On Feb 12, 5:58 pm, "Yuri Blanarovich" wrote:
Looks like W8JI type of response: I am right, I will not respond to
questions and arguments and my is last word!"

Well:

Point 1. What is wrong with a W8JI type of response?

Point 2. I already violated my promise to remain silent on this
thread, in the pointless belief that it might be possible to reason
with you. How wrong I am.


Fact 1. - When hairpin or beta match is inserted in the driven element or
any resonant half wave dipole, it has current flowing through it.

In my "lost" post, I addressed this. I said more-or-less, of course a
stub, hairpin, beta, or whatever you would like to call it has current
in it. If it didn't then it could be removed without effect. So
what? Also if the stub is 1/4 lambda and shorted then the current
is maximum at the short. Again, so what?


Fact 2. That element has to be shortened if resonance is to be maintained,
which means that it is a center inductive loading element, hairpin, beta
match is the part of the antenna standing wave circuit.

How about calling it a point rather than a fact? The real fact is
that when shortened, the DE is no longer resonant.

In an earlier post I said, and you apparently chose to ignore it or
were too busy formulating a response to read:

"Instead of the integral stub, which Yuri believes is part of the
antenna that is "folded back" along the boom, I will move the matching
system away from this location using an ideal ½ wavelength (34.7 foot)
transmission line with an ideal current balun at the antenna end. I
don't believe anyone would argue that the feedpoint impedance is not
replicated exactly at the input end of this line. "

Now when the hairpin, beta, or lumped inductor is 1/2 lambda away from
the DE and it acts *exactly* the same, are you telling me that it is
still part of the radiator and "loads" the DE?

If you are saying yes, then this discussion really is over. If you
will keep an open mind, let's say to keep the numbers easy, that the
DE feedpoint when shortened appropriately is 25 -j25.

At the input to an ideal 1/2 wavelength line, terminated with the DE,
we will also see 25 -j25. The parallel equivalent of this (I should
use admittences, but I'll leave it as impedance) is 50 || -j50.
Shunting this with an inductive reactance +j 50 results in 50 +j0, or
a perfect match.

Now I suppose it can be argued that this is "resonating" the DE,
although I would prefer to say the system is resonant.

I don't see how you can argue that the added inductor is "radiating"
or "subtracting" from the radiation of the DE, or whatever your point
is.

Nevertheless, you give it a try below. But I'm not sure Google is
going to post this anyway, so I'm stopping here. I think the OP's
original question has been answered: For all practical purposes, the
radiation characteristics are unaffected by the matching scheme.


Fact 3. There is current that would more, less belong to the antenna element
flowing in the hairpin, folded on itself and on right angle to driven
element, not contributing to overall current distribution. That can be seen
in my EZNEC model in the article. That portion of the current is not in line
with the rest of the element, it is missing from as to compared to full size
element and therefore efficiency is slightly lower (middle part of the
cosine wave is "missing" folded with hairpin). On typical 40m dipole, as I
modeled, it accounts for about 0.2 dB loss.

The original argument was is the beta match lossy or not.
My modeling of the real situation is ignored, also findings and experimental
verification of others who replaced hairpins with solenoid coils is ignored.
What you are "arguing" is that if you place the matching network in the
driven element (now you even lengthen it) that there is no loss.http://www.k3bu.us/beta_match.htm
Is my modeling of 40m dipole wrong? Where? On dipole it shows about 0.2 dB
loss, which gets bit more when used in multielement array.

If you are denying that there is a RF antenna current flowing in the Beta
match hairpin, then you are denying reality and I rest my case. No point of
arguing or discussing when you do not show where I am wrong, but you offer
"proof" by juggling matching components at the feedpoint. We are talking
about hairpin that is part of the antenna element, has current flowing in
the system and it has feedline attached at the feedpoint of element and
hairpin. No tricks with coils or capacitors inserted half wave somewhere.

73 Yuri, K3BU.us

We are not arguing loss due to loading, we are arguing - does hairpin beta
match contribute to the loss in the system or not.
I mentioned that there are better alternatives, if one choses to be less
lossy.

"Wes" wrote in message

oups.com...

On Feb 10, 7:27 pm, "Yuri Blanarovich" wrote:

My previous ISP sold me off to a new one. SInce that time I've had no
end of email problems and in addition the new guys don't provide
newsgroups, so I've been reduced to using Google Groups. Yuch.

I spent too much time composing a response to Yuri's last post and
then did not see it appear after posting. I will not waste further
time, but in a nutshell I offered another case using the self-
reactance of the DE as part of the matching network.

In this example, to avoid any "loss due to loading" resulting from
shortening the DE, I *lengthen* the DE to obtain an inductive
reactance for the series arm of an L-network and then use a shunt
capacitor for the second matching element.

The model is he

http://www.k6mhe.com/n7ws/N7WS_Yagi_Lengthened.EZ

Run the analysis, and add a shunt 195 pF and see what you get.

Regards,

Wes N7WS