The Utility of Catenaries to Electric Utilities

...A Graphing Calculator Approach to the Examples

We provide a graphing calculator approach to the solutions of the example problems in the catenary demo.  We illustrate the solutions using the TI-83 but most other graphing calculators have similar built-in features.

Example 1.  A flexible cable with length 150 feet is to be suspended between two poles with height 100 ft.  How far apart must the poles be spaced so that at its lowest point the cable is 50 feet off the ground?

We obtain the system of equations:

Let f(x) = c + a cosh(x/a).  Then f(-s) = f(s) = 100 so we have

c + a cosh(s/a) = 100.

(1)

The minimum value of f occurs when x = 0 and the minimum value of f at x = 0 is 50 so

c + a = 50

(2)

The length of the wire is 150 feet.  In the context of the catenary function, we can interpret the length of the wire as the arc length of the graph of f from x = -s to x = s.  This gives a third equation.

(3)

A bit of work gives the following:

which simplifies to

.

The integral can be evaluated directly

Thus, our problem is to solve a system of three equations in three unknowns:

c + a cosh(s/a) = 100

a + c = 50

2a sinh(s/a) = 150

for a, c, and s.

The system of three equations can be reduced to a system of two equations by setting
c = 50 - a.  With some rearrangement, we obtain

cosh(s/a) = (50+a)/a

sinh(s/a) = 75/a

Each equation can be solved for s (in terms of a) by using inverse functions:

We can now use a crossing graphs approach which will require that we use the intersect function on the TI-83

Step 1:  Associate a with x and s with y.  Define the functions.

Step 2:  Graph the functions.  You may need to experiment with the graphing window.  In this graph,

xmin = 0, xmax = 50
ymin = 0, ymax = 100

Step 3:  Adjust the window so that the intersection can be clearly seen. In this graph,

xmin = 25, xmax = 35
ymin = 45, ymax = 55

Step 4:  From the CALC menu, select intersect.

Step 5:  Enter the first curve.

Step 5:  Enter the second curve. 

Step 6.  Enter the guess.

Step 7.  Coordinates of intersection displayed.

Thus, the distance between the poles is about 100.6 feet.

Example 2. According to our electric utility, Excelsior Electric Membership Corp. (Excelsior EMC) in Metter, GA [2], due to terrain, easements, etc., the average distance between utility poles ranges from 325 to 340 feet.  The Georgia Department of Transportation (DOT) [3] states that the maximum height of a truck using interstates, national, and state routes is 13 feet and 6 inches.  However, special permits may be granted by the DOT for heights up to 18 feet.  With these restrictions in mind, Excelsior EMC maintains a minimum clearance of 20 feet under those lines it installs during cooler months because expansion causes lines to sag during warmer months.  For the obvious reason, Excelsior EMC prefers that the distance from its lines to the ground is greater than 18' 6 '' at all times.

Problem:  Find a and c so that f(x) = c + a cosh(x/a) models this situation.

We will assume that the distance between poles is 340 feet and that we want the minimum clearance to be 20 feet.

Proceeding as in the previous example, we require that

f(-170) = f(170) = 34 and f(0) = 20.

These two conditions give the equations

c + a cosh(170/a) = 34

c + a = 20

(3)

(4)

which can be reduced to the single equation

(20 - a) + a cosh(170/a) = 34 or

a cosh(170/a) - a = 14.  (5)

We can use a crossing graphs approach (the intersect function) of the calculator to obtain a numerical approximation for the value of a.  But by rewriting Eq. (5) below: 

a cosh(170/a) - a - 14 = 0

we put the equation into a form so that we can use the zero function.

Step 1.  Associate the variable a with x and enter the left hand side as y1.

Step 2.  Plot the function.  Adjust the scale as necessary so that you can see the x-intercept.  The window dimensions here were

xmin =900, xmax = 1100
ymin = -5,   ymax = 5

Step 3.  Select zero from the CALC menu.

Step 4.  Enter left bound.

Step 5.  Enter right bound.

Step 6.  Enter a guess.

Step 7.  The intersection is shown.

The x-coordinate of the intersection gives the value of a = 1034.4678.  We can compute c directly:  c = 20 - a, so for this example c = -1014.4678.  

With this information we can obtain the length of the wire between the poles by computing the arc length integral.

Step 1.  Enter the function to be integrated:

Step 2.  Graph the function.  Here the limits were

xmin = -170, xmax = 170
ymin = -5,     ymax = 3

Interesting, huh?  We'll look at this later.

Step 3.  Select the integration function.

Step 4.  Select lower limit of integration.

Step 5.  Select upper limit of integration.

Step 6.  The value of the definite integral is shown.  Note that the area under the curve is shaded.

The length of wire is about 341.53 feet. 

This graphical approach to the integration leads to an interesting discussion about the hyperbolic functions and their graphs.  Students should wonder why the graph in Step 2 appears to be linear. 

A closer look yields the following:

Thus, the arc length integral is equivalent to

This formulation of the integral still does not directly explain the flatness of the graph.  Since the integrand is a hyperbolic function, shouldn't we expect to have a curve?

In fact, we do.  Because the scale of the limits of integration relative to the denominator, the bending of the graph is obscured.  By changing the xmin and xmax dimensions of the graphing window to be of the same order as the denominator, we see that the graph of the integrand does have the typical shape of a hyperbolic cosine function. 

 

Our original graph only showed the relatively flat area of the hyperbolic cosine function.

LFR 12/31/2003