Cecil Moore wrote in message ...
Tom Bruhns wrote:
If you measure the voltage drop along the wire, it's essentially zero,
so along the wire the voltage between the end points of the dipole is
essentially the same as the voltage across the feedpoint.

Brain fart?

Just so we're clear on this, no, certainly not.

If you care why, consider the direction of the electric field adjacent
to the conductor, and integrate the component of that field parallel
to the conductor along the path of the conductor. You will in general
get a different answer than if you integrate along a path from the tip
of the antenna, out say a quarter wavelength, then parallel to the
antenna for a half wave, then back to the other end of the antenna.
There is no such thing as "the voltage" between the ends of your
excited dipole at an instant in time. There are infinitely many
potentials, as there are infinitely many paths you can follow through
the (time-varying) magnetic field.

Cheers,
Tom

Tom Bruhns wrote:
There is no such thing as "the voltage" between the ends of your
excited dipole at an instant in time.

Perhaps they meant the voltage 'across' the ends of the dipole. The
ends should always be an electrical half-wave out of phase, right?
There should only be two instants of time during a period when the
difference in potential from end to end is zero.

What are you saying exactly, Tom?

73, Jim AC6XG

Jim Kelley wrote in message ...
Tom Bruhns wrote:
There is no such thing as "the voltage" between the ends of your
excited dipole at an instant in time.

Perhaps they meant the voltage 'across' the ends of the dipole. The
ends should always be an electrical half-wave out of phase, right?
There should only be two instants of time during a period when the
difference in potential from end to end is zero.

What are you saying exactly, Tom?

I'm saying that if you measure the voltage between two points on a
good conductor, in a path along that conductor, it will be very small.
The electric field is always perpendicular to a perfect conductor at
the surface of that conductor. For a conductor with resistance, the
drop along it is I*R, and therefore the nearby electric field is in
general not quite perpendicular, but unless it's a darned inefficient
antenna, it's very nearly so.

I'm also saying that the voltage (potential) between two points
depends, in general, on the path you take between the two points. You
should be _especially_ aware of that fact when you're in the presence
of time-varying magnetic fields, such as you have around a powered
antenna.

As I said, if you measure the potential along a line perpendicular to
the antenna, it will be large (when the antenna is excited with some
power). I fully expect the electric field to be high near the wire,
but perpendicular to the wire, NOT parallel to it.

Cheers,
Tom

Tom Bruhns wrote:
I'm also saying that the voltage (potential) between two points
depends, in general, on the path you take between the two points. You
should be _especially_ aware of that fact when you're in the presence
of time-varying magnetic fields, such as you have around a powered
antenna.


In practice that will means that the voltage you measure between say the
end of a whip and ground will depend on how you choose to route the
connecting leads to the voltmeter, and how you connect to ground... and
above (below?) all on what you define "ground" to be.


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

Ian White, G3SEK wrote:
In practice that will means that the voltage you measure between say the
end of a whip and ground will depend on how you choose to route the
connecting leads to the voltmeter, and how you connect to ground... and
above (below?) all on what you define "ground" to be.

How about using an artificial ground at the measurement point?
--
73, Cecil, W5DXP

Cecil Moore wrote:
Ian White, G3SEK wrote:
In practice that will means that the voltage you measure between say
the end of a whip and ground will depend on how you choose to route
the connecting leads to the voltmeter, and how you connect to
ground... and above (below?) all on what you define "ground" to be.

How about using an artificial ground at the measurement point?

Define one, if you can! What are its properties, and how would you
achieve them?

It's rather like the early wireless users who "earthed" their receivers
to the aspidistra pot in the corner of the room - after all, it
contained earth, so why didn't it work?


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

How 'bout you model it with your concept of an "artificial ground", and
let us know the result? You can measure the voltage with EZNEC by
connecting the two points to be measured with a wire and inserting a
zero amplitude current source in the wire. The source will act like an
open circuit, and the voltage will be reported in the Source Data output.

After you've determined the voltage relative to your "artificial
ground", modify the "artificial ground" and note the effect on the
voltage. Then see if you can figure out what the voltage is between the
"artificial ground" and the Earth. Or, give us your justification for
assuming that it's zero. If it is zero, via what path? As Tom has been
saying, the voltage between two points depends on the path you take
between them.

Roy Lewallen, W7EL

Cecil Moore wrote:
Ian White, G3SEK wrote:

In practice that will means that the voltage you measure between say
the end of a whip and ground will depend on how you choose to route
the connecting leads to the voltmeter, and how you connect to
ground... and above (below?) all on what you define "ground" to be.


How about using an artificial ground at the measurement point?
--
73, Cecil, W5DXP

Roy Lewallen wrote:
After you've determined the voltage relative to your "artificial
ground", modify the "artificial ground" and note the effect on the
voltage. Then see if you can figure out what the voltage is between the
"artificial ground" and the Earth. Or, give us your justification for
assuming that it's zero. If it is zero, via what path? As Tom has been
saying, the voltage between two points depends on the path you take
between them.

Wow, you sure ASSume a lot from a simple question. Let's turn it around
and you guys prove that the voltage at the ends of a dipole is less
than or equal to the feedpoint voltage even though a florescent light
bulb is brighter at the ends.
--
73, Cecil http://www.qsl.net/w5dxp



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Tom Bruhns wrote:
I'm saying that if you measure the voltage between two points on a
good conductor, in a path along that conductor, it will be very small.

True for DC and RF traveling waves. Not true for standing waves. A
1/2WL dipole is a *standing-wave* antenna. What do you get when you
measure the voltage between the voltage maximum and voltage minimum
on a feedline with a 10:1 SWR? Exactly the same principle applies
to *standing-wave* antennas.
--
73, Cecil, W5DXP

Tom Bruhns wrote:
There is no such thing as "the voltage" between the ends of your
excited dipole at an instant in time.

Please reference Fig 1, page 2-2, in the 15th edition of the ARRL
Antenna Book. "Current and voltage distribution on a 1/2WL wire.

The RMS (or peak) values of the voltages at the ends of the dipole
are maximum and 180 degrees out of phase. The ratio of net voltage to
net current is the impedance anywhere along the wire.
--
73, Cecil http://www.qsl.net/w5dxp



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