#### SOME INTERESTING CURVES

**Professor John Barrow**

There was an interesting article in *The Independent* today about Mr Madoff, the biggest financial fraudster in history that we know about so far. The article revealed that really he had many, very strange, personal eccentricities, and one of them was an extreme dislike for curves. So in the buildings of his offices, he insisted on over-building any curved surfaces so they were all straight and square. So none of the window frames were allowed to be curved; none of the desktop surfaces were allowed to be curved. He also would not allow anybody to write with blue ink in his offices, only black ink, and there was a long list of further eccentricities. So curves are important and very influential - that is what I learnt from *The Independent* this morning!

Today, I am going to talk about only a small number of curves. There could be an entire lecture series about curves and shapes but I am going to pick on a few curves that will have been noticed by you, perhaps not appreciated in their mathematical depth, and I will try and show you how an understanding of curves plays an important role in many aspects of design and construction.

One curved building that will not have escaped your notice here in London is the Swiss Re Building, sometimes known as the Gherkin or 'the Towering Innuendo', which is my favourite title for it.

It was created by Foster and Partners and it was a project that had a very strong mathematical modelling input, and we will see why in a moment. The shape is very distinctive: the widest point of the building is about halfway up and it narrows towards the bottom and towards the top. It is about 180 metres high and it has a covered dome on the top. There are forty floors and it is at its widest at about Floor Sixteen. This building was created by mathematical modelling of a rather complex sort. There is not a simple equation that tells you the shape of the surface, although I think from overlaying, it really looks, until about halfway down, rather like a parabola, but the fact that it curves in makes it rather more complicated.

What has been done in creating this building, why it has the strange shape that it does, is the result of a combination of optimisations. Some are psychological and aesthetic, some are mathematical and some are aerodynamic. If you stand at the bottom of a building that curves inwards, then when you look at the sky, it does not blot out so much of the sky at the top because of the concavity. So this building does not seem so immense to people at ground level than if it was a conventional, rectangularly cross-sectioned skyscraper. It allows much more of the sky to be seen.

Having the thin bottom is aerodynamically important. If you walk around in Manhattan, in what seems like a calm day, you will turn a corner and suddenly run into a forty or fifty mile an hour wind. What is going on there is rather like when you put your finger over the tap when you turn the water on - you put your finger over the end and you get a high velocity jet. Because you are forcing the flow of water to go through a narrow constriction, the pressure goes up and the speed goes up. And so it is with winds channelled between high buildings: the wind speed will increase because the air has been channelled into a narrow nozzle. Therefore, by having the building go in at the bottom, you make more space available and you lower the wind speed that will be felt by people at ground level.

This Swiss Re building also has an apparent spiral shape, but this is only an illusion. Every six floors or so, there are great triangular vents that reach deep into the building to bring air into it; they create a natural venting situation. There is really an inner shell to this building, and a gap, which is like a super-cavity wall type of insulation, and so the building is twice as thermodynamically or energy efficient as a comparable building of the old design of a same size. These indentations that bring the air in have been slightly offset from one floor to another, and that creates this spiral pattern. So just by offsetting the circular slices, you create this circular spiral effect.

The other illusion that you have when you look at a building like this is that you think the surfaces are curved, which is an illusion created by the spiral effect. In fact, all of its plates of glass are flat, so there are no curved surfaces on the outside. The only curved surface is the small dome at the top. The smaller you make the little triangles and diamond shapes that you use, the better the matching will be when you join them together to create a curved surface. So all the curvature is at the angles where the flat surfaces meet.

So this building is a sort of triple or quadruple illusion. It is mathematically optimised with these different features in mind. The shape is to create a nice, smooth flow of air past the building on the outside, so you do not get a lot of drag, noise or a lot of other external factors.

This building was opened in 2003 and is 180 metres tall. It is not unique, by any means, in terms of its rough shape but it is a good example of why curves play a role in architecture. Later on, we are going to look in a bit more mathematical detail at some of the most important shapes that are used and try and understand why they are used. This hint of a parabola in this surface suggests we might like to think about parabolas a bit more.

There is one situation where parabolas play a key role, and that is if you throw something or if you launch a projectile of any sort, it is going to follow a parabolic trajectory.

There is one particular parabolic trajectory that I noticed about a year ago that is worth investigating. You probably remember, around Boxing Day, a year and a month or so ago, there was a tragic incident at San Francisco Zoo, when a tiger named Tatiana leapt out of its enclosure after being taunted by some visitors who seemed to have had rather too much to drink and various drugs. One of them was killed and I think two of them were seriously injured. A long investigation has since gone on - I think there is a court case at the moment. The issue is effectively whether you could have foreseen whether Tatiana could jump out of its enclosure, so was this a case of poor health and safety risk management, in modern speak, or was it something that could not be foreseen?

If we were to look at Tatiana's house from the air we would see that there is a a ground area, in front of which there is a deep ditch and then a fence. So, in order to escape, Tatiana has to leap the ditch and the fence to get to the outside:

We here need to just review what we know or remember about parabolas and projectiles. A typical projectile problem is shown here:

We launch something from a particular place, it follows a trajectory and then it comes back to the ground. We will measure the distance along the ground as x, and the height up in the air as y, and we will launch the projectile at some speed that has a horizontal component that we will call u, and a vertical component that is called little v. So it is the sort of thing you have to do at school. It is fairly easily to figure out what goes on.

So in the motion in the horizontal direction, the distance is just the speed times the time. But the vertical motion has got a force acting against it, a force of gravity, and so the height after time t is the vertical speed multiplied by t but minus a half gt^{2}, where g is the acceleration due to gravity. We can remove the time if we want, by substituting in x/u, and we have got an equation of y versus x, which is a parabola:

Launch velocity = (u, v)

x = ut

y = vt - ½ gt^{2}

= vx/u - ½ gx^{2}/u^{2}

This path is indeed a parabola, and it starts at x is zero and it ends at some maximum value of x.

You can work out interesting things from this. The maximum height is given by a half v squared over g: y_{max} = ½ v^{2}/g. Obviously the weaker the force of gravity - the smaller g - the greater the height you would reach, and the time it takes to maximum is v over g: t_{max} = v/g. So just by manipulating this equation, you can work out anything you might want to know about the trajectory of something that gets launched with a particular speed.

Well, what about our tiger? What we are interested in is that if the tiger launches itself with a speed v at some angle, which can be anything we like, if it is going to clear a height h, at a distance x away, how big does v have to be in order to do that? What we are interested in is particularly the case where h is 3.8 metres and x is 10 metres, because they are the measurements of the dry moat and the high fence in Tatiana's enclosure in San Francisco. The formula is really rather simple:

V^{2} = g (h +√(h^{2} + x^{2}))

This is the smallest speed that will do the trick. It depends on the square root of h squared plus x squared.

So, what do we know about tigers? If you look through your natural history books or look at tigerjumping.com or whatever it is on the internet, there is lots of information and films of tigers doing very alarming things. So on the flat, for fairly short distances, a tiger can run past you on the motorway at 50 miles an hour, or 22 metres per second. From just a five metre or so start in the enclosure, an average tiger would not have any difficulty launching at fourteen metres per second.

So the question is: what actually is the speed that is required, for the dimensions of Tatiana's enclosure, to leap over the top of the fence? Is it bigger or smaller than 14 metres per second? We can find this out very easily if we plug this into our formula, V^{2} = g (h +√(h^{2} + x^{2})), and we will find that, in fact, the speed that is needed is just twelve metre's per second, and that is ignoring the fact that a smart tiger does not need to leap all the way over the fence at all; it can just grab onto the top and clamber over. So we would not have expected, on the basis of using our simple school mathematics, that this tiger could be confined in this enclosure if it did not want to be. It ought to be able to sail straight over the top, with a lot of room to spare.

So it is an example of how knowing a little bit of simple mathematics can perhaps make life safer, and it is certainly something to bear in mind when you next visit the zoo!

The next curve I want to talk about has an interesting history. It is the curve you get when you suspend a chain from two points and allow the chain to hang. So this type of situation can be seen in many instances in the real world, including on the street outside:

It is a famous problem that was set in 1691 by one of the Bernoulli brothers as a challenge to the mathematicians of Europe, to work out the mathematical equation of a chain suspended, at same level at two ends, hanging under gravity. There are two winners of this competition, whose entries and analyses were published in what counted as the mathematical journal of the day, and they were rather famous people, the sort that you might expect would pop up and win the competition: one was Leibniz, co-inventor of the calculus; and the other was a great Dutch mathematician and physicist, Huygens, responsible for the theory in interference and much else in the study of light and the mathematics of the propagation of light.

Here are the drawings of their entries, which were transposed into the publication:

Leibniz has a lot of constructions in his diagram (on the left). He has calculated what the shape is, as we will see in a moment, as the sum of two exponential functions and then he has created a construction to add them together. Huygens has done it rather more cleanly, all in one go (on the right). The places he has hanging it from are slightly further apart, so it looks a little bit different.

The shape that the hanging chain forms is called catenaria; the line or the curve of catenaria. The English word, catenary, the hanging chain, curiously, was introduced by Thomas Jefferson, rather later, in a letter to Payne, another American revolutionary, which was about the shape of an arch of a bridge, when they were debating with each what the shape ought to be for an optimal arch. You will see why in a moment.

How does the mathematics of this work? Here is a summary:

The portion *AP* is in equilibrium under the horizontal tension *H* at *A*,

the tension *F* directed along the tangent at *P*, and the weight *W* of *AP*.

If the weight of the string is *w* per unit length and *s* is the arc *AP*, *W* = *ws*;

and from the force triangle, tan ψ = *ws*/*H* = *s*/*c*,

where *c* = *H*/*w* is called the parameter of the catenary is determined by dy/dx = s/c

With solutions

y = c cosh(x/c) s = c sinh(x/c)

In the diagram, the dotted line is the suspended chain. If we consider a section of it, there are three forces acting on that piece. There is gravity, which is the weight of that piece of the chain, going vertically downwards. At the top, there is a tension pulling upwards, in a direction that is tangent to the shape of the chain at that point, and then, at the bottom, there is another tension tangent, which is horizontal. So there is a triangle of forces here: tension acting diagonally upwards, tension acting horizontally to on side, and gravity acting downwards. By solving that little triangle, we can get the solutions we are looking for. The horizontal tension has the value is h and we can work that out in terms of the length along the curve, which we call s. The gradient of the curve is dy by dx, which is just the distance along the curve divided by a constant, which is given by the tension, and we can solve this equation rather simply. The solution is that the shape of this curve has y as equal to a constant multiplied by cosh of x divided by the same constant, and if you want to know the arc length, it is given by the sine. So there is a simple mathematical problem that works out the shape of the hanging chain, and we find this curve in all sorts of perhaps unexpected places.

If you turn the catenary curve upside down, it has all sorts of other rather nice structural properties. In a catenary arch there is almost no bending moment stress at all. It was Robert Hooke, in the late 17^{th} Century, who first noted this in a slightly encrypted statement - it was an anagram he created in Latin that was only unravelled after his death - where he remarked: 'As hangs the flexible chain, so inverted stand the touching pieces of an arch.' This is what Jefferson and Payne were in correspondence about: the matter of what the shape should be of an arch forming a bridge. A good example of such a bridge if the St. Louis Gateway Arch:

On the base of it is engraved the equation of this particular arch. Unfortunately it is in American Imperial units of feet, but it gives the dimensions as 630 ft x 630 ft, so y = -127.7 ft cosh(x/127.7ft) + 757.7ft. The 757.7ft is added at the end because the curve is inverted from how we had it before, so instead of the peak of the arch being at zero ft, it is actually at 757.5ft. When you then put x as zero, you get the height as 630 feet.

So these particular arches have robust, optimal structural properties. If you look at these types of figures around the world, where they often appear, you will find this inverted catenary, the cosh curve.

There is another unusual place where this appears, and it is all tied up with the question of whether you can ride a bicycle which has square wheels. But here you need to ask yourself what it means to ride a bicycle with square wheels? So what would a smooth ride correspond to? Because it could be that if the ground was sufficiently uneven, you would be able to move over it in such a way that your centre of gravity moved in a perfectly even straight line. So you want to ask: what would the shape of the road have to be in order for you to be able to ride in that way?

That is the sort of question that mathematicians like to answer, in particular this mathematician, Stan Wagon, who first posed this problem and solved it, and actually built the bicycle to demonstrate it in action:

The solution to this problem lies in the fact that he is moving on this curved surface, with his square-wheeled bike, so that the centre of his wheels moves in a perfectly straight line:

The shape of the road that makes this possible is the catenary. So if these arches here are the cosh inverted catenaries, then this motion becomes possible in a smooth way. All you have to decide is how far you should let the catenary curve run down before you come up to the next one.

If we think about what is going on here, we see that each corner of the square is of course a right angle - it is at 90 degrees - so to solve the problem, you have to find where this catenary intersects the next one at an angle that is equal to 90 degrees, and that tells you where you cut off the catenary curve.

First you have to check that this is possible at all. If you picked any old curve, like a sine or a cosine, it would not be possible at all, but with this function, it is. Then you can go a bit further: it turns out that if your wheel is a polygon, with any number of sides, you can also have a smooth ride. You have to cut off the catenary at a slightly different point, so the angle at the point at which the catenary curves meet is equal to the interior angle inside the polygon. You can do it for a triangle, but the triangle gets in the way of itself, in that if you try and do it with a triangle, the triangle cannot get out because the next bit of the road is in the way. So the triangle has to lay its own road as it goes along, but if it does, it can still do it!

So the next time you are interested in an unusual bike, this is the inverted catenary equation for the road: y = -Acosh(x/A). If we have a polygonal wheel that has got n sides to it, that is the angle inside: A = Rcot(p/n). So this tells you what the exact shape is of this oscillatory road surface.

In fact, one person who thought he had solved the problem of the hanging chain was Galileo. In his dialogues, he makes a remark that if you suspend a chain, the equation of the hanging chain is obviously that of a parabola. It was later shown to him by someone else that this was not true, but you cannot tell whether he was interested in exactly the same problem that we have been talking about, because there is another problem about which his answer would have been correct, and it is the problem of a suspension bridge.

This beautiful picture is of Brunel's fantastic Clifton Suspension Bridge, completed in 1865, a few years after Brunel himself died. This problem of the suspension bridge is different. This is because, the roadway is suspended below the arch of the bridge, which is a factor not present in the hanging chain problem. What matters here is not the weight along the curve, but the weight horizontally along the bottom. Mathematically, this is actually much easier, and it was why Galileo could probably work this out in his head, with a few lines of geometry.

Again, there are three forces acting here. For each part of the hanging support, there is a tension going up, there is a tension at the bottom acting horizontally due to the other and of the supporting arch being hung up on the other end, and there is a weight going downwards, which is the weight of all the road that is being supported underneath that section. So these three forces can be made into a triangle of forces:

The slope of the hypotenuse of the triangle can be represented as dy/dx, that is, the angle in the triangle, is equal to the weight going down divided by this tension k:

We can now integrate, if we let the weight per unit length be p, and we know how long the roadway is along the horizontally, let us call it x:

Therefore, the total weight is p times x, and so our equation dy by dx is just px over the constant tension. We can integrate that: y is just equal to x squared times a constant plus another constant. If we decide that the middle of the bridge is going to be where x is zero and y is zero, then the equation of the bridge is just y is a constant times x squared:

So it is our parabola again that we looked at with the tiger. So we can see that these suspension bridges are much simpler in their overall geometrical construction than simple hanging chains.

Another type of exotic structure with a complicated mathematical background are towers. A very good example of this is the Eiffel Tower.

Until the 1930s, this was the tallest manmade structure in the world. It was completed in 1899 by Gustav Eiffel, and his statement about it was, somewhat enigmatically: 'It was moulded in a way by the action of the wind itself.' So what did this poetic statement, translated from the French, actually mean?

What it meant was that, in deciding on the structure, again, there was some mathematical modelling. This is a structure that is experiencing a force of gravity; it is also standing up in the wind, and it is experiencing bending stresses moving it from side to side. It needs to be optimised so that those bending moment stresses are not going to make it break. So those types of stresses are the reasons why you do not see trees or flagpoles half a mile high. You cannot build a flagpole or grow a tree that is too tall, or else, when it is deflected in the wind, it will break. Indeed, things break much more easily by this bending moment, deflection, than they do by being compressed by the weight at their base. So if you break your arm, which I hope you don't but you may have done, you know you never break limbs by compression; you break them because you twist it and you bend it to the side. So things are often much more fragile to that type of bending breaking than they are to being compressed.

So Eiffel's tower is optimised in various ways to minimise the effect of this swaying and susceptibility to breaking. One way to do this is to have lots of holes in it, so that the wind goes through. The optimisation of the gravitational balance versus the internal stresses created by the bending, produce actually a rather complicated equation, which is both an integral equation and a differential equation:

f(x) òx f(t) dt = f ¢(x) òx (x - t) f(t) dt ® f(x) = Aebx

f(x) is the shape of the outside, so f is the vertical height, and x is the horizontal height. So this is quite a nasty equation to solve. You fully cannot solve it generally, but there is a particular solution which describes the shape almost exactly, and that solution is just an exponential. So this external shape of the Eiffel Tower is a simple exponential curve. Of course, it is concave, so it is not like the buildings that we saw - the Gherkin and so forth. It is optimised in a rather different way.

As an aside, having seen a picture of this Tower, I shall tell you a story from the town of my birth, which is Wembley. I come from Wembley and, like me, you will therefore know about Wembley Stadium, but what you may not know about Wembley Stadium is that if there was not a Wembley Stadium, there might have been an Eiffel Tower. Back at the turn of the century, a rather ambitious man called Sir Edward Watkin, who was Chairman of the Metropolitan Railway at the time, owned lots of the land around Wembley Park and that area of Wembley by the Triangle. He saw Eiffel's tower being constructed in France and he saw it at the Great Exhibition, and he thought 'I want one of those on my land - only bigger, much bigger!' So he approached Eiffel and said, 'I want you to build me a tower at Wembley, on my land, but I want it to be much bigger than the one you have got in Paris. I want it to be 1,200 feet high.' Eiffel said, 'Well, if I was to do this, I would be regarded as a traitor by the French; I am not going to do it, and I am not going to help you build a bigger tower.' So Watkin went off and recruited other people to work on this project, people who later would build the Blackpool Tower and other great engineering projects. By 1895, they had made the first stage.

But then, unfortunately, things started to go wrong. Suddenly, things slowly started to sag, and they realised that the foundations were shifting, that the land was much marshier and boggier in the winter than they had imagined when they began, so they had to stop building. They then used it as a slightly inelegant teashop for a while, and when the visitors' numbers rather fell off to basically be only Watkin and his wife. Then it was all closed down, declared unsafe seven years later, and eventually pulled down between 1904 and 1907. The price of iron was quite good in those years, so they decided to pull it down to at least make something from the materials. But this same site, owned by Watkin, he did alright in the end, because he sold the land to build Wembley Stadium, in the 1920s, ready for the first F.A. Cup Finals. But this would have been an exponentially shaped tower as well, if only it had got off the ground.

The last big collection of shapes we are going to look at is going to be rollercoasters and the shapes of their courses. You will find that these have something to do with the shapes of motorway junctions and their turnoffs.

What are the mechanics? What is going on in a rollercoaster? Basically, you get pulled up to the top, and then you are going to fall down under gravity, and then you go round usually in what we could call a circular curve. Therefore, in a rollercoaster experience two types of motion, two types of force: when you are falling vertically, you feel the force of gravity - forget the mass of the car, let us just think about your mass - so your weight, heading downwards towards the centre of the Earth; but you feel another force when you move around these part-circular curves. Schoolteachers get very agitated about the directions of these forces. If you want to move in a circle, you have to apply an inward force that points towards the centre of the circle, because, as Newton taught us, bodies acted upon by no forces act in a straight line, so if you want to coax something to move in a circle, you have got to keep sort of prodding it to move towards the centre of the circle you want it to move in. But if you are moving in a circle, by Newton's third law, you feel an equal and opposite reaction, so when you are moving in a circle, you feel an outward force pushing you away from the centre of the circle. This is the so-called centrifugal force. So when you are on the bus and it is going round a sharp corner, you get pushed away from the centre of the corner. This centrifugal force is the other force that you are going to feel when you go around these sharp bends, and the force you feel points radially outward, and it is equal to your mass times the square of your speed, as you move in the circle, divided by the radius of the circle that you move in: Mv^{2}/r. So if the radius of the circle is really huge, so the curve is very gentle, r on the bottom is very large, and this force will be small; if you fix the radius, as you increase the speed that you move around, the bigger the force will be. So this is quite intuitive. So we can see the rollercoaster as a battle of sum and difference between these two forces.

This is a typical type of set-up of what you might experience:

What is going on here in terms of the forces? If we think about the point when you are passing through the top, you will not stop, you will just keep going, but when you are instantaneously at this point at the very top of the loop, your weight is pushing downwards, but this centrifugal force is pushing upwards from the centre of this circle that you are moving in. So, as we will see in a moment, it better be that that centrifugal force is bigger than your weight, otherwise you will either fall out or you will be hanging from the straps. This is shown in the diagram by the green arrow, which is the centrifugal force, which must be greater than the force of gravity marked by the blue arrow. As you come down at the bottom, faster and faster, when you are at the bottom, the net force on you is the sum of your weight and this centrifugal force, both of which are now acting downwards. So we are going to look at how big these two forces are, and how fast we are going to move as we move around.

This is something of a schematic where I have put some numbers on. We are not going to use these actual numbers, but just this is a typical rollercoaster set-up:

Let us suppose we start from rest, so somebody just gently nudges the car off the high point. As you go down the slope from there, you gain more and more speed until you go up around the loop, and then off to the end. So you can see that there are a few critical factors here: that when you get to the bottom of the first slope, all of your potential energy from starting at the initial height will be converted into energy of motion. So if we know this starting height, we can work out our speed at the bottom. What we are interested in knowing, first of all, is how high does that have to be so that the speed at the bottom is big enough to get you around to the top of the loop, and then when you get up to the top, and you come back to the bottom. So we can ask: how fast will you be going at the bottom and how big will the force be on you at the bottom? In order to have enough speed at the bottom to get to the top, you have got to start pretty high up on the left, and that means you get to the top with a pretty good speed, and it means something about the force you are going to feel when you get to the bottom.

Let us see how this works. We are going to imagine that that loop that we are going around is a circle - let us forget about the join. So we are going to see what happens if you try to make a rollercoaster where the loop is a circle. Back at the turn of the century, at around 1900, the very first rollercoasters did indeed just use circles. It is a bit like the forces we talked about with the projectile problem. The potential energy of gravity, if your mass is m, and you start from a height h, is just your weight, mgh, mg times the height, h. Therefore, if you start from rest and you let it fall, then the speed with which this would hit the ground, if I let go of it from height h, is equal to half mV squared, where half mv squared is mgh.

½mV_{b}^{2} = mgh

So if you drop something from a height h under gravity, this is the speed with which it reaches the ground:

V_{b} =Ö2gh

So when it gets to the bottom, it is going to start moving in along the circular loop that has got radius r. If it is going to get all the way up to the top, what has it got to do? It has got to have enough potential energy to rise to a height equal to 2r, which is the diameter of the circle, so it needs potential energy 2mgr, and if when it gets there, it has got some speed - we will call it V_{t}, velocity top - then it is going to have some energy ½mV_{t}^{2}. So that is going to be the energy needed to get to the top.

We know what the energy is at the bottom: it is ½mV^{2}, which is mgh, so the energy you start off with is the energy you are going to have when you get to the top, assuming energy is not created or destroyed, and so we have a simple formula relating h to r and the speed at the top:

mgh = ½mV_{b}^{2} = 2mgr + ½mV_{t}^{2}

What we are here interested in is: what is the condition to stay in the car at the top and not to fall out? That condition is that the centrifugal push, the force upwards, is bigger than the weight, mg. So mV_{t}^{2}/r - mg, must be positive, or you will fall out: mV_{t}^{2}/r - mg > 0. That is a very simple condition, because we know what V_{t}^{2} is just mgh minus 2mgr divided by ½m. So all the 'm's cancel out and it works out that all that condition is is that V_{t}^{2} has to be bigger than gr. But we know that that is related to h, because h has to be bigger than 2.5 times the radius of the circle. So to stay in, you have to start out quite a bit higher than this loop. So 2r is the diameter of the loop, so it is around three radii up of the loop.

So let us suppose we start out at that height. What we are interested in is that we therefore get to the top, we come all the way back to the bottom. So what then is the force that we experience when we get to the bottom?

If we start off with a height, what is the speed when we get to the bottom? We know that the speed at the bottom is the square root of 2g times the height, and that height is bigger than 2.5 times the radius, and so we come to the bottom with a speed that is bigger than the square root of 5gr:

V_{b} = √(2gh) > √(2gx2.5r) = √(5gr)

So we can now work out the force that you feel at the bottom: it is your weight plus the centrifugal force. Where your weight is mg, the centrifugal force is m V squared over r - mV_{b}^{2} - where V_{b} is the speed at the bottom, and V^{2} is 5gr. So the force you feel at the bottom, if you have enough speed to get to the top and not fall out, exceeds 6mg, so you feel a force of 6g at the bottom of this rollercoaster:

mg + mV_{b}^{2} /r > mg + 5mg = 6mg

As it happens, that will probably kill you, or, at least, it would certainly render you seriously unconscious, unless you are an off-duty astronaut or you have g-suit on. So usually fairground rides try to avoid forces bigger than about two or three g. So this is a serious acceleration, and what would happen in the body were you to experience this would be that, under that force acting downwards, the blood in the heart would not be able to work against that downward force to be able to move blood - and therefore oxygen - into the brain. Therefore circular rollercoasters fail their risk analysis. If you want to have a semi-circular or a circular loop, the numbers do not add up and you will experience lethal forces if you attain enough speed to avoid falling out.

At this point you can see what the solution needs to be: in a rollercoaster you want to have the force at the top such that you do not fall out, so you want a big centrifugal force pointing outwards at the top of the loop in order to keep you in. So you want your mV^{2} /r to be big at the top, but to be small at the bottom, so that it does not push you down too strongly. So the answer is to create a different shape that is like a slightly distorted circle. This would be so that the 'r' at the top would be small, but that the 'r' at the bottom would be large.

At first, when people appreciated these sorts of factors by trial and error, they had a go with ellipses. So even in 1901, an elliptical trajectory was tried. This certainly helps, but from the 1970s, the optimal solution was found to this problem by a gentleman called Werner Stengel, who has a big company that designs and creates rollercoasters. There is a particular type of curve, known as the clothoid, which we will look at in a moment, which solved this problem rather nicely, and also gives an optimally smooth ride.

So here's the first clothoid curve rollercoaster that was ever used in public:

It is in Six Flags in Arlington, Texas, and you can see the idea. At the top, it is as though you are moving in a circle that has a very small radius of curvature. There is a strong force up that keeps you in. But when you come down to the bottom, you are going to be part of a very gentle curve, with a very large radius of curvature, and you feel a very small force.

What is the technicality, what is the definition of this curve; what is its special ingredient? Well, its mathematical equation is rather messy. If you wanted to plot it on a piece of graph paper, the whole curve looks rather like this:

It is a spiral which goes round one way and then goes back the other way; it is a bit like a musical cleft. The parametric equation for it, so there is a quantity here, which we have labelled as t, so you can define x and y in terms of t, and basically, t looks like an integral over cos of t squared and sine of t squared. So as you vary t, you change x and y following this pattern.

This is rather messy, so it must have some other simpler nice special property. Why would anyone have bothered to do this? As you do things to these formulae, you gradually unpick what the nice properties are. If you take the time derivatives of these quantities, you find the speeds, the components of the velocity in the x and y direction - dx by dt and db by dt.

So things are starting to look simpler. So the speed is just this constant 'a' times the sine, and the other component is the cosine.

If you take another time derivative, then you are going to get the acceleration. The differential of sine t squared just looks like t times sine t squared, so you get some 't's out the front, so here's the acceleration - it's proportional to this t factor:

If we want to measure the distance along the arc length - we will call it s, instead of the x or y position - then that distance is proportional to the acceleration, the root mean square of the acceleration components.

Therefore, if you take a squared, then you get a squared t squared, times cos squared plus sine squared, which is one. The rate of change of the arc length is just given by the acceleration.

So this curve has very special, simple properties: that the curvature is inversely proportional to the acceleration; the arc length is proportional to the acceleration; and the rate of change of the arc length is proportional to the acceleration.

What does this mean? Well, it is one of the reasons that the motion is so smooth, and so this curve is used in all sorts of construction purposes. When you are driving, for example, and you are going to change speed, from high speed to low speed, by taking a curved path, this particular curved path is optimal for various reasons.

To look at an example, here is an aerial picture of a complicated motorway junction:

Its four turnoffs are parts of the clothoid, just like the rollercoaster. So what is happening with this curve, as we have seen, is that its curvature changes; it is not like a circle, which has always got the same curvature. At the top of the rollercoaster curve, you had rather large curvature, and at the bottom, you had rather small one. So if you measure the distance along the curve from the bottom, then the curvature will grow in proportion to the length that you go along the curve. So when you have not gone very much, at the bottom of the rollercoaster, it is small, but as you get to the top, the curvature becomes large, and it is a direct proportionality. So what this curve then provides is that it is the smoothest way to link and extrapolate from a straight line to a circular curve.

As a note here, if you want to drive off one of the turning on the motorway we say earlier, whilst keeping at constant speed, you are going to end up moving with a constant angular velocity. What that means in practice is that if you move your steering wheel at a constant angular velocity, then if the bend is a clothoid shape, you do not need to change the rate at which you turn the wheel, if you keep the speed of the car constant. If the turnoff was a semi-circle or a part of a circle, you would have to keep changing the speed with which you move the steering wheel if you wanted to keep the speed of the car the same, or if you wanted to move the speed, with the steering wheel constant, you would have to change the speed of the car. So the clothoid curve has this very nice, optimally safe situation, where moving the steering wheel at constant angular speed, at a constant number of degrees per second, keeping the car speed constant, produces the required result. So clothoids sound rather odd and rather idiosyncratic, but we use them all the time in the construction of roads and not only the construction of rollercoasters.

Lastly, another curve that is quite fun is the one that Mr August Mobius created long ago. Here is the first time he ever drew it in his notebook.

That was in 1858, though it was not published until rather later. All he did was to take a strip of paper and join it together to make a cylinder. It has then got two sides: it has got an inside and an outside. But if you give it a twist into the shape shown in the diagram, it has only got one side. So if you were to colour it with a crayon, without taking the crayon off the surface of the paper, you would eventually colour the whole of the paper surface. This unusual shape, a closed curve, had all sorts of odd commercial uses.

In the 1920s, there was a great spate of patents which were placed in the US for things like conveyor belts, because people woke up to the fact that if you had a belting system where you gave a twist, so you made it into a Mobius curve, then it would last twice as long, because instead of just wearing on one side, which would happen if you had no twist, the wear would be shared over the two sides and it would last twice as long.

The other place that you see that curve, without perhaps realising it, is in the universal recycling symbol, which is not a trademark, as some people would like to make you think. This symbol that you see printed on just about everything, you can see is a strip which has been cut up into arrows to represent the sort of incestuous nature of the recycling idea:

In fact, this was created by a student at the University of Southern California back in 1970, when there was a design competition for logos. He created this symbol and won the competition with it through this representation.

So the Mobius strip is useful, and it is useful in advertising, but of course it is more famous for the role of this curve in various types of artistic representation. Escher did not very often draw Mobius strips or impossible figures, only a few, mostly tessellations, but this is famous one, with the ants climbing around the Mobius curve:

This is a picture of a sculpture by Robert Wilson, who was at one time the Director of the Fermilab, the particle physics laboratory in Illinois.

He was someone who was very active in sculpture, and the grounds of Fermilab still have many impressive works of sculpture in them, including this one of his, which you can see is a Mobius figure, made out of hollowed metal.

So the last contribution of curves that we looked at here are both functional and also artistic. So I hope you have had something of a whistle-stop tour here through a small number of curves that have a lot of unexpected applications: in architecture; in design; in the creation of structures and arches; the optimisation of structures, like rollercoasters, where forces of motion are present; and then also curves like the Mobius band, which are both useful and aesthetic.

©Professor John Barrow, Gresham College, 29January 2009