I know that many people think G3LHZ is a little bit off his rocker, but out of
curiosity... what he suggests on slide 15 he
http://frrl.files.wordpress.com/2009...heuristics.pdf -
- is that a valid approach to measuring antenna efficiency? -- Use a thermal
camera to note how much an antenna heats up with a given input power, find out
how much DC power it required to heat it to the same temperature (the
antenna's loss), and -- poof! -- antenna efficiency = (input power-loss)/input
power?

What are the significant loss mechanisms that he's not accounting for? (He
claims his matching network isn't getting at all hot.)

Thanks,
---Joel

On 23 mar, 02:24, "Joel Koltner" wrote:
I know that many people think G3LHZ is a little bit off his rocker, but out of
curiosity... what he suggests on slide 15 hehttp://frrl.files.wordpress.com/2009...f-small-an...-
- is that a valid approach to measuring antenna efficiency? -- Use a thermal
camera to note how much an antenna heats up with a given input power, find out
how much DC power it required to heat it to the same temperature (the
antenna's loss), and -- poof! -- antenna efficiency = (input power-loss)/input
power?

What are the significant loss mechanisms that he's not accounting for? *(He
claims his matching network isn't getting at all hot.)

Thanks,
---Joel

Hello Joel,

As with many questions, the answer to the temperature rise method can
be "yes" or "no".

For this discussion I assume an antenna as a device or system to emit
radio waves to certain directions. Mostly designers try to maximize
radiation intensity over electrical input ratio, or total radiated
power over electrical input power.

For situations where obstacles are wavelengths away from the antenna,
temperature rise can be a means of evaluating antenna efficiency. I
once used temperature rise to accurately measure efficiency of a high
efficiency RF amplifier.

In cases where obstacles are very close to the antenna, just
determining temperature rise of the metallic structure being the
antenna does not satisfy me. You will know the dissipated power inside
the antenna, but not inside the obstacle in the reactive field. When
this obstacle dissipates 90% of the electrical input power, overall
efficiency will not be high.

By using the temperature rise of the antenna only, you will notice
higher efficiency when the (loop) antenna is closer to an obstacle
(for example a thick wall). In case of a loop, the Q-factor drops,
resulting in less reactive currents, hence less dissipated power in
the loop and tuning capacitor. Of course more power is dissipated in
the wall.

When the antenna is close to metallic structures with certain
geometry, the real efficiency (so Prad/Pelec) may increase. The large
structure may extract energy from the loop and reradiate it (instead
of converting into heat). The extraction of energy from the loop
results in lower Q-factor, hence less heat loss in the loop and tuning
capacitor. Theoretically spoken, the temperature rise method is a
good one.

How will you relate temperature rise of arbitrary structures to
dissipation? If this question remains unanswered temperature rise
method will also not solve the antenna efficiency question.

Regarding temperature rise methods in general, it is good way to find
where losses are and whether it is worth to do some redesign to lower
losses.

Best regards,


Wim
PA3DJS
www.tetech.nl
PM will reach me, but don't forget to remove abc.

Joel Koltner wrote:
I know that many people think G3LHZ is a little bit off his rocker, but
out of curiosity... what he suggests on slide 15 he
http://frrl.files.wordpress.com/2009...heuristics.pdf
- - is that a valid approach to measuring antenna efficiency? -- Use a
thermal camera to note how much an antenna heats up with a given input
power, find out how much DC power it required to heat it to the same
temperature (the antenna's loss), and -- poof! -- antenna efficiency =
(input power-loss)/input power?

What are the significant loss mechanisms that he's not accounting for?
(He claims his matching network isn't getting at all hot.)

Thanks,
---Joel

A thermal camera is NOT a good way to do calorimetry. It's a fine way to
look for hot spots. Here are some of the potential problems: 1) the
thermal camera converts long wave IR brightness to temperature using
some assumptions about the emissivity of the surface; 2) convective and
radiative losses to the surroundings will change the surface
temperature; 3) surface temperature may or may not correlate well to
dissipated heat.

It's an RF device, so the physical distribution of the power dissipation
will be different than with DC. In a classic substitution RF power
measurement, a lot of effort is made to try and make sure that the
thermal properties are identical for the DC and RF dissipation cases
(well defined broadband load that is physically small, etc.).

In the subject case here, think of this: say you had a 2cm diameter
copper bar and you run 100 Amps of DC through it. The current is
distributed evenly, as is the power dissipation. Now run 1 MHz RF
through that same bar. The skin depth is about .065 mm, so virtually
ALL the RF current is contained within a layer less than 1/3 mm thick.
That's a very different heat and thermal distribution (sort of like the
difference between putting that thick steak in the 200F oven and
throwing it on the blazing hot grill).

One can calibrate for all this, but, still, it's tough.

A better way to do this measurement is to put the antenna in a suitable
far field test site, accurately measure the power flowing into it,
accurately measure the power flowing out of it (e.g. E & H field
strengths in the far field)

Now, finding a suitable site is difficult, particularly at lower
frequencies: you want to be "many" wavelengths away from the ground, for
instance.

How about hanging it from a balloon with a battery powered transmitter
(or receiver: I assume nobody is claiming that reciprocity doesn't work)
and have the field strength detector also hanging from a balloon.

On Mar 22, 9:24*pm, "Joel Koltner"
wrote:
I know that many people think G3LHZ is a little bit off his rocker, but out of
curiosity... what he suggests on slide 15 hehttp://frrl.files.wordpress.com/2009...f-small-an...-
- is that a valid approach to measuring antenna efficiency? -- Use a thermal
camera to note how much an antenna heats up with a given input power, find out
how much DC power it required to heat it to the same temperature (the
antenna's loss), and -- poof! -- antenna efficiency = (input power-loss)/input
power?

What are the significant loss mechanisms that he's not accounting for? *(He
claims his matching network isn't getting at all hot.)

With some feedlines and frequencies, feedline radiation can become an
issue. For example, using 4" ladder line at UHF.

I think his method, especially for physically compact antennas and
feed systems which tend to have very low radiation resistance at HF
frequencies, is a great check on theoretical calculations. There has
to be a meeting point between mathematical models/NEC and reality and
he is working at one such point. There are of course other points too
(e.g. near field and far field measurements).

Tim.

Hi Tim,

"Tim Shoppa" wrote in message
...
On Mar 22, 9:24 pm, "Joel Koltner"
wrote:
I think his method, especially for physically compact antennas and
feed systems which tend to have very low radiation resistance at HF
frequencies, is a great check on theoretical calculations. There has
to be a meeting point between mathematical models/NEC and reality and
he is working at one such point.

Agreed -- the controversy comes into play in that he ends up computing
electrically-small loop antennas as being upwards of 70-90% efficient, when
everyone "knows" that such antennas are typically 10% efficient. He even
goes after Chu/Wheeler/McLean/etc. in suggesting that the fundamental limits
for the Q of an ESA are orders of magnitude off (slide 47), and that's pretty
sacrosanct terriority (see, e.g., www.slyusar.kiev.ua/Slyusar_077.pdf -- even
the Ruskies buy into the traditional results :-) ).

Hence, while I don't really have the background to know precisely how much of
what Underhill promotes is true or not, it's definitely intriguing to me, and
I'm looking around for various rebuttals by those more skilled in the art than
I am.

One link I found: http://qcwa70.org/truth%20and%20untruth.pdf (but this was
written before the PowerPoint presentation I originally linked to).

---Joel

Hi Jim,

Thanks for the thoughts; I hadn't thought of many of the additional loss
mechanisms you mention.

"Jim Lux" wrote in message
...
In the subject case here, think of this: say you had a 2cm diameter copper
bar and you run 100 Amps of DC through it. The current is distributed
evenly, as is the power dissipation. Now run 1 MHz RF through that same bar.
The skin depth is about .065 mm, so virtually ALL the RF current is
contained within a layer less than 1/3 mm thick. That's a very different
heat and thermal distribution (sort of like the difference between putting
that thick steak in the 200F oven and throwing it on the blazing hot grill).

If you're just looking at surface temperature (i.e., with a thermal camera),
will it take more or power at 1MHz to obtain a given surface temperature
increase than at DC?

At DC, since you're heating up the entire bar, and the only way for the heat
to go is up "out" to the surface... I'm thinking... less power is needed for a
given rise?

That would certainly then overestimate antenna efficiency.

---Joel

On Mar 23, 12:13*pm, Tim Shoppa wrote:
On Mar 22, 9:24*pm, "Joel Koltner"
wrote:

I know that many people think G3LHZ is a little bit off his rocker, but out of
curiosity... what he suggests on slide 15 hehttp://frrl.files.wordpress.com/2009...ts-of-small-an...
- is that a valid approach to measuring antenna efficiency? -- Use a thermal
camera to note how much an antenna heats up with a given input power, find out
how much DC power it required to heat it to the same temperature (the
antenna's loss), and -- poof! -- antenna efficiency = (input power-loss)/input
power?

What are the significant loss mechanisms that he's not accounting for? *(He
claims his matching network isn't getting at all hot.)

With some feedlines and frequencies, feedline radiation can become an
issue. For example, using 4" ladder line at UHF.

I think his method, especially for physically compact antennas and
feed systems which tend to have very low radiation resistance at HF
frequencies, is a great check on theoretical calculations. There has
to be a meeting point between mathematical models/NEC and reality and
he is working at one such point. There are of course other points too
(e.g. near field and far field measurements).

Tim.

I can't see how the external fields come into it! That would
automatically be within the two vectors that supply acceleration, this
would be measure by the skin depth created by the displacement
current. The accelleration of charge is a constant dependent on the
conductor used. Where the particle goes when acceleration stops i.e.
after leaving the boundary is of no consequence.This would be seen in
the oscillation losses of the radiator
in the same way as with a pendulum

On Mar 23, 1:56*pm, Art Unwin wrote:
On Mar 23, 12:13*pm, Tim Shoppa wrote:



On Mar 22, 9:24*pm, "Joel Koltner"
wrote:

I know that many people think G3LHZ is a little bit off his rocker, but out of
curiosity... what he suggests on slide 15 hehttp://frrl.files.wordpress.com/2009...ts-of-small-an...
- is that a valid approach to measuring antenna efficiency? -- Use a thermal
camera to note how much an antenna heats up with a given input power, find out
how much DC power it required to heat it to the same temperature (the
antenna's loss), and -- poof! -- antenna efficiency = (input power-loss)/input
power?

What are the significant loss mechanisms that he's not accounting for? *(He
claims his matching network isn't getting at all hot.)

With some feedlines and frequencies, feedline radiation can become an
issue. For example, using 4" ladder line at UHF.

I think his method, especially for physically compact antennas and
feed systems which tend to have very low radiation resistance at HF
frequencies, is a great check on theoretical calculations. There has
to be a meeting point between mathematical models/NEC and reality and
he is working at one such point. There are of course other points too
(e.g. near field and far field measurements).

Tim.

I can't see how the external fields come into it! * That would
automatically be within the two vectors that supply acceleration, this
would be measure by the skin depth created by the displacement
current. The accelleration of charge is a constant dependent on the
conductor used. Where the particle goes when acceleration stops i.e.
after leaving the boundary is of no consequence.This would be seen in
the oscillation losses of the radiator
in the same way as with a pendulum

If you dont understand external fields then you dont understand
Maxwell's equations at all. Maxwell is all about fields. This pretty
much means you havent had a clue about anything you have ever said
about antennas.

Jimmie

On Mar 23, 1:10*pm, JIMMIE wrote:
On Mar 23, 1:56*pm, Art Unwin wrote:



On Mar 23, 12:13*pm, Tim Shoppa wrote:

On Mar 22, 9:24*pm, "Joel Koltner"
wrote:

I know that many people think G3LHZ is a little bit off his rocker, but out of
curiosity... what he suggests on slide 15 hehttp://frrl.files.wordpress.com/2009...ts-of-small-an...
- is that a valid approach to measuring antenna efficiency? -- Use a thermal
camera to note how much an antenna heats up with a given input power, find out
how much DC power it required to heat it to the same temperature (the
antenna's loss), and -- poof! -- antenna efficiency = (input power-loss)/input
power?

What are the significant loss mechanisms that he's not accounting for? *(He
claims his matching network isn't getting at all hot.)

With some feedlines and frequencies, feedline radiation can become an
issue. For example, using 4" ladder line at UHF.

I think his method, especially for physically compact antennas and
feed systems which tend to have very low radiation resistance at HF
frequencies, is a great check on theoretical calculations. There has
to be a meeting point between mathematical models/NEC and reality and
he is working at one such point. There are of course other points too
(e.g. near field and far field measurements).

Tim.

I can't see how the external fields come into it! * That would
automatically be within the two vectors that supply acceleration, this
would be measure by the skin depth created by the displacement
current. The accelleration of charge is a constant dependent on the
conductor used. Where the particle goes when acceleration stops i.e.
after leaving the boundary is of no consequence.This would be seen in
the oscillation losses of the radiator
in the same way as with a pendulum

If you dont understand external fields then you dont understand
Maxwell's equations at all. Maxwell is all about fields. This pretty
much means you havent had a clue about anything you have ever said
about antennas.

Jimmie

Jimmy
I am referring to the boundary laws which is energy in versus energy
out.
Maxwells laws finish with the completion of acceleration of charge.
The boundary laws are covered by this action and reaction per Newton.
The particle that is accellerated is the smallest known with respect
to mass and we know that it is accellerated to the speed of light
which is known for any particular medium.
Thus knowing the energy supplied we must equate it to the ejection
vector applied to the particles and the reaction force applied to the
radiating member.
NEC computer programs do just this and for such equations produce
arrays where each
element is resonant and tipped to oppose the two vectors of gravity
and the rotation of the Earth by supplying the array only where it is
in a state of equilibrium and all elements are resonant as a result of
the initial two vectors. The NEC computer programs do just this when
applying Newton's laws which are not mismanaged to reflect planar
forms.
Again I state that Newtons equations account for all vectors involved
in radiation and does not in any way reflect the fields that are
generated beyond acceleration and how the particles are dispersed
beyond this point. This way it represents all types of radiation that
the radiator is capable of and likewise makes it sensitive to all that
is thrown at the receiving end ie H,V, cw, ccw signals e.t.c. and all
the rest that is thrown at it which it converts to a useable current
signal for the radio.Now if you are still in a state of flux as to the
use of equilibrium in all the laws of the Universe then you are still
spitting into the wind.
Art Unwin KB9MZ....xg

On Mon, 22 Mar 2010 18:24:06 -0700, "Joel Koltner"
wrote:

What are the significant loss mechanisms that he's not accounting for? (He
claims his matching network isn't getting at all hot.)

Hi Joel,

Your source is thrashing against a number of conventions that he has
disproved by relying on Schopenhauer - what a wheeze! This is the
same argument that "revolutionary" thinkers appeal to forgetting that
their theories can be dismissed by the same mechanism.

Let's examine this heuristically (irony there). Two observations
provided by your source raise the temperature of
1. a can by 100 deg C with 100W;
2. a tube by 100 deg C with 150W.
Heat is always concerned with mass and surface area, and heat transfer
is expressed in Watts per square Meter. Temperature rise is expressed
in joules for specific heat capacity and mass.

The two examples provided by the source differ by -50/+100% in heat
transfer. How does this impact the development of a "new theory?"
Let's revisit the two examples normalized to transfer:
1. the can exhibits 127W/m²
2. the tube exhibits 57.7W/m²
Such discrepancies are meaningless? I assume so, because the source
doesn't respond to answering them.

There is, of course, a very simple explanation that heuristics reveals
in specific heat capacity (a term completely absent from the
discussion of heat in a radiator).

Let's consider the nearly 1m loop with a heating wire inside -
curiously unspecified for such a "scientific" and unbiased report. It
is, in fact, an 18ga wire of 80% nickel and 20% chromium (yes
nichrome) which by the current supplied heats up to 550 deg. C. These
details must be immaterial to the original source that ignores them.

Could one imagine a heater wire at 550deg. C heating the outer tube by
100 deg C only? Well, let's consider that light bulb in the can that
exhibits double the heat transfer. It's filament is running at
somewhere between 2200 and 3000 deg C for the same temperature rise.
Nature must have some curious law of heat plateau in this "new
theory."

If we were to rely on heat reportings alone to carry the logic, then
this vast wobbly range of -50/+100% reported values is not very
compelling. Intuition is often appealed to in these pages (some call
it heuristics); and it would seem that the heated wire, with 150W and
supporting 550 deg C would eventually melt the antenna components.
Where else is the heat at that temperature going? It is entirely
contained by the copper tube. Yet our source does not reveal this -
and ironically points out that would be the fate if RF could raise
temperatures.

However, temperature is NOT power. How much power would it take to
raise that specific 83 cm diameter looped 10 mm copper tube 100 deg C?
The answer is 2.16 W-Hours. The 550 deg C hot wire is
running at 150 W continuous and only pumping
up the temperature a paltry 100 deg C.
That is pretty pathetic.

The answer is ALSO 7800 W-Seconds.
Clearly 150W for several seconds wouldn't
make a noticeable heat boost.
That is pretty pathetic too.

Obviously heat rise is a product of time (the term "rise" demands
this) and yet throughout this source's "scientific" and unbiased
report time measurement for these heat readings are wholly absent. How
many in this forum recognize the duty cycle of a typical QSO in this?
How many antenna failures follow from heat, how many QSOs still suffer
from poor efficiency?

For those who care, the formula is simple:
Q = m · C · (Tf - Ti)
where
Q is joules
m is mass in grams
Tf is final temperature (K or C)
Ti is initial temperature (K or C)
C is specific heat capacity in
J/(K · g)
or
J/(C · g)
and for copper is 0.39

For comparison, water's specific heat capacity is 4.18, more than ten
times higher and thus more power is required to raise water that same
100 deg. and this is why water is used in radiators. Mineral oil runs
about half that (and folks struggle to obtain transformer oil for
their dummy loads when water is superior).

An aluminum tube at 0.90 is thus going to run cooler than copper when
we put a heater wire inside (for the same, unstated, time interval).

Given the inordinate number of facts missing, undisclosed time
intervals, no discussion of heat capacity (in a heat report) and the
obvious cynicism cloaked in "heuristics, science, and unbias" -
further reading on other topics has every chance of being equally
blighted.

To put icing on the cake, I am astonished that the original source can
only eke out low 90s percent efficiency in the 20M band! To employ
the original source's arguments: Put 450W to that sucker and tell us
where the 20 to 30W were lost. As it takes only 2W-Hr to raise that
antenna 100 deg C, that air cooled dummy load must be boiling rain and
burning fog.

--- and then again, maybe I did the math wrong.....

73's
Richard Clark, KB7QHC