"Roy Lewallen" wrote in message
...
Jerry Martes wrote:
Roy

Since I consider antenna efficiency difficult to measure accurately,
I'd
be willing to try to measure it as carefully as *practical*. If that
required that the referance heater had a similar "thermal mass" to that
of
the antenna under test, I could probably make a heater that did have a
thermal mass equivalent to that of the antenna.
But, I would propose that there is no easier way to actually measure
an
antenna's efficiency than to measure the amount of power it turns into
heat.
Thats a statement I cant back up, and I know it.

What is a better way to measure an antenna's efficiency??


I believe it would usually be easier and more accurate to measure the
amount of power it turns into radiation. That is, measure the strength
of the radiated field compared with an antenna of known efficiency and
preferably having a similar pattern.

Neither is easy, and a method that's "practical" from your point of view
might well yield results which are so grossly inaccurate as to be
useless. But go ahead, do a careful estimate of the accuracy you expect,
do some measurements of various simple antennas whose loss is easy to
calculate and compare the measured results, and see how well you can do.
It's probably within the reach of a very careful amateur to make
measurements which are accurate enough to be useful. It just isn't easy,
and requires knowing the relationship between heat and temperature,
where the heat is going, sources of error and the amount they can
contribute, and a whole lot of care and attention to detail. And that's
to get even crudely accurate results.

For myself, I'd model it as Reg suggests (although I'd use a program
rather than the back of an envelope as he would) and easily believe that
the model results are more accurate than any measurement I'd be able to
make. However, neither calculation or measurement results are likely to
alter the claims made by manufacturers of miracle antennas and their fans.

Roy Lewallen, W7EL

Roy

If I had your credentials, I'd measure antenna efficiency the way you
prescribe.
It is only because I have no faith in my abiliuty to convince anyone what
the pattern of a "New-Revolutionary is, that I'd get a thermometer out to
chech my predictions that the antenna has *no* loss, like this guy does.

As I read this thread, there is one guy who thinks measuring the
temperature of an antenna for getting information on its efficiency. And
all the other posts consider that guy to be 'on the wrong path'. I'd advise
doing it some way *other than* measuring heat of the antenna.

Jerry

Yuri

I realize you are a smart and important guy who knows alot about
electronucs. But, when you enter in a thread at this time with statements
about defining "3 DB" I wonder if you think everyone else is supid except
you. Whats the point of defining "3DB??
I used to work with some very good engineers who happened to think
measuring measuring antenna heating was a fairly decent way of getting
"loss" data on an antenna. I happen to think it is a good way to get some
preliminary info on antenna loss.
So, when you tell me to "forget the igloo", you seem to have placed
yourself in a position where you think you know how to evaluate antenna
efficiency and that I dont. I dont think you are *that* smart.

Jerry


"Yuri Blanarovich" wrote in message
...

I sure disagree about being able to determine an antenna's efficiency
by
either resistance measurements or by measuring received signa;l strength
at
any one point in space.

From practical, engineer's point of view, and we are looking at shortened
(loaded) antennas, you (or at least I) want to know how efficient the
loaded
antenna is vs. equivalent (same pattern producing) antenna. You want to
compare
oranges with oranges. The ultimate indicator is how much is one better
(worst)
than the other producing more (less) transmit (or receive) signal. That is
the
ultimate parameter that we are looking for, that's what you want to
measure and
compare. We have dB as a unit for that. 3dB means you gain (lose) double
(power) signal. If you lose 50% in the heat, you will see corresponding
loss in
signal strength. So forget the igloo!


But, you are certainly a much smarter fuy than I
ever was. So, I am not equipped to get into a news group contest.

How did you arrive at that without putting me in the styrofoam igloo, or
that I
am Fuy? :-)

It is my contention that the antenna under test's complete radiation
pattern and field strength would have to be measured and integrated if
field strength is used to determine it's efficiency. That could get
coplicated.

NOT! That's why you want to use the same pattern producing antennas for
comparison (apples to apples) i.e. quarter wave (electrical) shortened
(loaded)
vertical vs. full size quarter wave vertical made of same material
(tubing).
You can make measurements at the same point (properly chosen) and compare
signal levels while swapping the antennas at the same test site.

It would have seemed to me that the loss in any "R" is dependent on the
current flowing in it. And, I'd be concerned that any resonances could
have
higher circulating currents at the operating frequency.


Circulating in what?

If you can point out the errors in my convictions, I'd like to read
about
where I'm wrong.


You can be "convicted" in anything you like. But you have to look at the
problems and see what are you trying to achieve. I always try to make
antenna
to produce the maximum signal in the desired direction, pattern. That's
what
you try to measure, evaluate. Anything else is just contributing factor
that
gets included in the final parameter - signal strength. You can fart with
heat,
resistances, etc., I do not use antennas for heating, I use them for
producing
or extracting signals and that's what I am interested in and want to
quantify.
You need proper "standard" and use proper parameters to compare your
"miracle"
against.

If Mr. Vincent "discovers" that his shortened antenna is more broadband
than
full size (same electrical length) radiator, than he has some serious
resistors
"broadening" the response. My dummy load is perfect broadband "antenna"
and
almost 100% "efficient" - turns almost 100% of power into a heat, but
radiates
almost nothing.

The point is, you can measure other things like heat generated by the
loses,
but you are neglecting other parameters that come to play, leading you
astray,
while neglecting the most important parameter - the result you are (or I
am)
after - the signal strength!
Does that "convict" you? Otherwise I rest my case.
Jerry


Yuri, K3BU.us

On Mon, 16 Aug 2004 11:42:19 -0700, Roy Lewallen
wrote:

It's not clear to me how you could get quantitative data from
measurements in a styrofoam container. Let's say you put 100 watts into
the antenna for five minutes and the temperature rise (of the coil?
helical winding? air?) was 10 degrees C. How would you calculate the
loss or efficiency from that information?


First of all, the test should not be run for 5 minutes, but until a
thermal equilibrium inside the container has been established,
possibly mixing the air inside the container with a fan. This solves
the thermal mass problems when equilibrium has been established.
Measure the power fed to the antenna.

The temperature outside of the container should remain stable during
the test. Then the temperature difference across the container wall is
known, as well as the wall surface area and thickness and hopefully
also the thermal conductivity for styrofoam (from the manufacturer).
Now the thermal resistance of the container as well as the temperature
difference is known and the power dissipation can be calculated. This
is very similar to calculating the heatsink requirements for
transistors.

If the thermal conductivity for the styrofoam is not known, run a
predefined amount of power to the antenna until equilibrium has been
reached and measure the temperature. Inside the container, disconnect
the antenna, attach the feedline to a dummy load (possibly attaching
it to some large metallic part of the antenna to act as a heatsink).
Close the container and feed RF-power into the dummy load at an
adjustable power level, until the same stable temperature is reached
as with the actual antenna. Measure the power fed into the dummy load.
This will eliminate the feedline conductivity problem as well as
offset errors in the power meter (but not linearity errors).

This way the power dissipated by the dummy load is the same as the
power dissipated in the antenna losses in the first place and thus,
the efficiency can be calculated.

To avoid any power meter linearity problems, reconnect the antenna
again, feed in the original full power and verify that there is a
thermal equilibrium at the original temperature and outside
temperature. Insert a calibrated step attenuator ahead of the power
meter and adjust it until it reads the same value as in the dummy load
test.

You can then read directly from the step attenuator, how many dBs the
total full transmitter power was compared to the dummy load power
which is also the antenna thermal dissipation value. A 10 dB setting
would indicate that the thermal losses are only 10 % of the total
power or 90 % antenna efficiency, a 3 dB reading would indicate that
half of the power is radiated and the other half is dissipated in the
antenna, thus 50 % efficiency. However, you would need a step
attenuator with 0.1 steps to measure the worst antennas :-).

Paul OH3LWR

So, when you tell me to "forget the igloo", you seem to have placed
yourself in a position where you think you know how to evaluate antenna
efficiency and that I dont. I dont think you are *that* smart.

Jerry


You win, I ain't so smart as I thought. :-)
I thought that antenna is supposed to radiate RF in the direction and with
pattern one desires. That's what I want to measure, how effective radiator it
is, how much RF it produces at the point of interest.
If you want to know how well it works as a heater, more power to you.
I am not in a ****ing contest who is "smarter", I was trying to convey some
practicality and what I do and what I am interested in.
Over and out!

Back to DR1 (www.computeradio.us)

Yuri, K3BU

On 17 Aug 2004 11:12:46 GMT, oUsama (Yuri Blanarovich)
wrote:

I thought that antenna is supposed to radiate RF in the direction and with
pattern one desires. That's what I want to measure, how effective radiator it
is, how much RF it produces at the point of interest.

Of course that is what everyone would want to measure, but unless you
are working in the millimeter bands in a non-echoic chamber,
determining the true (absolute) radiation pattern can be quite
problematic.

You could fly around in a plane with a calibrated field strength meter
in trying to determine the true radiation pattern of the antenna.
However, the variations in ground conductivity may create a large
variations if the measurement is repeated at different places,
especially with unbalanced antennas.

Thus, it would help a lot, if the efficiency could be determined
separately from the radiation pattern, since much of the calibration
problems could be eliminated. The calorimetric method of measuring the
heat dissipated by the losses is one usable method to get the
efficiency.

If some manufacturer claims extraordinary _gain_ figures for some
LF/MF/HF antenna, you should publicly ask which plane was used for the
measurement and when were these measurements made. You can then check
the local airports, if that plain was in the air at that time.

Paul OH3LWR

First of all, the test should not be run for 5 minutes, but until a
thermal equilibrium inside the container has been established,
possibly mixing the air inside the container with a fan. This solves
the thermal mass problems when equilibrium has been established.
Measure the power fed to the antenna.

The temperature outside of the container should remain stable during
the test. Then the temperature difference across the container wall is
known, as well as the wall surface area and thickness and hopefully
also the thermal conductivity for styrofoam (from the manufacturer).
Now the thermal resistance of the container as well as the temperature
difference is known and the power dissipation can be calculated. This
is very similar to calculating the heatsink requirements for
transistors.

If the thermal conductivity for the styrofoam is not known, run a
predefined amount of power to the antenna until equilibrium has been
reached and measure the temperature. Inside the container, disconnect
the antenna, attach the feedline to a dummy load (possibly attaching
it to some large metallic part of the antenna to act as a heatsink).
Close the container and feed RF-power into the dummy load at an
adjustable power level, until the same stable temperature is reached
as with the actual antenna. Measure the power fed into the dummy load.
This will eliminate the feedline conductivity problem as well as
offset errors in the power meter (but not linearity errors).

This way the power dissipated by the dummy load is the same as the
power dissipated in the antenna losses in the first place and thus,
the efficiency can be calculated.

To avoid any power meter linearity problems, reconnect the antenna
again, feed in the original full power and verify that there is a
thermal equilibrium at the original temperature and outside
temperature. Insert a calibrated step attenuator ahead of the power
meter and adjust it until it reads the same value as in the dummy load
test.

You can then read directly from the step attenuator, how many dBs the
total full transmitter power was compared to the dummy load power
which is also the antenna thermal dissipation value. A 10 dB setting
would indicate that the thermal losses are only 10 % of the total
power or 90 % antenna efficiency, a 3 dB reading would indicate that
half of the power is radiated and the other half is dissipated in the
antenna, thus 50 % efficiency. However, you would need a step
attenuator with 0.1 steps to measure the worst antennas :-).

Paul OH3LWR


Paul, simple approaches to calorimetry are, as you know, very difficult. Having
done them extensively, I don't recommend radio amateurs try it, as the
likelihood of missing even a second-order issue/factor is high.

Knowledge of Netwon's equations is essential vis a vis heating and cooling
rates and equilibrium. (again, you know this; others might not.)

Since the DLM/DLD is a monochromatic antenna, it readily fits into a Wheeler
cap approach for efficiency measurement. There was no data nor discussion of
such an attempt in this veign.

There are no valid efficiency measurements of the DLM from the data I saw. If
someone sees otherwise, kindly let us know and enlighten us.

73,
Chip N1IR
..

On Tue, 17 Aug 2004 13:10:19 +0300, Paul Keinanen
wrote:

First of all, the test should not be run for 5 minutes, but until a
thermal equilibrium inside the container has been established,
possibly mixing the air inside the container with a fan.

Hi Paul,

First, you show that this is not an intractable problem, merely one
that requires care. Good work.

However, as to the statement above, and the presumptions that follow,
there is no equilibrium to observe as the test implementation has been
described. You are pouring calories into an insulated environment
which can only raise temperature without end (short of destruction, of
course).

As I described elsewhere, this is called a caloric bomb, and as such
should be limited in time for all comparisons. What you are
describing requires a steady flow of coolant through the chamber where
you measure the inlet and exhaust temperatures and use THAT
differential. Care must be given to not exhaust into the inlet, and
to maintain the inlet with a body of constant temperature air (to
simplify matters).

The problem is not really that difficult - it only takes attention to
detail such as your response indicates.

73's
Richard Clark, KB7QHC

On Tue, 17 Aug 2004 15:50:05 GMT, Richard Clark
wrote:

However, as to the statement above, and the presumptions that follow,
there is no equilibrium to observe as the test implementation has been
described. You are pouring calories into an insulated environment
which can only raise temperature without end (short of destruction, of
course).

As I described elsewhere, this is called a caloric bomb, and as such
should be limited in time for all comparisons.

You seem to assume that no thermal power will flow through the
styrofoam walls of the box.

Your statement is true if you insert a huge amount of power into the
box, in which case the temperature would climb, until the styrofoam
would melt. However, the measurement can be done at much lower power
levels that are compatible with the thermal conductivity of the
styrofoam.

If styrofoam would be an ideal isolator, you could put some deep
freeze food into a styrofoam box, move to your summer cottage and hope
that the food would still be eatable after a week. Unfortunately this
is not true :-(.

When doing antenna efficiency measurements, the power levels should be
set to a level, in which the temperature increase is manageable (below
the melting point of the styrofoam).

To give an example of the power levels required, assume that
you insert a 1 kg (1 liter) bottle of drink at 0 C into the test
container and after 10000 seconds (about 3 hours) the drink
temperature is at +10 C, the energy needed is about 40 kJ, thus the
leakage through the container walls would be about 4 W. With the
outside temperature of 20 C, the average temperature difference would
be 15 C and the thermal conductivity about 4 C/W. This is what you
usually get with some power transistors with a heat sink :-).

If you build an igloo around the "miracle" antenna, you should be able
to measure the heat generated (and thus efficiency) with power levels
well below the legal limit in most countries and with bad antennas
10-100 W should be enough.

Of course, you should make sure that the styrofoam own RF dissipation
is sufficiently small at the test frequency.

Paul OH3LWR

OH3LWR:

Of course that is what everyone would want to measure, but unless you
are working in the millimeter bands in a non-echoic chamber,
determining the true (absolute) radiation pattern can be quite
problematic.


We are looking at shortened, loaded verticals, monopoles. So you stick the
"standard" - quarter wave radiator in the test point, measure it at proper
point, then you replace it with "miracle" antenna and do the same, feeding them
with the same power. The difference is in overall efficiency. Lets not lose
sight of what are we talking about and what the real "problem" is.
How do you build igloo for 4 square? Comon guys, this is old stuff, has been
done for 100 years already. Lets not fly off on tangents.
Is anyone out there who "calorimetered" the antenna?

Yuri, K3BU

Yes, the temperature will rise until the rate of heat leaving the
container equals the power (rate of energy) entering the box. As the
interior of the box gets hotter (more exactly, as the inside-outside
temperature differential increases), heat exiting the box will increase.
So at some internal temperature an equilibrium will be reached. You'd
want to use a power level high enough to make this temperature well
above ambient so that small variations in ambient temperature won't
badly skew the results, but low enough that you don't cook the box's
innards.

The method you propose seems workable, except I didn't see any mention
of heat leaving the box via thermal conduction along the wires.
Depending on the box's insulating property and the wires, this could be
a significant contributor to the total heat loss from the box. Therefore
it's very important to either insure that this loss is negligible
compared to the loss through the styrofoam, or else to manage it (most
easily by using exactly the same wires and wire orientation during
calibration and test).

As I mentioned earlier, it's probably within the means of an amateur to
make useful measurements, but it would take a lot of care and attention
to detail. There may well be other factors none of us have considered
(as examples, the presence of variable drafts in the vicinity of the
box, which could impact the heat transfer rate in a major way, or
nonlinearity of convective cooling), so measurements of various bodies
of known dissipation would have to be measured to confirm that all
significant factors are accounted for before there will be any
confidence in the results.

Roy Lewallen, W7EL

Paul Keinanen wrote:

On Tue, 17 Aug 2004 15:50:05 GMT, Richard Clark
wrote:


However, as to the statement above, and the presumptions that follow,
there is no equilibrium to observe as the test implementation has been
described. You are pouring calories into an insulated environment
which can only raise temperature without end (short of destruction, of
course).

As I described elsewhere, this is called a caloric bomb, and as such
should be limited in time for all comparisons.


You seem to assume that no thermal power will flow through the
styrofoam walls of the box.

Your statement is true if you insert a huge amount of power into the
box, in which case the temperature would climb, until the styrofoam
would melt. However, the measurement can be done at much lower power
levels that are compatible with the thermal conductivity of the
styrofoam.

If styrofoam would be an ideal isolator, you could put some deep
freeze food into a styrofoam box, move to your summer cottage and hope
that the food would still be eatable after a week. Unfortunately this
is not true :-(.

When doing antenna efficiency measurements, the power levels should be
set to a level, in which the temperature increase is manageable (below
the melting point of the styrofoam).

To give an example of the power levels required, assume that
you insert a 1 kg (1 liter) bottle of drink at 0 C into the test
container and after 10000 seconds (about 3 hours) the drink
temperature is at +10 C, the energy needed is about 40 kJ, thus the
leakage through the container walls would be about 4 W. With the
outside temperature of 20 C, the average temperature difference would
be 15 C and the thermal conductivity about 4 C/W. This is what you
usually get with some power transistors with a heat sink :-).

If you build an igloo around the "miracle" antenna, you should be able
to measure the heat generated (and thus efficiency) with power levels
well below the legal limit in most countries and with bad antennas
10-100 W should be enough.

Of course, you should make sure that the styrofoam own RF dissipation
is sufficiently small at the test frequency.

Paul OH3LWR