Article 32
Catenary of Uniform Strength.

If the area of the normal section at any point be made proportional to the tension at that point, there will then be a constant tension per unit of area, and the tendency to break will be the same at all points. To find the equation of the curve of equilibrium under gravity, consider the equilibrium of an element PP whose length is ds, and whose weight is gρωds, where ω is the section at P, and ρ the uniform density. This weight is balanced by the difference of the vertical components of the tensions at P and P, hence

d(Tsinφ) = gρωds, d(Tcosφ) = 0;

therefore Tcosφ = H, the tension at the lowest point, and T = Hsecφ. Again, if ω0 be the section at the lowest point, then by hypothesis ω ω0 = T H = secφ, and the first equation becomes

Hd(secφsinφ) = gρω0 secφds,

or

cdtanφ = secφds,

where c stands for the constant H gρω0, the length of string (of section ω0) whose weight is equal to the tension at the lowest point; hence,

ds = csecφdφ,s c = gd1φ,

the intrinsic equation of the catenary of uniform strength.

Also

dx = dscosφ = cdφ,dy = dssinφ = ctanφdφ;

hence

x = cφ,y = clogsecφ,

and thus the Cartesian equation is

y c = logsec x c,

in which the axis of x is the tangent at the lowest point.

Prob. 95.
Using the same data as in Art. 31, find the parameter c and the depth of the lowest point. (The equation x c = gd s c gives 10 c = gd 15 c , which, by putting 15 c = z, becomes gd z = 2 3z. From the graph it is seen that z is nearly 1.8. If f(z) = gd z 2 3z, then, from the tables of the gudermanian at the end of this chapter,

f(1.80) = 1.2432 1.2000 = +.0432, f(1.90) = 1.2739 1.2667 = +.0072, f(1.95) = 1.2881 1.3000 = .0119, whence, by interpolation, z = 1.9189 and c = 7.8170. Again, y c = log esec x c ; but x c 10 c = 1.2793; and 1.2793  radians  = 731755; hence y = 7.8170 ×.41914 ×2.3026 = 7.5443, the required depth.)
Prob. 96.
Find the inclination of the terminal tangent.
Prob. 97.
Show that the curve has two vertical asymptotes.
Prob. 98.
Prove that the law of the tension T, and of the section ω, at a distance s, measured from the lowest point along the curve, is
T H = ω ω0 = cosh c h;

and show that in the above numerical example the terminal section is 3.48 times the minimum section.

Prob. 99.
Prove that the radius of curvature is given by ρ = ccosh s f. Also that the weight of the arc s is given by W = Hsinh s f, in which s is measured from the vertex.