From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: Catenary
Date: 5 Nov 1999 14:33:22 -0500
Newsgroups: sci.math
Keywords: fitting a catenary to match a suspended cable

See a file below; observe that you will have to solve one transcendental
equation whose solution has no known elementary formula .

Cheers, ZVK(Slavek).

In article <3823AAF7.2789@net.hcc.nl>,
Cornelis Wessels  <Keez.Wessels@net.HCC.nl> wrote:
>Hello there,
>
>Given two suspension points A (ax, ay ) and B (bx, by )
>and a cable length L. Now I want to determine the parameters
>
>a, b, c and d of the function:
>
>y = a * cosh ( b * x + c ) + d
>
>to fit the appropriate catenary.
>
>Does somebody have the solution for my engineering problem?
>
>
>Bye,
>-- 
>Cornelis Wessels van der Val
>The Netherlands
>Office : K.Wessels@boskalis.nl
>Private: Keez.Wessels@net.HCC.nl

Upright catenary:  y - y_0 = a * cosh((x - x_0)/a) ,  a > 0.
The vertex is (x_0, y_0+a). 
The parameter a turns out to be the radius of curvature at the vertex.
Remark: The radius of curvature at a point above x is
   a * (cosh((x-x_0)/a)^2.

Problem: Parameters a, x_0, y_0 to be found so that the catenary arc passes
through (x_1, y_1), (x_2, y_2) and has length L between these points
(provided L > sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2) ):

The equations to be solved are (after some symmetrizing manipulation
of the arclength integral)

      y_1 - y_0 = a * cosh ((x_1 - x_0)/a)
      y_2 - y_0 = a * cosh ((x_2 - x_0)/a)
     2*a * cosh((x_1 + x_2 - 2*x_0)/(2*a)) * sinh((x_2 - x_1)/(2*a)) = L

and it can be reduced to solving for one unknown at a time.

 First, we introduce an auxiliary unknown z which is to satisfy

     sinh(z) / z = sqrt(L^2 - (y_2 - y_1)^2) / abs(x_2 - x_1)  ,   z > 0

(the only transcendental non-elementary equation)

and then the unknowns pop out:

   a = abs(x_2 - x_1) / (2*z)

  x_0 = (1/2)*(x_1 + x_2 - a * ln ((L + (y_2 - y_1)) / (L - (y_2 - y_1)))

  y_0 =  (y_1 + y_2)/2 - (L/2) * coth(z).