The lions gate bridge in vancouver BC, is a suspension bridge that spans 1516m. Large cables are attached to the tops of the towers, 50 m above the road. the road is suspended from the large cables with small vertical cables, the smallest one being 2m, find a quadratic equation to model the large cable shape.

2. Originally Posted by sureshrju
The lions gate bridge in vancouver BC, is a suspension bridge that spans 1516m. Large cables are attached to the tops of the towers, 50 m above the road. the road is suspended from the large cables with small vertical cables, the smallest one being 2m, find a quadratic equation to model the large cable shape.
Maybe it's because I'm in Civil Engineering but this is a terrible question. There is no way to create a singular model to map the behavior of the cables for the entire span of the bridge. Well there is, but it certainly cannot be done with 1 quadratic function.

There are 3 distinct parts: http://upload.wikimedia.org/wikipedi..._Vancouver.jpg

So we will assume they are talking about the middle part (from left tower to right tower).

If the cable is of constant density (which it is) then the positioning in the cable can be treated liked the deflection of a beam with a point load in the middle. Thus, the max deflection happens at Mid-Span and the height here is 2. Having said that, if the cables are held down via smaller wires the engineer could have positioned it differently, but this makes no sense because a symmetric structure will transfer the loads better (more evenly and efficiently) so we will say that this structure is symmetric.

Treat this like a graph, so at the left we have the point (0,50) and on the right we have (1516,50). Of course, your analysis will depend on where you take your co-ordinate axis. You could take mid-span to be the origin whereas I am taking the left tower to be at the origin.

Let us say

$\displaystyle F(x)=Ax^2 + bx + C$

$\displaystyle F(0)=50=C$

$\displaystyle F(x)=Ax^2 + bx + 50$

$\displaystyle F(1516/2)=2=A(1516/2)^2 + b(1516/2) + 50$

And

$\displaystyle F(1516)=50=A(1516)^2 + b(1516) + 50$

Solve the 2 equations for the 2 unknowns.

3. Hello, sureshrju!

The description is not clear,
but I can guess the design of the bridge.

The Lion's Gate bridge in Vancouver, BC, is a suspension bridge that spans 1516m.
Large cables are attached to the tops of the towers, 50 m above the road.
The road is suspended from the large cables with small vertical cables, the smallest one being 2m.
Find a quadratic equation to model the large cable shape.

Place the bridge on coordinates axes with the center of the bridge at the Origin.

Code:
              :
*       :       *(758,50)
|       :       |
|*      :      *|
| *     :     * |
|   *   :   *   |
|       *       |
|     (0,2)     |
|       :       |
|       :       |
- - * - - - + - - - * - -
-758      0      758
:

The general form of this parabola is: .$\displaystyle f(x) \:=\:ax^2 + c$

We have two point on the parabola: .$\displaystyle (0,2),\;(758,50)$

$\displaystyle f(0) = 2:\;\;0^2a + c \:=\:2 \quad\Rightarrow\quad c \;=\;2$

$\displaystyle f(758,50):\;\;758^2a + 2 \:=\:50 \quad\Rightarrow\quad a \:=\:\frac{12}{143,\!641}$

Therefore, the equation is: .$\displaystyle f(x) \;=\;\frac{12}{143,\!641}\,x^2 + 2$