d=wl^2/8T - Where d = wire sag; w = weight of the
wire per unit length; l = horizontal span between
supports; and
T = the horizontal tension in the wire.
The 0.016 inch diameter steel wire weighs about 0.0007
pounds per foot. So for a 10 foot span between level
supports, and 30 pounds tension at a given temperature,
the sag at the low point (mid point of the span) is d=wl^2/8T
= 0.0003 feet, or about 3/1000 of an inch. For a 100
foot span with that same tension, d = wl^2/8T = 0.03
feet (about 3/8 of an inch). Now to get the wire
deflections at other points in the span, let's take the
100 foot span case, using the parabolic approximation
y=ax^2 (letting the low point of the curve be at origin
(0,0)), then you can solve for 'a' using the condition
that y=.03 when x = 50, and get a = .03/2500 = 0.000012;
so now, for the 100 foot span case, you have y =
.000012x^2, which defines the shape of the curve, and
where y is the value measured up from the low point. For
example, at x= 0 (low point) y = 0, implying a sag of
0.03 -y = .03'; or at the 1/4 points, where x = 25, y=
.0075, and the deflection at that point is .03 -y = .03
- .0075 = .0225 feet.
Note also that if the wire is subject to temperature
variations, it's tension and sag will change (more sag ,
less tension, when hot; less sag, more tension, when
cold.).
That just seems like too much work. Let's just
print out the pages below and save our strength.
