On Apr 7, 1:55*am, wrote:
On Apr 6, 11:41*pm, Owen Duffy wrote:



Owen Duffy wrote :

... Here is a deck for 8 halfwaves at 435MHz with a single stub tuner.

...

Ouch, that had some remanents of a matching scheme using RG62. Here is a
better deck.

CM 8 half waves coaxial collinear on 435MHz
CM Assumes lossline TL, VF=0.667, effective choke at bottom of array.
CM Matched to 50 ohms with single stub tuner.
CM Owen Duffy 2009/04/05
CE
GW 2 20 0 0 -0.229885 0 0 0 0.0045
GW 3 20 0 0 -0.45977 0 0 -0.229885 0.0045
GW 4 20 0 0 -0.689655 0 0 -0.45977 0.0045
GW 5 20 0 0 -0.91954 0 0 -0.689655 0.0045
GW 6 20 0 0 -1.149425 0 0 -0.91954 0.0045
GW 7 20 0 0 -1.37931 0 0 -1.149425 0.0045
GW 8 20 0 0 -1.609195 0 0 -1.37931 0.0045
GW 9 10 0 0 -1.83908 0 0 -1.609195 0.0045
GW 200 1 -0.01 0 -2.25308 0.01 0 -2.25308 0.001
GW 201 1 -0.01 0 -2.35308 0.01 0 -2.35308 0.001
GE 0
GN -1
EK
EX 0 200 1 1 0
TL 3 1 2 1 -50 0.3448276 0 0 0 0
TL 4 1 3 1 -50 0.3448276 0 0 0 0
TL 5 1 4 1 -50 0.3448276 0 0 0 0
TL 6 1 5 1 -50 0.3448276 0 0 0 0
TL 7 1 6 1 -50 0.3448276 0 0 0 0
TL 8 1 7 1 -50 0.3448276 0 0 0 0
TL 200 1 8 1 50 0.471 0 0 0 0
TL 200 1 201 1 50 0.0855 0 0 1e99 0
FR 0 0 0 0 435 0
RP 0 361 1 1000 -180 0 1

The coax half wave sections are exactly a half wave (electrically).

Owen

Isn't there an inherent problem with this design when using coax
sections wich have a velocity factor which differs from that of free
space ?

The alternating sections rely on radiation from the outer (common
mode) and a phase shift occuring along the inner (differential mode
subject to the coax VF).

Because of this mismatch the cumulative phase error along the length
of the antenna will result in it only being close to the required
phase shifts over the first few sections. Hence the gain reduction as
more sections are added.

To work properly the coax sections would need to be air spaced.

UKM

So, consider this: with lossless transmission line and each section
of transmission line being an electrical half-wave long, the VOLTAGE
between adjacent ends of any two sections is identically the same
(including being in phase). So every feedpoint -- that is, every gap
between sections -- is fed with the same in-phase voltage. Practical
feedlines come pretty close to that lossless ideal, short as they
are. (You can model this quite accurately to see just what the
variation in feedpoint voltages is along the array.)

That does not guarantee that the currents on all sections are in
phase, nor does it guarantee that they are the same magnitude, but the
simulations I've run tell me that they are pretty close to being in
phase. It's really not so important that they be all the same
magnitude. Were it not for mutual impedances among the elements,
having identical feedpoint voltages would yield the same current on
each section (except the very end sections, which don't have a
feedpoint connection at their outer ends).

I note that Owen has reported some different results (larger current
phase differences than I recall seeing), and if I can find time, I'd
like to explore those with him, but at the moment I'm tied up with
other things.

Cheers,
Tom

K7ITM wrote in
:

....
I note that Owen has reported some different results (larger current
phase differences than I recall seeing), and if I can find time, I'd
like to explore those with him, but at the moment I'm tied up with
other things.

Tom,

The models I offered (and they are very similar) do have fairly good
cophase operation.

My comment earlier was that some designs aren't nearly as good, although it
seems to degrade gain by only a small amount (though giving rise to more
and narrower lobes).

In your own time...

73
Owen

Hi Owen

I would like to put a question regarding a previous discussion.

Not having your e-mail address, I have to catch you on another thread.

I report below what you wrote at that time:

QUOTE
From TLLC, the matched line loss in dB of LMR400 (a foam coax of similar OD to
RG213) is
3.941e-6*f^0.5
+1.031e-11*f
The first term is due to R and the second due to G.

At 144MHz, the percentage of power lost per meter due to R is
(1-10^-(3.941e-6*f^0.5)/10)*100 is 1.08%. If you do similar for G, the loss is
0.034%, so loss in R is more than 30 times loss in G
UNQUOTE

I put those formulas on a spreadsheet, but I only obtain the 0.034% figure if I
change the second formula into +1.031e-5*f (instead of +1.031e-11*f)

Any comment?

Thanks and 73

Tony I0JX
Rome-Italy

"Antonio Vernucci" wrote in
:

Hi Owen

I would like to put a question regarding a previous discussion.

Not having your e-mail address, I have to catch you on another thread.

I report below what you wrote at that time:

QUOTE
From TLLC, the matched line loss in dB of LMR400 (a foam coax of
similar OD to RG213) is
3.941e-6*f^0.5
+1.031e-11*f
The first term is due to R and the second due to G.

At 144MHz, the percentage of power lost per meter due to R is
(1-10^-(3.941e-6*f^0.5)/10)*100 is 1.08%. If you do similar for G, the
loss is 0.034%, so loss in R is more than 30 times loss in G
UNQUOTE

I put those formulas on a spreadsheet, but I only obtain the 0.034%
figure if I change the second formula into +1.031e-5*f (instead of
+1.031e-11*f)

Any comment?

Thanks and 73

Tony I0JX
Rome-Italy



Ok, firstly, there was a mistake in my formula... it is missing a pair of
parentheses, and should be (1-10^(-(3.941e-6*f^0.5)/10))*100.

That correctly finds 1.08%/m for R loss.

From my spreadsheet check, =(1-10^(-(0.00000000001031*f)/10))*100
correctly calculates 0.034%/m. Note the exponent of f is 1 in the G case.

Apologies for the parenthesis omission. I wrote down what I 'did' on an
RPN calculator rather than copying an expression that evaluated properly.

Does this answer your question?

Owen

On Apr 5, 6:50*pm, Owen Duffy wrote:
wrote in news:a706ece1-d519-47d0-8ffe-8e76cf03ff66
@z16g2000prd.googlegroups.com:

I always assumed having more 1/2 elements in a collinear was best,
because that obviously raises gain and lowers angle. So from that
point of view, I was thinking that it would actually be better to use
a low VF coax, since that would give you shorter length elements, thus
being able to fit more elements in a shorter space. But I've also read
that having the element lengths closer to actual 1/2 length (longer)
is actually more efficient than having more elements at a shorter (low
VF) length. Just wondering what people's opinion is on this... So lets
say you have a choice between using the insulated center conductor of
a VF 78 coax through brass tube outer elements and having room for a
few extra elements in a given length, verses using just an insulated
wire that has a 99 VF through brass tube outer elements, thereby
allowing slightly less elements because they're longer, but the
elements you do have are closer to actual 1/2 length. Which would be
best? And this is basically various UHF bands we're talking about.

Dave,

The popular explanation for these things constructed with reversing coax
sections is that the currents all along the vertical are (exactly) in
phase.

That explanation doesn't seem consistent with the nominal half wave
elements being in fact a halfwave at VF=0.67, ie about two thirds the
length. Nor does it deal with the fact that the element ends are
connected to each other, ie no charge difference permitted.

I have been playing with an NEC model of eight half wave
elements using RG213. The currents are certainly not exactly in phase,
not nearly, and the gain in freespace at 6dBi is less than I see claimed
for this antenna. People seem to justify a claim of dB gain as 3dB for 2
elements, 6dB for 4 elements, 9dB for 8 elements... in which case they
must mean gain to be wrt a dipole, and therefore gain of 8 half waves
would be 11.2dBi... very sus, too simplistic.

You asked specifically about efficiency. The efficiency in my NEC model
(which includes internal loss in the RG213) is very good, 99%.

Confusing part is that some of the designs have a half wave sticking out
the top, and others a quarter wave, similarly different treatments at the
bottom, and they aren't consistent about whether the coax breaks at or
near voltage maxima or minima.

I am still working on this, I suspect the antenna doesn't quite work as
often explained, and not nearly as good as claimed.

Owen

Owen you make a very good point by alluding that the zero cross over
point is not 50% of a the time for a full period but 0.67 % of a
period.Thus if a tank circuit is inclusive of a radiation period this
represent a difference in speed or time with respect to charging and
discharging of lumped loads both of which must be included
to achieve equilibrium in the absence of friction ( perpetual motion)
Thus the peak
amplitude of the resulting occillation at a particular frequency
determines the characteristic impedance when resonance occurs at that
frequency. In the real world on Earth occillation without friction is
impossible thus frequency losses are in parabolic or additive form
with the change in the number of periods which equates to frequency.
Thus the 0.67 cross over point represents the difference of time
between kinetic and potential energy accomodation which produces a
curve that deviates from a true sinosoidal curve. Since NEC is formed
on the condition that the exchange time between potential and kinetic
exchange are equal i.e sinosoidal then the results obtained have an
inherrent error ie assumtion is different from the factual. ie not
truelly sinosoidal.
Regards
Art

Ok, firstly, there was a mistake in my formula... it is missing a pair of
parentheses, and should be (1-10^(-(3.941e-6*f^0.5)/10))*100.

That correctly finds 1.08%/m for R loss.

From my spreadsheet check, =(1-10^(-(0.00000000001031*f)/10))*100
correctly calculates 0.034%/m. Note the exponent of f is 1 in the G case.

Apologies for the parenthesis omission. I wrote down what I 'did' on an
RPN calculator rather than copying an expression that evaluated properly.

Does this answer your question?

Owen,

I simply pasted and copied your formulas in my Excel.

For f=144:

- the first formula gives me 0.00108893 that is 0.108%, which is 10 times lower
than your figure
- the second formula gives me 3.41851e-08 that is 0.0000034%, which is 10,000
times lower than your figure

I did that three times, same results.....

I cannot understand what can be wrong....

Tony I0JX

"Antonio Vernucci" wrote in
:

Ok, firstly, there was a mistake in my formula... it is missing a
pair of parentheses, and should be (1-10^(-(3.941e-6*f^0.5)/10))*100.

That correctly finds 1.08%/m for R loss.

From my spreadsheet check, =(1-10^(-(0.00000000001031*f)/10))*100
correctly calculates 0.034%/m. Note the exponent of f is 1 in the G
case.

Apologies for the parenthesis omission. I wrote down what I 'did' on
an RPN calculator rather than copying an expression that evaluated
properly.

Does this answer your question?

Owen,

I simply pasted and copied your formulas in my Excel.

For f=144:

- the first formula gives me 0.00108893 that is 0.108%, which is 10
times lower than your figure
- the second formula gives me 3.41851e-08 that is 0.0000034%, which is
10,000 times lower than your figure

I did that three times, same results.....

I cannot understand what can be wrong....

Tony I0JX



Ok, you have calculated for 144Hz. Try f=144e6 and don't format the cells
with %.

Owen

Ok, you have calculated for 144Hz. Try f=144e6

Yes it works fine now

Thanks for help.

Tony I0JX