i.e. in equilibrium.
* Define 'equilibrium' in terms of normal physics and cite a reference.
Now
cast your mind back to the Gauss extension where radiators and
particles lie in a closed static field. Now you should see that
equilibrium must reign for a closed static field. Now you blithely
mentioned an electrostatic field with nary a mention as to how it is
formed and how it fits into the whole picture.
* Recent discussion has been about the electric field produced by the
voltage that appears between the ends of a dipole - that's what I was
writing about, and so were you. How can you state 'with nary a mention as
to how it is formed'?
Why? Because the books do not provide an unbroken trail that fully
describes radiation
( books admit that) as I have done where everything dove tails into
the existing laws of the Universe.
* Kraus provides a trail that's as unbroken as can be comprehended by most
engineers. Physicists may wish to take it further but there's no evidence
of physicists in this newsgroup.
And then I gave you a bonus with
respect to the weak field that Einstein searched for in vain. Einstein
looked at the package presented by the Big Bag but omitted to keep
that which it was wrapped into.
* Is the universe enclosed in a Big Bag?
That was the arbitrary boundary around
which were forces or vectors that were equal and opposite when a
smallest of smallest of particles edged out towards the border. Yes it
was of a weak force but for all of that it broke the equilibrium
boundary as the forces at that point was not now equal. The breakage
was one where the opposing forces were offset to each other thus
providing a torque force that provided spin. Now we come to our own
Earth encased in a arbitrary border and outside the border we have the
same conditions of equilibrium that must equal the forces of the Big
Bang which means the outside has two vectors, gravity vector which is
straight and a vector denoting spin i.e. rotation of the Earth. On the
other side of the border you again have two vectors a straight vector
and that curly one you don't like me to call eddy current.
* Call it what you wish. Capitalising the word within a sentence, as you
did before, is usually reserved for phenomena named after the people who
discovered them - but that's an engineer thing.
Yup. Everything falls together nicely thank you, when you study
radiation from first principles instead of binding yourself to books
that readily admit to not understanding the radiation process.
* If you believe so fervently in your own version of all this then why don't
you submit it to peer review at sci.physics or sci.physics.research? It's
rather unfair to expose this only to an amateur radio newsgroup when what
you are doing is apparently re-writing physics in such a major way. Those
newsgroups would be more appropriate, considering the depth to which you are
going (i.e. well beyond amateur radio) - wouldn't you agree.
Now
this is not being arrogant when one has applied all principles
* My dictionary defines 'arrogant' as: 'having or showing an exaggerated
opinion of one's own importance , merit, ability, etc.' from the Latin
'arrogare': 'to claim as one's own'. Your re-writing of physics conforms to
this definition precisely.
but I
do think it is arrogant of you and others to asasinate the character
of "S" purely because of his english and spelling and not to help one
that wants to learn.
* He is showing no inclination of wanting to learn - he asks a question, and
then when answered responds with non-physical contradiction. His treatement
here is a direct consequence of that behaviour, like yours.
Just think about it all. I have shown how a particle moves in space in
a straight line trajectory such that the particle maintains a straight
line without the parabolic force of gravity driving it down to the
Earth. Go back to your books and show just how the electrostatic field
came about and what was the borders that it was contained in
Art
via applied spin and where gravity
* As I say, you should present your theory to sci.physics and
sci.physics.research if you have any interest in checking whether it is
correct, and not limit its exposure to this group. Do let us know when you
have posted there.
Chris
Cecil Moore
> Now we should work out
> the consensus on which part of the radiator radiate the radio waves.
Here's what you are missing: RF is AC. At some point
in the radiation cycle, the instantaneous radiation has
to be zero at the zero-crossing time. That is when the
magnetic field energy is essentially zero and close to
100% of the energy is contained in the electric fields
at the ends of the dipole.
You cannot have it both ways. You cannot have a radiation
peak at a net current maximum time and also have a radiation
peak at a net voltage maximum time since they are ~90 degrees
out of phase. If what you are saying were really happening,
an antenna would radiate two times the applied frequency,
but it obviously doesn't.
Or did you not realize that when the electric field is
at its maximum amplitude, the current is close to zero-
crossing all up and down the standing wave antenna? The
phase of the feedpoint current is within a couple of
degrees of the phase of the current all up and down
the antenna.
That's why the current on a standing wave antenna cannot
be used to measure the delay through a wire or a loading
coil. All such "measurements" are bogus.
Richard Fry
> It apply to 1/2WL dipole. But what radiate in 0.05WL dipole? There the
> "maximum radiation " is in the transmission line (1/4WL from the end).
> The feed point is also in the transmissing line.
There is no radiation from a transmission line as long as the currents
in it are balanced. Attaching a transmission line to a 0.05WL center-
fed dipole does not change that balance.
The feedpoint of such a short dipole has very high capacitive
reactance, so getting much current to flow into it would be difficult
without proper impedance matching to the transmission line.
But whatever the impedance match, all of the radiation occurs from the
0.05WL dipole itself, and maximum radiation is generated at the
feedpoint, where the r-f current is greatest.
RF
Richard Clark
wrote:
>> Which has absolutely nothing to do with the failure of your swampy
>> metaphor - EXCEPT to demonstrate its stagnation into a cesspool by
>> being completely ignored by you.
>>
>> It's amusing to see you wading out there tho'. ;-)
>
>For me it is important that at the end the voltage is doubled
Yes, we can all tell what is important to you. Unfortunately it does
not equate with what is important for you. No doubt that distinction
is lost in translation and you will continue to fumble on.
I'm glad to see you shed that nonsense about hydraulics. As you
understood that topic far less than RF (which is in itself on very
shaky ground), it wouldn't have done to poison the well.
This poor understanding is quite obvious by your last comment above.
73's
Richard Clark, KB7QHC
Richard Clark
<chris...@btinternet.com> wrote:
>Chris
Could you turn your newsreader's quoting mechanism back on? It is
very confusing to have to fumble with Art's babbling interleaved with
your "special" editorial marks.
Richard Clark
wrote with abysmal quoting:
>It will not be easy to work out here all details.
>S*
>
>
>Regards
>Art
You guys need to get a room.
Szczepan Białek
"Richard Fry" <rf...@adams.net> wrote
news:f9e080ad-4116-424f...@d34g2000vbm.googlegroups.com...
On Sep 22, 2:21 am, Szczepan Bia�ek <sz.bia...@wp.pl> wrote:
>> It apply to 1/2WL dipole. But what radiate in 0.05WL dipole? There the
> "maximum radiation " is in the transmission line (1/4WL from the end).
> The feed point is also in the transmissing line.
>There is no radiation from a transmission line as long as the currents
in it are balanced. Attaching a transmission line to a 0.05WL center-
fed dipole does not change that balance.
>The feedpoint of such a short dipole has very high capacitive
reactance, so getting much current to flow into it would be difficult
without proper impedance matching to the transmission line.
The feed point is the part of the transmissing line and not radiate.
>But whatever the impedance match, all of the radiation occurs from the
0.05WL dipole itself, and maximum radiation is generated at the
feedpoint, where the r-f current is greatest.
See above.
Dipole 0.05 is probably the straight. How long are the folded dipoles and
the loop antennas? Are there the short version?
S*
Szczepan Białek
"Cecil Moore" <nos...@w5dxp.com> wrote
news:uc3um.2236$lR3....@newsfe25.iad...
> Szczepan Bia�ek wrote:
>> Now we should work out the consensus on which part of the radiator
>> radiate the radio waves.
>
> Here's what you are missing: RF is AC. At some point
> in the radiation cycle, the instantaneous radiation has
> to be zero at the zero-crossing time. That is when the
> magnetic field energy is essentially zero and close to
> 100% of the energy is contained in the electric fields
> at the ends of the dipole.
>
> You cannot have it both ways. You cannot have a radiation
> peak at a net current maximum time and also have a radiation
> peak at a net voltage maximum time since they are ~90 degrees
> out of phase. If what you are saying were really happening,
> an antenna would radiate two times the applied frequency,
> but it obviously doesn't.
We do not have the both. But we have the Luxembourg effect. Each dipole
antena radiate two times the applied frequency, The pulses from the ends
are 180 degrees apart.
>
> Or did you not realize that when the electric field is
> at its maximum amplitude, the current is close to zero-
> crossing all up and down the standing wave antenna?
> phase of the feedpoint current is within a couple of
> degrees of the phase of the current all up and down
> the antenna.
>
> That's why the current on a standing wave antenna cannot
> be used to measure the delay through a wire or a loading
> coil. All such "measurements" are bogus.
S*
Richard Fry
> But what radiate in 0.05WL dipole? There the
> "maximum radiation" is in the transmission line (1/4WL from the end).
> The feed point is also in the transmissing line.
Then later the same day he wrote:
> The feed point is the part of the transmissing line and not radiate.
Pick one of the above comments, only, Szczepan.
The feed points are terminals of the antenna. On center-fed dipoles
that are 1/2WL or less in length, antenna current is highest at those
terminals.
> How long are the folded dipoles and the
> loop antennas? Are there the short version?
They can be any length, but some lengths have better input
characteristics and/or more useful radiation patterns than others.
RF
Cecil Moore
> Each dipole
> antenna radiate two times the applied frequency, ...
Sorry, 2*sin(2wt) <> sin(wt)
Here's the question: Is the radiated RF wave in phase
with the standing wave current or in phase with the
standing wave voltage? The radiated RF wave cannot be
in phase with both since they are 90 degrees out of
phase on the standing wave antenna.
Art Unwin
Easily done. Boundary laws are based on Newtons laws where every
action has an equal and opposite reaction. So a subject under
discussion is placed inside an arbitrary border
where the forces of the contents must be equal to the sum of the
forces outside the boundary. Where does Krauss deviate from from
Maxwells laws? well for one thing his radiator differed from a full
WL or period so immediately he has strayed. Another thing is by
straying from a full WL he was unable to contain the electrostatic
field within a border
such as a quad. And it goes on especially with regards to his windings
pitch theory.
Thus the contents of the arbitrary boundary is now in a state of
equilibrium. this is true whether a unit of time has been added or not
to the equation.
>
> To do that you must think in terms of
> wavelength. After all with respect to science all revolve around
> boundary laws of the Universe and you blindly ignore that fact.
>
> * How is there a boundary to the universe? It is unbounded - the universe
> is known to be expanding into empty space.
Hogwash. Nobody knows the extent of the Universe boundary. Every day
discoveries are made beyond our Universe so we are nowhere near the
point we can ascertain what forces
are required to contain a boundary.
Again I am surprised you didn't know that!
>
> Now back to radiation which applies spin to a p.
>
> * No it doesn't - cite a reference.
Well look at how salvage yards sort out metals into different
enclosures.
They apply a displacement current to a conveyor where each piece of
metal is elevated with spin such that it lands in the appropriate
enclosure which is dependent on the resistivity of the metal elevated.
This method of elevating scrap for recovery has been used for years
and it is the same action that is applied to particles for radiation.
Why would you need a citation for a practice that is well known and in
use? On top of that a citation is only an instrument to poin out who
and where the statement was made. This has its use for those who
easily believe what they see in print and thus save them from effort.
>
> Nowarticle where as
> with a crt no spin is applied as it is heat that separates the resting
> particle
>
> * ... separates it from what?
Now you are being silly. first the particle was resting. Then heat was
applied to its butt.
Now think about it. What fields are in place to take control of the
free particle that now has a increase in its potential energy?
>
> Now earlier you refered to a electrostatic field that
> according to what you stated did not have a border and surely you know
> that just can't be unless it is in equilibrium which requires a closed
> circuit.
>
> * Fields in unbounded space tend to be unbounded. The fields around a
> dipole are bounded by the surface of the dipole, which is the boundary
> condition used by NEC, and usually in practice by the earth.
Tend, tend. You mean you don't know? NEC is per the laws of Maxwell.
Equilibrium is a must as is the encirclement of a electrostatic field
just like a quad or loop antenna. Since the programmer did not follow
Maxwell in its entirety and frankly did many modifications such that
it would solve planar radiators, who knows what license the
programmers took to get the job done where it could foolmost of the
people most of the time?
>
> For this to come to fruition you go back to the boundary laws where
> the arbitrary border is one that is closed i.e. in equilibrium.
>
> * Define 'equilibrium' in terms of normal physics and cite a reference.
Hmm I thought I had done that before!
>
> Now
> cast your mind back to the Gauss extension where radiators and
> particles lie in a closed static field. Now you should see that
> equilibrium must reign for a closed static field. Now you blithely
> mentioned an electrostatic field with nary a mention as to how it is
> formed and how it fits into the whole picture.
>
> * Recent discussion has been about the electric field produced by the
> voltage that appears between the ends of a dipole - that's what I was
> writing about, and so were you. How can you state 'with nary a mention as
> to how it is formed'?
I have concentrated on radiator that were in equilibrium and of a
closed circuit called a tank circuit
. You are mixing things up. I explained for somebody else what a
charge does for a radiator that is NOT in equilibrium because others
were unable to describe it adequately.
> Why? Because the books do not provide an unbroken trail that fully
> describes radiation
> ( books admit that) as I have done where everything dove tails into
> the existing laws of the Universe.
> * Kraus provides a trail that's as unbroken as can be comprehended by most
> engineers. Physicists may wish to take it further but there's no evidence
> of physicists in this newsgroup.
No Krauss did no such thing. He expanded a yagi style radiator where
the elements ere made of loops. All he did was to connect them into a
continuous conductor. When experimenting with this he saw evidence
that field did not follow the axis of the windings
( rotation of the earth again) so he adjusted the pitch of the
windings to obtain a satisfactory pattern. He did not improve on the
existing knowledge of radiators as his radiators were not in
equilibrium and thus strayed from Maxwell.
>
> And then I gave you a bonus with
> respect to the weak field that Einstein searched for in vain. Einstein
> looked at the package presented by the Big Bag but omitted to keep
> that which it was wrapped into.
>
> * Is the universe enclosed in a Big Bag?
If you wish to put it into a three dimensional border be my guest. It
would be interesting to know what your starting point is bearing in
mind your position of a huge surrounding void.
>
> That was the arbitrary boundary around
> which were forces or vectors that were equal and opposite when a
> smallest of smallest of particles edged out towards the border. Yes it
> was of a weak force but for all of that it broke the equilibrium
> boundary as the forces at that point was not now equal. The breakage
> was one where the opposing forces were offset to each other thus
> providing a torque force that provided spin. Now we come to our own
> Earth encased in a arbitrary border and outside the border we have the
> same conditions of equilibrium that must equal the forces of the Big
> Bang which means the outside has two vectors, gravity vector which is
> straight and a vector denoting spin i.e. rotation of the Earth. On the
> other side of the border you again have two vectors a straight vector
> and that curly one you don't like me to call eddy current.
>
> * Call it what you wish. Capitalising the word within a sentence, as you
> did before, is usually reserved for phenomena named after the people who
> discovered them - but that's an engineer thing.
Big deal
>
> Yup. Everything falls together nicely thank you, when you study
> radiation from first principles instead of binding yourself to books
> that readily admit to not understanding the radiation process.
>
> * If you believe so fervently in your own version of all this then why don't
> you submit it to peer review at sci.physics or sci.physics.research? It's
> rather unfair to expose this only to an amateur radio newsgroup when what
> you are doing is apparently re-writing physics in such a major way. Those
> newsgroups would be more appropriate, considering the depth to which you are
> going (i.e. well beyond amateur radio) - wouldn't you agree.
Not really, espeilly if the groups has the same stature as this one.
>
> Now
> this is not being arrogant when one has applied all principles
>
> * My dictionary defines 'arrogant' as: 'having or showing an exaggerated
> opinion of one's own importance , merit, ability, etc.' from the Latin
> 'arrogare': 'to claim as one's own'. Your re-writing of physics conforms to
> this definition precisely.
So it is arrogant to do things for your self from first principles to
confirm whether printed matter is correct or not? Why would anybody
broaden the outlook of others work without determining that its
foundations was of rock or sand? Trust but verify seems very apt here.
>
> but I
> do think it is arrogant of you and others to asasinate the character
> of "S" purely because of his english and spelling and not to help one
> that wants to learn.
>
> * He is showing no inclination of wanting to learn - he asks a question, and
> then when answered responds with non-physical contradiction. His treatement
> here is a direct consequence of that behaviour, like yours.
Could be but he is not alone. He is surrounded by pseudo experts in
all fields who can prove nothing for themselves and relying on the
printed matter of others. Might just as well close the doors of a
library for a year to determine which book came out on top. Why on
earth would somebody turn him away from hydrolics when it like
electric current has laminar flow
>
> Just think about it all. I have shown how a particle moves in space in
> a straight line trajectory such that the particle maintains a straight
> line without the parabolic force of gravity driving it down to the
> Earth. Go back to your books and show just how the electrostatic field
> came about and what was the borders that it was contained in
> Art
> via applied spin and where gravity
>
> * As I say, you should present your theory to sci.physics and
> sci.physics.research if you have any interest in checking whether it is
> correct, and not limit its exposure to this group. Do let us know when you
> have posted there.
No, that is not the choice I have made. I decided to merge a paper
aproach with that of a patent request . You have read one patent
request and you have to wait for the PTO to print out the concluding
application. I am sharing it with industry and not the boneheads
who bunch themselves into secret rooms away from those outside who
cannot possibly provide anything of interest. They have the common
interest that if it doesn't come from them..........!!!!!! Pretty much
the same as this group. We shall see
Art
>
> Chris
Mike Coslo
> On Tue, 22 Sep 2009 10:48:43 +0200, Szczepan Bia?ek <sz.b...@wp.pl>
> wrote with abysmal quoting:
>
>> It will not be easy to work out here all details.
>> S*
>>
>>
>> Regards
>> Art
>
> You guys need to get a room.
I think that might already be the case.
- 73 de Mike N3LI -
Art Unwin
> > wrote with abysmal quoting:
>
> >> It will not be easy to work out here all details.
> >> S*
>
> >> Regards
> >> Art
>
> > You guys need to get a room.
>
> I think that might already be the case.
>
> - 73 de Mike N3LI -
Mike I am certainly not wired the same as Clark as I'm married and
have great grand children. It maybe normal for him to go into a room
with a man but why would he paint such a picture of "S" to infer that
he is as sick as him? I like women like any other real man
and certainly do not condone the practices that Clark and possibly you
would aproove of
with respect to males or airport bathroom manners
Richard Fry
> > If what you are saying were really happening,
> > an antenna would radiate two times the applied frequency,
> > but it obviously doesn't.
>
> We do not have the both. But we have the Luxembourg effect. Each dipole
> antena radiate two times the applied frequency, The pulses from the ends
> are 180 degrees apart.
So then, Szczepan, should transmissions using such antennas, and
expecting to be received on frequency "X" transmit on frequency "X" /
2 ? Note that such is not the reality.
Some low-level radiation from the transmit antenna may exist at twice
the carrier frequency, but in almost all cases it arises from
insufficient suppression of the 2nd harmonic of, and in the
transmitter.
And in NO case is it produced as you describe above.
RF
Registered User
<arthu...@gmail.com> wrote:
>
>Well look at how salvage yards sort out metals into different
>enclosures.
>They apply a displacement current to a conveyor where each piece of
>metal is elevated with spin such that it lands in the appropriate
>enclosure which is dependent on the resistivity of the metal elevated.
This isn't exactly how such systems work. Abstractly the system is a
metal detector and a sorting table hanging off a CAN. A controller at
the other end of the CAN 'reads' the discriminator and 'writes' to the
sorter. The writes open and close ejector nozzles. These are the magic
devices that cause the material to 'elevate with spin'.
>This method of elevating scrap for recovery has been used for years
>and it is the same action that is applied to particles for radiation.
>Why would you need a citation for a practice that is well known and in
>use?
Because you might be wishing your agenda into how you propose things
work. Who'da thunk that!
christofire
"Richard Clark" <kb7...@comcast.net> wrote in message
news:v44ib5dm14lno81rg...@4ax.com...
Richard,
It's not something I have control over using Outlook Express, and it has
been commented upon by others before. Whether OE maintains the '>'s before
lines of quoted text seems to depend on something in the text I'm quoting.
I gather it is a known issue, and it's the reason why I sometimes use a '*'
before the first line of new paragraphs of my response. I have been told I
should ditch OE but I don't want to use webmail; I could use Agent but I
haven't so far.
Chris
Roy Lewallen
>
> Richard,
>
> It's not something I have control over using Outlook Express, and it has
> been commented upon by others before. Whether OE maintains the '>'s before
> lines of quoted text seems to depend on something in the text I'm quoting.
> I gather it is a known issue, and it's the reason why I sometimes use a '*'
> before the first line of new paragraphs of my response. I have been told I
> should ditch OE but I don't want to use webmail; I could use Agent but I
> haven't so far.
>
> Chris
I highly recommend Mozilla Thunderbird.
Roy Lewallen, W7EL
christofire
On Sep 22, 6:51 am, "christofire" <christof...@btinternet.com> wrote:
> "Art Unwin" <arthurun...@gmail.com> wrote in message
- - snip - -
> * As I say, you should present your theory to sci.physics and
> sci.physics.research if you have any interest in checking whether it is
> correct, and not limit its exposure to this group. Do let us know when you
> have posted there.
No, that is not the choice I have made. I decided to merge a paper
aproach with that of a patent request . You have read one patent
request and you have to wait for the PTO to print out the concluding
application. I am sharing it with industry and not the boneheads
who bunch themselves into secret rooms away from those outside who
cannot possibly provide anything of interest. They have the common
interest that if it doesn't come from them..........!!!!!! Pretty much
the same as this group. We shall see
Art
* OK, at an appropriate juncture I'll invite some of them to come over and
take a look at what you write here (crossposting would probably be frowned
upon). It might be enlightening to receive the views of some physicists.
Chris
Art Unwin
> "Art Unwin" <arthurun...@gmail.com> wrote in message
>
> news:9eaf5504-1d95-48b7...@f33g2000vbm.googlegroups.com...
> On Sep 22, 6:51 am, "christofire" <christof...@btinternet.com> wrote:
>
> > "Art Unwin" <arthurun...@gmail.com> wrote in message
>
> - - snip - -
>
> > * As I say, you should present your theory to sci.physics and
> > sci.physics.research if you have any interest in checking whether it is
> > correct, and not limit its exposure to this group. Do let us know when you
> > have posted there.
>
> No, that is not the choice I have made. I decided to merge a paper
> aproach with that of a patent request . You have read one patent
> request and you have to wait for the PTO to print out the concluding
> application. I am sharing it with industry and not the boneheads
> who bunch themselves into secret rooms away from those outside who
> cannot possibly provide anything of interest. They have the common
> interest that if it doesn't come from them..........!!!!!! Pretty much
> the same as this group. We shall see
> Art
>
> * OK, at an appropriate juncture I'll invite some of them to come over and
> take a look at what you write here (crossposting would probably be frowned
> upon). It might be enlightening to receive the views of some physicists.
>
> Chris
If you know of any I would welcome their views. There are many retired
educated people in this world today that turn to that which they had
an interest with when young. Now it is difficult to get up to speed in
different sciences because various journals get the rights of various
papers from Universities e.t.c which are then denied to libraries and
the public.
This is a resource the country should assist because its costs are low
and where all have
large experience obtained thru their working years. Imagine
professionals who when retired
have twenty or more years of experience be allowed to follow and
contribute in areas where an interest has laid dormant for so long.
Today's efforts are applied to computers where data comes out in
bundles which have to be sorted to determine if anything good is being
offered by using a mish mash of arithmetic formulae that are merged
with similar formulae from different functions. Sad, sad, sad.
tom
I second that recommendation.
tom
K0TAR
christofire
* It's true that a lot of effort is put into the areas that yield the
greatest profit, and computing in one form or another does seem to have a
grip at the moment. However, it is enlightening to take a look from time to
time at news groups like the two I named to see the sorts of things they are
discussing, and the _unbounded_ nature of the universe (which is what I
wrote) is one of them. They too appear to have input from ex-professionals.
Have you ever tried to obtain access to a technical library in a university
or one of the engineering institutions? You might be surprised how easy or
inexpensive it turns out to be. As a member of the general public I have
access to the IET library in London to read as much as I wish, and to
photocopy.
Chris
Art Unwin
> On Tue, 22 Sep 2009 13:24:17 -0700 (PDT), Art Unwin
>
Interesting. Can you point to an article or something on the web that
describes what you
say. For myself I have only run into articles by special purpose
machine manufacturers
who deal with sorting machines for scrap yards which deals with many
materials including plastics , glass etc as well as different metallic
materials. This sorting aproach that you mention sounds rather
interesting if they are relying on magic or voodoo!
Richard Fry
after major clippage:
> Today's efforts are applied to computers where data comes out in
> bundles which have to be sorted to determine if anything good is being
> offered by using a mish mash of arithmetic formulae that are merged
> with similar formulae from different functions. Sad, sad, sad.
However as far as NEC is concerned, your sad-sad-sadness applies only
to those (possibly even yourself) using computer software without
sufficient understanding of how it should be used, and applied.
Note that the U.S. FCC has endorsed the use of NEC in the licensing
process of many types of directional MW broadcast arrays -- given that
such NEC analysis was performed by someone with provable and
acceptable competence.
RF
Art Unwin
Yes, some university libraries allow access to the public but not for
copies. These must
come from journals at quite high prices. Here you can be a member of a
professional society say IEEE but to get the journals of say antennas
and propagation then you must pay a couple of $100 to have access to
them. This is on top of the fees for the institution and the group
that you are personally a member of which also requires fees. But the
U.S. is not like being in London where you can take the tube to any
where such as the patent office library or visit the one on Birdcage
Walk ( Royal Institution of Mechanical Engineers in the old days)
I am not aware of this IET that you mentioned. Here in Illinois which
is the size of the UK plus has a population that London sees every
day of the week! One library I would like to get into is on
Whitechapel road in Stepney ( Queen Mary college) where extensive work
is done on antennas. Anyway what I do is to start right at the
beginning ie first principles
and with antennas stuck in a rut for so long it was a good one for me
as a retired person
to fiddle with as it was nice to talk to my buddies at BAC St Albans
and nearby towns when radio itself was a hobby for me but most have
now passed away. Now I have finished my personal antenna studies and I
will have to turn to the honey doos that have piled up over the last
few years even tho I have had a handy man come in regularly even so I
go thru periods where every thing that I own is broken and I must turn
away from my hobbies.
Dave
On Sep 21, 7:33 pm, "christofire" <christof...@btinternet.com> wrote:
<snip crap>
>No it does not appear in Kraus book. He never followed Maxwells laws
>with respect to equilibrium.
<snip more crap>
and he had a good reason not to... because its all made up by you art, and
its totally WRONG!
Dave
"Szczepan Bialek" <sz.b...@wp.pl> wrote in message
news:h99mvm$9hm$1...@node1.news.atman.pl...
>
> "Dave" <no...@nowhere.com>
> news:LETtm.2963$Jd7....@nwrddc02.gnilink.net...
>>
>> "Szczepan Bialek" <sz.b...@wp.pl> wrote in message
>> news:h980ba$tt$1...@node1.news.atman.pl...
>>>>>>
>>>>>>>In the Gas Analogy the monopole antena is exactly like the Kundt's
>>>>>>>tube.
>>>>>>
>>>>> Heaviside did the Hydraulic Analogy. All is exactly the same like in
>>>>> the fluids mechanics.
>>>>> Next the electrons were discovered. Automatically Heaviside is a
>>>>> history and the Gas Analogy is in power.
>>>>>
>>>>> But you, radio people, are very close to waves and should be easy for
>>>>> you to work out the answer for the Question:
>>>>> Which Analogy is right?
>>>>
>>>> neither analogy is 'right'. they are useful in limited circumstances
>>>> to demonstrate some basic pressure wave physics to young students. but
>>>> neither one properly reproduces electromagnetic waves.
>>>
>>> "electromagnetic waves" are paper waves. Radio waves are real waves. Now
>>> we must not know what the waves are like. Now we should estabilish from
>>> "At the open circuited ends of a resonant antenna there is almost double
>>> the forward voltage but zero total current due to cancellation of the
>>> dorward and reflected currents at the open circuit. At the open circuit
>>> in the wire, all the energy in the wave is transferred to the electric
>>> field."
>>> S*
>>>
>>>
>> we know what they are like, you just have to understand the mathematics.
>
> I understand. Without that it was be impossible to know that Maxwell
> proposed the displacement current to save the incompressible electric
> fluid. In Maxwell times AC current was known. To pass the incompressible
> fluid through a capacitor the displacement current is necessary. I prefere
> the compressible electrons. They compress in the plates and nothing flow
> between them. The polarization is not the macro flow.
>
>> and yes, richard's statements are true, but a bit too restrictive, it
>> doesn't HAVE to be resonant. Voltage doubles and current=0 at the end of
>> any wire fed with a time varying current, it doesn't even have to be a
>> sine wave... note the effect of sending square waves from a time domain
>> reflectometer down an open circuited wire.
>
> Yes. But antennas are in resonance.
> S*
>>
antennas don't have to be 'in resonance'.... a very short dipole radiates
almost as well as one 1/2 wavelength long... its all in the fields.
Art Unwin
> "Art Unwin" <arthurun...@gmail.com> wrote in message
So you are back David ! have you built that four poster antenna yet,
of steel I presume,
for the top band? Hopefully the system of yours is in a state of
equilibrium so you can tell us good things about it. Did you have to
make a ground plane system? They are not needed for a system in
equilibrium so you may have displaced a lot of moles from their
habitat for no good reason. Haven't heard you mention anymore about
that book you were writing on antennas. I assume you do not have a
chapter about equilibrium as yet. And that problem you had about the
legality of turning a static field into a dynamic field, have you made
any progress on that yet ( just pulling your tail)
tom
>> Yes. But antennas are in resonance.
>> S*
>>>
> antennas don't have to be 'in resonance'.... a very short dipole
> radiates almost as well as one 1/2 wavelength long... its all in the
> fields.
As I tried to make Art understand with respect to yagis, it's not
necessarily resonant (or even usually), it's matched. By the matching
network.
It was like releasing excess nitrogen containing waste into the wind,
like all things are with Art.
See wall, hit with head, repeat.
tom
K0TAR
tom
> So you are back David ! have you built that four poster antenna yet,
> of steel I presume,
> for the top band? Hopefully the system of yours is in a state of
> equilibrium so you can tell us good things about it. Did you have to
> make a ground plane system? They are not needed for a system in
> equilibrium
Wow! No ground system needed for top band. The AM broadcasters are
going to be all over this. Art, you are about to become very very rich.
This overturns every measurement ever made on AM broadcast antenna
arrays. You should be very proud that hundreds of engineers and their
measurements were wrong.
Again, congratulations.
tom
K0TAR
Art Unwin
And you think I didn't know that like it was a personal secret of
yours?
I don't even remember discussing yagis with you. I moved away from
those years ago when I was building antennas with 80 foot booms and a
dozen or more elements all made of fishing poles with aluminum foil
surfaces. Now I am interested in antennas the size that Chip works
with but not with the same aproach. In fact I am committed to having
antennas made that are small enough for those with small gardens to
use, where how much money or land that you have allow you to run over
others is not a measure of ones skills. As for you you just sit on the
side lines with nothing to offer but insults since you appear to live
alone berift of friends. Try smiling instead of just growling and get
a life.
tom
>> As I tried to make Art understand with respect to yagis, it's not
>> necessarily resonant (or even usually), it's matched. By the matching
>> network.
>>
>> It was like releasing excess nitrogen containing waste into the wind,
>> like all things are with Art.
>>
>> See wall, hit with head, repeat.
>>
>> tom
>> K0TAR
>
> And you think I didn't know that like it was a personal secret of
> yours?
> I don't even remember discussing yagis with you. I moved away from
> those years ago when I was building antennas with 80 foot booms and a
> dozen or more elements all made of fishing poles with aluminum foil
> surfaces. Now I am interested in antennas the size that Chip works
> with but not with the same aproach. In fact I am committed to having
> antennas made that are small enough for those with small gardens to
> use, where how much money or land that you have allow you to run over
> others is not a measure of ones skills. As for you you just sit on the
> side lines with nothing to offer but insults since you appear to live
> alone berift of friends. Try smiling instead of just growling and get
> a life.
Well, let me refresh your memory on yagis and resonance. It was a LONG
time ago, September 2 2009. Here's the copy.
---------------------------------
>> It is possible to have a resonance at the match frequency, but unusual.
>> Run a couple yagis through Eznec and check the driven element impedence.
>> For instance, the NBS yagi that comes with Eznec. At 50.125 it's
>> 11.7+j5.9, not very resonant. I have a 222 yagi that has a DE longer
>> than the reflector, also unusual, but it works fine.
>>
>> tom
>> K0TAR
>
> My goodness. This is the silliest thing I have heard in years.
> A yagi has a bandwidth based around a resonant frequency which is the
> change over point
> around which the impedance is not measured by the resistance alone.
> If the feed radiator is devoid of any proximitry effect then its
> material length differs from that of an array where the proximetry
> actions exists. Thus a yagi driven element is resonant and different
> from a driven element alone because of the proximitry effect. If your
> radiator impedance is 11.7 +j5.9 then you are not at the point of
> resonance which must be solely resistive. Modify the proximetry effect
> to make it resonant.
> This is exactly what you did by using a short length reflector at a
> abnormal distance from the driven element. Methinks you regard only 50
> ohms as being resonant where anything different from that is not
> resonant. Tom you are nuts!
Well, let's take a look at some yagis from known companies, the US
National Bureau of Standards and some well known hams. This is a small
subset of the yagis I have models for. Almost all are HF or low VHF so
we don't have to get into an argument over proper boom correction
factors. Yes, something else you've never heard of, Art.
Cushcraft
Model Impedance at design frequency
215 23.9+j6.4
17B2 27.4+j8.3
20-3CD 17.9-12.0
20-4CD 18.7-j17.7
617-6B 36.4+j11.5
A147-11 24.9+j6.1
A50-5 28.6-j20.1
A50-5S 36.7-j10.9
A50-6S 25.2-j19.0
US NBS
NBS122 20.0-j4.8
NBS32 25.2-j7.6
HyGain
HG105BA 33.4-j26.0
HG155BA 20.1-j7.6
HG175CA 21.4-j27.0
HG204BA 25.8-j20.7
HG205CA 20.7-j22.4
HG215DX 27.7-j29.2
M Squared (or M^2)
2M5WL 24.9-j33.9
2M7 31.6-j10.4
6M2.5WL 40,2+j9.5
DL6WU (note this is the grandfather of the great 2m/70cm designs
like K2RIW, K1FO, and M^2 among others)
DL6WU 34.4-11.4
W1JR
W1JR3170 15.8-j17.7
W1JR66 29.2+j24.6
Note that none of these are particularly close to resonance at the
design frequency.
Art, you don't know what you're talking about. As usual.
-----------------------------------
Remember now Art?
tom
K0TAR
Art Unwin
I really don't know what the point is that you are trying to describe.
Statements like the "design frequency "of the above antenna don't
really make sense if you do not provide what the intended design
frequency was in the first place.
On top of that, all the antennas that you point to could all be made
resonant in the right type of environment. Really there is more
important things than just getting rid of that which is to get rid of
your SWR meter since bandwidth, gain and many other things force one
into different judgements. All that you show may well be very good
designs but it is surely not a recommendation to avoid radiators that
are resonant in favor of those that are not exactly resonant as your
list shows
Is it possible that your definition of resonance means a resistive
impedance of fifty ohms?
You have made similar errors in the past is why I ask.
All antenna requirements have to be weighted to achieve what the
designer is aiming for, but nowhere is the degree of achievements
shown compared to the designers needs. So what exactly are your
judgement parameters of a good antenna and are they considered
universal for all environments?
I have given you a good opportunity to show off your talents with
respect to antennas Let's see how you use it or will you decline for
fear of exposure to the group
and fall back on a insult or two as you normally do ?
Art Unwin
No congratulations necessary. The methods will be available to all
amateurs when the PTO print out comes about. All of this represents a
different aproach than that taken before me. If it is 90 % in error
where only 10% is useful that's fine by me as it represents an advance
supplied by an amateur who wasn't detered from experimenting by
others.
If I feel that something that is taken for granted is in error just
because an author had it printed in a book I will always step forward
if I feel the emperer does not wear any clothes.
Which of you have the courage to stand along side me knowing full well
your character will be dismantled before the true facts become known?
Jeff Liebermann
<arthu...@gmail.com> wrote:
>Yes, some university libraries allow access to the public but not for
>copies. These must
>come from journals at quite high prices. Here you can be a member of a
>professional society say IEEE but to get the journals of say antennas
>and propagation then you must pay a couple of $100 to have access to
>them. This is on top of the fees for the institution and the group
>that you are personally a member of which also requires fees.
Ummm... IEEE full membership is now $175/year. Membership in the
Antennas and Progagation group is $24/year.
<http://www.ieee.org/web/membership/Cost/dues.html>
That's about 4000 pages for the extra $24.
If you are a retired former IEEE member, currently unemployed,
disabled, or are working for only peanuts or stock options, you can
get up to a 50% discount on dues.
<http://www.ieee.org/web/membership/Cost/special_circumstances.html>
You can see the articles in each past issue at:
<http://ieeexplore.ieee.org/xpl/RecentIssue.jsp?puNumber=8>
If you're a cheap tightwad non-member, like me, you can buy individual
articles ala carte for $29 each:
<http://ieeexplore.ieee.org/guide/g_tools_apo.jsp>
If you don't mind just older papers, a member can buy the DVD with
everything from AP-S from 1952-2000:
<http://www.ict.csiro.au/aps/cdrom.htm>
for $100.
If you're a non-member, you really pay hansomly for the printed
publications.
<http://www.ieee.org/portal/cms_docs_iportals/iportals/publications/subscriptions/info/IEEE_Sub_Price_List_2010.pdf>
For example, the Antenna and Propagation IEEE Transactions for a year
(12 issues) costs $1,200.
--
Jeff Liebermann je...@cruzio.com
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558
Szczepan Białek
"Richard Clark" <kb7...@comcast.net> wrote
news:5o3ib5172vqk81so7...@4ax.com...
>
> I'm glad to see you shed that nonsense about hydraulics. As you
> understood that topic far less than RF (which is in itself on very
> shaky ground), it wouldn't have done to poison the well.
>
> This poor understanding is quite obvious by your last comment above.
I was absent. You wrote:
">All is exactly the same like in the fluids mechanics.
"Exactly" makes it very, very easy to show how an analogy fails:
Describe the laminar flow in terms of
the Reynolds number for
the interface between RF and a Biconical Antenna
and
the interface between RF and a thin wire Antenna."
Maxwell's math is the same as for fluid:
"Maxwell's equations are simply a re-arrangement of relationships worked
out by Faraday in respect of charge and only verified at low speed.
Maxwell discovered that the relationships could be arranged in a form
which mirrored the mathematical description of a fluid"
In fluid are whirls. The magnetic field is like whirl. But not for all
scientists.
"Maxwell's equations" were wrote by Heaviside.
S*
Szczepan Białek
"Dave" <no...@nowhere.com> wrote
news:fMeum.3083$Jd7....@nwrddc02.gnilink.net...
>>>>>>>>In the Gas Analogy the monopole antena is exactly like the Kundt's
>>>>>>>>tube.
>>>>
>>>
> antennas don't have to be 'in resonance'.... a very short dipole radiates
> almost as well as one 1/2 wavelength long... its all in the fields.
In my antenna radiate the end where the voltage is doubled. In your
something alse. What?
S*
>
Szczepan Białek
"Richard Fry" <rf...@adams.net> wrote
news:ddc5f438-c0e6-429f...@o13g2000vbl.googlegroups.com...
On Sep 22, 3:34 am, Szczepan Bia�ek <sz.bia...@wp.pl> wrote:
>> Up to now the acoustic analogy is fully applicable.
>Not if one understands the physics of radiation.
Physics of radiation is unknown. Antennas are the nice apparatus to analyse
it.
>> But it is experimentally proved. Stationary charge - electric field,
> Moving charge - magnetic field.
>Untrue, and I challenge you to cite any credible experimental data
that you think proves your belief.
>Far-field EM radiation is produced only by the current flow on the
antenna, and that radiation contains BOTH the electric and the
magnetic fields.
For me the magnetic field is the illusion.
>You may have missed the accurate description posted by Chris, and
pasted below.
"The acceleration of charge in an antenna results almost entirely from
the
applied potential difference at its terminals. The radiated fields
result
from the alternating current effectively passing through the radiation
resistance, and all the other, reactive, fields have no direct effect
on the
radiation resistance, or the component of the current that passes
through it
in phase with the voltage that is developed across it, which together,
of
course, represent the radiated power. The reactive fields affect the
terminal impedance and a large imaginary part can upset the device
trying to
send power into the antenna, but that is more of a system issue. The
alternating current that passes through the radiation resistance is
composed
of charge that moves in time with each RF cycle, accelerating and
decelerating accordingly. The electrostatic field developed between
the
ends of a half-wave dipole reaches its maximum value a quarter of a
cycle
later than the voltage at the drive point so any effect it has on the
charge
in the antenna elements during each cycle must be reactive, and it
doesn't
affect the radiation resistance or the radiated wave."
My description is shorter:
The supply unit sends the voltage pulses (in opposite phase) in the
transmissing line. If such pulses collide the voltage is doubled and the
strong radiation take place. In straight radiator the forward pulse collides
with the reflected. In folded dipoles with that from the other wire.
S*
Szczepan Białek
"Cecil Moore" <nos...@w5dxp.com> wrote
news:nq9um.68047$u76....@newsfe10.iad...
> Szczepan Bia�ek wrote:
>> Each dipole antenna radiate two times the applied frequency, ...
>
> Sorry, 2*sin(2wt) <> sin(wt)
>
> Here's the question: Is the radiated RF wave in phase
> with the standing wave current or in phase with the
> standing wave voltage? The radiated RF wave cannot be
> in phase with both since they are 90 degrees out of
> phase on the standing wave antenna.
In the Gas analogy radiate only the doubled voltage. The doubling take place
in the ends of the two radiators. The two spherical waves are radiated. So
some receiver antennas can work on doubled frequency (Luxembourg effect).
S*
Szczepan Białek
"Richard Fry" <rf...@adams.net> wrote
news:7c13262e-ed2c-44bd...@e18g2000vbe.googlegroups.com...
On Sep 22, 1:44 pm, Szczepan Bia�ek <sz.bia...@wp.pl> wrote:
>> > If what you are saying were really happening,
> > an antenna would radiate two times the applied frequency,
> > but it obviously doesn't.
>
>> We do not have the both. But we have the Luxembourg effect. Each dipole
> antena radiate two times the applied frequency, The pulses from the ends
> are 180 degrees apart.
>So then, Szczepan, should transmissions using such antennas, and
expecting to be received on frequency "X" transmit on frequency "X" /
2 ? Note that such is not the reality.
It happened in 1930. The Luxembourg LW were received on MW radio sets.
>Some low-level radiation from the transmit antenna may exist at twice
the carrier frequency, but in almost all cases it arises from
insufficient suppression of the 2nd harmonic of, and in the
transmitter.
The Luxembourg effect is only possible if the both ends of the dipole are
"visible". The mast was on the tip top.
>And in NO case is it produced as you describe above.
It is easy to check. Now no vertical LW masts. But everybody has a
horizontal dipole.
S*