On Sat, 26 Feb 2011 07:30:00 -0800 (PST), Wimpie
wrote:

Hi Wimpie,

This may
require another approach then you should use in a professional
environment. If you prefer that, Edaboard.com (just an example) is a
more suitable place.

Now the result is a professional reaction of Norbert:

Curious combination of conflicting sentiments, there. What is
suitable, and how should we recognize it?

Radiation resistance (no coupling with other objects) will be about 1
mOhm.

There are many source for computation, I chose one that closely agrees
with several at hand. Perhaps I made an entry error, so I will take
the opportunity to examine that possibility he
Rr = 80 · pi² · (dl/lambda)²
80 · 9.87 · (2/80)²
790 · (0.025)²
790 · 0.0006
0.49 Ohm

Of course, the possibility of mis-entry remains, and cross checking is
helpful given an in dependant validation. If I examine my text
further it uses as an example a smaller loop at a lower frequency
dl = 1m
F = 1MHz
(lambda = 300)
resulting in
Rr = 0.0084 Ohm
which is roughly 10 times your computed radiation resistance for a
larger loop at a smaller wavelength.

Now, having said that, and examining my text for further possibilities
of error, I find that, yes, I made an error. My computation was based
for an electric dipole, not a loop. Let us examine the Rr for a loop
from the equation from the same source:
Rr = 320 · pi^6 · (r/Lambda)^4
320 · 961 · (1/80)^4
307,645 · 2.44^-8
0.0075 Ohm
This, too, is very different from your calculation, but certainly that
error is eclipsed by my own first reckoning. However, what does this
say about efficiency based upon the original design (but computed for
another)?

However, I did first ask Norbert for the equation used and the
parameters entered. Testing those results did not appear to be
appealing in the face of contradicting testimonial. It should come as
no surprise that many testimonials are tested here. Testimonials
stand or fall in such tests, and those tests are retested (as has
given rise to this and your response).

Curiously we entered into this with how the loop has superior
qualities over the standard dipole, and then the same loop is cited as
being very inefficient. How such contradictions are held within the
space of a short thread is certainly a denial of engineering
professionalism, but denial is not the standard of merit that is
typically lauded in this forum. A hearty defense of wounded ego
raises suspicion even further.

One consequence of that demurral brings us to a rather remarkable
insight in comparing the radiation resistance of the electric dipole
to the loop within the same spread of the loop (and in certainly a
smaller volume of space). The electric dipole enjoys 60 times more
radiation resistance that certainly impacts efficiency to the same
degree. This, of course, presumes no further errors in computation or
application.

73's
Richard Clark, KB7QHC

On 2/27/2011 11:03 PM, Richard Clark wrote:
On Sat, 26 Feb 2011 07:30:00 -0800 (PST),
wrote:

Hi Wimpie,

This may
require another approach then you should use in a professional
environment. If you prefer that, Edaboard.com (just an example) is a
more suitable place.

Now the result is a professional reaction of Norbert:

Curious combination of conflicting sentiments, there. What is
suitable, and how should we recognize it?

Radiation resistance (no coupling with other objects) will be about 1
mOhm.

There are many source for computation, I chose one that closely agrees
with several at hand. Perhaps I made an entry error, so I will take
the opportunity to examine that possibility he
Rr = 80 · pi² · (dl/lambda)²
80 · 9.87 · (2/80)²
790 · (0.025)²
790 · 0.0006
0.49 Ohm

Of course, the possibility of mis-entry remains, and cross checking is
helpful given an in dependant validation. If I examine my text
further it uses as an example a smaller loop at a lower frequency
dl = 1m
F = 1MHz
(lambda = 300)
resulting in
Rr = 0.0084 Ohm
which is roughly 10 times your computed radiation resistance for a
larger loop at a smaller wavelength.

Now, having said that, and examining my text for further possibilities
of error, I find that, yes, I made an error. My computation was based
for an electric dipole, not a loop. Let us examine the Rr for a loop
from the equation from the same source:
Rr = 320 · pi^6 · (r/Lambda)^4
320 · 961 · (1/80)^4
307,645 · 2.44^-8
0.0075 Ohm
This, too, is very different from your calculation, but certainly that
error is eclipsed by my own first reckoning. However, what does this
say about efficiency based upon the original design (but computed for
another)?

However, I did first ask Norbert for the equation used and the
parameters entered. Testing those results did not appear to be
appealing in the face of contradicting testimonial. It should come as
no surprise that many testimonials are tested here. Testimonials
stand or fall in such tests, and those tests are retested (as has
given rise to this and your response).

Curiously we entered into this with how the loop has superior
qualities over the standard dipole, and then the same loop is cited as
being very inefficient. How such contradictions are held within the
space of a short thread is certainly a denial of engineering
professionalism, but denial is not the standard of merit that is
typically lauded in this forum. A hearty defense of wounded ego
raises suspicion even further.

One consequence of that demurral brings us to a rather remarkable
insight in comparing the radiation resistance of the electric dipole
to the loop within the same spread of the loop (and in certainly a
smaller volume of space). The electric dipole enjoys 60 times more
radiation resistance that certainly impacts efficiency to the same
degree. This, of course, presumes no further errors in computation or
application.

73's
Richard Clark, KB7QHC


Wimpie is right, Richard.

Please go back to your laboratory and speak to someone who understands
your dumb-ass dialect. Also, please don't discourage those who are
trying to contribute their experiences here. Try to be positive for a
change.

John

On Sun, 27 Feb 2011 23:14:34 -0600, John - KD5YI
wrote:

Wimpie is right, Richard.

I presume Wimpie can speak for himself. As he offered musings that
were done on the back of a handy envelope, there is every chance he is
not right. I offered a similar chance that I was not right either,
but I offered complete (two in fact) equations that no one has
disputed, and none have faulted for computation. I admitted a
misapplication of one - which also passed without comment.

Considering Wimpie's work was not done for the antenna under
consideration (the size of his being much smaller where radiation
resistance varies by the FOURTH POWER of size) - what does "right"
mean?

73's
Richard Clark, KB7QHC

On 28 feb, 20:53, Richard Clark wrote:
On Sun, 27 Feb 2011 23:14:34 -0600, John - KD5YI
wrote:

Wimpie is right, Richard.

I presume Wimpie can speak for himself. *As he offered musings that
were done on the back of a handy envelope, there is every chance he is
not right. *I offered a similar chance that I was not right either,
but I offered complete (two in fact) equations that no one has
disputed, and none have faulted for computation. *I admitted a
misapplication of one - which also passed without comment.

Considering Wimpie's work was not done for the antenna under
consideration (the size of his being much smaller where radiation
resistance varies by the FOURTH POWER of size) - what does "right"
mean?

73's
Richard Clark, KB7QHC

Hello Richard,

Your formulas can be disputed:

When using (from http://www.ece.msstate.edu/~donohoe/ece4990notes5.pdf):

Rr_loop = 320*(pi)^4*A^2/lambda^4

for f = 3.6 MHz, Dloop = 1.27m (so A = 1.27 m^2),

Rr_loop = 0.001 mOhm.

This result agrees the number in my previous calculation (for the same
situation).

From the same source, but for a dipole of 1.27m with large end-
plates,

Rr_dipole = 80*(pi)^2*le^2/lambda^2 = 0.18
Rr_dipole = 0.045 Ohm (without large end-plates).

This is roughly a factor 45 or 180 more (for the dipole).

Maybe somebody can confirm the above calculations.

The actual efficiency depends on the required (space consuming)
reactive component to cancel the capacitive (dipole) or inductive
(loop) behavior.

The advantage of the loop (especially for reception) is that you need
a variable capacitor instead of a variable loop, and matching / balun
function can be made easily. He also mentioned the vertical radiation
component (NVIS operation) together with the nulls in the horizontal
plane.

Regarding claims, Norbert didn't make claims about the high
efficiency. Please read his conclusion that starts with "despite the
low efficiency of 3%….". His stated 3% reasonably agrees with my 3%
(though you think that the calculation may be wrong). The claim with
regards to performance comparable to a half wave or vertical antenna
is for higher frequencies (where the loop's efficiency increases
significantly).

Of course I have serious doubts about the conclusions regarding
general noise cancelling properties, but the conclusions can be right
for that special RF-environment. Whether they apply for another
situation, can be food for the radio amateur experimenter (or
professional?).


With kind regards,

Wim
PA3DJS
www.tetech.nl

The link should be:
http://www.ece.msstate.edu/~donohoe/ece4990notes5.pdf

Wim
PA3DJS
www.tetech.nl

On 2/28/2011 1:53 PM, Richard Clark wrote:
On Sun, 27 Feb 2011 23:14:34 -0600, John -
wrote:

Wimpie is right, Richard.

I presume Wimpie can speak for himself. As he offered musings that
were done on the back of a handy envelope, there is every chance he is
not right. I offered a similar chance that I was not right either,
but I offered complete (two in fact) equations that no one has
disputed, and none have faulted for computation. I admitted a
misapplication of one - which also passed without comment.

Considering Wimpie's work was not done for the antenna under
consideration (the size of his being much smaller where radiation
resistance varies by the FOURTH POWER of size) - what does "right"
mean?

73's
Richard Clark, KB7QHC


I didn't mean Wimpie was right about his technical response. I meant he
was right about a part of his message which you cut:

"I agree with you that several statements on Norbert's site will not
hold when scientifically reviewed. However I think the way you
respond will likely not result in better statements."

"As the name of the newsgroup indicates; this is a radio amateur group
and Norbert site starts with "Dutch amateur radio station". This may
require another approach then you should use in a professional
environment. If you prefer that, Edaboard.com (just an example) is a
more suitable place."

John

On Mon, 28 Feb 2011 15:36:27 -0600, John - KD5YI
wrote:

I didn't mean Wimpie was right about his technical response. I meant he
was right about a part of his message which you cut:

I selectively quote to make the response specific to the point being
responded to (like I am right now). It saves room, is not ambiguous,
and serves the technical community by confining discussion to
technical matters.

73's
Richard Clark, KB7QHC

On 2/28/2011 7:24 PM, Richard Clark wrote:
On Mon, 28 Feb 2011 15:36:27 -0600, John -
wrote:

I didn't mean Wimpie was right about his technical response. I meant he
was right about a part of his message which you cut:

I selectively quote to make the response specific to the point being
responded to (like I am right now). It saves room, is not ambiguous,
and serves the technical community by confining discussion to
technical matters.

73's
Richard Clark, KB7QHC

Then I'll just have to put the "point" back in

Quote Wim
I agree with you that several statements on Norbert's site will not
hold when scientifically reviewed. However I think the way you
respond will likely not result in better statements.

As the name of the newsgroup indicates; this is a radio amateur group
and Norbert site starts with "Dutch amateur radio station". This may
require another approach then you should use in a professional
environment. If you prefer that, Edaboard.com (just an example) is a
more suitable place.
/Quote

On 28 feb, 20:53, Richard Clark wrote:
On Sun, 27 Feb 2011 23:14:34 -0600, John - KD5YI
wrote:

Wimpie is right, Richard.

I presume Wimpie can speak for himself. *As he offered musings that
were done on the back of a handy envelope, there is every chance he is
not right. *I offered a similar chance that I was not right either,
but I offered complete (two in fact) equations that no one has
disputed, and none have faulted for computation. *I admitted a
misapplication of one - which also passed without comment.

Considering Wimpie's work was not done for the antenna under
consideration (the size of his being much smaller where radiation
resistance varies by the FOURTH POWER of size) - what does "right"
mean?

73's
Richard Clark, KB7QHC

Hello Richard,

you used r = 1m (as you have r in your formulas), that is D = 2m,
6.28m circumference.

I used D = 1.27m (4m perimeter), that is r = 0.635 m.

Quote from Norbert's site:
"When a magnetic loop antenna is used for 3.5 MHz with a perimeter of
4 meter (13.3 foot) , it has an efficiency of approximately 3%."

Maybe this helps you to explain the difference between your and my
result,


Wim
PA3DJS
www.tetech.nl
Don't forget to remove abc in case of PM.

This has been a good thread, I have little room for an antenna, a mag loop may be just the ticket for my small Tampa QTH ?