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Old May 12th 09, 03:44 PM posted to rec.radio.amateur.antenna
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Default wave polarisation

On Tue, 12 May 2009 09:56:18 +0200, Szczepan Bia?ek
wrote:


"Richard Clark" wrote
.. .
On Mon, 11 May 2009 19:21:32 +0200, Szczepan Bia?ek
wrote:


Now we know that
electrons have mass and are compressible.


Nonsense.


Millikan measured it in XIX centuary. Latelly Wim wrote: "Hello Szczepan,

You are right, charge is compressible. The charge that is required to
charge (for example) a sphere seems to break the coninuity equition as
is used for incompressible fluid in hydraulics".


Now we are into utter nonsense.

So at the ends appear and
disappear the huge charges. In that case an electric impulse is send when
the current flows to (from one end) , and the next when the current flows
fro (from the other end).


Nonsense.


I was told that the electrons want to escape from the ends of a dipole (
balls are a remedy).


Then you were told nonsense. At least study something with accurate
fundamentals.

Let assume that something is radiated from the end


This alone explains why you have a very poor understanding of the
dynamics. An antenna radiates in ALL directions from EVERYPOINT of
the antenna. No one has to make special "assumptions" that are
limiting and with the appearance of being special knowledge.

when the charge
(compressed electrons) appears. There are the two ends. How many pulses will
be send in space ?


Radiation is not pulses.

Does your measurements distinguish radiation from the spakrks from that
from
the plates?
S*


Of course they do. More the question: can you measure them too?


No.



Then you have very limited resources and absolutely no basis for
discussion of the topic.

The only instrument I have is the comb. But it was enough to verify the
second meaning of word "polarisation".


That adds nothing to the topic except you are working at very crude
levels with very limited knowledge about a vastly more complex issue.

If you cannot, then this explains the nonsense.


Something is radiated from the ends of the dipole. Can you detect it?


I already said I could, and you said you couldn't. Why do you ask
again?

73's
Richard Clark, KB7QHC
 
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