[TowerTalk] Long Wire Sag

[jimlux]

On 3/10/19 9:29 AM, dwkanepe at 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 at 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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