(Dave Platt) wrote in
news
....

I think you'll need to run a a simple calculation, based on the
frequencies you actually want to use.

What you'll want, is a wire whose length is not particularly close to
any multiple of 1/2 wavelength, on any frequency you want to use. A
wire which would match easily on 80 meters (e.g. 1/4 wavelength long)
would be a bad choice if you want to work on 40 meters as well, as
it'd be 1/2 wavelength long.

A simple program or spreadsheet ought to be able to do the necessary
calculations... try every wire length from 66 feet to 132 feet and see
if you can find a length which is a comfortable percentage away from
an even multiple of 1/2 wavelength on each frequency. Or,

You probably mean't any integral multiple of a half wave.

alternatively, iterate through each frequency, calculate the
1/2-wavelength multiples, and "blacklist" every possible length which
is too close to these multiples.

I have done just that, and searched for "sweet" wire lengths that aren't
within say 5% of band edge for all HF bands. It sounds like a solution to
the problem doesn't it. (5% implies that you have a pretty determinate
scenario, which is a big assumption. IIRC 10%+ will not give a practical
result on the higher bands.)

Problem is that it probably unecessarily constrains the solution.

The input impedance and the feed point voltage of an end fed wire at its
higher parallel resonances falls, so that whilst you might want to avoid
the first such resonance, the impedance (and feed point voltage) at the
third or higher resonance might well be low enough to not worry about it.

To demonstrate that life isn't simple, the antenna efficiency (Rr/Rtot at
the feedpoint) improves closer to those parallel resonances that everyone
wants to avoid.

Owen

Hi Owen. 5% was just a number picked out at random by me for clarification
of my question.
So you wouldn't want to share your 'sweet' findings, would you??
thanks!
W


"Owen Duffy" wrote in message
...




I have done just that, and searched for "sweet" wire lengths that aren't
within say 5% of band edge for all HF bands. It sounds like a solution to
the problem doesn't it. (5% implies that you have a pretty determinate
scenario, which is a big assumption. IIRC 10%+ will not give a practical
result on the higher bands.)
Owen

"Woody" wrote in news:hYdli.29709$BT3.8034@trnddc06:

Hi Owen. 5% was just a number picked out at random by me for
clarification of my question.
So you wouldn't want to share your 'sweet' findings, would you??
thanks!

W,

I have had a quick look for the spreadsheet and Perl scripts, but haven't
found them. I am sure I have them, but haven't filed them in an ordered
way apparently, I think that technically is lost... or galloping
senility.

The reason I didn't publish them as is at the time is is that they are an
incomplete picture.

Have a look at the article at
http://www.vk1od.net/InvertedL/InvertedL.htm which describes an InvertedL
at approximately one of those "sweet" lengths (~26m). In fact, the length
was juggled to avoid excessive feedpoint voltage on all bands.

Another of the "sweet" lengths is half that at 13m, and the voltage plot
is rougly scaled proportionately in frequency.

You will see from Fig 1 that the voltage peaks at higher frequency
parallel resonances are less an issue than the first and second
resonances (you can't see from the graphs, but ~10kV and 3kV
respectively).

Of course, none of this discussion addresses the pattern issues at the
higher frequencies.

As far as the earth system goes, it impacts efficiency of the system. It
is my view that the earth only needs to be good enough that its loss is
an acceptably low portion of the transmitter power. The shorter the
radiator in wavelengths, the lower its Rr and therefore the lower the
earth resistance for comparable efficiency. If you look at Fig 4 of the
article, you will see that Rr is 100 ohms or greater above 5MHz, so the
loss of an earth system resistance of say 30 ohms is near insignificant,
but at 80m where the length is relatively short, you need a better earth
for good efficiency.

Don't agonise over it too much, and treat the Rules of Thumb as ROT until
you understand the underlying assumptions and caveats.

Owen

W


"Owen Duffy" wrote in message
...




I have done just that, and searched for "sweet" wire lengths that
aren't within say 5% of band edge for all HF bands. It sounds like a
solution to the problem doesn't it. (5% implies that you have a
pretty determinate scenario, which is a big assumption. IIRC 10%+
will not give a practical result on the higher bands.)
Owen

"Woody" wrote in news:hYdli.29709$BT3.8034@trnddc06:

Hi Owen. 5% was just a number picked out at random by me for
clarification of my question.
So you wouldn't want to share your 'sweet' findings, would you??
thanks!

I had a dig around through a few thousand spreadsheets... and found it.

9.05m, 12.6m, 18.3m, 26.4m, 33.3m, 44.9m.

If you want to cover 60m, leave the 26.4 out of the list.

Did someone mention 60' (18.3m), it is in the list, but it is more
critical than 26.4 or 12.6.

As noted in my earlier post, if you want to be 10% or more away from n
half waves on all bands, there is no solution.

These calcs don't consider proximity of objects that may shift the
tuning.

My advice still stands that the higher order parallel resonances aren't
such an issue, so this method unnecessarily constrains the solution..

Owen

Hey, I'm just learning and absorbing so anything new is a place for me to
start. Thanks again!
woody

"Owen Duffy" wrote in message
...
"Woody" wrote in news:hYdli.29709$BT3.8034@trnddc06:

Hi Owen. 5% was just a number picked out at random by me for
clarification of my question.
So you wouldn't want to share your 'sweet' findings, would you??
thanks!

I had a dig around through a few thousand spreadsheets... and found it.

9.05m, 12.6m, 18.3m, 26.4m, 33.3m, 44.9m.

If you want to cover 60m, leave the 26.4 out of the list.

Did someone mention 60' (18.3m), it is in the list, but it is more
critical than 26.4 or 12.6.

As noted in my earlier post, if you want to be 10% or more away from n
half waves on all bands, there is no solution.

These calcs don't consider proximity of objects that may shift the
tuning.

My advice still stands that the higher order parallel resonances aren't
such an issue, so this method unnecessarily constrains the solution..

Owen

In article ,
Owen Duffy wrote:

A simple program or spreadsheet ought to be able to do the necessary
calculations... try every wire length from 66 feet to 132 feet and see
if you can find a length which is a comfortable percentage away from
an even multiple of 1/2 wavelength on each frequency. Or,

You probably mean't any integral multiple of a half wave.

You're right... integral multiple of a half-wavelength, or even
multiples of a quarter-wavelength are two alternative ways for stating
the lengths to be avoided (high feedpoint Z).

alternatively, iterate through each frequency, calculate the
1/2-wavelength multiples, and "blacklist" every possible length which
is too close to these multiples.

I have done just that, and searched for "sweet" wire lengths that aren't
within say 5% of band edge for all HF bands. It sounds like a solution to
the problem doesn't it. (5% implies that you have a pretty determinate
scenario, which is a big assumption. IIRC 10%+ will not give a practical
result on the higher bands.)

Problem is that it probably unecessarily constrains the solution.

Entirely possible!

The input impedance and the feed point voltage of an end fed wire at its
higher parallel resonances falls, so that whilst you might want to avoid
the first such resonance, the impedance (and feed point voltage) at the
third or higher resonance might well be low enough to not worry about it.

And, even if it was a bit high, you might not need to be more than a
couple of percent away from it to get it down to a length that might
work.

Odds are that some amount of experimentation is going to be required,
at any given installation, to find a wire length which tunes up well
with these ATUs. The orientation of the wire (vertical, inverted-L,
etc.), height about ground, presence of trees and metallic objects,
and (perhaps most importantly) the details of the ATU's grounding
system, are likely to change the impedances around enough to make the
"textbook" answers less than completely effective.

--
Dave Platt AE6EO
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