# The Museum of Mathematics of Catalonia (MMACA)

Last Wednesday I went to MMACA (Museum of Mathematics of Catalonia) with some of my students. This museum is located in Mercader Palace in Cornellà de Llobregat (near Barcelona) since February and we enjoyed a very interesting “mathematical experience”.

The museum is not so big but you can “touch” and discover Mathematics in all its rooms. I think that there are enough experiences to enjoy arithemtical and geometrical properties, simmetries, mirrors, impossible tessellations, Stadistics,…

For example, students could check the validity of theorem of Pythagoras in two ways. First of all, they coud weigh wooden squares and check that the square constructed on the hypotenuse of a right triangle weighs the same as the two squares constructed on the other two sides of the triabgle. Later, they discovered that the first square could be divided in some pieces of Tangram with which they could construct the other two squares. So the visitors demonstrated the theorem in a very didactic way: playing with balances and playing with tangram.

Students also learnt some properties of the cycloid and they could check its brachistochronic characteristic. I imagine Galileo or some of Bernoulli brothers in the 17th century doing the same experiments with a similar instrument. What a wonderful curve! The ball always reaches the central point in the same time and its initial position doesn’t matter!

Another of the studied curves is the catenary which is one of the emblematic mathematical symbols of Antoni Gaudi’s architecture in Barcelona.

Of course, polyhedra are very important in the exhibition and visitors can play with them so they discover some of their most important properties. For example, which is the dual polyhedron of the dodecahedron? Playing with it the students could see that the hidden polyhedron is a… You must visit MMACA and discover it!

Another example: look at these three wooden pieces…

The dodecahedron has an ortonormal symmetry and we can check it with an ortonormal set of mirrors:

There are more mirrors and more wooden pieces to play and construct other different Platonic and Archimedian polyhedra.

And… did you know that it’s possible to draw a right line playing with two circles? If the red circle rotates within the black one… what figure is described by the yellow point?

In the 13th century, the great Nasîr al-Dîn al-Tûsî had to build one similar instrument to improve the astronomical geometrical systems with his “Al-Tûsî’s pair”:

Rotating a circle within another one, he could move a point in a right line without denying Aristotelian philosophy. This dual system was used by al-Tûsî in his* Zîj-i Ilkhanî* (finished in 1272) and Nicolas Copernicus probably read this innovation together with other Arabic astronomical models. Thinking about them, he began to improve the astronomical system of his *De Revolutionibus* (1543). Al-Tûsî’s pair was very famous until the 15th century.

In Erathostenes Room there are some Sam lloyd’s puzzles, games about tesselations, Stadistics, Probablility and this quadric:

I didn’t know that it could be described only with a multiplication table! Is its equation *z = xy*? Yes, of course! My students also played to build the famous Leonardo’s bridge and they could see that there isn’t necessary any nail to hold a bridge.

Ah! And I can’t forget to say that if you visit MMACA with a person that don’t like Maths, he/she can always admire this beautiful XIX century Mercader Palace:

Furthermore, one of the rooms of the palace is decoratd by a chess lover!

So… you must go to MMACA and enjoy Mathematics in a way ever done!

# Clifton Suspension Bridge

Clifton Suspension Bridge was designed in 1830 by Isambard Kingdom Brunel (1806-1859). Its construction began in 1836 but was interrupted in 1843 through lack of funds. It was not until 1864 five years after Brunel’s death that the bridge was completed as a monument to his fame, the chains used being those from the hungerford bridge designed and erected by him in 1843.

Can you see the wonderful catenary between its pillars?s

**Location**: Clifton Suspension Bridge (map)

# The catenary resists

And now it’s tim for the catenary:

To admire the form of this curve, all we have to do is attach a chain, a string or a wire to two points in a constant gravity field. The wire or similar will adopt the form in which it only supports its own weight and no other additional tension. This is the situation in which it is most at rest, that of minimum rigidity. The urban landscape is full of catenaries, but the most interesting are, without doubt, the inverted catenaries that Gaudí used in many arches. The most notable difference between a Gothic cahedral and the Sagrada Familia in Barcelona is that Gaudí’s church rises to the same height without the need for buttresses. Question: are shells and the skeletons of large animals inverted catenaries?

Another example of the use of the catenary in the architecture is the modernist Masia Freixa (1907-1910) by Lluís Muncunill:

The first man who tried to find the solution to the problem of the hanging chain was probably Leonardo da Vinci (1452-1519), who draw some similar situations in his papers. Galileo Galilei (1564-1642) studied the problem and said that the chain should adopt a parabolic form in his *Discorsi e Dimostrazioni Matematiche, intorno a due nuove scienze attenenti alla meccanica & i movimenti locali *(1638):

SALVIATI: In this case of the rope then, Sagredo, you cease to wonder at the phenomenon because you have its demonstration; but if we consider it with more care we may possibly discover some correspondence between the case of the gun and that of the string. The curvature of the path of the shot fired horizontally appears to result from two forces, one (that of the weapon) drives it horizontally and the other (its own weight) draws it vertically downward. So in stretching the rope you have the force which pulls it horizontally and its own weight which acts downwards. The circumstances in these two cases are, therefore, very similar. If then you attribute to the weight of the rope a power and energy [possanza ed energia] sufficient to oppose and overcome any stretching force, no matter how great, why deny this power to the bullet?Besides I must tell you something which will both surprise and please you, namely, that a cord stretched more or less tightly assumes a curve which closely approximates the parabola. This similarity is clearly seen if you draw a parabolic curve on a vertical plane and then invert it so that the apex will lie at the bottom and the base remain horizontal; for, on hanging a chain below the base, one end attached to each extremity of the base, you will observe that, on slackening the chain more or less, it bends and fits itself to the parabola; and the coincidence is more exact in proportion as the parabola is drawn with less curvature or, so to speak, more stretched; so that using parabolas described with elevations less than 45° the chain fits its parabola almost perfectly.

SAGREDO: Then with a fine chain one would be able to quickly draw many parabolic lines upon a plane surface.

In 1669, the German mathematician Joachim Jungius (1587-1657) demonstrated that the form adopted by the chain wasn’t a parabola and one year later, Jakob Bernoulli (1654-1705) proposed a contest looking for the first mathematician who could find out the real forma of a hanging chain. The problem was solved by Johann Bernoulli (1667-1748), Christiann Huygens (1629-1695) and Gottfried W. Leibnitz (1646-1717) each independently. All three solutions were published in the German *Acta Eruditorum* in 1691. One year later, Johann Bernoulli wrote his *Lectures on the Integral Calculus* compiling his lessons to Guillaume Marquis de L’Hôpital (1661-1704). In this work we can read:

The importance of the problem of the catenary in Geometry can be seen from the three solutions in the

Actaof Leipzig of last year (1691), and especially from the remarks that the renowned Leibniz makes there. The first to consider this curve, which is formed by a free-hanging string, or better, by a thin inelastic chain, was Galileo. He, however, did not fathom its nature; on the contrary, he asserted that it is a parabola, which it certainly is not. Joachim Jungius discovered that it is not a parabola, as Leibniz remarked, through calculation and his many experiments. However, he did not indicate the correct curve for the catenary. The solution to this important problem therefore remained for our time. We present it here together with the calculation, which was not appended to the solution in theActa.There are actually two kinds of catenaries: the common, which is formed by a string or a chain of uniform thickness, or is of uniform weight at all points, and the uncommon, which is formed by a string of non-uniform thickness, which therefore is not of uniform weight at all points, and certainly not uniform in relation to the ordinates of any given curve. Before we set about the solution, we make the following assumptions, which can easily be proven from Statics:

1) The string, rope, or chain, or whatever the curve consists of, will be assumed to be flexible and inelastic at all of its points, that is, it undergoes no stretching as a result of its weight.

2) If the catenary ABC is held fixed at any two points A and C, then the necessary forces at points A and C, are the same as those which support a weight D, that is equal to the weight of the chain ABC and is located at the meeting point of two weightless strings AD and CD, that are tangent to the curve at points A and C. The reason for this is clear.

Because the weight of the chain ABC exerts its action at A and C in one direction at each point, namely, in the directions of the tangents AD and CD, and the pull of the same or equal weight D at A and C likewise goes in the directions of AD and CD. Therefore the necessary forces at points A and C must also in both cases be the same. Accordingly, one obtains the necessary force at the lowest point B, when one seeks the force that the weight E [Figure 2] exerts at the same point, when it is held by two weightless strings, one of which is tangent to the curve at B, and therefore is horizontal, while the other is tangent curve at B, and therefore is horizontal, while the other is tangent to the curve at point A:

3) When a chain fastened at points A and C is then fastened at any other point F, so that one could remove the portion AF, the curve represented by the remaining piece of chain FBC does not change, that is, the remaining points will stay in the same position as before the fastening at F. This needs no proof, because Reason advises it and experience lays it daily before our eyes.

4) If we retain the previous assumptions, then before and after the fastening at F, the same (that is, the original) force must obtain at particular positions on the curve, or, what amounts to the same thing, a point will be pulled with the same force after the fastening [at F] as before it. This is nothing but a corollary of the preceding number. Consequently, as one lengthens or shortens the chain BFA, that is, wherever one chooses the fastening point F, the force at the lowest position B neither increases nor decreases, but always remains the same.

5) The weight P, which is held by any two arbitrarily situated strings AB and CB, exerts its forces on the points A and C in such a relation, that the necessary force at A is to the necessary force at C (after drawing vertical line BG), as the sine of angle CBG is to the sine of angle ABG, and the force of the weight P is to the force at C as the sine of the whole angle ABC is to the sine of the opposite angle ABG. This is proven in every theory of Statics.

With these assumptions, we find the common catenary curve in the following manner. Let BAa be the desired curve; B, its deepest point; the axis or the vertical through B, BG; the tangent at the deepest point, which will be horizontal, BE; and let AE be the tangent at any other point A. Draw the ordinate AG and the parallel EL to the axis:

Let x = BG, y = GA, Gg = dx and dy = Ha, and the weight of the chain, or, since it is of uniform thickness, the length of the curve BA = s. Since at point B, an ever constant force will be required (by assumption 4), whether the chain be lengthened or shortened, that force, or the segment C = a expressing it, will therefore be a constant.3 Imagine now that the weight of the chain AB is concentrated at and hangs at the meeting point E of the tangent strings AE, BE; then (by assumption 2) the same force is required at point B to hold the weight E as was required to hold the chain BA. But the weight E (by assumption 5) is to the force at B, as the sine of the angle AEB, or as the sine of its complementary angle EAL is to the sine of angle AEL, that is, as EL is to AL. Wherever on the curve one chooses the fixed point A (the curve always remains the same, by assumption 3), the weight of the chain AB is to the force at B [which force equals the constant a], as EL is to AL, that is:

s : a = EL : AL = AH : Ha = dx : dy or dy : dx = a : s

Hence it follows that the catenary BA is the same as that curve whose construction and nature we have given above, by the method of inverse tangents, where we first converted the proportion dy : dx = a : s to the following:

dy= a · dx / √(2ax + x

^{2})at which point the curve was constructed through the rectification of the parabola as well as through the quadrature of the hyperbola.

Then, Johann Bernoulli explained how to find dy:

To find the nature of the curve so created that DC : BC = E : AD:

Let AC = x, CD = y, AD = s. By assumption, dy : dx = a : s [we have a constant segment E = a]. Therefore: dy = a dx : s.

However, to be able to eliminate the letter s (which is always necessary in the determination of curves), one must proceed thus:

dy

^{2}= a^{2}· dx^{2}: s^{2}Therefore:

ds

^{2 }= dx^{2 }+ dy^{2}= (s^{2}· dx^{2}+ a^{2}· dx^{2}): s^{2 }⇒ ds = [dx · √(s^{2}+ a^{2})] : sTherefore:

dx = (s · ds) : √(s

^{2}+ a^{2})and the integral thereof:

x = √(s

^{2}+ a^{2})From this is obtained s = √(x

^{2}– a^{2}) and ds = (x · dx) : √(x^{2}– a^{2}) = √(dx^{2}– dy^{2}). If the equation is simplified one obtains:x

^{2}dx^{2}– a^{2}dy^{2}= a^{2}dx^{2}And finally: dy = (a · dx) : √(x

^{2}– a^{2}).

We can say that Johann Bernoulli calculated the equation of the catenary in a such original way. Nowadays, we write:

*y* =* a* cosh (*x/a*) =* a* (*e ^{x/a} + e^{-x/a}*)/2

I think that that’s enough for today. Thanks Leibnitz, thanks Johann and Jakob Bernoulli and thanks to all the men who made possible this beautiful curve!