Article 31
The Catenary

A flexible inextensible string is suspended from two fixed points, and takes up a position of equilibrium under the action of gravity. It is required to find the equation of the curve in which it hangs.

Let w be the weight of unit length, and s the length of arc AP measured from the lowest point A; then ws is the weight of the portion AP. This is balanced by the terminal tensions, T acting in the tangent line at P, and H in the horizontal tangent. Resolving horizontally and vertically gives

Tcosφ = H,Tsinφ = ws,

in which φ is the inclination of the tangent at P; hence

tanφ = ws H = s c,

where c is written for H w , the length whose weight is the constant horizontal tension; therefore

dy dx = s c,ds dx = 1 + s2 c2 ,dx c = ds s2 + c2, x c = sinh1s c,sinh x c = s c = dy dx,y c = cosh x c,

which is the required equation of the catenary, referred to an axis of x drawn at a distance c below A.

The following trigonometric method illustrates the use of the gudermanian: The “intrinsic equation,” s = ctanφ, gives ds = csec2φdφ; hence dx = dscosφ = csecφdφ; dy = dssinφ = csecφtanφdφ; thus x = cgd1φ,y = csecφ; whence y c = secφ = secgd x c = cosh x c ; and s c = tangd x c = sinh x c .

Numerical Exercise.—A chain whose length is 30 feet is suspended from two points 20 feet apart in the same horizontal; find the parameter c, and the depth of the lowest point.

The equation s c = sinh x c gives 15 c = sinh 10 c , which, by putting 10 c = z, may be written 1.5z = sinhz. By examining the intersection of the graphs of y = sinhz, y = 1.5z, it appears that the root of this equation is z = 1.6, nearly. To find a closer approximation to the root, write the equation in the form f(z) = sinhz 1.5z = 0, then, by the tables,

f(1.60) = 2.3756 2.4000 = .0244, f(1.62) = 2.4276 2.4300 = .0024, f(1.64) = 2.4806 2.4600 = +.0206;

whence, by interpolation, it is found that f(1.6221) = 0, and z = 1.6221, c = 10 z = 6.1649. The ordinate of either of the fixed points is given by the equation

y c = cosh x c = cosh 10 c = cosh1.6221 = 2.6306,

from tables; hence y = 16.2174, and required depth of the vertex = y c = 10.0525 feet.14

Prob. 91.
In the above numerical problem, find the inclination of the terminal tangent to the horizon.
Prob. 92.
If a perpendicular MN be drawn from the foot of the ordinate to the tangent at P, prove that MN is equal to the constant c, and that NP is equal to the arc AP. Hence show that the locus of N is the involute of the catenary, and has the property that the length of the tangent, from the point of contact to the axis of x, is constant. (This is the characteristic property of the tractory).
Prob. 93.
The tension T at any point is equal to the weight of a portion of the string whose length is equal to the ordinate y of that point.
Prob. 94.
An arch in the form of an inverted catenary15 is 30 feet wide and 10 feet high; show that the length of the arch can be obtained from the equations cosh z 2 3z = i, 2s = 30 z sinh z.