On Sep 27, 7:45 am, John Ferrell wrote:
On Sun, 23 Sep 2007 12:46:30 -0400, "Rick (W-A-one-R-K-T)"

wrote:

I just read the following on one of the mailing lists I subscribe to:

"Quarter wave ground mounted radials are a waste of wire and a hold over
from the olden days. Check the antenna handbook, the new philosophy is
more and shorter. The thing is that the bulk of the energy from the
vertical antenna is near the base of the antenna and this is what you are
trying to capture. A quarter wave radial sounds logical but the planet
will detune it so a quarter wave means nothing to the current."

What say you all?

Read this and see if you really want to know more.

http://www.bencher.com/pdfs/00361ZZV.pdf

A challenge to all you experts out the
Can you find anything you disagree with in this document?

John Ferrell W8CCW
"Life is easier if you learn to
plow around the stumps"

That seems a very practical ap note. But since you issued the
challenge, I'll say I disagree with the wording of an early sentence,
where it says that ground return currents are "greatly attenuated" if
they come through lossy earth. Clearly, the current is not
attenuated; the current is what it is. However, the current through
lossy ground causes power dissipation (and loss of radiated power) in
the ground.

I think the meaning is clear, but the wording would not pass muster
with a good technical editor.

Cheers,
Tom

On Fri, 28 Sep 2007 20:21:23 -0700, K7ITM wrote:

On Sep 27, 7:45 am, John Ferrell wrote:
On Sun, 23 Sep 2007 12:46:30 -0400, "Rick (W-A-one-R-K-T)"

wrote:

I just read the following on one of the mailing lists I subscribe to:

"Quarter wave ground mounted radials are a waste of wire and a hold over
from the olden days. Check the antenna handbook, the new philosophy is
more and shorter. The thing is that the bulk of the energy from the
vertical antenna is near the base of the antenna and this is what you are
trying to capture. A quarter wave radial sounds logical but the planet
will detune it so a quarter wave means nothing to the current."

That seems a very practical ap note. But since you issued the
challenge, I'll say I disagree with the wording of an early sentence,
where it says that ground return currents are "greatly attenuated" if
they come through lossy earth. Clearly, the current is not
attenuated; the current is what it is. However, the current through
lossy ground causes power dissipation (and loss of radiated power) in
the ground.

I think the meaning is clear, but the wording would not pass muster
with a good technical editor.

Cheers,
Tom
Good catch. It could have been more precisely stated.
In defense of the the point, Dictionary.com offers this definition of
"attenuated"- to weaken or reduce in force, intensity, effect,
quantity, or value.

OTH, lossy earth is a limiting factor to the current component of the
equation.

I think it unlikely to be misinterpretd so I would be inclined to
leave it alone. Of course, I am not a Technical Editor.

John Ferrell W8CCW
"Life is easier if you learn to
plow around the stumps"

"K7ITM" wrote
... I'll say I disagree with the wording of an early sentence,
where it says that ground return currents are "greatly attenuated"
of they come through lossy earth. Clearly, the current is not
attenuated; the current is what it is. However, the current through
lossy ground causes power dissipation (and loss of radiated
power) in the ground.
___________

But without a low-loss r-f ground for a monopole such as provided by a good
buried radial system, those returning r-f currents ARE greatly attenuated
before they can enter into the ground terminal of the antenna system.

That ground resistance is in series with the radiation resistance of the
monopole, and so will reduce the current that will flow on the monopole --
hence the field it will radiate.

RF

Richard Fry wrote:
... those returning r-f currents ARE greatly
attenuated before they can enter into the ground terminal of the antenna
system.

Richard, too many people, including some of the gurus,
are thinking DC circuits. The only difference between
the voltage equation and the current equation is a
division by Z0, i.e. the current is attenuated exactly
by the same factor as the voltage.
--
73, Cecil http://www.w5dxp.com

K7ITM wrote:
I'll say I disagree with the wording of an early sentence,
where it says that ground return currents are "greatly attenuated" if
they come through lossy earth. Clearly, the current is not
attenuated; the current is what it is.

There is some confusion between DC circuits and
distributed RF networks. In a DC circuit, the
current is the same throughout the circuit. In
a distributed RF network, the current is usually
*NOT* the same throughout the network. One can
put one amp of current into a buried radial and
measure zero amps from some point outward.

In particular, when dealing with RF EM waves, the
H-field to which the current is proportional in
a transmission line is attenuated by the same
attenuation factor as is the E-field to which
the voltage is proportional. Please reference
the transmission line equations to verify that
fact.

Such is easy to see. If one has a flat Z0=50 ohm
transmission line with 100 watts in and 50 watts
out, there is 1.414 amps in and 1.0 amp out because
the ratio of voltage to current is fixed at 50 ohms.
The traveling-wave current in an EM wave in a
transmission line (or in a radial in lossy earth)
is attenuated exactly as much as the voltage.

And don't feel too ignorant about that fact of physics.
Some of the gurus on this newsgroup make a similiar
mistake about RF EM wave current through a loading coil.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

There is some confusion between DC circuits and
distributed RF networks.

No comment.

In a DC circuit, the
current is the same throughout the circuit.

That's true only for a simple series circuit. And it's true for RF
as well.

In
a distributed RF network, the current is usually
*NOT* the same throughout the network.

The same is true for a DC network.

One can
put one amp of current into a buried radial and
measure zero amps from some point outward.

In particular, when dealing with RF EM waves, the
H-field to which the current is proportional in
a transmission line is attenuated by the same
attenuation factor as is the E-field to which
the voltage is proportional. Please reference
the transmission line equations to verify that
fact.

Such is easy to see. If one has a flat Z0=50 ohm
transmission line with 100 watts in and 50 watts
out, there is 1.414 amps in and 1.0 amp out because
the ratio of voltage to current is fixed at 50 ohms.
The traveling-wave current in an EM wave in a
transmission line (or in a radial in lossy earth)
is attenuated exactly as much as the voltage.

Note: according to Ohms law, current scales directly with voltage and
inversely with resistance.

ac6xg

Jim Kelley wrote:
In a DC circuit, the
current is the same throughout the circuit.

That's true only for a simple series circuit. And it's true for RF as
well.

A foot of wire with reflections at one GHz has
the same current throughout the circuit?
--
73, Cecil http://www.w5dxp.com

On Oct 1, 9:09 pm, Cecil Moore wrote:
Jim Kelley wrote:
In a DC circuit, the
current is the same throughout the circuit.

That's true only for a simple series circuit. And it's true for RF as
well.

A foot of wire with reflections at one GHz has
the same current throughout the circuit?
--
73, Cecil http://www.w5dxp.com

A simple series circuit can be expected to behave as a simple series
circuit. Other circuits can be expected to behave differently. Which
do you think applies?

ac6xg

Jim Kelley wrote:
Cecil Moore wrote:
A foot of wire with reflections at one GHz has
the same current throughout the circuit?

A simple series circuit can be expected to behave as a simple series
circuit. Other circuits can be expected to behave differently. Which
do you think applies?

An *ordinary prudent man* would think that one foot
of wire is a "simple series circuit" and it is in a
DC circuit. If as you say, the current in an RF circuit
is the same throughout, why does the current vary
every inch in a circuit with reflections?
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Jim Kelley wrote:

Cecil Moore wrote:

A foot of wire with reflections at one GHz has
the same current throughout the circuit?


A simple series circuit can be expected to behave as a simple series
circuit. Other circuits can be expected to behave differently. Which
do you think applies?


An *ordinary prudent man* would think that one foot
of wire is a "simple series circuit" and it is in a
DC circuit.

Most ordinary prudent men that I know wouldn't characterize a one foot
length of wire as a series circuit.

ac6xg