Missouri

103

Gateway Arch, St. Louis, MO

38° 37′ 28.81″ N, 90° 11′ 5.82″ W

Gateway to the West

The Gateway Arch in St. Louis, Missouri, stands at 192 meters, and is the tallest monument in the U.S. (Figure 103-1). It was opened in 1967, and the view from the top is spectacular. The top is reached by riding in small five-person trams that travel up the legs of the arch to the observation deck at the top. While you’re up there, you can contemplate the perfect shape of the arch—it’s a catenary.

Figure 103-1. The Gateway Arch; courtesy of Pat Dye (gobucks2)

Even though the arch is man-made, there’s something natural-looking about it, because a catenary is a shape made simply by the force of gravity. Take a short piece of rope (or string, or a necklace of uniform thickness) and hold one end in each hand. With your hands level, the rope falls into a graceful curve under its own weight pulled down by gravity. That shape is a catenary (a term coined by Thomas Jefferson based on the Latin word for chain), and the weight of the rope is transmitted by tension through the rope itself.

Invert a catenary and build it out of something rigid, and you’ve got a catenary arch. The Gateway Arch is one example, but catenary arches are very common. They’ve been built since antiquity because they are stable—the weight of the arch’s components acts downward through the arch’s legs.

The arch follows the shape of a catenary curve and is constructed from stacked metal equilateral triangles. The centroid of each of the triangles is aligned with the curve being followed; the triangles start out at the bottom with 16-meter sides that narrow to 5 meters at the top. This narrowing explains why the trams are so small—you should be grateful that they’re there at all, though, because the original plan was to have visitors walk 1,076 steps to the top.

The exterior of the Gateway Arch is made of plates of stainless steel that cover 90 meters of reinforced concrete, topped by a steel structure that completes the shape. Because the arch curves, the trams are mounted on gimbals that allow them to stay level as they climb the curved interior of the legs.

The arch is as high as it is wide, and it weighs about 15,600 tonnes. It sways in the wind up to a maximum of close to 50 centimeters in each direction; on very windy days, the observation area is closed.

At the base of the arch is a visitor center that explains its construction and a little of the mathematics behind its shape. There’s also the Museum of Westward Expansion, which explains the Lewis and Clark expedition that helped open up the western U.S., and also covers the Louisiana Purchase and the debate over slavery. The arch is sometimes called the Gateway to the West.

The arch is part of the Jefferson National Expansion Memorial, a 37-hectare park along the Mississippi River. Once you’ve spent a couple of hours seeing the arch inside, and the associated museum, the park makes an ideal spot for a walk, lunch, or bike ride. It’s also possible to take helicopter tours of St. Louis for an aerial view of the arch from all angles.

Practical Information

Visiting information (and, I promise, no calculus—unlike the upcoming sidebar) is available from http://www.gatewayarch.com/.

The Catenary

A telephone cable strung between a pair of telephone poles hangs down in a curve called a catenary. The curve looks a bit like a parabola, but it is not—it’s actually described by a hyperbolic cosine function. What follows is probably the hardest bit of mathematics in this book; it involves a bit of calculus, but try to stick with it if you can!

Start by imagining a cable hanging under its own gravity, with a y-axis passing vertically through the cable’s lowest point; call that point A. Pick another point on the cable to the right and call it B. The line from the origin, O, to the point B makes an angle of a with the x-axis. (See Figure 103-2.)

Figure 103-2. A catenary

Three forces act on the segment AB—there’s the horizontal tension, T, at point A; the tension along the cable, F, at point B; and the weight of the segment AB acting downward, W. Since the cable is not moving, these three forces are in equilibrium and can be redrawn into a triangle of forces (Figure 103-3).

Figure 103-3. Triangle of forces

Basic trigonometry shows that F sin a = W and F cos a = T. The weight of the segment can be expressed by taking the linear density of the cable (call that D) and multiplying it by the segment’s length (call that l) to obtain W = Dl. Dividing the two trigonometric equations eliminates the force F to express the angle (which gives the slope of the cable) in terms of the cable weight and horizontal tension (Equation 103-1).

Equation 103-1. Basic catenary equation

The coordinates of B are (x, y), and some basic trigonometry gives that dx/dl = cos a and dy/dl = sin a. Differentiating the equation above gives dl/da = T/D sec2 a. So it’s possible to express everything in terms of differentiating by a (see Equation 103-2).

Equation 103-2. Catenary differential equations

Now integrate those two equations to get equations for x and y (Equation 103-3).

Equation 103-3. x and y expressed in terms of a

Both the constants are in fact 0. c0 is 0 because the angle a is 0 when x is 0 (see the diagram back in Figure 103-2). c1 can be made 0 by placing the x-axis at position T/D.

Now flip the equation for x around and raise it to the power of e. Inverting the resulting equation, plus a little trigonometry (recalling that tan2 a + 1 = sec2 a), gives the two exponential equations shown in Equation 103-4.

Equation 103-4. Catenary exponential equations

Add those together, and you’ve got the equation for y based on x (which gives the shape of the curve) shown in Equation 103-5.

Equation 103-5. The catenary equation