On Sep 22, 11:34*am, tom wrote:
Art Unwin wrote:
On Sep 21, 9:23 pm, tom wrote:
Art Unwin wrote:
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

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

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.


------------ Current laws of physics ---------------

Wowl he got something right for a change


Jaro

"Cecil Moore" ...
Szczepan Bia³ek wrote:

"Cecil Moore" 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*

"christofire" wrote
...

"Szczepan Bia³ek" wrote in message
...


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/Sear...nnas&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...diation%20Lab/

S*

U¿ytkownik "Richard Clark" napisa³ w wiadomo¶ci
...
On Mon, 21 Sep 2009 18:53:15 +0200, Szczepan Bia?ek
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'. ;-)

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*

"Dave"
...

"Szczepan Bialek" wrote in message
...

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
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*

"Richard Fry" wrote
...
On Sep 21, 12:23 pm, Szczepan Bia³ek 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

"Art Unwin" wrote
...
On Sep 21, 12:23 pm, Szczepan Białek wrote:
 "tom"
news:4ab41e80$0$42842$8046368 ...

Szczepan Bia³ek wrote:

God forbid that you should actually do some research! Â 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.

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.

It will not be easy to work out here all details.
S*


Regards
Art

On Sep 22, 3:13*am, Szczepan Bia³ek wrote:

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

Szczepan Bia³ek wrote:
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

On Sep 22, 3:34*am, Szczepan Bia³ek wrote:

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.

"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."

RF