It might be helpful to elaborate a bit more about radiation resistance.

Consider an antenna that has no loss. If we apply 100 watts, say, to
this antenna, the radiation resistance "consumes" that 100 watts. That
is to say, all 100 watts is radiated.

In the case of a resonant lossless quarter wave vertical, for example,
the current at the base will be about 1.67 amps if 100 watts is being
radiated. We can solve for Rr at the base from P = I^2 * Rr, where I is
the base current, to get Rr = P / I^2, with the result that Rr is about
36 ohms. This is the radiation resistance referred to the base -- it
"consumes" the 100 watts. (We could have calculated Rr at some other
point along the antenna, where I is different, and gotten a different
value. But P still has to equal I^2 * Rr, where I is the current at the
point to which Rr is being referred.)

Now, what determines the current we get at the base, for a given applied
power and radiator length? The answer is the current distribution --
that is, the way the current varies along the length of the conductor.
(I'm only considering a simple single wire antenna here. When other
conductors are involved, mutual coupling between conductors also plays a
role.) Putting a loading coil at the bottom of the antenna doesn't
change the current distribution, it only changes the feedpoint
reactance. So it doesn't change the feedpoint current for a given power
input, so the radiation resistance doesn't change. But if you put a
loading coil part way up the antenna, the current distribution does
change. This alters the base current for the same power input and
therefore the radiation resistance changes. Remember that Rr = P / I^2,
where Rr and I are measured at the same point (in this case the base
feedpoint), so if I changes, Rr changes. Likewise, top loading alters
the current distribution and consequently the radiation resistance. For
people who would like to see this graphically, the demo version of EZNEC
is adequate. Just look at the View Antenna display after running a
pattern, source data, or current calculation, and you'll see how the
current varies along the antenna. If you set a fixed power level in the
Options menu (Power Level selection), you can also see, by clicking Src
Dat, exactly how the current at the source changes as the current
distribution does.

If you have a fixed amount of loss, say at the base of the antenna due
to ground system loss, the amount of power dissipated in that loss as
heat is Ploss = I^2 * Rloss, where I is the current flowing through that
loss, in this case the current at the antenna base. So for a given
amount of applied power, you minimize the power lost when you minimize
the base current. This is exactly equivalent to saying you're raising
the radiation resistance referred to the base. That's why mobile antenna
users consider higher radiation resistance a virtue -- it means lower
feedpoint current for a given power input, and therefore less power lost
in the necessarily imperfect ground system.

While the principles are all the same, wire loss has to be treated a bit
differently because altering the current distribution changes the amount
of wire loss (which is usually combined into a single loss resistance
referred to the feedpoint, or the same point where the radiation
resistance is referred). Also, changing the wire length alters the wire
loss, as does the total current in the wire which increases as the wire
is shortened for a given power input. All these can be dealt with
analytically or with a modeling program, but it's easy to lose track of
exactly what's happening when all these factors are present at the same
time. Fortunately, wire loss is insignificant for the vast majority of
typical amateur applications. With modeling, it's easy to determine when
it is and isn't significant, simply by turning wire loss on and off and
observing how much the results change.

Roy Lewallen, W7EL

I'm sorry, but I can't make any sense out of your response, so I'm
unable to answer your question.

Roy Lewallen, W7EL

Richard Clark wrote:

On Fri, 05 Nov 2004 13:06:40 -0800, Roy Lewallen
wrote:


What makes you think that the Rr isn't changing?


Hi Roy,

With what?

I have offered a very ascetic report of very simple actions from very
simple terms. You have chosen to change those terms to fit your own
answer (I do not choose to put this into the context of a perfect
world). Please offer effort in kind, or point out what error I've
made instead of what error I might have made if something were changed
from my model.

73's
Richard Clark, KB7QHC

On Fri, 05 Nov 2004 15:20:26 -0800, Roy Lewallen
wrote:

I'm sorry, but I can't make any sense out of your response, so I'm
unable to answer your question.

What makes you think that the Rr isn't changing?
With what?

Loading.

You wrote:

So much for theories of Rr being modified by loading.

Being a simple-minded, non-liberally educated person, I took this to
mean that you'd shown that Rr isn't modified by loading, disproving
theories that Rr is modified by loading. (Otherwise, I figured, you
would have said "So much for theories of Rr not being modified by loading.")

Did you mean instead that there are theories that Rr isn't modified by
loading? If so, who on earth is proposing those theories? As a basic
understanding of Rr shows, and as your modeling results also show,
there's no basis for such a "theory". What's next, a "theory" that V = I/R?

Roy Lewallen, W7EL

Richard Clark wrote:
On Fri, 05 Nov 2004 15:20:26 -0800, Roy Lewallen
wrote:


I'm sorry, but I can't make any sense out of your response, so I'm
unable to answer your question.


What makes you think that the Rr isn't changing?

With what?

Richard Clark wrote:

On Fri, 05 Nov 2004 13:37:27 -0800, Jim Kelley
wrote:


Point being that antenna gain has spatial implications which Rr by
itself could not provide in the solutions.


Hi Jim,

Not sure where this is going, so I will stand with my own statements.

I wouldn't have it any other way, Richard.

The examination of the underlying theory taking place in this discussion
resolves in different ways. Any other 'work in kind' would depend
largely on which way one resolves the theory. Horse first, then cart.

Reasonable questions have been posed here. Some of the points being
raised remain to an extent, unreconciled with certain aspects of
conventional treatments. Obviously, thanks in no small part to
empirical methods, this poses no greater hinderance to the design of
efficient antennas than does the lack of a precise understanding of the
nature of turbulent flow to the design of efficient fluid pumps. I find
any contention that no further investigation is necessary to be less
than convincing. On the contrary, the work done by the oft cited E&M
engineers clearly indicates there is more to the story - a point that
said engineers would undoubtedly concede without hesitation.

On the other hand, if the impedance of the
coil is not constant from one end to the other, and it in fact does have
some real physical and/or electrical length, then I think the radiation
resistance of the antenna would have to be effected by its presence in
the circuit. That is, if indeed Ro is the integral of disributed r
along the entire physical and/or electrical length of the antenna
(credit for that formula going to an esteemed contributor to this
newsgroup earlier today). Or perhaps more concisely put, if the loading
coil itself contributes to the field radiating from the antenna, then it
should likewise have a Rr associated with it. The converse would of
course still be true.

This returns us to matters of degree. By simple observation comparing
the standard full sized radiator to ANY of the iterations, it is
obvious that nothing significant can be said of the physicality of the
load, much less its contribution. In other words, is there some
magical coil or magical placement that would create a short
super-radiator that exceeds the performance of the standard
quarterwave? [no here can recognize a eh/cfa claim?]

Let's put a handicap on this. Is there some magic combination of
coil/position that would even EQUAL the performance of the standard
quarterwave?

Absent a thorough understanding of the phenomena, one might be inclined
to conjecture about magical things. But I'm inclined toward taking
satisfaction from persuing a thorough understanding of the phenomena
rather than from strutting the pretense of already having one.

73, Jim AC6XG

Concerning current distribution -- at least on a monopole above a perfectly
conducting ground. I have noticed that in any antenna, of any length,
inductively loaded, or not, the: Integral of I(z)dz (where the units of dz
are in fractions of a wavelength, and I(z) the current distribution) is
virtually a constant. Assuming the same input power in all cases. Suppose
it seems pretty obvious since the total radiated power from any structure is
also essentially constant. I must admit I have not seen gain increasing,
with decreasing size, but then I have not seriously tried.

Regards,

Frank


"Roy Lewallen" wrote in message
...
It might be helpful to elaborate a bit more about radiation resistance.

Consider an antenna that has no loss. If we apply 100 watts, say, to this
antenna, the radiation resistance "consumes" that 100 watts. That is to
say, all 100 watts is radiated.

In the case of a resonant lossless quarter wave vertical, for example, the
current at the base will be about 1.67 amps if 100 watts is being
radiated. We can solve for Rr at the base from P = I^2 * Rr, where I is
the base current, to get Rr = P / I^2, with the result that Rr is about 36
ohms. This is the radiation resistance referred to the base -- it
"consumes" the 100 watts. (We could have calculated Rr at some other point
along the antenna, where I is different, and gotten a different value. But
P still has to equal I^2 * Rr, where I is the current at the point to
which Rr is being referred.)

Now, what determines the current we get at the base, for a given applied
power and radiator length? The answer is the current distribution --
that is, the way the current varies along the length of the conductor.
(I'm only considering a simple single wire antenna here. When other
conductors are involved, mutual coupling between conductors also plays a
role.) Putting a loading coil at the bottom of the antenna doesn't change
the current distribution, it only changes the feedpoint reactance. So it
doesn't change the feedpoint current for a given power input, so the
radiation resistance doesn't change. But if you put a loading coil part
way up the antenna, the current distribution does change. This alters the
base current for the same power input and therefore the radiation
resistance changes. Remember that Rr = P / I^2, where Rr and I are
measured at the same point (in this case the base feedpoint), so if I
changes, Rr changes. Likewise, top loading alters the current distribution
and consequently the radiation resistance. For people who would like to
see this graphically, the demo version of EZNEC is adequate. Just look at
the View Antenna display after running a pattern, source data, or current
calculation, and you'll see how the current varies along the antenna. If
you set a fixed power level in the Options menu (Power Level selection),
you can also see, by clicking Src Dat, exactly how the current at the
source changes as the current distribution does.

If you have a fixed amount of loss, say at the base of the antenna due to
ground system loss, the amount of power dissipated in that loss as heat is
Ploss = I^2 * Rloss, where I is the current flowing through that loss, in
this case the current at the antenna base. So for a given amount of
applied power, you minimize the power lost when you minimize the base
current. This is exactly equivalent to saying you're raising the radiation
resistance referred to the base. That's why mobile antenna users consider
higher radiation resistance a virtue -- it means lower feedpoint current
for a given power input, and therefore less power lost in the necessarily
imperfect ground system.

While the principles are all the same, wire loss has to be treated a bit
differently because altering the current distribution changes the amount
of wire loss (which is usually combined into a single loss resistance
referred to the feedpoint, or the same point where the radiation
resistance is referred). Also, changing the wire length alters the wire
loss, as does the total current in the wire which increases as the wire is
shortened for a given power input. All these can be dealt with
analytically or with a modeling program, but it's easy to lose track of
exactly what's happening when all these factors are present at the same
time. Fortunately, wire loss is insignificant for the vast majority of
typical amateur applications. With modeling, it's easy to determine when
it is and isn't significant, simply by turning wire loss on and off and
observing how much the results change.

Roy Lewallen, W7EL

On Fri, 05 Nov 2004 07:45:16 GMT, Richard Clark
wrote:

Good day Richard,

Sorry this isn't too timely, but I've been off visiting a dear friend
and soon to be silent key.

|On Thu, 04 Nov 2004 20:38:35 -0700, Wes Stewart
|wrote:
|
|Or, model a short lossless monopole over perfect ground and determine
|the feedpoint R. In this case, R is totally due to radiation loss,
|i.e. "radiation resistance." Add a lossless loading inductance
|somewhere in the middle and see what happens to R.
|
|Hi Wes,
|
|The difference between the two (perfect/real) insofar as Z is hardly
|remarkable.

True.
|
|First I will start with a conventionally sized quarterwave and by
|iteration approach the short antenna and observe effects. I am using
|the model VERT1.EZ that is in the EZNEC distribution and modifying it
|by turns. For instance, I immediately turn on the wire loss.
|
|40mm thick radiator 10.3 meters tall:
| Impedance = 36.68 + J 2.999 ohms
|which lends every appearance to expectation of Rr that could be
|expected from a lossless perfect grounded world.
|Best gain is
| -0.03dBi
|
|next iteration:
|
|cut that sucker in half:
| Impedance = 6.867 - J 301 ohms
|which, again, conforms to most authorities on the basis of Rr.
|best gain
| 0.16dBi
|How about that! More gain than for the quarterwave (but hardly
|remarkable). This makes me wonder why any futzing is required except
|for the tender requirements of the SWR fearing transmitter (which, by
|the way, could be as easily taken care of with a tuner).
|
|next iteration:
|
|load that sucker for grins and giggles:
| load = 605 Ohms Xl up 55%
| Impedance = 13.43 + J 0.1587 ohms
|Did I double Rr? (Only my hairdresser knows.)
|best gain
| 0.13dBi
|Hmm, losing ground for our effort, it makes a pretty picture of
|current distribution that conforms to all the descriptions here (sans
|the balderdash of curve fitting to a sine wave). I am sure someone
|will rescue this situation from my ineptitude by a better load
|placement, so I will leave that unfinished work to the adept
|practitioners.
|
|next iteration:
|
|cut that sucker down half again (and remove the load):
| Impedance = 1.59 - J 624.6 ohms
|Something tells me that this isn't off the scale of the perfect
|comparison.
|best gain:
| 0.25dBi
|Hmm, the trend seems to go counter to intuition.
|
|next iteration:
|
|-sigh- what charms could loading bring us?
| load = 1220 Ohms Xl up 55%
| Impedance = 3.791 + J 1.232 ohms
|more than doubled the Rr?
|best gain:
| 0.23dBi
|
|Now, all of this is for a source that is a constant current generator;
|we've monkeyed with the current distribution and put more resistance
|(Rr?) into the equation with loading; and each time loading craps in
|the punch bowl.
|
|So much for theories of Rr being modified by loading. I would
|appreciate other effort in kind to correct any oversights I've made
|(not just the usual palaver of tedious "explanations" - especially
|those sophmoric studies of current-in/current-out).

Sure. Your calculations are impeccable so far (although it would be
better to use more than 10 segments IMHO).

I believe it is your contention that loading to resonance with an
arbitrarily positioned inductor, or not loading at all, does not
affect the gain, and the radiation resistance is not the same as the
changing feedpoint resistance.

I am in the other camp, along with Hansen, Devoldere, et. al. who say
that the current distribution does affect the radiation resistance
(and in the real world, the gain/efficiency).

I hope you would agree that the normalized gain would be a good proxy
for efficiency.

For example if we use the lossless 1/4 wavelength monopole over
perfect ground as a reference, then gain with respect to that (5.15
dBi) would be an indicator of efficiency.

I believe that you will agree that the efficiency can be determined
by:

Rr
eta = ------------- Eq.1
Rr + Rg + Rl

where Rr = radiation resistance
Rg = ground resistance
Rl = all other resistances (conductor, etc)

I think you would also agree that for the full-sized monopole over
perfect ground the feedpoint resistance of ~36 Ohm = radiation
resistance.

As an old (sorry [g]) metrologist, you're very familiar with
substitution, so let's set Rl = 0 (lossless case) and eta to 0.5 (-3
dB). Per Eq. 1, Rg = Rr.

So in our model, if I add a simulated ground resistance, Rg, that
reduces the gain by 3 dB, I have by substitution, determined the
radiation resistance.

Sure enough, if I add a 36 Ohm load at the bottom of the perfect 1/4
wave monopole, the gain drops to 2.14 dBi, and the feedpoint
resistance doubles.

I will let you try this with the other cases. I trust you will find
that the radiation resistance does decrease with shorter radiators
and/or lower loading points.

I too I would appreciate other effort in kind to correct any
oversights I've made.

Wes

On Sat, 06 Nov 2004 01:31:40 GMT, "Frank"
wrote:

Assuming the same input power in all cases.

Hi Frank,

What would you say to "assuming the same current in all cases" and
either/both the Rr and drivepoint Z have changed to no particular
evidence in the far field?

I have been posited with
a "theory" that V = I/R?
to which I would wryly note that nothing has been said of V (and
certainly no interest has been shown), I is known and observations of
P (as evident in the far field, which has occurred to me may be the
issue - had to say it for myself, didn't I?) for a changing R
(somewhere through the iterations).

73's
Richard Clark, KB7QHC

Ah, yes, that's one of those simple observations that lures people into
making false generalizations. I think you'll find that your observation
isn't true for any length, but only for lengths of a quarter wavelength
(for a monopole) and shorter. And only for an antenna consisting of a
single straight wire. 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. The maximum field strength is, therefore, proportional to the sum
of the fields from the individual segments which, in turn, is
proportional to the integral of the currents on the segments. Since the
maximum field strength (or gain) doesn't change much from a very short
wire to a quarter wavelength one, the integral of the current stays
pretty constant. But don't, for heaven's sake, think you've discovered a
rule that applies for all antennas. Not even all straight, single wire
monopoles.

Roy Lewallen, W7EL

Frank wrote:
Concerning current distribution -- at least on a monopole above a perfectly
conducting ground. I have noticed that in any antenna, of any length,
inductively loaded, or not, the: Integral of I(z)dz (where the units of dz
are in fractions of a wavelength, and I(z) the current distribution) is
virtually a constant. Assuming the same input power in all cases. Suppose
it seems pretty obvious since the total radiated power from any structure is
also essentially constant. I must admit I have not seen gain increasing,
with decreasing size, but then I have not seriously tried.

Regards,

Frank

Oh, I would never make such an assumption that it applies to all antennas --
well maybe it did cross my mind!. My guess was that probably everybody knew
it anyway, but thought it was interesting. I have also noticed that the
phase of the current is constant over thin wire monopoles of less than a 1/4
wave

Regarding Richard's comments "Assuming the same currents". Not sure I
understand, since the feed-point current varies with constant power -- as
does the voltage. Also confused by the expression V=I/R.

73,

Frank



"Roy Lewallen" wrote in message
...
Ah, yes, that's one of those simple observations that lures people into
making false generalizations. I think you'll find that your observation
isn't true for any length, but only for lengths of a quarter wavelength
(for a monopole) and shorter. And only for an antenna consisting of a
single straight wire. 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. The maximum field strength is, therefore, proportional to the sum of
the fields from the individual segments which, in turn, is proportional to
the integral of the currents on the segments. Since the maximum field
strength (or gain) doesn't change much from a very short wire to a quarter
wavelength one, the integral of the current stays pretty constant. But
don't, for heaven's sake, think you've discovered a rule that applies for
all antennas. Not even all straight, single wire monopoles.

Roy Lewallen, W7EL

Frank wrote:
Concerning current distribution -- at least on a monopole above a
perfectly conducting ground. I have noticed that in any antenna, of any
length, inductively loaded, or not, the: Integral of I(z)dz (where the
units of dz are in fractions of a wavelength, and I(z) the current
distribution) is virtually a constant. Assuming the same input power in
all cases. Suppose it seems pretty obvious since the total radiated
power from any structure is also essentially constant. I must admit I
have not seen gain increasing, with decreasing size, but then I have not
seriously tried.

Regards,

Frank