of the _common catenary_, as it is called (fig. 56). The omission of the additive arbitrary constants of integration in (8) is equivalent to a special choice of the origin O of co-ordinates; viz. O is at a distance a vertically below the lowest point ([psi] = 0) of the curve. The horizontal line through O is called the _directrix_. The relations s = a sinh x/a, y² = a² + s², T = T0 sec [psi] = wy, (10) [Illustration: FIG. 56.] which are involved in the preceding formulae are also noteworthy. It is a classical problem in the calculus of variations to deduce the equation (9) from the condition that the depth of the centre of gravity of a chain of given length hanging between fixed points must be stationary (§ 9). The length a is called the _parameter_ of the catenary; it determines the scale of the curve, all catenaries being geometrically similar. If weights be suspended from various points of a hanging chain, the intervening portions will form arcs of equal catenaries, since the horizontal tension (wa) is the same for all. Again, if a chain pass over a perfectly smooth peg, the catenaries in which it hangs on the two sides, though usually of different parameters, will have the same directrix, since by (10) y is the same for both at the peg. As an example of the use of the formulae we may determine the maximum span for a wire of given material. The condition is that the tension must not exceed the weight of a certain length [lambda] of the wire. At the ends we shall have y = [lambda], or x [lambda] = a cosh ---, (11) a and the problem is to make x a maximum for variations of a. Differentiating (11) we find that, if dx/da = 0, x x --- tanh --- = 1. (12) a a It is easily seen graphically, or from a table of hyperbolic tangents, that the equation u tanh u = 1 has only one positive root (u = 1.200); the span is therefore 2x = 2au = 2[lambda]/sinh u = 1.326[lambda], and the length of wire is 2s = 2[lambda]/u = 1.667 [lambda]. The tangents at the ends meet on the directrix, and their inclination to the horizontal is 56° 30´. [Illustration: 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 [psi] the inclination to the horizontal at A or B, we have 2T[psi] = W, AB = 2[rho][psi], approximately, where [rho] is the radius of curvature. Since 2k[rho] = (½AB)², ultimately, we have k = (1/8)W·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.