# Parabolic Bridges

Real Life Parabolas
Field: Algebra
Image Created By: Aaron Logan
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Real Life Parabolas

Parabolas are very well-known and are seen frequently in the field of mathematics. Their applications are varied and are apparent in our every day lives. For example, the main image on the right is of the Golden Gate Bridge in San Francisco, California. It has main suspension cables in the shape of a parabola.

# Basic Description

For a detailed overview of parabolas, see the page, Parabola. However, we will provide a brief summary and description of parabolas below before explaining its applications to suspension bridges.

#### Basic Definition

You may informally know parabolas as curves in the shape of a "u" which can be oriented to open upwards, downwards, sideways, or diagonally. But to be more mathematical, a parabola is a conic section formed by the intersection of a cone and a plane. Below is an image illustrating this.

When you were first introduced to parabolas, you learned that the quadratic equation, $y= a(x-h)^2+ k$ is its algebraic representation (where $h$ and $k$ are the coordinates of the vertex and $x$ and $y$ are the coordinates of an arbitrary point on the parabola.

#### Suspension Bridges

Suspension Bridges are the most commonly built bridges. Known for their long spans, these bridges feature a deck with vertical supports, from which long wire cables hang above. These cables are made up of hangers that run vertically downwards to hold the cable up. The suspension cables hang over the towers until they are anchored on land by the ends of the bridges. Notably, the way these cables are hung resemble the shape of a parabola.

# Usefulness of Suspension Bridges

Due to their elegant structure, suspension bridges are used to transport loads over long distances, whether it be between two distant cities or between two ends of a river. Suspension bridges are able to work efficiently because of their cables, which are interesting from a mathematical perspective.

Since the bridge’s deck spans a long distance, it must be very heavy in weight by its own, not to mention all the weight of the heavy load of traffic that it must carry. Because of all this weight, this results in two active forces: compression and tension. The cable’s parabolic shape results in order for it to effectively address these forces acting upon the bridge. For instance, the deck sags from all the weight of the traffic because of compression forces, which travels upwards the cables. The cables then transfer those compression forces downwards the vertical towers, down into the foundations buried deep within the earth. However, the cables receive the brunt of the tension forces, as they are supporting the bridge’s weight and its load of traffic, being stretched by the anchors' ends on-land.

Overall, the suspension bridge does its job with minimal material (as most of the work is accomplished by the suspension cables), which means that it is economical from a construction cost perspective.

# A conceptual explanation

This links to other page, Catenary. But we shall explain the differences between parabola and catenary with more emphasis on the parabola.

Why is that the main suspension cables hang in a shape of a parabola, and not in a catenary, a similar ‘u-shaped’ curve?

Despite their visual similarities, catenaries and parabolas are two very different curves, both conceptually and mathematically.

A catenary curve is created by its own weight, pulling down because of gravity. The parabolic curves of the suspension cable are not created by gravity alone, but also by other forces: compression and tension acting on it. Also the weight of the suspension cable is negligible compared to that of the deck, but it is also supporting the weight of the deck. This is also another conceptual reason why the suspension cables hang in a parabolic curve.

# A More Mathematical Explanation

Note: understanding of this explanation requires: *Calculus, Physics

Some basic differential calculus is needed to derive an equation for the suspension cables, which giv [...]

Some basic differential calculus is needed to derive an equation for the suspension cables, which gives us a parabolic equation.

From the previous sections, we explained the forces active in the bridge: tension and compression. Well, by looking at only an interval of the cable, let’s examine these related forces.

The following three forces are active on the cable:

Tension( $\vec{T}$)= horizontal direction coming from the left because it’s an opposing force.

Weight( $\vec{W}$)= gravitational pull downwards of the cable. $\vec{F_x}$=Force tangent to the cable at point of the cable, arbitrarily labeled $x$. This is a force coming from the right that corresponds to the direction of the suspension cable.

In physics, these three forces can be visualized in the form of a free body diagram.

(Image of a triangle underneath an interval of a cable with its lowest point conveniently positioned at the origin-so one of our known points of this curve is $(0,0)$)

The arrows indicate the direction that the forces are going in. Overall, the net force is $\vec{0}$ because the segment of the cable is motionless and as such, has no acceleration. Therefore, when drawing the free body diagram, all three vectors’ heads and tails must meet up where the head of one the vectors meets up at the tail of another vector. This forms a right triangle.

Slope of $F_x$ is equivalent to the slope of the cable, which we are looking for in order to find an equation that best describes the cable.

From the diagram, we see that the slope of $F_x = {W/T}$

Note: That this is not a vector quantity, but rather a magnitude quantity.

But what exactly is $W$? Well, weight is distributed evenly throughout the deck below the cables, so at this interval of the cable, the interval of the deck below it must have uniform weight and as such, uniform linear density $u$. Let the length of the deck be defined as the distance from $(0,0)$ to the point $x$. So the weight is equal to $W= ux$

So the slope of $F_x$ can also be rewritten $F_x= {ux/T}$, which is also the slope of the cable.

Now with our knowledge of the slope of the cable, an equation for the curve containing the above slope can be derived with the tools of basic integration.

Integrating the slope of $F_x$ with respect to $x$, we get: $\int {ux\over T} dx = {u\over2T}(x^2) + C$

Plugging our known point, $(0,0)$ into this result, we get the equation: $y = {u\over 2T}(x^2)$

Which describes a parabola.

So optimal shape of the suspension cables is the parabola.

#### A Real World Application

Back to the Golden Gate Bridge, statistics show that the main span of the bridge is approx. $l=4200 ft.$while its tower height is approx. $h=500 ft$ At the middle of the bridge, $T=W$, which defines the most efficient use of the suspension cable.