spruce, EFGH (fig. 595.). Draw the diagonal BD, and about B make Bd equal to one righth 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, baring 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 marie 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 B В Fig.CO. Fig. 6C1. Fig. 602. enelosing 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.). 1516. 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 ottained by the simple rule of three ; for, knowing the three sides of the triai g e ABD. similar to the smaller triangle Bde, we have BD : Bd::AD: ed. Thus, suppose th length of wall represented by AD=28 feet, and its height AB=12 feet, we shall have th 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 mar be referred to in the section “ Trigonometry,” and from thence here applied. P. 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 angle into as many triangles as it has sides. In fig. 601., on one of these triangles let fall fron C(which is the vertex of each triangle) a perpendicular CD on the base or side A B whic is supposed horizontal. The area of this triangle is equal to the product of DB (hall AB by CD, or to the rectangle DCFB. Making DB=x, CD=y, and the arva given = P, w all have, For the equilateral triangle, xxy 3=p, or xy = For the hexagon, xy x6 =p, or ry= 1519. In the equilateral triangle this proportion is as the sine of 60 degrees to the sine o 30 degrees; that is, using a table of sines, as 86603 : 50000, or 83 :5, or 26 : 15, wheno r:y::26 : 15, and 15x=26y; whence y= 26p '=45.6, and the side AB=91.2. 80902.r 58799 and x= V 58779 x 720 58779 which makes r= 92.87, and the side AB =45-74. For the hexagon, z: y::sin. 30° : sin. 60°, or as 50000 : 86603::5: 83, whence the valu This value, substituted in the equation ry= =600; when 600 x 15 ; lastly, therefore, r=1346·15=18.61, and the side AB=37•22. 152 26 we have 1572 P 3' and I= 26p 45 45 45 2 3600 80902 : of y= 263 will give 15 26 Geometrically. 1520. Suppose the case that of a pentagon (fig. 601.) one of whose equal triangles i: ACB. Let fall the perpendicular CD, which divides it into two equal parts; whence it: 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 g draw parallels to GE and GB, which will determine on AB the points e and h, and 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 circuin ference in which it is inscribed. This is manifest because of the similar triangles EGI and egb, from which BD: DE :: 6D: De. 1522. From the truth that the sides of similar figures are to each other as the squarı 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 rool of the greater area, and the chord of the arc, which determines the size of the angle equa B Fig. ses. to the square root of the smaller area. Let, for instance, the larger area =1156, and that of the smaller, to which the figure 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 leagth equal to the square root 23 of 529, set out Bg; through g draw Ag, which will be the angle of reduction gAB, by means of wbich the figure may be reduced, transferring all the measures 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 (933.), being but a polygon of an infinite number of sides, it would follow that a circular enclosure would be stable with an infnitely 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. 1525. 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 wali 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 Bd 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 144 inches. 1527. By calculation. Add the square of the height to that of half the radius, that is, f 18 =324, and of 14=196 (=520). Then extract the square root of 520, which will be found =22-8, and this will lie the value of the diagonal BD. Then we have the follow. ing 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 224 ft. high. In applying to it the preceding rule, we shall find the diagonal of the rectangle, whose base sould be the side of a polygon, equal to half the radius and 224 ft. high, would be 7494 x 494 +221 x 22}=543%. Then, using the proportion 54:37 : 49.5 :: : 2A. 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 tbe area will be about 78540000, as before; we have in such a case a length of valling = 44310. 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 toof, supposing that to be well constructed. Some of the larger edifices, such as the ancient basilicæ 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 of 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 theiu together. D D |