SIMP. Before proceeding further I should like to have one of my difficulties removed. Up to this point you have not taken into consideration a certain other kind of resistance that, it appears to me, diminishes as the solid grows longer, and this is quite as true in the case of pulling as of bending. For example, in the case of a rope we observe that a very long one is less able to support a large weight than a short one. Thus, I believe, a short rod of wood or iron will support a greater weight than if it were long, provided that the force be always applied longitudinally and not transversely, and provided also that we take into account its own weight, which increases with its length.
SALV. I fear, Simplicio, that in this particular you are making the same mistake as many others, if I correctly catch your meaning; that is, if you mean to say that a long rope, one of perhaps forty cubits, cannot hold up so great a weight as a shorter length, say one or two cubits, of the same rope.
SIMP. That is what I meant, and as far as I see the proposition is highly probable.
SALV. On the contrary, I consider it not merely improbable but false; and I think I can easily convince you of your error. Let AB represent the rope, fastened at the upper end A. At the lower end, attach a weight C whose force is just sufficient to break the rope. Now, Simplicio, point out the exact place where you think the break ought to occur.
[162] SIMP. Let us say D.
SALV. And why at D?
SIMP. Because at this point the rope is not strong enough to support, say, one hundred pounds, made up of the portion of the rope DB and the stone C.
SALV. Accordingly, whenever the rope is stretched with the weight of one hundred pounds at D, it will break there.
SIMP. I think so.
SALV. But tell me, if instead of attaching the weight at the end of the rope, B, one fastens it at a point nearer D, say, at E; or if instead of fixing the upper end of the rope at A, one fastens it at some point F, just above D; will not the rope, at the point D, be subject to the same pull of one hundred pounds?
SIMP. It would, provided you include with the stone C the portion of rope EB.
SALV. Let us therefore suppose that the rope is stretched at the point D with a weight of one hundred pounds. Then, according to your own admission, it will break. But FE is only a small portion of AB. How can you therefore maintain that the long rope is weaker than the short one? Give up then this erroneous view which you share with many very intelligent people, and let us proceed.
We have already demonstrated that in the case of prisms and cylinders of constant thickness, the moment of force tending to produce fracture varies as the square of the length; and likewise we have shown that when the length is constant and the thickness varies, the resistance to fracture varies as the cube of the side or diameter of the base; so let us go on to the investigation of the case of solids that simultaneously vary in both length and thickness. Here I formulate Proposition 5: Prisms and cylinders that differ in both length and thickness offer resistances to fracture that are directly proportional [163] to the cubes of the diameters of their bases and inversely proportional to their lengths.
Let ABC and DEF be two such cylinders; then the resistance of the cylinder AC bears to the resistance of the cylinder DF a ratio that is the product of the cube of the diameter AB divided by the cube of the diameter DE, and the length EF divided by the length BC. Make EG equal to BC; let H be the third proportional to the lines AB and DE; let I be the fourth proportional; and let I be to S as EF is to BC.
Now, since the resistance of the cylinder AC is to that of the cylinder DG as the cube of AB is to the cube of DE, that is, as the length AB is to the length I; and since the resistance of the cylinder DG is to that of the cylinder DF as the length FE is to EG, that is, as I is to S; it follows, by equidistance of ratios, that the resistance of the cylinder AC is to that of the cylinder DF as the length AB is to S. But the line AB bears to S a ratio that is the product of AB/I and I/S. Hence the resistance of the cylinder AC bears to the resistance of the cylinder DF a ratio that is the product of AB/I (that is, the cube of AB to the cube of DE) and I/S (that is, the length EF to the length BC). This is what I meant to prove.
This proposition having been demonstrated, let us next consider the case of prisms and cylinders that are similar. Concerning these we shall show Proposition 6: In the case of similar cylinders and prisms, the compound moments, namely, those produced by their own weight and length (which latter acts as a lever arm), bear to each other a ratio that is the three-halves power of the ratio between the resistances of their bases.
In order to prove this, let us consider the two similar cylinders AB and CD. Then I say that the moment of the cylinder AB, opposing the resistance of its base B, bears to the moment of CD, opposing the resistance of its base D, a [164] ratio that is the three-halves power of the ratio between the resistance of the base B and the resistance of the base D. For, the solids AB and CD are effective in opposing the resistances of their bases B and D in proportion to both their weights and the mechanical advantages of their lever arms; and the advantage of the lever arm AB is equal to the advantage of the lever arm CD (this is true because, in virtue of the similarity of the cylinders, the length AB is to the radius of the base B as the length CD is to the radius of the base D); so it follows that the total moment of the cylinder AB is to the total moment of the cylinder CD as the weight alone of the cylinder AB is to the weight alone of the cylinder CD, that is, as the volume of the cylinder AB is to the volume CD; but these are as the cubes of the diameters of their bases B and D; and the resistances of the bases, being to each other as their areas, are to each other consequently as the squares of their diameters; therefore, the moments of the cylinders are to each other as the three-halves power of the resistances of their bases.
SIMP. This proposition strikes me as both new and surprising. At first glance it is very different from anything which I myself should have guessed. For since these figures are similar in all other respects, I should have certainly thought that the moments and the resistances of these cylinders would have borne to each other the same ratio.
SAGR. This is the proof of the proposition to which I referred, at the very beginning of our discussion, as one imperfectly understood by me.
SALV. For a while, Simplicio, I used to think, as you do, that the resistances of similar solids were similar. But a certain casual observation showed me that similar solids do not exhibit a strength that is proportional to their size, the larger ones being less fitted to undergo rough usage just as tall men are more apt than small children to be injured by a fall. [165] And as we remarked at the outset, a large beam or column falling from a given height will go to pieces when under the same circumstances a small scantling or small marble cylinder will not break. It was this observation that led me to the investigation of the fact which I am about to demonstrate to you: it is a very remarkable thing that, among the infinite variety of solids that are similar one to another, there are no two whose moments are related in the same ratio to their own resistances.
SIMP. You remind me now of a passage in Aristotle’s Questions of Mechanics14 in which he tries to explain why it is that a wooden beam becomes weaker and can be more easily bent as it grows longer, notwithstanding the fact that the shorter beam is thinner and the longer one thicker. And, if I remember correctly, he explains it in terms of the simple lever.
SALV. Very true. But, since this solution seemed to leave room for doubt, Monsignor di Guevara,15 whose truly learned commentaries have greatly enriched and illuminated this work, indulges in additional clever speculations with the hope of thus overcoming all difficulties. Nevertheless, even he is confused as regards this particular point, namely, whether, when the length and thickness of these solid figures increase in the same ratio, their strength and resistance to fracture, as well as to bending, remain constant. After much thought upon this subject, I have reached the following results. First I shall show that (Proposition 7): Among prisms or cylinders that are similar and hav
e weight, there is one and only one which, under the stress of its own weight, lies just on the limit between breaking and not breaking, such that every larger one is unable to carry the load of its own weight and breaks, while every smaller one is able to withstand some additional force tending to break it.
Let AB be a prism, the longest possible that will just sustain its own weight, so that if it be lengthened the least bit it will break. Then, I say, this prism is unique among all similar prisms—infinite in number— in occupying [166] that boundary line between breaking and not breaking; so that every larger one will break under its own weight, and every smaller one will not break but will be able to withstand some force in addition to its own weight. Let the prism CE be similar to, but larger than, AB; then, I say, it will not remain intact but will break under its own weight. Lay off the portion CD, equal in length to AB. Since the resistance of CD is to that of AB as the cube of the thickness of CD is to the cube of the thickness of AB, that is, as the prism CE is to the similar prism AB, it follows that the weight of CE is the utmost load which a prism of the length CD can sustain; but the length of CE is greater; therefore the prism CE will break. Now take another prism FG smaller than AB, and let FH equal AB. Then it can be shown in a similar manner that the resistance of FG is to that of AB as the prism FG is to the prism AB, provided the distance AB (that is, FH) is equal to the distance FG; but AB is greater than FG; therefore the moment of the prism FG applied at G is not sufficient to break the prism FG.
SAGR. The demonstration is short and clear; while the proposition which, at first glance, appeared improbable is now seen to be both true and inevitable. In order therefore to bring this prism into that limiting condition that separates breaking from not breaking, it would be necessary to change the ratio between thickness and length either by increasing the thickness or by diminishing the length. An investigation of this limiting state will, I believe, demand equal ingenuity.
SALV. Nay, even more; for the question is more difficult. This I know because I spent no small amount of time in its discovery which I now wish to share with you. Proposition 8: Given a cylinder or prism of the greatest length consistent with its not breaking under its own weight, and given a greater length, to find the thickness of another cylinder or prism of this greater length that shall be the only and largest one capable of withstanding its own weight.
Let BC be the largest cylinder capable of sustaining its own weight; and let DE be a length greater than AC. The problem is to find the thickness of the [167] cylinder which, having the length DE, shall be the largest one just able to withstand its own weight. Let I be the third proportional to the lengths DE and AC; let the diameter FD be to the diameter BA as DE is to I; draw the cylinder FE; then, I say, among all cylinders having the same proportions, this is the largest and only one just capable of sustaining its own weight. Let M be the third proportional to DE and I; also let O be the fourth proportional to DE, I, and M; lay off FG equal to AC. Now since the diameter FD is to the diameter AB as the length DE is to I, and since O is the fourth proportional to DE, I, and M, it follows that the cube of FD is to the cube of BA as DE is to O; but the resistance of the cylinder DG is to the resistance of the cylinder BC as the cube of FD is to the cube of BA; hence the resistance of the cylinder DG is to that of cylinder BC as the length DE is to O. And since the moment of the cylinder BC is held in equilibrium by its resistance, we shall accomplish our end (which is to prove that the moment of the cylinder FE is equal to the resistance located at FD), if we show that the moment of the cylinder FE is to the moment of the cylinder BC as the resistance DF is to the resistance BA, that is, as the cube of FD is to the cube of BA, or as the length DE is to O. But the moment of the cylinder FE is to the moment of the cylinder DG as the square of DE is to the square of AC, that is, as the length DE is to I; and the moment of the cylinder DG is to the moment of the cylinder BC as the square of DF is to the square of BA, that is, as the square of DE is to the square of I, or as the square of I is to the square of M, or as I is to O. Therefore, by equidistance of ratios, it results that the moment of the cylinder FE is to the moment of the cylinder BC as the length DE is to O, that is, as the cube of DF is to the cube of BA, or as the resistance of the base DF is to the resistance of the base BA. This is what was being sought.
SAGR. This demonstration, Salviati, is rather long and difficult to keep in mind from a single hearing. Will you not, therefore, be good enough to repeat it?
SALV. As you like. But I would suggest instead a more direct and a shorter proof. This will, however, necessitate a different figure.
[168] SAGR. The favor will be that much greater. Nevertheless, I hope you will oblige me by putting into written form the proof just given, so that I may study it at my leisure.
SALV. I shall gladly do so. Now, let A denote a cylinder of diameter DC and the largest capable of sustaining its own weight; the problem is to find a larger cylinder that shall be at once the maximum and the unique one capable of sustaining its own weight. Let E be such a cylinder, similar to A, having an assigned length, and having the diameter KL; let MN be the third proportional to the two lengths DC and KL; let MN also be the diameter of another cylinder, X, having the same length as E; then, I say, X is the cylinder sought. For the resistance of the base DC is to the resistance of the base KL as the square of DC is to the square of KL, that is, as the square of KL is to the square of MN, or, as the cylinder E is to the cylinder X, that is, as the moment of E is to the moment of X; but the resistance of the base KL is to the resistance of the base MN as the cube of KL is to the cube of MN, that is, as the cube of DC is to the cube of KL, or, as the cylinder A is to the cylinder E, that is, as the moment of A is to the moment of E; hence it follows, by perturbed equidistance of ratios, that the moment of A is to the moment of X as the resistance of the base DC is to the resistance of the base MN; therefore, moment and resistance are related to each other in prism X precisely as they are in prism A.
Let us now generalize the problem. Then the proposition will read as follows: Given a cylinder AC in which moment and resistance are related in any manner whatsoever, and given that DE is the length of another cylinder, then determine what its thickness must be in order that the relation between its moment and resistance shall be identical with that of the cylinder AC.
Using again the penultimate figure and almost in the same manner, we may say the following. Since the moment of the cylinder FE is to the moment of the portion DG as the square of ED is to the square of FG, that is, as the length DE is to I; and since the moment of the cylinder FG is to the moment of the cylinder AC as the square of FD is to the square of AB, or, as the square of ED is to the square of I, or, as the square of I is to the square of M, [169] that is, as the length I is to O; it follows, by equidistance of ratios, that the moment of the cylinder FE is to the moment of the cylinder AC as the length DE is to O, that is, as the cube of DE is to the cube of I, or, as the cube of FD is to the cube of AB, that is, as the resistance of the base FD is to the resistance of the base AB. This is what was to be proven.
From what has been demonstrated so far, you can plainly see the impossibility of increasing the size of structures to vast dimensions either in art or in nature. Thus, it would be impossible to build ships, palaces, or temples of enormous size in such a way that their oars, masts, beams, iron bolts, and, in short, all their other parts will hold together. Nor could nature produce trees of extraordinary size, because the branches would break down under their own weight. Likewise it would be impossible to build up the bony structures of men, horses, or other animals so as to hold together and perform their normal functions; for these animals would have to be increased enormously in height and this increase could be accomplished only by employing a material that is harder and stronger than usual, or by enlarging the size of the bones, thus changing their shape until the form and appearance of the animals would be monstrous. This is perhaps what our wise poet had in mind, when he said, in describing a huge giant: “Impossible it
is to reckon his height / So beyond measure is his size.”16
To illustrate briefly, I have sketched a bone whose natural length has been increased three times and whose thickness has been multiplied until, for a correspondingly large animal, it would perform the same function which the small bone performs for its small animal. From the figures here shown you can see how out of proportion the enlarged bone appears. Clearly then if one wishes to maintain in a great giant the same proportion of limb as that found in an ordinary man, one must either find [170] a harder and stronger material for making the bones, or one must admit a diminution of strength in comparison with men of medium stature; for if his height be increased inordinately, he will fall and be crushed under his own weight. On the other hand, if the size of a body be diminished, the strength of that body is not diminished in the same proportion; indeed the smaller the body the greater its relative strength. Thus a small dog could probably carry on his back two or three dogs of his own size; but I believe that a horse could not carry even one of his own size.
SIMP. This may be so. But I am led to doubt it on account of the enormous size reached by certain fish, such as the whale which, I understand, is ten times as large as an elephant; yet they all support themselves.
SALV. Your question, Simplicio, suggests another principle, one that had hitherto escaped my attention and that enables giants and other animals of vast size to support themselves and to move about as well as smaller animals do. This result may be secured by increasing the strength of the bones and other parts intended to carry not only their weight but also the superincumbent load. But there is another way: keeping the proportions of the bony structure constant, the skeleton will hold together in the same manner or even more easily provided one diminishes, in the proper proportion, the weight of the bony material, of the flesh, and of anything else which the skeleton has to carry. It is this second principle that is employed by nature in the structure of fish, making their bones and muscles not merely light but entirely devoid of weight.
The Essential Galileo Page 42