[TowerTalk] "Magic" length from tower to first insulator
jimlux
jimlux at earthlink.net
Wed Mar 16 13:49:45 EDT 2016
On 3/16/16 9:01 AM, Larry Loen wrote:
> Thanks, Jim. A little too much "magic" in your formula, though. I'm
> trying to understand this as well as "cook-book" it.
>
> So, I did the "obvious" thing and constructed my own table based on
> multiples of 1/2 wavelength. Took the classic 468 and divided it by the
> upper and lower bounds of each frequency band.
That's my 3.28 * 299.7/f -> feet/meter * meters/sec / freq
to compute the free space wavelength in feet. (a half wavelength would
be 491/f)
( I used that rather than the "resonant dipole" formula of 468, which is
about 5% lower, figuring that later, I'd be comparing with a 10%
tolerance anyway)
Then I "deducted" 10 per
> cent from the shorter length (higher frequency) and "added" 10 per cent
> to the longer length (higher frequency).
So, for each frequency,
> multiple of 1/2 wavelength, I calculate an upper and lower bound
> (distinances in-between are also on the "bad idea" list).
That works.. I figured I'd put band centers, or upper and lower in the
list and look.
I just take the guy length and divide it by the wavelength: so a 1/2
wavelength winds up as 0.5, a 3/2 wavelength is 1.5, etc.
if you take x-floor(x) that just removes everything to the left of the
decimal point, so you get numbers between 0-1.
I then compare against 0.45 to 0.55 (which is 10%)
>
> That produced this little table:
>
> 0.5 wlength 1.0
> wlength 1.5 wlength
> 3.50 4.00 147.09 105.30 294.17
> 210.60 441.26 315.90
> 7.00 7.30 73.54 57.70 147.09
> 115.40 220.63 173.10
> 10.10 10.15 50.97 41.50 101.94
> 83.00 152.91 124.49
> 14.00 14.30 36.77 29.45 73.54
> 58.91 110.31 88.36
> 18.06 18.16 28.50 23.19 57.01
> 46.39 85.51 69.58
> 21.00 21.45 24.51 19.64 49.03
> 39.27 73.54 58.91
> 24.89 24.99 20.68 16.85 41.37
> 33.71 62.05 50.56
> 28.00 30.00 18.39 14.04 36.77
> 28.08 55.16 42.12
> 50.00 54.00 10.30 7.80 20.59
> 15.60 30.89 23.40
>
> First off, it tells me that if you try and account for 6 meters, you're
> going to end up with a very short initial cable length. Nobody has told
> me yet whether there is a minimally practicable minimum size for that
> first length of guy. I guess I will have to hope that at 6m, at least,
> that any antenna will be high enough up the mast as to not matter (which
> should be true according to our plans).
>
> This still gives me acceptable results, or should, if I pick around 10
> feet as my "consensus" length. It might disturb the pattern on the
> upper end of 15 meters, but it should be well "out of synch" with
> everyone else. At least that's what my all-too-quick study of this
> suggests.
>
> I don't know if there is a big structural difference between, say, 10
> and 20 feet, and perhaps even 3 feet wouldn't matter (except for the
> sheer logistics of getting the cable tied off), but 10 feet looks like
> the overall safest best. So, absent further discussion from those far
> more learned than I, that's what I am going to pick. I know what the
> initial configuration is, but I'm trying to account for every use.
> Unless this becomes a 15 meter monoband tower someday (very doubtful)
> and the upper end of 15 becomes important (actually, kind of expendable
> given my interests if _something_ has to go), then this looks good to me.
>
> Comments? Dissents? What did I overlook?
>
> On Wed, Mar 16, 2016 at 5:25 AM, jimlux <jimlux at earthlink.net
> <mailto:jimlux at earthlink.net>> wrote:
>
> On 3/16/16 12:31 AM, Larry Loen wrote:
>
> A friend of mine told me that someone, somewhere had calculated
> a "magic"
> length for guy cable. If you went down 10 or 11 or 13 feet (or
> whatever it
> is) off of your tower and installed your first insulator there,
> your guys
> would be sufficiently 'non-resonant' so that they wouldn't
> interfere with
> the pattern of the antenna(s) above.
>
> What is this magic length and where is it documented?
>
>
> There's a table in the ARRL Antenna Book of "good" guy segment lengths..
>
> I do it by setting up an excel spreadsheet that calculates the
> number of wavelengths for my frequencies of interest for a given
> length, and then just making sure they're not close to a multiple of 0.5
>
> if cell C3 is the frequency (in MHz) and cell A5 is the length (in ft)
> C4 = 3.28*299.7/C3 wavelength in feet
>
> C5 = $A5/C$4 - floor($A5/C$4) frac part of guy length in wavelengths
>
> Then conditional format C5 to be red if it's in the range 0.45-0.55
> (or whatever tolerance you want)
>
> I put a sample sheet out on google docs
>
> https://docs.google.com/spreadsheets/d/1rXeWjPMej_TlzN0h8OfzEFAU2mwKGiar_QaxK94tozA/pubhtml
>
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