On Apr 9, 11:36 am, "Antonio Vernucci" wrote:
Hi Tony,

I would be happy to have a look at the file and add the hairpin if you
wish. But I can also guide you through doing it yourself. It should
be quite easy. Simply click on the " Trans Lines" button, the second
one below Sources in the main window. Specify the first end the
same as the source--perhaps 50% along wire 2, assuming the D.E. is
wire 2. That end will then be in parallel with the source. You can
type "S" in for the other end's wire #, or if you click in the End 2
Wire # box, you should see a list of "open" and "short" and you can
select "short" there. Then enter the length, Z0, velocity factor (1)
and loss. Reverse or normal doesn't matter since only one end is
connected.

Thanks on you guidance. My hairpin is a U of aluminum tube connected directly to
the antenna feedpoints. The U center is connected to the boom.

The hairpin tube diameter is about 0.8 cm. The three sides of the U are
approximately 35 cm, 10 cm and 35 cm. I presume that I shall enter 35 cm as the
transmission line length, but how to calculate Z0? Moreover what about the
velocity factor, perhaps 0.98 for a tube in open air?

Would you have a formula at hand?

Thanks and 73

Tony I0JX

Hi Tony,

Since the spacing is so large (10cm), it may be better to enter the
hairpin as wires in the model. It might be interesting to compare the
results between wires and a transmission line. First the wires:

I assume that the D.E. is one wire in the model you have now. I
suppose it has an odd number of segments, so that the source can go
just in the middle. I'm assuming the hairpin parallel tube segments
are center-to-center 10cm apart. I will assume the D.E. is currently
250cm long, all at some x value, and y extending from -1.25m to
+1.25m, as one "wire" of some diameter. To be able to connect the
hairpin at the right places, we need wire ends at -.05m and +.05m, so
we can change the D.E. wire to be one segment between y=-.05m and +.
05m, and then add 2 wires the same diameter as the D.E., one from
y=-1.25m to -.05m, and one from y=+.05m to +1.25m, and each with about
half as many segments as the original D.E., all at the original x
value. (Or exchange x and y if I have guessed wrong.) So if the
original D.E. had 11 segments, put 5 segments on each of the two new
wires. Now add three new wires to represent the hairpin U. Assuming
the hairpin is oriented perpendicular to the plane of the antenna, it
will be all at the same x value as the D.E. I'll assume the antenna
is at z=0. So the wires will have ends at x,-.05,0;x,-.05,.35 and x,-.
05,.35;x,+.05,.35 and x,+.05,.35;x,+.05,0 -- and all with diameter 8mm
or .008m.

For the transmission line model: the diameter (8mm) and spacing
(100mm center to center) tells me the line impedance is about 380
ohms. It should be OK to assume VF=1.00, if there is no insulation
around the tube. One problem is that the short across the end is not
really a short--it is more an inductance. We could go to the trouble
of calculating the inductance and putting that as a load on the end of
the transmission line instead of just a short, but it should be a good
approximation to just lengthen the line by half the length of the
"short"; I would enter the line length just as 40cm instead of 35. (I
won't be surprised if Owen pops in here with either a confirmation or
a better suggestion!)

Cheers,
Tom