On Sep 18, 12:55*pm, "Dave" wrote:
"Art Unwin" wrote in message

...

With "David" being barred from the group there should be no obstacle
to closing this thread

who is barred?? *i've just been busy with real antennas getting ready for
winter. *i will be gone in another week or so because verizon caved in and
will be dumping usenet. *maybe i'll resurrect myself with some other
service, but none that i have to pay for, you aren't that much fun.

It really was all for his benefit so he could provide insults and now
there is no need for it.

awww, i'm sure there are others who get as much of a laugh at your rantings
and random phrases as i do! *keep going!!

Your profile states you are barred on all of your past postings!

"Art Unwin" wrote in message
...

Your profile states you are barred on all of your past postings!
profile?? what profile?? we don't need no stinking profile! this is
usenet, i post via a news server, it has no profile of me, it just accepts
postings and sends them out to the world.

On Sep 18, 3:18*pm, "Dave" wrote:
"Art Unwin" wrote in message

...

Your profile states you are barred on all of your past postings!

profile?? *what profile?? *we don't need no stinking profile! *this is
usenet, i post via a news server, it has no profile of me, it just accepts
postings and sends them out to the world.

Well that is good news. I await your next insult

Cecil, W5DXP wrote:
"The forwerd and the reflected current sre in phase at the feedpoint and
thus radiation is maximum at that maximum current point."

Exactly. The reflected current has made s round trip of 180 degrees plus
the reflection has added another 180 degrees for a total of 360 degrees
or back in phase with the forward current at the feedpoint.

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.

The relationship between the input power of a given antenna and the
strength of the far electromagnetic field which it will produce depends
among other factors the length and shape of the antenna.

Best regards, Richard Harrison, KB5WZI

Dave wrote:
i will be gone in another week or so because verizon caved
in and will be dumping usenet. maybe i'll resurrect myself with some
other service, but none that i have to pay for, you aren't that much fun.

Groups.google.com is free if one can tolerate the clumsy interface.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

"Richard Harrison" wrote
...
Cecil, W5DXP wrote:
"The forwerd and the reflected current sre in phase at the feedpoint and
thus radiation is maximum at that maximum current point."

Exactly. The reflected current has made s round trip of 180 degrees plus
the reflection has added another 180 degrees for a total of 360 degrees
or back in phase with the forward current at the feedpoint.

At the open circuited ends of a resonant antenna there is almost double
the forward voltage

Could you tell us if the "almost double voltage" is the measured of
theoretically predicted?

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.

The relationship between the input power of a given antenna and the
strength of the far electromagnetic field which it will produce depends
among other factors the length and shape of the antenna.

Best regards, Richard Harrison, KB5WZI
S*

Szczepan Bia³ek wrote:
Could you tell us if the "almost double voltage" is the measured of
theoretically predicted?

Since a voltage reference point is difficult to achieve
at the ends of a dipole, we rely on the conservation of
energy principle. Since the current is zero at the ends
of a dipole, all the energy must be contained in the
electric field. With that knowledge, the voltage can
be estimated.

Such is easy to understand by taking voltage measurements
on a 1/4WL open-circuit stub. If one uses resistance wire
for the stub, one can simulate radiation loss in a dipole.

The following is a transmission line simulation of a 1/4WL
monopole designed to run on the free demo version of EZNEC
available from www.eznec.com

http://www.w5dxp.com/stub_dip.EZ

The user defined resistivity of the wire is what causes
the 35 ohm feedpoint resistance akin to a 1/4WL monopole.
The 10 megohm load allows us to look at the voltage
across the "open" end of the stub. It is 1033 volts for
a 100 watt input. We can assume the forward voltage and
the reflected voltage at the end of the stub are equal
at 516 volts.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

"Cecil Moore" wrote
...
Szczepan Bia³ek wrote:
Could you tell us if the "almost double voltage" is the measured of
theoretically predicted?

Since a voltage reference point is difficult to achieve
at the ends of a dipole, we rely on the conservation of
energy principle. Since the current is zero at the ends
of a dipole, all the energy must be contained in the
electric field. With that knowledge, the voltage can
be estimated.

Such is easy to understand by taking voltage measurements
on a 1/4WL open-circuit stub. If one uses resistance wire
for the stub, one can simulate radiation loss in a dipole.

The following is a transmission line simulation of a 1/4WL
monopole designed to run on the free demo version of EZNEC
available from www.eznec.com

http://www.w5dxp.com/stub_dip.EZ

The user defined resistivity of the wire is what causes
the 35 ohm feedpoint resistance akin to a 1/4WL monopole.
The 10 megohm load allows us to look at the voltage
across the "open" end of the stub. It is 1033 volts for
a 100 watt input. We can assume the forward voltage and
the reflected voltage at the end of the stub are equal
at 516 volts.

The acoustic analogy predict it: "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."
S*

--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

Szczepan Bia³ek wrote:
In antennas is electron gas.

Your "electron gas" is a cloud of photons emitted
by electrons. Those photons are infinitely
"compressible".

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."

Again, it is impossible for the voltage maximums and
current maximums to both radiate since such would
double the radiated frequency which it doesn't. We
usually transmit and receive on the same frequency.

You must decide whether the peak radiation comes
from the current maximum or the voltage maximum since
they do not occur at the same time. Until you make
that choice, rational discussion is not possible.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

"Szczepan Bialek" wrote in message
...

"Richard Harrison" wrote
...
Cecil, W5DXP wrote:
"The forwerd and the reflected current sre in phase at the feedpoint and
thus radiation is maximum at that maximum current point."

Exactly. The reflected current has made s round trip of 180 degrees plus
the reflection has added another 180 degrees for a total of 360 degrees
or back in phase with the forward current at the feedpoint.

At the open circuited ends of a resonant antenna there is almost double
the forward voltage

Could you tell us if the "almost double voltage" is the measured of
theoretically predicted?

both