### Bezier Bridge

Anyone who has been to Jerusalem in the last couple of years could not have missed the relatively new monstrosity built at the city’s gates: the Chords Bridge. This mammoth of a bridge is designed to carry the weight of two trains, and is suspended by powerful metal chords connected to a 120 meter tall tower. Most probably, the first thought that sprang into the minds of those who gazed upon the monument was “Why, look at how neat those chords are! I wonder what shape their edges make.” We shall try to say something interesting about that shape in the following paragraphs.
Let us take an axes system, (X, Y) orthogonal to each other, and draw lines from the X axis to the Y axis. The coordinate (0, 0) is analogous to the base of the bridge, the X axis is analogous to the bridge itself, and the Y axis is analogous to the gigantic tower from which it hangs. The lines are drawn so that they intersect both the X axis and the Y axis somewhere in the range [0, 1], in the following manner: if a line intersects the X axis at point (t ,0), then it intersects the Y axis at point (0, 1-t). The lines we draw are uniformly spaced across the range [0, 1], so we get something like this: Let us look at a line, which intersect the X axis at a certain point t. Its function, yt(x), is 0 at t, and is 1-t at 0. $y_{t}(t) = 0$ $y_{t}(0) = 1-t$

The general equation for such a line is: $y_{t}(x) = (1-t) - x \frac{1-t}{t}$

Now, let us look at two lines: one that intersects the X axis at t, and another that intersects it at t+Δt, where Δt is really really small. As Δt approaches 0, the point in which the two lines intersect approaches a point on our “envelope”. In other words, in order to find out the shape that the edges of all the lines make, we shall look at the intersection between two very close lines; so close, in fact, that we can actually treat the whole ordeal as a continuum, and not as a collection of lines. Anyway, our goal is to find out which x values satisfy: $y_{t}(x) = y_{t+\Delta t}(x)$

A little bit of basic math: $(1-t) - x \frac{1-t}{t} = (1 - t - \Delta t) - x \frac{1 - t - \Delta t}{t + \Delta t}$ $-x \frac{1-t}{t} = - \Delta t - x \frac{1 - t - \Delta t}{t + \Delta t}$ $-x(1 - t)(t + \Delta t) = - x\cdot t (1 - t - \Delta t) - \Delta t \cdot t (t + \Delta t)$ $x = \Delta t \cdot t^{2} + \frac{t \cdot \Delta t ^{2}}{\Delta t}$ $x = t^{2} + t \cdot \Delta t$

And as Δt → 0, we get $x = t^{2}$

Putting this in yt(x), we get: $y_{t}(t^{2}) = (1-t) - t^{2}\frac{(1-t)}{t}$ $y_{t}(t^{2}) = (1-t)^{2}$

So, we now know that our curve is defined as all the points $(x,y) = (t^{2}, (1-t)^{2})$

for any t in [0, 1].
How very interesting “That doesn’t seem to be any shape I know”, one can say. And indeed it requires another trick to see how what this really is. One can look at the points he received, and try to see if there is anything of value to tell about them. After playing around a bit, the following equation can be reached: $2(x+y) = (x-y)^{2} + 1$

Substituting R = x+y, S = x-y, we get: $2R = S^{2} + 1$

Which is a parabola! only not in the same axes that we previously chose, but in a new coordinate system, (S, R), which is also orthogonal but is tilted at 450 relative the the one we chose. However, it is a parabola nonetheless!
That’s quite neat. However, our work here is not yet finished – we have only shown what the shape is for a set of lines which intersect the axes somewhere in the range [0, 1], and we must wonder how this relates to the actual Chords Bridge. First and foremost, it is quite obvious that this can be extended to any range, [0, p]. However, what if we want different p’s for the different axes? Suppose we say that the lines may intersect the X axis somewhere in [0, x1], and the Y axis somewhere in [0, y1]. It will not be shown here, but one can repeat the previous calculations for any arbitrary x1,y1 (also, one can simply elongate or compress one or more of the new axes in the second coordinate system).
Another problem arises – the cables in the Chord Bridge do not start at the base of the “axes”, but somewhere in the middle, so we actually draw lines that intersect the X axis somewhere in [x0, x1], and the Y axis somewhere in [y0, y1]. This is indeed a problem for our model, since it affects the “density” of the lines being drawn. We cannot just say, “let us extend x0, y0 to 0 and everything will be ok”, because that would change the position of the lines, they would connect in a different manner. So, in reality, the shape of the Chord Bridge is not really parabolic, and has a deformation caused by the fact that the cables are not uniformly spread throughout [0, p]. One can still perform work similar to what was described above, but will not arrive at the same results.
Lastly, there is the fact that the Chords Bridge’s tower and bridge are not actually orthogonal to each other, but are rather tilted at an awful angle. This can still be solved, and we can still get a parabola (although, in a new coordinate system), but I would like to address another interesting topic here: Bezier Curves. Wikipedia has an absolutely excellent article on the subject, filled with diagrams and animations, and I highly recommend you read it. Generally, Bezier curves are defined by several points, and can be constructed using either equations or a recursive definition. The edge described in the previous mathematical analysis is actually a third order Bezier Curve, also known as a quadratic curve. It takes the form of a parabola, and can be drawn either by a quadratic equation, or by using two linear equations, as shown in the following animation.
All in all, the next time you see a bridge, do take some time to appreciate its beauty, be it architectural, or mathematical.

A note for the artistic and wild among us: an orthogonal coordinate system and uniform distribution is just one (dull) way to connect lines to one another. What’s way more exciting, is to try and experiment with what happens when you take any two curves, define density functions, and start connecting lines from one curve to the other. The results can be very appeasing to the eye, and can easily be replicated using the given parameters, meaning that one can generate a high quality and pretty picture, while using very little disk space (but a lot of computations, most probably).