On Fri, 28 Nov 2008 17:08:06 -0800 (PST), K7ITM wrote:

On Nov 27, 2:46*pm, Jeff Liebermann wrote:
...
Good luck, but first a little math. *What manner of tolerance do you
thing you need to cut your coax pieces? *Let's pretend you wanted to
get the center frequency accurate to 1Mhz. *At 418MHz, one wavelength
is:
* *wavelength(mm) = 300,000 / freq(mhz) * VF
* *wavelength = 3*10^5 / 418 * 0.83 = 596 mm
That works out to:
* *596 / 418 = 1.4 mm/MHz
So, if you want the center frequency accurate to within +/- 1MHz, you
gotta cut it to within +/- 1.4 mm. *Good luck. *Like I previously
ranted, you'll need a cutting fixture. *A steady hand, good eye,
quality coax, and plenty of patience are also helpful.

But why would you care to try to get it within 1MHz?

I don't. I wanted a number to show how accurate the cut would need to
be if he wanted the minimum VSWR point to be accurate to within 1MHz.
I picked 1MHz because the tolerance is easily scaled to other
bandwidth and accuracy numbers. My main point was that a fixture of
some sort was necessary to obtain that level of accuracy.

With only four
radiating elements, the beam 3dB width will be roughly 8 degrees if
the bottom of the antenna is a wavelength above ground (30 degrees in
freespace...). There's not much point in putting a lot of effort into
get closer than perhaps 4 electrical degrees along the line, and I
don't believe even that is necessary to get good performance. That's
several mm, and should be easy with such short lengths.

Well, at 418MHz, one wavelength (electrical) is about 600 mm.
4 degrees is:
600 mm * 4/360 = 6.7 mm
Yeah, that's fairly loose and could be done with diagonal cutters and
a tape measure. Normally, I would punch the numbers into an NEC
model, but I couldn't figure out how to model a radiating coax cable
section as an antenna element (using 4NEC2). I sorta faked it with
wire segments, but got stumped on what to do with the dielectric and
it's velocity factor.

Here's one way to build a coax cable colinear (for 2.4GHz):
http://www.nodomainname.co.uk/Omnicolinear/2-4collinear.htm
Note the measurements in Fig 1 for where to measure the half
wavelength sections. I'm suspicious. Of course, at 418MHz, it's less
critical.

Using foam-
Teflon coax makes it easy to do: the insulation doesn't melt when you
solder things together.

I've only played with the RG6/u CATV flavor, where everything is
crimped. I never have tried to solder the stuff.

RG8/u with foam teflon dielectric:
http://www.westpenn-cdt.com/pdfs/coaxial_spec_pdfs/50%20ohm%20cables/25810.pdf
Looks nice.

I cut the sections to matched lengths, use a
little jig to trim the layers to the same lengths on each, and then
put a wrapping of 30AWG or so silver plated wire (wire-wrap wire)
around each joint to hold it while soldering. That makes it easy to
adjust before soldering, and solid after.

See photos of the jig at bottom of:
http://www.nodomainname.co.uk/Omnicolinear/2-4collinear.htm

Incidentally, since the top 1/4 wave element represents something
close to perhaps 50 ohms, it would be interesting to measure the
amount of RF that isn't radiated and actually gets to the top section
of the antenna. *If my analysis of the antenna is correct, the first
section (near the coax connector) radiates 1/2 the power. *The next
section 1/4th. *After that 1/8th, etc. *By the time it gets to the top
of the antenna, there won't be much left. *However, that's theory,
which often fails to resemble reality. *It would interesting if you
stuck a coax connector on the top, and measured what comes out.

There's very little loss in a half wave of decent coax at 450MHz.
That means that the voltage across the lowest junction between
sections is echoed up the antenna at each other junction. In
freespace, by symmetry, the currents will be very nearly the same
going down from the top as going up from the bottom. My model over
typical ground (bottom a wavelength above the ground) shows current
symmetry within a percent or so, assuming equal voltages driving each
of the three junctions. If you wish, you can use the parameters of
the line you're actually using to figure the differences among the
feedpoint voltages, based on the loads at each junction. When I've
done that in the past, the differences are practically negligible.
You can iterate, feeding those voltages back into the model to find
new load impedances, etc., repeating till you're happy that the models
have converged. Recent versions of EZNEC even let you put the
transmission line into the model, along with its loss.

I hate to admit that I made a mistake, but as you and Roy Lewallen
point out, my explanation of how this antenna operates is almost
certainly wrong. I'll do a fast measurement tomorrow to satisfy my
curiousity, but from your explanation and Roy's, I've erred big time.
With a constant current distribution along the length of the antenna,
and a constant voltage at the various feed points, it's a fair
conclusion that the power radiatated around each of these feed points
are equal. I goofed(tm).

The supporting tube certainly will affect the feedpoint impedance, but
in my experience, it does not materially affect the pattern. I deal
with the impedance through a matching network; it's no trouble to
adjust for a low enough reflection that I don't worry about it.
Decoupling is the more interesting problem, to me.

Cheers,
Tom

Gone sulking...

--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558