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...."
____________

Below is a link to a calculator on the FCC website giving some
insight into this. It restricts inputs to values allowed for AM broadcast
stations, but still might be of some value to amateurs.

For one example, it shows a 1/4-wave monopole using 120 x
1/4-wave buried radials as generating a groundwave field of
about 306 mV/m at 1 km for 1 kW of applied power.

If the number of radials is reduced to 90, and their length is reduced
to 0.153 wavelength, the groundwave field is reduced to 267 mV/m.

The difference in radiated power then is (267/306)^2, or about
24%, which value is dissipated by heating the earth.

Whether or not that reduction is important to amateurs is a judgment call.
But an antenna system producing 267 mV/m for these conditions would be
unusable by a "regional" AM broadcast station -- which per their station
license must produce a groundwave rms field of at least 282 mV/m at 1 km
for 1 kW of applied power.

http://www.fcc.gov/mb/audio/bickel/figure8.html

RF

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

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

Richard Fry wrote:

Below is a link to a calculator on the FCC website giving some
insight into this. It restricts inputs to values allowed for AM broadcast
stations, but still might be of some value to amateurs.

For one example, it shows a 1/4-wave monopole using 120 x
1/4-wave buried radials as generating a groundwave field of
about 306 mV/m at 1 km for 1 kW of applied power.

If the number of radials is reduced to 90, and their length is reduced
to 0.153 wavelength, the groundwave field is reduced to 267 mV/m.

The difference in radiated power then is (267/306)^2, or about
24%, which value is dissipated by heating the earth.

Whether or not that reduction is important to amateurs is a judgment
call. But an antenna system producing 267 mV/m for these conditions
would be unusable by a "regional" AM broadcast station -- which per
their station license must produce a groundwave rms field of at least
282 mV/m at 1 km for 1 kW of applied power.

http://www.fcc.gov/mb/audio/bickel/figure8.html

RF

This is a good example of why we shouldn't assume that what's suitable
or optimum for AM broadcasting or some other service is necessarily the
best solution for amateur applications. A reduction in radiated power of
24% is just about 1 dB. While this amount of attenuation makes the
system unsuitable for AM broadcasting, it would be difficult to even
detect that amount of difference except just perhaps in the most
demanding amateur communication -- right at the noise level -- and it
would go completely unnoticed in the vast majority of cases.

Roy Lewallen, W7EL

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

Jim Kelley wrote:

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


Come on now guys, let's get series!

- 73 de Mike KB3EIA -