Prisms
Table of Contents
Since the prism AB grows constantly stronger and more resistant to the pressure of its load at points which are more and more removed from the middle, we could in the case of large heavy beams cut away a considerable portion near the ends which would notably lessen the weight, and which, in the beam work of large rooms, would prove to be of great utility and convenience.
Fig. 32
What is the proper shape to give a solid in order to make it equally resistant at every point so that a load placed at the middle would not produce fracture more easily than if placed at any other point?
Let DB represent a prism.
Its resistance to fracture [bending strength] at the end AD, owing to a load placed at the end B, will be less than the resistance at CI in the ratio of the length CB to AB.
This same prism to be cut through diagonally along the line FB so that the opposite faces will be triangular; the side facing us will be FAB.
Such a solid will have properties different from those of the prism.
For, if the load remain at B, the resistance against fracture [bending strength] at C will be less than that at A in the ratio of the length CB to the length AB.
This is easily proved: for if CNO represents a cross-section parallel to AFD, then the length FA bears to the length CN, in the triangle FAB, the same ratio which the length AB bears to the length CB.
Therefore, if we imagine A and C to be the points at which the fulcrum is placed, the lever arms in the two cases BA, AF and BC, CN will be proportional [simili].
Hence the moment of any force applied at B and acting through the arm BA, against a resistance placed at a distance AF will be equal to that of the same force at B acting through the arm BC against the same resistance located at a distance CN.
But now, if the force still be applied at B, the resistance to be overcome when the fulcrum is at C, acting through the arm CN, is less than the resistance with the fulcrum at A in the same proportion as the rectangular cross-section CO is less than the rectangular cross-section AD, that is, as the length CN is less than AF, or CB than BA.
Fig. 33
Consequently the resistance to fracture at C, offered by the portion OBC, is less than the resistance to fracture at A, offered by the entire block DAB, in the same proportion as the length CB is smaller than the length AB.
By this diagonal saw-cut we have now removed from the beam, or prism DB, a portion, i. e., a half, and have left the wedge, or triangular prism, FBA.
We have 2 solids possessing opposite properties.
One body grows stronger as it is shortened while the other grows weaker.
This being so it would seem not merely reasonable, but inevitable, that there exists a line of section such that, when the superfluous material has been removed, there will remain a solid of such figure that it will offer the same resistance [strength] at all points.
Evidently one must, in passing from greater to less, encounter equality.
But now the question is what path the saw should follow in making the cut.
By sawing the prism along the diagonal line and removing half of the material, the remainder acquires a property just the opposite to that of the entire prism, so that at every point where the latter gains strength the former becomes weaker, then it seems to me that by taking a middle path, i. e., by removing half the former half, or one-quarter of the whole, the strength of the remaining figure will be constant at all those points where, in the two previous figures, the gain in one was equal to the loss in the other.
You have missed the mark, Simplicio.
The amount which you can remove from the prism without weakening it is 1/3, not 1/4.
What is the path the saw must travel?
This must be a parabola demonstrated by the following lemma:
If the fulcrums are so placed under two levers or balances that the arms through which the forces act are to each other in the same ratio as the squares of the arms through which the resistances act, and if these resistances are to each other in the same ratio as the arms through which they act, then the forces will be equal.
Let AB and CD represent two levers whose lengths are divided by their fulcrums in such a way as to make the distance EB bear to the distance FD a ratio which is equal to the square of the ratio between the distances EA and FC. Let the resistances located at A and C be to each other as EA is to FC.
The forces which must be applied at B and D in order to hold in equilibrium the resistances at A and C are equal. Let EG be a mean proportional between EB and FD.
Then we shall have BE:EG=EG:FD=AE:CF. But this last ratio is precisely that which we have assumed to exist between the resistances at A and C. And since EG:FD=AE:CF, it follows, permutando, that EG:AE=FD:CF.
Seeing that the distances DC and GA are divided in the same ratio by the points F and E, it follows that the same force which, when applied at D, will equilibrate the resistance at C, would if applied at G equilibrate at A a resistance equal to that found at C.
Fig. 34
But one datum of the problem is that the resistance at A is to the resistance at C as the distance AE is to the distance CF, or as BE is to EG. Therefore the force applied at G, or rather at D, will, when applied at B, just balance the resistance located at A. q. e. d.
Draw the parabola FNB in the face FB of the prism DB.
Let the prism be sawed along this parabola whose vertex is at B.
The portion of the solid which remains will be included between the base AD, the rectangular plane AG, the straight line BG and the surface DGBF, whose curvature is identical with that of the parabola FNB.
This solid will have, I say, the same strength at every point. Let the solid be cut by a plane CO parallel to the plane AD. Imagine the points A and C to be the fulcrums of two levers of which one will have the arms BA and AF.
The other BC and CN.
Then since in the parabola FBA, we have BA:BC=AFmacronsup2: CNmacronsup2, it is clear that the arm BA of one lever is to the arm BC of the other lever as the square of the arm AF is to the square of the other arm CN.
Since the resistance to be balanced by the lever BA is to the resistance to be balanced by the lever BC in the same ratio as the rectangle DA is to the rectangle OC, that is as the length AF is to the length CN, which two lengths are the other arms of the levers, it follows, by the lemma just demonstrated, that the same force which, when applied at BG will equilibrate the resistance at DA, will also balance the resistance at CO.
The same is true for any other section.
Therefore this parabolic solid is equally strong throughout.
Fig. 35
It can now be shown that, if the prism be sawed along the line of the parabola FNB, one-third part of it will be removed; because the rectangle FB and the surface FNBA bounded by the parabola are the bases of two solids included between two parallel planes, i. e., between the rectangles FB and DG; consequently the volumes of these two solids bear to each other the same ratio as their bases.
But the area of the rectangle is one and a half times as large as the area FNBA under the parabola; hence by cutting the prism along the parabola we remove one-third of the volume. It is thus seen how one can diminish the weight of a beam by as much as thirty-three per cent without diminishing its strength; a fact of no small utility in the construction of large vessels, and especially in supporting the decks, since in such structures lightness is of prime importance.
The advantages derived from this fact are so numerous that it would be both wearisome and impossible to mention them all; but leaving this matter to one side, I should like to learn just how it happens that diminution of weight is possible in the ratio above stated.
I can readily understand that, when a section is made along the diagonal, one-half the weight is removed; but, as for the parabolic section removing 1/3 of the prism, this I can only accept on the word of Salviati who is always reliable; however I prefer first-hand knowledge to the word of another.
You would like then a demonstration of the fact that the excess of the volume of a prism over the volume of what we have called the parabolic solid is one-third of the entire prism.
This I have already given you on a previous occasion; however I shall now try to recall the demonstration in which I remember having used a certain lemma from Archimedes’ book On Spirals, namely, Given any number of lines, differing in length one from another by a common difference which is equal to the shortest of these lines; and given also an equal number of lines each of which has the same length as the longest of the first-mentioned series; then the sum of the squares of the lines of this second group will be less than three times the sum of the squares of the lines in the first group. But the sum of the squares of the second group will be greater than three times the sum of the squares of all excepting the longest of the first group.
Inscribe in the rectangle ACBP the parabola AB.
The mixed triangle BAP whose sides are BP and PA. Its base is the parabola BA, is a third part of the entire rectangle CP.
If this is not true it will be either greater or less than a third.
Suppose it to be less by an area which is represented by X. By drawing lines parallel to the sides BP and CA, we can divide the rectangle CP into equal parts;
If the process be continued we shall finally reach a division into parts so small that each of them will be smaller than the area X; let the rectangle OB represent one of these parts and, through the points where the other parallels cut the parabola, draw lines parallel to AP.
Let us now describe about our “mixed triangle” a figure made up of rectangles such as BO, IN, HM, FL, EK, and GA; this figure will also be less than a third part of the rectangle CP because the excess of this figure above the area of the “mixed triangle” is much smaller than the rectangle BO which we have already made smaller than X.
Fig. 36
More slowly, please; for I do not see how the excess of this figure described about the “mixed triangle” is much smaller than the rectangle BO.
Does not the rectangle BO have an area which is equal to the sum of the areas of all the little rectangles through which the parabola passes? I mean the rectangles BI, IH, HF, FE, EG, and GA of which only a part lies outside the “mixed triangle.”
Have we not taken the rectangle BO smaller than the area X?
Our opponent might say the triangle plus X is equal to a third part of this rectangle CP.
The circumscribed figure, which adds to the triangle an area less than X, will still remain smaller than a third part of the rectangle, CP.
But this cannot be because this circumscribed figure is larger than a third of the area. Hence it is not true that our “mixed triangle” is less than a third of the rectangle.