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Catenary Curve


Date: 03/30/99
From: Anonymous
Subject: Catenary Curve

I know what the general equation for a catenary curve is, but a friend 
is trying to find where the vertex of the curve is if the poles on 
each side are at a different height because they lie on a slope. Is 
there a variation of the formula that would allow us to know where the 
vertex is (making sure it is far enough above the ground)?


Date: 03/30/99
From: Doctor Anthony
Subject: Re: Catenary Curve

We have supports at different heights.  Let one support be at 
(x1,y1) and the other at (x2,y2) with y1 > y2.

To avoid negatives I am measuring x1 positive to the right (from
the lowest point on the curve) and x2 positive to the left. 
This is okay provided we are consistent. It means that x1+x2 is 
the distance between the posts and can be measured.  We also 
know s1+s2, the length of the cable L.

The equation of a catenary curve is:

       y = a.cosh(x/a)  

and the equation for arc length from the lowest point is:
 
       s = a.sinh(x/a),

where a is the distance below the lowest point to the origin of 
coordinates. When x = 0,  y = a and this is the lowest point.

We can combine the following equations

  y1 - y2  = a[cosh(x1/a) - cosh(x2/a)]
  
           = 2a.sinh((x1 + x2)/2a).sinh((x1 - x2)/2a)  ....(1)


  s1 + s2 = a.sinh(x1/a) + a.sinh(x2/a)
  
          = 2a.sinh((x1 + x2)/2a).cosh((x1 - x2)/2a)

to get

   y1-y2
   -----  = tanh[(x1-x2)/2a]
   s1+s2

We know all the values on the left, which means we can find the value
of (x1-x2)/2a.  So we can find 

  sinh[(x1-x2)/2a]  =  k1 (say)

With x1+x2 = L, from equation (1) we have

   y1-y2 = 2a.sinh(L/2a).k1

So   

  2a.sinh(L/2a) = (y1-y2)/k1

This equation has 'a' as the only unknown and can be solved by an 
iterative method.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   


Date: 03/30/99
From: Doctor Rob
Subject: Re: Catenary curve

If the catenary is y = a*(e^x + e^[-x])/2 and the line is y = m*x + b, 
then the point on the catenary of minimum distance from the line is
(x0, y0), where

   x0 = ln([m + sqrt(m^2 + a^2)]/a)

   y0 = sqrt(m^2 + a^2)

The catenary is above the line if and only if

   y0 > m*x0 + b

This was found by using the formula for the distance from the point
(x,y) to the line y - m*x - b = 0:

   d = (y - m*x - b)/sqrt(1 + m^2)

I substituted for y the catenary formula, differentiated with respect
to x0, set the result equal to zero, and solved for x = x0. Then, I
used the catenary formula to find y0.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus
College Euclidean Geometry
College Trigonometry
High School Calculus
High School Euclidean/Plane Geometry
High School Functions
High School Trigonometry

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