2s = 2λ/u = 1.667 λ.
The tangents at the ends meet on the directrix, and their inclination to the horizontal is 56° 30′.
| Fig. 57. |
The relation between the sag, the tension, and the span of a wire (e.g. a telegraph wire) stretched nearly straight between two points A, B at the same level is determined most simply from first principles. If T be the tension, W the total weight, k the sag in the middle, and ψ the inclination to the horizontal at A or B, we have 2Tψ = W, AB = 2ρψ, approximately, where ρ is the radius of curvature. Since 2kρ = (1⁄2AB)2, ultimately, we have
k = 1⁄8W · AB/T.
(13)
The same formula applies if A, B be at different levels, provided k be the sag, measured vertically, half way between A and B.
In relation to the theory of suspension bridges the case where the weight of any portion of the chain varies as its horizontal projection is of interest. The vertical through the centre of gravity of the arc AP (see fig. 55) will then bisect its horizontal projection AN; hence if PS be the tangent at P we shall have AS = SN. This property is characteristic of a parabola whose axis is vertical. If we take A as origin and AN as axis of x, the weight of AP may be denoted by wx, where w is the weight per unit length at A. Since PNS is a triangle of forces for the portion AP of the chain, we have wx/T0 = PN/NS, or
y = w · x2/2T0,
(14)