Roy Lewallen wrote:
This is indeed an interesting result, even though it's small. As a
dipole or monopole gets shorter than a resonant length, the current
distribution changes from sinusoidal to triangular.

Actually, it just moves to a straighter portion of the sinusoidal
curve. The cosine curve from 75 degrees to 90 degrees does resemble
a triangle but it is not a straight line. Assuming a triangle from
75 deg to 90 deg is a simplified shortcut that introduces almost a
1% error at 82.5 deg.
--
73, Cecil http://www.qsl.net/w5dxp


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Roy Lewallen wrote:
Under those conditions, the phase of the current
is nearly constant along the wire, so the fields from the various parts
of the antenna add together in phase at a distant point broadside to the
wire.

And remember, that's the net current, i.e. the current in the standing
wave. It, as a net stand alone entity, is not flowing anywhere. It is
merely the phasor sum of two traveling waves traveling in opposite
directions. The phase of the forward current and the phase of the
reflected current are changing in the normal traveling wave manner.
But since their phases are rotating in opposite directions, their
net phasor sum results in very little phase shift. In fact, Kraus
shows zero phase shift for the standing wave current on an
infinitesimally thin wire.
--
73, Cecil http://www.qsl.net/w5dxp


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Almost 1%! Holy moly!

You're correct that the distribution is only approximately triangular.
But since a difference between a full quarter sine wave and a true
triangular distribution results in less than a 0.5 dB difference in
field strength, a 1% difference between a true triangular distribution
and the end of a sine wave will make a difference in gain, efficiency,
or feedpoint impedance that would be entirely negligible and in fact
impossible to measure. For that matter, the current distribution on a
wire of finite diameter isn't really a sine wave anyway, but a close
approximation (although very possibly not within 1%). So the current
distribution on a short antenna is triangular for any practical purpose.

Roy Lewallen, W7EL

Cecil Moore wrote:

Roy Lewallen wrote:

This is indeed an interesting result, even though it's small. As a
dipole or monopole gets shorter than a resonant length, the current
distribution changes from sinusoidal to triangular.


Actually, it just moves to a straighter portion of the sinusoidal
curve. The cosine curve from 75 degrees to 90 degrees does resemble
a triangle but it is not a straight line. Assuming a triangle from
75 deg to 90 deg is a simplified shortcut that introduces almost a
1% error at 82.5 deg.

Wes Stewart wrote:
for the extremely short antennas that were (are) the subject of the
"shootout" business, the current actually increases between the ends
of the inductor.

Heh, heh, the helix feature in EZNEC shows the same thing. So here's
a question for all the "constant current" gurus. How can the current
in the middle of the loading coil be a greater magnitude than the
current at either end?

Hint: it happens all the time in distributed networks. I can design
an antenna with a loading coil that has one amp at one end and zero
amps at the other end. The forward current and reflected current
are simply 180 degrees out of phase at the end where the net current
is zero and thus a current node (minimum) is developed.

A naive person would say one amp is flowing into one end and zero
amps is flowing out the other end. But standing wave current doesn't
flow. This is W8JI's basic mistake in his explanation on his web page.
The standing wave current is an artifact caused by the superposition
of two traveling waves, traveling in opposite directions at the speed
of light. For a 1/4WL antenna, there is a standing wave current antinode
at the feedpoint and a standing wave current node at the tip of the antenna.
--
73, Cecil http://www.qsl.net/w5dxp


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Roy Lewallen wrote:
Almost 1%! Holy moly!

Aren't you the same Roy Lewallen who raked me over the coals
for using a simplified shortcut? (The devil made me do it. :-)
--
73, Cecil http://www.qsl.net/w5dxp


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On Sat, 06 Nov 2004 00:05:49 -0600, Cecil Moore
wrote:
Yep, it changed "supress" to "suppers".
An expensive Bank Shot....

I hope most readers can tell the difference between an approximation or
simplification that we understand and that produces negligible and
unmeasurable errors, from one that's based on invalid premises and leads
to major errors.

Roy Lewallen, W7EL

Cecil Moore wrote:

Roy Lewallen wrote:

Almost 1%! Holy moly!


Aren't you the same Roy Lewallen who raked me over the coals
for using a simplified shortcut? (The devil made me do it. :-)

Now I understand the confusion. I am not using EZNEC, and am not familiar
with the program. My analysis is written in NEC code. I do have access to
some graphical results with the use of NEC-Win Pro, but have taken the NEC
output file and imported it to an Excel spread sheet (NEC-Win Pro's
simplified data entry formats do not recognize cards such as 'GH" and "GM"
so am forced to resort to direct code entry). For a constant power input I
therefore have to read the input impedance results, then calculate the
required "Peak" Voltage for the desired power -- note that NEC requires all
voltages and currents to be peak values.

If anybody is interested I have uploaded the models to my wife's business
web site (she will probably kill me when she finds out!). The models show
the current distribution of two 84" monopoles; one of which is loaded with a
6" long, 12 turn helix, of 2.5" diameter. The other is unloaded. The
ground is defined as "Perfect". The input power is set at 100 W. Just
looking at the curves led me to the (mostly) wrong conclusion (verified with
numerical integration of the Excel data) that the Integral of I(z)dz is a
constant. The results are posted at www.carolyns-creations.com/ve6cb

Regards,

Frank


"Roy Lewallen" wrote in message
...
I'd be glad to help clarify any confusion, but I can't for the life of me
figure out what the confusion is about.

Regarding EZNEC's sources, you have the option of using a constant-current
source, constant-voltage source, or having a fixed power applied to the
model. If a fixed power level is chosen, multiple sources will have the
same ratio of voltages and currents as you've specified. For example, if
your model has two sources, one of 1 amp and the other 2 amps, and you
choose a fixed power level, the current from the second source will always
be twice the current from the first (and at the specified relative phase),
but both will be modified as necessary to produce the total power you
specified.

Fixed power is selected in the Options menu. While the relative source
currents (2 amps and 1 amp in the example) are saved with the model, the
Options menu choices aren't. They're applied to all models. Changes you
make in the Options menu remain effective until you end the program. If
you choose "Save as Default" from the Options menu, the current settings
will remain in effect even after you end the program, until you change
them.

I don't personally use a fixed power level very often, since I'm more
often interested in things like currents on wires relative to a source, so
it's convenient for me to set the source to 1 amp. The fixed power level
is useful whenever you need an absolute value of voltage or current,
though, or want to see how the voltage, current, or load power dissipation
changes as you modify the model. It can be helpful in understanding
radiation resistance, by illustrating the relationship among feepoint
power, current, and resistance.

Just what is it that's confusing about radiation resistance and EZNEC
results? Richard's EZNEC results are exactly what I'd expect, and they're
consistent with theory. (I'm talking about established theory found in
textbooks, not ones cooked up by people who don't understand basic
principles or common nomenclature.) The small gain change due to the
ground reflection reaction to altered current distribution was admittedly
a surprise, but it makes perfect sense after a moment's thought. If
someone can summarize what seems to be wrong, I'll do my honest best to
explain it. Is the problem that the gain doesn't go up with Rr? Of course
it doesn't, in the nearly lossless antennas of Richard's models. Why
should it? I tried to explain why in my earlier posting, but I'll try
again if that's what the problem is.

Roy Lewallen, W7EL

Richard Clark wrote:

On Sat, 06 Nov 2004 02:34:32 GMT, "Frank"
wrote:


Regarding Richard's comments "Assuming the same currents". Not sure I
understand, since the feed-point current varies with constant power


Hi Frank,

That is quite simple. The file VERT1.EZ has as a source, a constant
current generator - a fact I pointed out in the summary of my results.
Your presumption of constant power is a natural one for the sake of
common discussion, but it is not even the default source for EZNEC.
[However, having said that, a curiosity of program design has found
that the concept of a "new file" has been orphaned. Any time the
application is opened, it is opened with LAST.EZ and I cannot recall
upon initial acquisition if EZNEC ever started with a blank slate.
Given that you cannot have less than 1 wire nor 1 source, then there
is on way to force a blank file. Hence the concept of a default
resides in the last file opened.]

In other words, each and every iteration of antenna, irrespective of
its presumed or actual Rr or drivepoint Z had the same current applied
to it, 1 Ampere.

Now, if ANY resistance had changed, it then follows that the POWER
would have changed too (which presents us with that puzzle

confused by the expression V=I/R.

that came out of the blue) at a linear rate (1² = 1).

Note, a doubling of drivepoint Z by the addition of a load does not
result in 3dB gain over the former design. I have seen I * I * R
bandied about as an "explanation" and yet at least 3dB is remarkably
absent in the results. What R is this that everyone speaks of?
Certainly not the real component of drivepoint Z. What about the Rr
that must've changed? Copper loss absorb it? Ground loss?

As I offered, I must've done something wrong, taken the wrong turn,
interpreted the modeler in error, -ahem- not read the help file....

I will bet it was that last one - which only reveals no one has. ;-)

[aside]
So, Wes, I will amend my ways and delve into that treasure of
knowledge before returning your work in kind (no point in retread
effort) sometime tomorrow morning. [to the audience]
C'mon folks, this has to be an especially simple resolution - I'm glad
that so many are just as flummoxed (even Reg is uncharacteristically
silent in this regard :-)

73's
Richard Clark, KB7QHC

As mentioned in an earlier post. I have analyzed an 84" monopole, at 21.3
MHz, with and without loading coils. To view the graphical results of
current distribution go to www.carolyns-creations.com/ve6cb These data are
placed without comment, since I cannot intelligently add to the ongoing
arguments of current distribution.

Regards,

Frank


"Cecil Moore" wrote in message
...
Wes Stewart wrote:
for the extremely short antennas that were (are) the subject of the
"shootout" business, the current actually increases between the ends
of the inductor.

Heh, heh, the helix feature in EZNEC shows the same thing. So here's
a question for all the "constant current" gurus. How can the current
in the middle of the loading coil be a greater magnitude than the
current at either end?

Hint: it happens all the time in distributed networks. I can design
an antenna with a loading coil that has one amp at one end and zero
amps at the other end. The forward current and reflected current
are simply 180 degrees out of phase at the end where the net current
is zero and thus a current node (minimum) is developed.

A naive person would say one amp is flowing into one end and zero
amps is flowing out the other end. But standing wave current doesn't
flow. This is W8JI's basic mistake in his explanation on his web page.
The standing wave current is an artifact caused by the superposition
of two traveling waves, traveling in opposite directions at the speed
of light. For a 1/4WL antenna, there is a standing wave current antinode
at the feedpoint and a standing wave current node at the tip of the
antenna.
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil Moore wrote:


The decrease (drop) in current across a loading coil installed in
a standing-wave antenna does NOT in any way violate Kirchhoff's current
law.

True.

One can imply from Kirchhoff's current law that there is no current
decrease (drop) across a point.

By 'point' I believe you are referring to what is commonly referred to
as a node. Kirchoff's current law stipulates that charge may not
accumulate at nodes. Therefore by definition, any feature of the system
where charge accumulation needs to be considered is not a node. This
makes sense since charge must accumulate on a surface (surface charge
density) (coul / meter squared) or in a dielectric volume (volume charge
density) (coul / meter cubed). Both concepts require some sort of
area or volume which is inconsistant with the notion of a node.

I don't know anyone who disagrees with
that so any argument is just a straw man. Kirchhoff never said the
current at one point in a network had to equal the current at another
point in the network.

The currents through two nodes connected in series, without branches, is
identical. I think that fact was established before Kirchoff but it's
certainly stipulated in circuit theory.

Many patches have been added to the DC circuit model to try to adapt
it to RF networks.

Many? Seems to me that the concept of electric displacement introduced
by Maxwell provides everything needed to extend DC theory all the way
through classical electromagnetics. What am I missing?

Some function after a fashion and some fail utterly.

Like what?

We all need to be able to recognize the difference. For EM waves, the
E-field and H-field are often affected in the same way.

Huh?

Saying that
the E-field voltage drops but the H-field current doesn't drop is
simply nonsense.

Saying the electric field voltage drops is nonsense. Voltage is the
scalar potential defined as the electric potential difference between
two points in space.

The electric field is vector field, characterized as having a field
strength in volts per meter dependant on spatial location, direction,
and perhaps time.

I don't understand what the term 'E-field voltage drop' could mean.
Same with 'H-field current drop'.

Likewise, saying that the H-field current flows and
the E-field voltage doesn't flow is nonsense.

H-field current flows?

The field H (amps per meter), is the so called magnemotive field. It
doesn't flow anymore than voltage flows through a resistor, and is
associated with the generation of magnetic flux. The magnetic flux
density, B, has the units of webers per meter squared and can be
integrated over an arbitrary surface to evaluate the total magnetic flux
passing through that surface. Magnetic flux is somewhat analogous to
current but H is not at all.

The E-field and H-field
are usually inseparable.

In the classical electromagnetic model, E & H are completely separable.
They are coupled via Faraday's law, and Maxwell's so called
displacement current. At steady state (DC) no coupling exists. When
one field quantity _varies_ in time, so will the other in accordance
with the curl equations. The coupling described by the time varying
part of the curl equations only involves the time varying components.

When determining the analysis method used to gather insight into a
physical system, one of the first considerations is to determine if the
time varying field components need to be considered, and if so, which
ones. For example, analysis of a 60 Hz power supply choke, or electric
motor, usually ignores the electric field in the air gap arising from
the time varying magnetic flux density. It's not important in the gap,
but is the driver of undesirable eddy currents in the core laminations.

bart
wb6hqk