Clifton Suspension Bridge

For my project, I have decided to analyze the Clifton Suspension Bridge located in Bristol, England.

image from Wikipedia

This bridge goes between Clifton, a town in Bristol, England, and Leigh Woods in North Somerset, England. It goes over the Avon Gorge, which is a 1.5 mile part of the River Avon. It is considered a landmark of the area. The Clifton Suspension Bridge opened for use in 1864. It took many, many years to build, due to funding and rioting problems. The bridge was designed by Isambard Kingdom Brunel, who won a contest for the best bridge design in 1831. One interesting fact is that the bridge has been used in many British TV shows and commercials.

The Clifton Suspension Bridge consists of two towers on each side with a parabola-shaped suspension cable between them. The towers rise 86 feet above the deck of the bridge. The span is 702 feet. There are actually three sets of chains that make up the cabling. Eighty-one iron rods suspended from the chains hold the bridge deck in place. These rods range in size from 65 feet at the ends to 3 feet in the middle. The bridge is 245 feet above the water level. According to Wikipedia, the bridge weighs about 1,500 tons and supports 4 million vehicles each year.

this parabola is a little tricky. I will let the roadway be the x-axis. The y-axis will be the first tower. According to my research, the height of each tower is 86 feet. So then, the point (0,86) is on my parabola. Since the span is 702 feet, the point at the top of the other tower is (702,86). I just need one more point to get my equation. The shortest rod going from the road to the cables is 3 feet, at the center of the parabola. This must be exactly halfway between the towers, so by taking half of 702, it must be 351 feet from each of the towers. Therefore, the point (351,3) is on my graph -- and it's also my vertex! Here is the graph I created using Microsoft Excel. I used Quadratic Regression to make the actual parabola:


There are actually several parabolas present on the bridge in addition to the suspension cabling. As you can see at right, each tower includes a parabola that opens down as part of its decoration. There are also cables leading up to each tower from the land that form parts of parabolas. For this project, though, I will focus on the large parabola that forms the suspension between the two towers.

I used quadratic regression to find an equation for my parabola. I did this with a spreadsheet program and got y = 0.0007x^2 - 0.4729x + 86. This equation is in standard form. So here are the three forms for my equation:

  • Standard Form: y = 0.0007x^2 - 0.4729x + 86
  • Vertex Form: y - 3 = 0.0007 (x - 351)^2
  • Factored form: not possible!

To find vertex form, I only needed find the a value, since I already knew the vertex (351,3). Here's my work:

  • I know the equation will be like this: y - 3 = a (x - 351)^2
  • Since I know another point on the graph, I can fill in values for x and y and solve for a. I'll use the point (0,86).
  • So now I have 86 - 3 = a (0 - 351)^2, which simplifies to 83 = 123,201a.
  • Dividing both sides by 123,201 gives a = 0.0007.
  • So the equation in vertex form is y - 3 = 0.0007 (x - 351)^2
I could have also used completing the square to find the vertex form, but it's tough with all the decimals:
  • Standard Form: y = 0.0007x^2 - 0.4729x + 86
  • Subtract 86 from both sides: y - 86 = 0.0007x^2 - 0.4729x
  • Divide both sides by 0.0007: y/0.0007 - 122,857.143 = x^2 - 675.571429x
  • To complete the square, add half of b, squared (that's about 114,099.1888) to both sides: y/0.0007 - 122,857.143 + 114,099.1888 = x^2 - 675.571429x + 114,099.1888
  • Simplify the left side and rewrite the right side of the equation as the product of two binomials: y/0.0007 - 8,757.95422 = (x - 337.7857145)^2
  • Multiply both sides by 0.0007: y - 6.130568 = 0.0007(x - 337.7857145)^2. As you can see, this is a little off from the actual vertex. This is probably due to rounding, so we'll use the vertex form I found using the first method above.

As we said earlier, the vertex is (351,3). It is a minimum, since the graph opens up. I can tell the graph opens up because the a value is positive. The vertex represents the lowest point of the bridge's cabling span.

The y-intercept of the graph is the point (0,86). This represents the height of the first tower on the bridge. We could have found it by substituting 0 for x in the standard form equation.

The discriminant of my equation is b^2 - 4ac = (-0.4729)^2 - 4(0.0007)(86) = 0.22363441 - 0.2408 = -0.01716559. This is a negative number. Therefore, the equation cannot be factored over the integers. This also tells me that there are NO x-intercepts, and that if we tried to use the quadratic formula to find the x-intercepts, there would be no solutions. I could also tell just by looking at the graph that there are no x-intercepts, since the parabola never crosses the x-axis. If there were x-intercepts, they would mean that the bridge cable touches or goes past the level of the roadway.

Now let's use the quadratic formula to find some solutions. Let's say that an iron rod was being replaced on the bridge, but we don't know exactly where. When the new rod arrives, we measure it to be 12 feet long. Where could this rod go? In other words, when is the cable 12 feet from the roadway?


  • Well, since the rod is 12 feet long, let's let y = 12 in our standard form equation.
  • Then we have this equation: 12 = 0.0007x^2 - 0.4729x + 86.
  • In order to use the quadratic formula, we need the left side to be 0, so we will subtract 12 from both sides: 0 = 0.0007x^2 - 0.4729x + 74.
  • Now we know our a, b, and c values: 0.0007, -0.4729, and 74, respectively.
  • Let's use the quadratic formula to solve for x:
    • -b +- sqrt(b^2 - 4ac) all over 2a
    • 0.4729 +- sqrt(0.4729^2 - 4(0.0007)(74)) all over 2(0.0007)
    • 0.4729 +- sqrt(0.22363441 - 0.2072) all over 0.0014
    • 0.4729 +- 0.128197 all over 0.0014
    • (0.4729 + 0.128197)/ 0.0014 = 429.3548 and
    • (0.4729 - 0.128197)/ 0.0014 = 246.2164
  • The two solutions are about 429 and 246. So, the cable could go at one of those distances from the towers. Those are the two points at which the height of the cable will be 12 feet.
  • The graph above shows our parabola with a line at y = 12. The places where this line crosses the parabola are the points (429,12) and (246,12).

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