Towertalk
[Top] [All Lists]

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 09:04:45 -0700
List-post: <mailto:towertalk@contesting.com>
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.

_______________________________________________



_______________________________________________
TowerTalk mailing list
TowerTalk@contesting.com
http://lists.contesting.com/mailman/listinfo/towertalk

_______________________________________________



_______________________________________________
TowerTalk mailing list
TowerTalk@contesting.com
http://lists.contesting.com/mailman/listinfo/towertalk

<Prev in Thread] Current Thread [Next in Thread>