He's truly a master, isn't he?

Roy Lewallen, W7EL

Cecil Moore wrote:
Tdonaly wrote:

What causes the light to light up, Cecil, an E field, or a V field?


According to quantum electrodynamics, fields don't exist. So are
you talking about photons or virtual photons above? In either case,
voltage can still be measured by a voltmeter.

Cecil, I'm sorry you don't understand that in the presence of
time-varying fields, the potential between two points depends on the
path you take. Grasping that concept can be very empowering in
understanding what's going on in antennas, and in transmission lines,
and in inductors and transformers. I can only hope that some lurkers
have benefitted from the discussion.

Cheers,
Tom

Cecil Moore wrote in message ...
Tom Bruhns wrote:
There is very little voltage ALONG the conductors, ...

Depends upon how long the conductors are. The difference along a
1/4WL conductor is known to be minimum VS maximum assuming a
minimum at one end. Are you saying that EZNEC doesn't display
the current distribution on an antenna when I press the 'i' key?

Tdonaly wrote:
I see you've gone into objection-stopper mode: write something
whether it makes sense or not.

Sorry, my neighbor forced some Amaretto upon me yesterday. The
fact that the voltage is hard to measure doesn't prove that it
doesn't exist. Simply loop the ends of the dipole close together
and measure the voltage with an RF voltmeter. It's a no brainer.
These are copper wires. No need to discuss fields.

During 1/2 cycle, there is an excess of electrons on one end.
During the next 1/2 cycle, there is an excess of electrons on
the other end. A voltmeter measures excesses of electrons between
two wires very well.

Or take a look at the end feedpoint impedance of a 1/2WL
monopole. It's a no brainer to discover that the voltage is high
and the current is low. So what is the feedpoint voltage for a
1/2WL end-fed monopole? EZNEC says it is around 3500 ohms giving
a feedpoint voltage of about 550v for 100w compared to a 1/4WL
monopole's feedpoint voltage of about 60v. Do you not understand
that the voltage to current ratios vary from high to low at 1/4WL
intervals along a wire antenna?

So please tell us how photons manage to light up a florescent
light bulb.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Roy Lewallen wrote:
He's truly a master, isn't he?

Ask an irrelevant question - obtain an irrelevant answer. :-)

Tdonaly wrote:
What causes the light to light up, Cecil, an E field, or a V field?

According to quantum electrodynamics, fields don't exist. So are
you talking about photons or virtual photons above? In either case,
voltage can still be measured by a voltmeter.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Tom Bruhns wrote:
Cecil, I'm sorry you don't understand that in the presence of
time-varying fields, the potential between two points depends on the
path you take.

I know that, Tom, but we are talking about measuring the RF voltage
between two copper wires one inch apart. The path is well defined.
It is a no brainer. There is no need for obfuscation. The measurement
proves the voltages at the ends of a dipole to be at least a magnitude
higher than the voltage at the feedpoint. Are you not aware of how
the ratio of voltage to current varies over 1/4WL of a wire antenna?
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Cecil wrote,


Tdonaly wrote:
I see you've gone into objection-stopper mode: write something
whether it makes sense or not.

Sorry, my neighbor forced some Amaretto upon me yesterday. The
fact that the voltage is hard to measure doesn't prove that it
doesn't exist. Simply loop the ends of the dipole close together
and measure the voltage with an RF voltmeter. It's a no brainer.
These are copper wires. No need to discuss fields.

During 1/2 cycle, there is an excess of electrons on one end.
During the next 1/2 cycle, there is an excess of electrons on
the other end. A voltmeter measures excesses of electrons between
two wires very well.

Or take a look at the end feedpoint impedance of a 1/2WL
monopole. It's a no brainer to discover that the voltage is high
and the current is low. So what is the feedpoint voltage for a
1/2WL end-fed monopole? EZNEC says it is around 3500 ohms giving
a feedpoint voltage of about 550v for 100w compared to a 1/4WL
monopole's feedpoint voltage of about 60v. Do you not understand
that the voltage to current ratios vary from high to low at 1/4WL
intervals along a wire antenna?

So please tell us how photons manage to light up a florescent
light bulb.
--
73, Cecil http://www.qsl.net/w5dxp


After all this time and all these posts, you're still trying to use
circuit theory to explain electromagnetic phenomena in
situations where length and time, as parameters, just can't
be ignored. Secondly, you haven't even demonstrated that you
took time to digest and understand Tom's original post, which
you responded to with a knee-jerk "Brain fart." That's o.k. for
you. If you really want to believe in a simplified view of
electronics that's fine. But, for the rest of us, all the books,
the research, the mathematics, and the thinking of people
from Faraday to Feynman, actually mean something.
73,
Tom Donaly, KA6RUH

Tom Bruhns wrote:
Cecil, I'm sorry you don't understand that in the presence of
time-varying fields, the potential between two points depends on the
path you take.

I know that, Tom, but we are talking about measuring the RF voltage
between two copper wires one inch apart. The path is well defined.

That's one path.

It is a no brainer. There is no need for obfuscation.

Universally recognized principles of electromagnetics are obfuscation?

The measurement
proves the voltages at the ends of a dipole to be at least a magnitude
higher than the voltage at the feedpoint.

You changed the geometry. But even if you hadn't, you might be
able to say the changing electrical fields are greater at the ends of
a dipole, but not the voltages, because the voltages
aren't uniquely defined.

Are you not aware of how
the ratio of voltage to current varies over 1/4WL of a wire antenna
--
73, Cecil http://www.qsl.net/w5dxp


There's not much point in arguing with you Cecil, since you don't want
to countenance the more sophisticated ideas of some of the other
posters to explain what's going on at the ends of a dipole. That's
too bad. You'll give some people the impression that things are
as simple as you say they are when things are not simple
at all. If they were, even an old hick like me could become an
engineer, and the job wouldn't pay much at all.
73,
Tom Donaly, KA6RUH

Tdonaly wrote:
But, for the rest of us, all the books,
the research, the mathematics, and the thinking of people
from Faraday to Feynman, actually mean something.

I don't recall Faraday or Feynman ever saying that the impedance
of an antenna wire is constant up and down the wire.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Tdonaly wrote:
Universally recognized principles of electromagnetics are obfuscation?

Complicating a simple measurement task beyond belief is obfuscation.

You changed the geometry. But even if you hadn't, you might be
able to say the changing electrical fields are greater at the ends of
a dipole, but not the voltages, because the voltages
aren't uniquely defined.

When a dipole is bent into an open loop, the relative voltage between
the ends is uniquely defined just like the voltage across a transmission
line is uniquely defined. If I poke two wires through two holes in a
faraday cage and ask you to measure the 10 MHz voltage between them
with 10% accuracy, would you say it can't be done?

You'll give some people the impression that things are
as simple as you say they are when things are not simple
at all.

The measurement may be extremely challenging, but the *concepts* are simple.
All you need to do is note the similarity of the transmission line impedances
on an SWR circle to a wire antenna. If the spacing on a transmission line is
an appreciable percentage of a wavelength, the transmission line will radiate.
That's all a center-fed wire antenna is - a transmission line with large
spacing between the conductors and it radiates.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Cecil wrote,


Tdonaly wrote:
But, for the rest of us, all the books,
the research, the mathematics, and the thinking of people
from Faraday to Feynman, actually mean something.

I don't recall Faraday or Feynman ever saying that the impedance
of an antenna wire is constant up and down the wire.
--
73, Cecil http://www.qsl.net/w5dxp

Interesting. Still trying to divert the argument to your own
frame of reference. Nice try.
Hasta la vista, Cecil - at least until you write something else
I disagree with.
73,
Tom Donaly, KA6RUH