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Re: [TowerTalk] Long Wire Sag

To: towertalk@contesting.com
Subject: Re: [TowerTalk] Long Wire Sag
From: jimlux <jimlux@earthlink.net>
Date: Sun, 10 Mar 2019 10:27:53 -0700
List-post: <mailto:towertalk@contesting.com>
On 3/10/19 9:29 AM, dwkanepe@aol.com wrote:
For information, NFPA-70 (National Electrical Code) Table 810.52 requires minimum 10 AWG (Hard Drawn Copper) or 12 AWG (Copper Clad Steel) for spans greater than 150 feet.

Or various clad steel..


Yeah - but I'd venture that most amateur antenna installations are not fully NEC compliant. Bonding to ground systems, the feedthrough into the structure, clearances from structures and trees, etc.

Part of it is a safety argument - if the antenna breaks and falls down, is a hazard created (i.e. is it going to wind up hitting a power line). The NEC rules are derived from the same rules as current carrying conductors between poles and are all about "it stays up regardless". Those rules also date from the days of King Spark when long wire antennas were a commercial thing.

I note that commercial wire antennas from companies like TCI use Alumoweld (Model 548 Log Periodic), and I've also seen Stainless Steel (in the infamous Terminated folded dipoles.. it's a lossy antenna already, so using a lossy conductor isn't a big problem)


One can also argue that the long wire antenna made with magnet wire is almost by definition a "temporary" installation and perhaps not subject to NEC.






Don Kane
WB2BEZ

In a message dated 3/10/2019 12:04:56 PM Eastern Standard Time, jimlux@earthlink.net writes:

    On 3/10/19 8:25 AM, Gedas wrote:
     > I am planning to put up a long inverted v antenna with it's
    feedpoint at
     > 85' using 600' total wire (300' on each leg). The ends will be
    near the
     > ground, only 20-25 feet high.
     >
     > My question is given that each leg of this antenna will be 300'
    long am
     > I better off going with a lighter weight #14 THHN insulated stranded
     > wire or some heavier #12 THHN stranded? I am not going to purchase a
     > different wire that would be better suited like copper-weld etc
    since I
     > have plenty of these other two and want to try something today.
     >
     > I realize there is going to be a _lot_ of sag in either case but
    I am
     > not sure of the breaking strengths of either #12 or #14 and in
    the end
     > which will help keep the wire up higher with less sag. Any ideas?

    You can calculate it:

    There's a spreadsheet called catenary.xls with my name (Jim Lux) you
    can
    download if you google it.
    I'd put the URL, but Earthlink is having problems displaying it off my
    website..

    Here's the equations..

    We assume that the origin is at the center of the span.

    Total span = L
    Sag in the cable = h
    So, the coordinates of the endpoints are (+/- L/2,h).

    The weight per unit length = w

    Total length of wire/cable = S

    Length along the cable from the origin = s

    Fh is the horizontal force component everywhere, and is equal to half
    the tension at the center.

    Equations

    The horizontal force, in terms of total cable length and sag is

    Fh = w / (8*h) * (S^2 - 4*h^2)

    The y coordinate (height) of any point in terms of the horizontal force

    y= Fh / w * (cosh(w * x / Fh) - 1 )

    (Change suggested by Stephen Argles, 24 Nov 2003)

    The span, given horizontal force, weight, and length of cable

    L = (2 * Fh / w ) * arcsinh(S * w / (2 * Fh))

    The total cable length, given span and horizontal force (useful for
    computing how long a span can be supported)

    S = (2 * Fh / w ) * sinh( w * L / (2 * Fh))

    The arc length from origin (center):

    s = Fh / w * sinh(w * x / Fh)



    Tension

    It's also useful to know the peak tension in the cable, which occurs at
    the end points. The Vertical force at support is

    Fv = w * S / 2,

    i.e. the total weight of the cable divided by two. And, the Horizontal
    force, computed above, is Fh. So the tension is simply the combination
    of the two:

    T^2 = Fh^2 + (w * S / 2)^2

    The minimum tension is, of course, Fh, at the center point where the
    cable doesn't support any of it's own weight. If you need the
    tension in
    between, you just need to compute the vertical force at a given point,
    which is equal to the weight of the cable from that point to the center
    (i.e. s*w).







     >
     > Gedas, W8BYA
     >
     > Gallery at http://w8bya.com
     > Light travels faster than sound....
     > This is why some people appear bright until you hear them speak.
     >
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