ve2pid wrote in news:cd290c79-c0c7-4308-af41-
:

Let's suppose that we use to identical length of coax, and on each
one, we connect the shield to the inner conductor at each of its ends.
If we use these two independant (?) sections as a substitute for an
open-wire line, could we use the usual formula Z=276*log(2S/D) to
compute the impedance of that line?

In that case, the characteristic impedance Zo of the coax could be
neglected. And what about the matched losses in dB/100' of the
resultant line? Twice the one of the coax?

73 de Pierre VE2PID


Sounds kinda messy and unstable, to me.... but do-able I suppoes. I'd
rather search for a few short pieces of 25 ohm coax and do the coaxial
balanced line thing as normally done.

Ed K7AAT

ve2pid wrote:
In the ARRL's Antenna Book 21st ed page 24-21, we see that if we
connect the two shields of the coax cables together, we obtain
'Shielded parallel Lines' . In that case, the resultant impedance is
simply the sum of the characteristic impedances of each coax.

So, there is quite a difference between the two independent coax I
mentioned in my first message (we connect the shield to the inner
conductor at each of its ends) (A) and the 'Shielded Parallel Lines'
case (B).

I am trying to understand why and it is the reason I posted my first
message...

In (A), the Z=276*log(2S/D) applies, so the Zo of each coax does not
matter.. but in (B), Z=Zo1+Zo2, so the value of each Zo matters.


I am sure that in case B it is as for resistors in parallel, ie 2 50ohm
cables in parallel give you 25ohms Zo.


Jeff

Jeff wrote in news:hm5dj4$4ls$1
@speranza.aioe.org:

I am sure that in case B it is as for resistors in parallel, ie 2 50ohm
cables in parallel give you 25ohms Zo.

Jeff, you you offer more explanation that just that your are "sure".

If you can't explain it, it speaks of whether you are sure.

Owen

Owen Duffy wrote in
:

steveeh131047 wrote in
:
The application is no secret - the first paragraph in my linked page
explains it.

Steve,

Another thought comes to mind, and that is a pair of rectangular
conductors, preformed to the parabolic like shape and screwed togeter with
insulated fixings.

You could achieve the Zo you seek with say 20mm wide flat with manageable
spacing... however it would be a water trap to some extent (just like LP
feeds using twin booms).

Owen

Owen Duffy wrote:
Jeff wrote in news:hm5dj4$4ls$1
@speranza.aioe.org:

I am sure that in case B it is as for resistors in parallel, ie 2 50ohm
cables in parallel give you 25ohms Zo.

Jeff, you you offer more explanation that just that your are "sure".

If you can't explain it, it speaks of whether you are sure.

Owen

Well having just tried it for real on a network analyser, and simulated
it on Ansoft designer I am now convinced rather than being sure!!

Jeff

On Feb 25, 4:51*pm, Owen Duffy wrote:
Owen Duffy wrote :

steveeh131047 wrote in
:
The application is no secret - the first paragraph in my linked page
explains it.

Steve,

Another thought comes to mind, and that is a pair of rectangular
conductors, preformed to the parabolic like shape and screwed togeter with
insulated fixings.

You could achieve the Zo you seek with say 20mm wide flat with manageable
spacing... however it would be a water trap to some extent (just like LP
feeds using twin booms).

Owen

Owen,

Thanks for the suggestion - water shouldn't be a problem because this
"band interconnect" TL can be installed within the hexbeam centre post
where it would be protected. One chap has already fabricated a "square
coaxial" version - second photo, he

http://lema.epfl.ch/images/stories/L...HB9MCZ_wb.html

but that will be beyond the capability of many constructors!

It's not a major issue - most constructors simply use RG213; but some
find the physical terminations "fiddly", especially if you try to
assemble it inside the centre-post for neatness and weather
protection. I'm pretty sure that twin line using the right gauge ecw
with teflon tubing on one of the wires would be about the correct Zo
and easy to fabricate. It's just another thing on my "to do" list - a
list which seems to get longer rather than shorter

73,
Steve G3TXQ

steveeh131047 wrote in
:

....
Thanks for the suggestion - water shouldn't be a problem because this
"band interconnect" TL can be installed within the hexbeam centre post
where it would be protected. One chap has already fabricated a "square
coaxial" version - second photo, he

http://lema.epfl.ch/images/stories/L...exbeam/hexbeam
_HB9MCZ_wb.html

Seeing the construction, I think a pair of rectangular tubes would be
easier to construct, and raise less water retention issues.

I didn't realise that the line section is straight, I had visualised it
taking the shape of one of the spreaders if you understand what I mean.

Anyway, my recommendation would be like in the pic, but using two side by
side rectangular tubes, say 20mm with 4mm spacing.

Owen

Jeff wrote in
:

Owen Duffy wrote:
Jeff wrote in news:hm5dj4$4ls$1
@speranza.aioe.org:

I am sure that in case B it is as for resistors in parallel, ie 2
50ohm cables in parallel give you 25ohms Zo.

Jeff, you you offer more explanation that just that your are "sure".

If you can't explain it, it speaks of whether you are sure.

Owen

Well having just tried it for real on a network analyser, and
simulated it on Ansoft designer I am now convinced rather than being
sure!!

That is not an explanation at all.

Your confirmation might just be confirmation of a wrong interpretation of
the B configuration.

Owen

Owen Duffy wrote in
:

....

Anyway, my recommendation would be like in the pic, but using two side
by side rectangular tubes, say 20mm with 4mm spacing.

I should have added that I would attach the coax at the bottom, and use a
ferrite cored broadband balun.

Owen

Owen Duffy wrote:
Jeff wrote in
:

Owen Duffy wrote:
Jeff wrote in news:hm5dj4$4ls$1
@speranza.aioe.org:

I am sure that in case B it is as for resistors in parallel, ie 2
50ohm cables in parallel give you 25ohms Zo.
Jeff, you you offer more explanation that just that your are "sure".

If you can't explain it, it speaks of whether you are sure.

Owen
Well having just tried it for real on a network analyser, and
simulated it on Ansoft designer I am now convinced rather than being
sure!!

That is not an explanation at all.

Your confirmation might just be confirmation of a wrong interpretation of
the B configuration.

Owen

and you just might be trolling.

Jeff