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### Finding the Arc Length of a Hanging Catenary

```
Date: 07/23/97 at 19:19:43
From: Anonymous
Subject: Finding the arc length of a hanging  catenary

I have been assigned a project in my class to find the arc length of a
catenary suspended between two equal poles at equal height. The
distance between the poles is 400 feet, and it sags in the center 40
feet. Can anyone out there help me?
```

```
Date: 07/23/97 at 21:12:50
From: Doctor Anthony
Subject: Re: Finding the arc length of a hanging  catenary

Take the x axis horizontal and the y axis vertical, through (but not
at) the lowest point.  Let s be the length of arc measured from C,
the lowest point, to any point, P, and let w be the weight per unit
length of the chain.  We let 'phi' be the angle the tangent at P makes
with the horizontal.

Consider the equilibrium of the portion CP. The forces on it are the
tension T at P along the tangent at P, the horizontal tension T' at C,
and the weight ws. Resolving horizontally and vertically,

T cos(phi) = T'   and   T sin(phi) = ws

By division  ws/T' = tan(phi) = dy/dx

If we write T' in the form wa, i.e. if the tension at the lowest point
is equal to the weight of a length 'a' of the chain, we have

dy/dx = ws/wa  =  s/a

Now   ds/dx = sqrt[1 + (dy/dx)^2]  = sqrt[1 + s^2/a^2]

sqrt(a^2 + s^2)
=  ---------------
a

ds               dx
--------------  =  -------
sqrt(a^2 + s^2)       a

Integrating:    sinh^(-1)(s/a)  =  x/a  + const

at x = 0,  s = 0,   so const = 0

So we get      s/a = sinh(x/a)

But we had   dy/dx =  s/a =  sinh(x/a)  so integrating

y = a cosh(x/a) + const.

Now when x = 0,  y = a cosh(0) + const

y = a + const,

and it is convenient to choose the origin so that the const. is zero.
So if y = a when x = 0, we get

y = a cosh(x/a)

and 'a' is the distance of the origin of coordinates below the lowest
point.

Now, considering the problem as set in the question, when x = 200, the
value of y is  a+40, so we have

a+40 = a cosh(200/a)

Trial and error shows a = 507 (approx.).  With this value, the left
hand side is 547, and the right hand side 546.962

So using this value in the equation  s = a sinh(x/a)  for x = 200 we
get

s =  507 sinh(200/507)

= 507 sinh(.39447)

= 507 x 0.40479

= 205.22

This is for half the span, so the total length of chain is 2 x 205.22

Length of chain  =  410.455 feet.

If we carry forward the solution of the equation for a to 3 places of
decimals, we get
a =  506.489

So we get         s = 506.489 sinh(200/506.489)

= 205.238,  so  the total length  = 410.476 feet.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Functions
High School Geometry
High School Trigonometry

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