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 Acta of 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 the Acta.
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 + x2)
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:
dy2 = a2 · dx2 : s2
ds2 = dx2 + dy2 = (s2 · dx2 + a2 · dx2): s2 ⇒ ds = [dx · √(s2 + a2)] : s
dx = (s · ds) : √(s2 + a2)
and the integral thereof:
x = √(s2 + a2)
From this is obtained s = √(x2 – a2) and ds = (x · dx) : √(x2 – a2) = √(dx2 – dy2). If the equation is simplified one obtains:
x2 dx2 – a2 dy2 = a2 dx2
And finally: dy = (a · dx) : √(x2 – a2).
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 (ex/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!