Fig. 591. Fig. 592. case (fig. 592.) the wall G F is opposed to the force MN, so that only the triangle of it HIF can be detached; lastly, in fig. 593. the force MN would only be effective against Fig. 595. the triangle CGH, which would, of course, be greater in proportion to the increased distance of the walls CD, HI. 1506. In the first case, the unequal settlement of the soil or of the construction may produce the effect of the force MN. The wall will fall on the occurrence of an horizontal disunion between the parts. 1507. In the second case the disunion must take place obliquely, which will require a greater effort of the power MN. 1508. In the third case, in order to overturn the wall, there must be three fractures through the effort of MN, requiring a much more considerable force than in the second case. 1509. We may easily conceive that the resistance of a wall standing between two others will be greater or less as the walls CD, HI are more or less distant; so that, in an extreme approximation to one another, the fracture would be impossible, and, in the opposite case, the intermediate wall approaches the case of an isolated wall. 1510. Walls enclosing a space are in the preceding predicament, because they mutually tend to sustain one another at their extremities; hence their thickness should increase as their length increases. 1511. The result of a vast number of experiments by Rondelet, whose work we are still using, will be detailed in the following observations and calculations. 1512. Let ABCD (fig. 594.) be the face of one of the walls for enclosing a rectangular space, EFGH (fig. 595.). Draw the diagonal BD, and above B make Bā equal to one eighth part of the height, if great stability be required; for a mean stability, the ninth or tenth part; and, for a light stability, the eleventh or twelfth part. If through the point d a parallel to AB be drawn, the interval will give the thickness to be assigned to the great walls EF, GH, whose length is equal to AD. 1513. The thickness of the walls EG, FH is obtained by making AD' equal to their length, and, having drawn the diagonal as before, pursuing the same operation. 1514. When the walls are of the same height but of different lengths, as in fig. 596., the operation may be abridged by describing on the point B (fig. 597.) as a centre with a radius equal to one eighth, one tenth, or one twelfth, or such other part of the height as may be considered necessary for a solid, mean, or lighter construction, then transferring their lengths, EF, FG, GH, and HE from A to D, D', D", and D’”; and having made the rectangles AC, AC, AC", and AC", draw from the common point B the diagonals BD, BD, BD", and BD”, cutting the small circle described on the point B in different points, through which parallels to AB are to be drawn, and they will give the thickness of each in proportion to its length. 1515. In figs. 598. to 602. are given the operations for finding the thicknesses of walls Fig.600. Prior enclosing polygonal areas supposed to be of the same height; thus AD represents, the side of the hexagon (fig. 602.); AD that of the pentagon (fig. 601.); AD” the side of the square (fig. 599.); and AD” that of the equilateral triangle (fig. 600.). 15iG. It is manifest that, by this method, we increase the thicknesses of the walls in proportion to their heights and lengths; for one or the other, or both, cannot increase or diminish without the same happening to the diagonal. 1517. It is obvious that it is easy to calculate in numbers the results thus geometrically obtained by the simple rule of three; for, knowing the three sides of the triangle ABD, similar to the smaller triangle Bale, we have BD : Ba:: AD : ed. Thus, suppose the length of wall represented by AD=28 feet, and its height AB = 12 feet, we shall have the length of the diagonal =30 feet 54 inches; and, taking the ninth part of AB, or 16 inches, as the thickness to be transferred on the diagonal from B to d, we have 30 ft. 6 in. : 16 in. :: 28 ft. : 14 in. : 8 lines (ed). The calculation may also be made trigonometrically; into which there is no necessity to enter, inasmuch as the rules for obtaining the result may be referred to in the section “Trigonometry," and from thence here applied. Method of enclosing a given Area in any regular Polygon. 1518. It is manifest that a polygon may be divided by lines from the centre to its angles into as many triangles as it has sides. In fig. 601., on one of these triangles let fall from c (which is the vertex of each triangle a perpendicular CD) a vertical on the base or side AB which is supposed horizontal. The area of this triangle is equal to the product of DB (half AB) by CD, or to the rectangle DCF B. Making DB =r, CD=y, and the area given = p, we shall have, For the equilateral triangle, r x yx 3=p, or ry= %; For the square, ry x 4 = p, or a y = *: For the pentagon, ryx 5 = p, or ry= }; For the hexagon, ry x 6 = p, or ry= #. Each of these equations containing two unknown quantities, it becomes necessary to ascertain the proportion of r to y, which is as the sines of the angles opposite to the sides DB and CD. 1519. In the equilateral triangle this proportion is as the sine of 60 degrees to the sine of 30 degrees; that is, using a table of sines, as 86603 : 50000, or 83 : 5, or 26 : 15, whence a : y :: 26 : 15, and 15+ =269; whence y='. Supposing the area given to be 3600, we shall therefore have 1520. Suppose the case that of a pentagon (fig,601.) one of whose equal triangles is ACB. Let fall the perpendicular CD, which divides it into two equal parts; whence its area is equal to the rectangle CDBF. 1521. Upon the side AB, prolonged, if necessary, make DE equal to CD, and from the middle of BE as a centre describe the semi-circumference cutting CD in G, and GD will be the side of a square of the same area as the rectangle CDBF. The sides of similar figures (Geometry, 961.) being as the square roots of their areas; find the square root of the given area and make Dg equal to it. From the point 9 draw parallels to GE and GB, which will determine on AB the points e and b, which will give on one side Db equal to one half of the side of the polygon sought; and, on the other, the radius De of the circumference in which it is inscribed. This is manifest because of the similar triangles EGB and egb, from which BD : DE::b1) : De. 1522. From the truth that the sides of similar figures are to each other as the square roots of their areas we arrive at a simple method of reducing any figure to a given area. Form an angle of reduction (fig. 603.) one of whose sides is equal to the square root of the greater area, and the chord of the arc, which determines the size of the angle equal to the square root of the smaller area. Let, for instance, the larger area = 1156, and that of the smaller, to which the figure g is to be reduced, = 529. Draw an indefinite line, on which make AB =34, the square root of 1156. Lastly, from the point A, as a centre, having described an indefinite arc, with a length equal to the square root 23 of 529, set out Bg ; through g draw Ag, which will be the angle of reduction g AB, by means of which the figure may be reduced, transferring all the mea- b sures of the larger area to the line AD, with which arcs are to be described whose chords will be the sides sought. 1523. If it be not required to reduce but to describe a figure whose area and form are given, we must make a large diagram of any area larger than that sought, and then reduce it. 1524. The circle, as we have already observed in a previous subsection (93.3.), being but a polygon of an infinite number of sides, it would follow that a circular enclosure would be stable with an infinitely small thickness of wall. This property may be easily demonstrated by a very simple experiment. Take, for instance, a sheet of paper, which would not easily be made to stand while extended to its full length, but the moment it is bent into the form of a cylinder it acquires a stability, though its thickness be not a thousandth part of its height. §3. But as walls must have a certain thickness to acquire stability, inasmuch as they are composed of particles susceptible of separation, we may consider a circular enclosure as a regular polygon of twelve sides, and determine its thickness by the preceding process. Or, to render the operation more simple, find the thickness of a straight wall whose length is equal to one half the radius. 1526. Suppose, for example, a circular space of 56 ft. diameter and 18 ft. high, and the thickness of the wall be required. Describe the rectangle ABCD (fig. 594.), whose base is equal to half the radius, that is, 14 ft., and whose height AB is 18 ft. ; then, drawing the diagonal BD, make Ba equal to the ninth part of the height, that is, 2 ft. Through d draw ad parallel to the base, and its length will represent the thickness sought, which is 143 inches. 1527. By calculation. Add the square of the height to that of half the radius, that is, of 18 = 324, and of 14-196=520. Then extract the square root of 520, which will be found = 22:8, and this will be the value of the diagonal BD. Then we have the following proportion: 22:8: 14 :: 2 ft. ( the height): 14-74. 1528. The exterior wall of the church of St. Stefano Rotondo at Rome (Temple of Claudius) incloses a site 198 feet diameter. The wall, which is contructed of rubble masonry faced with bricks, is 2 ft. 4 in. (French) thick, and 22 ft. high. In applying to it the preceding rule, we shall find the diagonal of the rectangle, whose base would be the side of a polygon, equal to half the radius and 22 ft. high, would be w/49, X 49,422, x 22}=54%. Then, using the proportion 54:37:49:5 :: * : 2 ft. 3 in. and 4 lines, the thickness sought, instead of 2 ft. 4 in., the actual thickness. We may as well mention in this place that a circle encloses the greatest quantity of area with the least quantity of walling; and of polygons, those with a greater number of sides more than those with a lesser: the proportion of the wall in the circle being 31416 to an area of 78540000; whilst in a square, for the same area, a length of wall equal to 35448 would be required. . As the square falls away to a flat parallelogram, say one whose sides are half as great, and the others double the length of those of the square, or 17724 by 4431, in which the area will be about 78540000, as before; we have in such a case a length of walling = 44310. Fig. 603. On the Thickness of Walls in Buildings not vaulted. 1529. The walls of a building are usually connected and stiffened by the timbers of the roof, supposing that to be well constructed. Some of the larger edifices, such as the ancient basilica at Rome, have no other covering but the roof; others have only a simple ceiling under the roof; whereas, in palaces and other habitations, there are sometimes two or more floors introduced in the roof. 1530. We will begin with those edifices covered with merely a roof of carpentry, which are, after mere walls of enclosure, the most simple. 1531. Among edifices of this species, there are some with continued points of support, such as those wherein the walls are connected and mutually support each other; others in which the points of support are not connected with each other, such as piers, columns, and pilasters, united only by arcades which spring from them. 1532. When the carpentry forming the roof of an edifice is of great extent, instead of being injurious to the stability of the walls or points of support, it is useful in keeping them together. 1533. Many edifices exist wherein the walls and points of support would not stand without the aid of the carpentry of the roofs that cover them. 1534. The old basilica of St. Paolo fuori le murd at Rome was divided into five naves formed by four ranks of columns connected by arcades, which carried the walls whereon the roof rested; the centre nave 734 ft. (French) wide, and 93 ft. 10 in. high. The walls of it are erected on columns 31 ft. 9 in high, and their thickness is a little less than 3 ft., that is, only 3, part of their height. 1535. At Hadrian's Villa the most lofty walls, still standing, were but sixteen times their thickness in height, and 51 ft. 9 in. long. The walls were the enclosures of large halls with only a single story, but assisted at their ends by cross walls. And we may therefore conclude that if the walls of the basilica above mentioned were not kept in their places by the carpentry of the great roof they would not be safe. It is curious that this supposition, under the theory, was proved by the fire which destroyed the church of St. Paolo in 1823. The walls which form the nave of the church of Santa Sabina are raised on columns altogether 52 ft. high ; they are 145 ft. long, and somewhat less than 2 ft., that is, a part of their height, in thickness. They are, therefore, not in a condition of stability without the aid of the roof. In comparing, however, the thickness of these walls with the height only of the side aisles, in the basilica of St. Paolo the thickness is #, and at Santa Sabina * In the other basilicae or churches with columns, the least thickness of wall is l, of greater proportion unconnected with the nave, as at Santa Maria Maggiore, Santa Maria in Trastevere, St. Chrysogono, St. Pietro in Vincolà, in Rome; St. Lorenzo and St. Spirito, in Florence; St. Filippo Neri, at Naples; St. Giuseppe and St. Dominico, at Palermo. 1536. We must take into account, moreover, that the thickness of walls depends as much on the manner in which they are constructed, as on their height and the weight with which they are loaded. A wall of rough or squared stone 12 inches thick, wherein all the stones run right through the walls in one piece, is sometimes stronger than one of 18 or 20 inches in thickness, in which the depth of the stones is not more than half or a third of the thickness, and the inner part filled in with rubble in a bad careless way. We are also to recollect that stability more than strength is ofttimes the safeguard of a building; for it is certain that a wall of hard stone 4 inches thick would be stronger than would be necessary to bear a load equal to four or five stories, where a thickness of 18 inches is used; and yet it is manifest that such a wall would be very unstable, because of the narrowness of the base. 1537. From an examination which Rondelet made of 280 buildings in France and Italy, ancient as well as modern, he found that in those covered with roofs of two inclined sides and constructed in framed carpentry, with and without ceilings, and so trussed as not to act at all horizontally upon the walls, the least thickness in brick or rough stones was 2, of the width in the clear. 1538. In private houses, divided into several stories by floors, it was observed, generally, that the exterior walls ran from 15 to 24 inches, party walls 16 to 20 inches, and partition walls 12 to 18 inches. 1539. In buildings on a larger scale, exterior walls 2 to 3 feet thick, party walls 20 to 24 inches, partition walls 15 to 20 inches. 1540. In palaces and buildings of great importance, whose ground floors are vaulted, the exterior walls varied from 4 to 9 feet, and the partition walls from 2 to 6 feet. In many of the examples which underwent examination, the thicknesses of the walls and points of support were not always well proportioned to their position, to the space they enclosed, nor to the loads they bore. In some, great voids occur, and considerable loads were supplied with but slender walls and points of support; and in others, very thick walls enclosed very small spaces, and strong points of support had but little to carry. 1541. For the purpose of establishing some method which in a sure and simple manner would determine the thickness of walls in buildings which are not arched, we have considered, says Rondelet, that the tie-beams of the trusses of carpentry whereof the roofs are composed, being always placed in the direction of the width, as well as the girders and leading timbers of floors, serve rather to steady and connect the opposite walls; but, considering the elasticity and flexibility of timber, it is found that they strain the walls which support them in proportion to the widths of the spaces enclosed, whence it becomes often the better plan to determine the thickness of the walls from the width and height of the apartments requisite. Hence the following rules. First Rule. 1542. In buildings covered with a simple roof, if the walls are insulated throughout, their height up to the under side of the tie-beams of the trusses, being as shown in fig. 604. Having drawn the diagonal BD and thereon made Bb and Dd, equal to the twelfth part of the height AB, then through the points b and d, draw lines parallel to BA and DC, which will bound the thickness of the walls required. 1543. If the height AB and width AD be known, the thickness Ac may be calculated, |