the parabola; and from C" to G, m, i on one side, and on the other to G, m, i, for the hyperbola. Each of these is represented in figs. 506. and 507. DEVELOPEMENT OF BODIES OR SOLIDS WHOSE SURFACES HAVE A DOUBLE CURVATURE. 1184. The developement of the sphere and other bodies whose surface has a double flexure would be impossible, unless we considered them as consisting of a great number of plane faces or of simple curvatures, as the cylinder and the cone. Thus a sphere or spheroid may be considered, I. As a polyhedron of a great number of plane faces formed by truncated pyramids whose base is a polygon, as fig. 508. II. By truncated cones, forming zones, as in fig. 509. III. By parts of cylinders cut in gores, forming flat sides that diminish in width, shown by fig. 510. 1185. In reducing the sphere or spheroid to a polyhedron with flat faces, the developement may be accomplished in two ways, which differ only by the manner in which the faces are developed. 1186. The most simple method of dividing the sphere 10 reduce it to a polyhedron is that of parallel circles crossed by others perpendicular to them, and intersecting in two opposite points, as in the common geographical globes. If, instead of the circle, the polygons are supposed of the same number of sides, a polyhedron A will be the result, similar to that represented by fig. 508., whose half ADB shows the geometrical elevation, and AEB the plan of it. 1187. For the developement, produce A1, 12, 23, so as to meet the produced axis CP in order to obtain the summits P, q, r, D of the truncated pyramids which form the semi-polyhedron ADB; then from the points P, q, r, with the radii PA, Pl, ql, q2, r2, r3 and Ds, describe the indefinite arcs AB', 1b', 1b", 2f', 2f", 3g', and Sg, upon which, after having transferred the divisions of the demi-polygons AEB, 1e6"", 2e'5"", Se", 4", from all the transferred points, as A, 4', 5', 6', 7, 8, 9, B', for each truncated pyramid draw lines to the summits PqrD, and other lines which will form inscribed polygons in each of the arcs A B', 1b', 1b", &c. These lines will represent for each band or zone the faces of the truncated pyramids whereof they are part. 1188. We may arrive at the same developement by raising upon the middle of each side of the polygon AEB indefinite perpendiculars, upon which must be laid the height of the faces of the elevation in 1, 2, 3, 4; through which points draw parallels to the base, upon which transfer the widths of each of the faces taken on the plan, whereby trapezia will be formed, and triangles similar to those found in the first developement, but ranged in another manner. This last developement, which is called in gores, is more suitable for geographical globes; the other method, for the formation of the centres, moulds, and the like, of spherical vaults. 1189. The developement of the sphere by conic zones (fig. 509.) is obtained by th same process as that by truncated pyramids, the only difference being, that the develope ment of the arrisses AB', 16', 2f', 3g, are arcs of circles described from the summits cones, instead of being polygons. 1190. The developement of the sphere reduced to portions of cylinders cut in gores fig. 510.) is conducted in the second manner, but instead of joining with lines the points h. i, k, d, (fig. 508.) they must be united by a curve. This last method is useful in Irawing the caissons or pannels in spherical or spheroidal vaults. OF THE ANGLES OF PLANES OR SURFACES BY WHICH SOLIDS ARE BOUNDED. 1191. In considering the formation of solids, we have already noticed three sorts o angles, viz. plane angles, solid angles, and the angles of planes. The two first have beer treated of in the preceding sections, and we have now to speak of the third, which must not be confounded with plane angles. Of these last, we have explained that they are formed by the lines or arrisses which bound the faces of a solid; but the angles of planes, whereof we are about to speak, are those formed by the meeting of two surfaces joining in an edge. D 1192. The inclination of one plane ALDE to another ALCB (fig. 511.) is measured by two perpendiculars FG, FH raised upon each of these planes from the same point F of the line or arris AL formed by their union. A L H B C 1193. It is to be observed, that this angle is the greatest of all those formed by lines drawn from the point F upon these two planes; for the lines FG, FH being perpendicular to AL, common to both the planes, they will be the shortest that can be drawn from the point F to the sides ED, BC, which we suppose parallel to AL; thus their distance GH will be throughout the same, whilst the lines FI, FK will be so much the longer as they extend beyond the perpendiculars FG, FH, and we shall always have KI equá to GH, and consequently the angle IFK so much smaller than GFH as it is more distant. Fig. 511. 1194. Thus, let a rectangular surface be folded perpendicularly to one of its sides so that the contours of the parts separated by the fold may fall exactly on each other. If we raise one of them, so as to move it on the fold as on a hinge, and so as to make it form all degrees of angles, we shall see that each of the central extremities of the moveable part is always in a plane perpendicular to the part that is fixed. 1195. This property of lines moving in a perpendicular plane, furnishes a simple method of finding the angles of planes of all sorts of solids whose vertical and horizontal projec tions or whose developements are known. 1196. Thus, in order to find the angles formed by the tetrahedron or pyramid on a tri angular base (fig. 477.), we must for the angles of the base with the sides, let fall from the angles ABC perpendiculars to the sides ac, cb, and ab, which meet at the centre of the base in D. It is manifest from what has just been said on this subject, that if the three triangles are made to move, their angles at the summit A, B, C will not be the vertical planes shown by the lines AD, DB, DC, and that they will meet at the extremity of the vertical, passing through the intersection of these planes at the point D. Thus we obtain for each side a rectangular triangle, wherein two sides are known, namely, for the side cb, the hypothenuse ed, and the side eD. Thus raising from the point D an indefinite perpendicular, if from the point e with eB for a radius an arc is described cutting the perpendicular in d, and the line de be drawn, the angle de D will be that sought, and will be the same for the three sides if the polyhedron be regular; otherwise, if it is not, the operation must be repeated for each. 1197. These angles may be obtained with great accuracy by taking de, or its equal e B, for the whole sine; then de: eD::sine: sine 19° 28', whose complement 70 32' will, if the polyhedron be regular, be the angle sought. In this case, all the sides being equal, and each being capable of serving as base, the angles throughout are equal. In respect of the cube (figs. 479. and 483.) whose faces are composed of equal squares, and whose angles are all right angles, it is evident that no other angles can enter into their combination with each other. 1198. To obtain the angle formed by the faces of the octahedron (fig. 480.) from the points C and D with a distance equal to a vertical dropped upon the base of one of the triangles of its developement (fig. 484.), describe arcs crossing each other in F; and the angle CFD will be equal to that formed by the faces of the polyhedron, and will be found by trigonometry to be 70° 32'. In the dodecahedron (fig. 481.), the angle formed by the faces will be found by drawing upon its projection the lines DA, and producing the side to E, determined by an are made from the point D with a radius equal to BA. angle sought will be found to be 108 degrees. The 1199. For the icosahedron (fig. 482.), draw the parallels Aa, Bb, Cc, and after having made be parallel and equal to BC, with a radius equal thereto, describe an arc cutting in a the parallel drawn from the point A; the angle abc will be equal to that formed by the sides of the polygon, which by trigonometry is found to be 108 degrees, as in the dodeca bedron. 1200. For the pyramid with a quadrangular base (fig. 487.) the angle of each face with the base is equal to PAB or PBA, because this figure, which represents its vertical projection, is in a plane parallel to that within which will be found the perpendiculars dropped from the summit on the lateral faces of the base. 1201. In order to obtain the angles which the inclined sides form with one another, draw upon the developement (fig. 488.) the line ED, which, because the triangles PEC, equal and isoceles, will be perpendicular to the line PC, representing one of the arrisses which are formed. Then from the point D with a radius equal to DF, and PCD are from the point C with a radius equal to the diagonal AD (of the square representing the square of the base) describe arcs intersecting each other. The angle FDG will b the angle sought. We may suppose it taken along the line BC traced in fig. 487. 1202. In order to obtain the angles formed by the faces of an oblique pyramid (fig. 489.) through some point q of the axis draw the perpendicular mo, showing the base oqmq'oʻo the right pyramid mpo, whose developement is shown in fig. 490., by the portion of the polygon a, b, c", e", d", a'F. 1203. By means of this base and the part developed, proceeding as we have already ex plained for the right pyramid, we shall find the angles formed by the meeting of the faces and they will differ but little from those of the little polygon oqmq'o'. 1204. In respect to the angles formed by the faces inclined to the base, that of the face answering to the side De of the base is expressed by the angle ADP of the vertical projec tion, fig. 489. 1205. As to the other faces, for instance, that which corresponds to the side AE of the base (fig. 490.), through any point g draw of perpendicular to it, meeting the line AF, showing the projection of one of the sides of the inclined face; upon the developement of this face, expressed by A′′E F, raise at the same distance from the point E' another perpendicular g'm', which will give the prolongation of the line shown on the base by Af. If we transfer A"m of the developement upon Am, which expresses the inclination of the arris represented by this line, we shall have the perpendicular height mf of the point m above the base, which, being transferred from fm" upon a perpendicular to gf, we shall have the two sides of a triangle whose hypothenuse gm" will give m'of, the angle sought. 1206. In the oblique prism (fig. 491.), the angles of the faces are indicated by the plane of the section perpendicular to the axis, represented by the polygon hiklmn. 1207. Those of the sides perpendicular to the plan of the inclination of the axis are expressed by the angles Ddb, Abd of the profile in the figure last named. 1208. In order to obtain the angles formed with the other sides, for instance Cc Dd and Cc Ab, draw the perpendiculars csbt, whose projection in plan are indicated by s'c' and b't', then upon fc, drawn aside, raise a perpendicular c"c" equal to cs of the profile, fig. 491. Through the point c"" draw a line parallel to fc, upon which, having transferred c's' of the projection in plan (fig. 492.), draw the hypothenuse s'c", and it will give the angle s'e"ƒ formed by the face Cc Dd with the inferior base. 1209. To obtain the angles of the face Cc Ab, raise upon Fb”, drawn on one side, a perpendicular b't''', equal to bt (fig. 492.), and drawing as before a parallel to b" through the point t", transfer bit of fig. 492. to tt"; and drawing t"b", the angle t'b"F is that required. 1210. As the bases of this prism are parallel, these faces necessarily form the same angles with the superior base. 1211. An acquaintance with the angles of planes is of the greatest utility in the preparation of stone, as will be seen in chap. iii., and a thorough acquaintance with it will well repay the architectural student for the labour he may bestow on the subject. SECT. VI. MENSURATION. 1212. The area of a plane figure is the measure of its surface or of the space contained within its extremities or boundaries, without regard to thickness. This area, or the content of the plane figure, is estimated by the number of small squares it contains, the sides of each whereof may be an inch, a foot, a yard, or any other fixed quantity. Hence the area is said to consist of so many square inches, feet, yards, &c., as the case may be. 1213. Thus if the rectangle to be measured be ABCD (fig. 512.), and the small square E, whose side we will suppose to be one inch, be the measuring D 1214. PROBLEM I. To find the area of a parallelogram, whether it be a square, a rectangle, a rhombus, or a rhomboid. Multiply the length by the perpendicular breadth or height, and the product will be the area. Example 1. Required the area of a parallelogram whose length is 12-25 feet, and height 8.5 feet. 12.25 x 8.5=104-125 feet, or 104 feet 11 inches. Example 2. Required the content of a piece of land in the form of a rhombus whose length is 6:20 chains, and perpendicular height 5:45. Recollecting that 10 square chains are equal to a square acre, we have, 6-20 x 5·45=33.79 and =3.379 acres, which are equal to 3 acres, 1 rood, 20, perches. 33.79 Example 3. Required the number of square yards in a rhomboid whose length is 57 feet, and breadth 5 feet 3 inches (= 5.25 feet). Recollecting that 9 square feet are equal to 1 square yard, then we have 37 × 5-25=194-25, and 19425-21-584 yards. 9 1215. PROBLEM II. To find the area of a triangle. Rule 1. Multiply the base by the perpendicular height, and take half the product for the area. Or multiply either of these dimensions by half the other. The truth of this rule is evident, because all triangles are equal to one half of a parallelogram of equal base and altitude. (See Geometry, 904.) Example 1. To find the area of a triangle whose base is 625 feet, and its perpendicular height 520 feet. Here, 625 x 260=162500 feet, the area of the triangle. Rule 2. When two sides and their contained angle are given: multiply the two given sides together, and take half their product; then say, as radius is to the sine of the given angle, so is half that product to the area of the triangle. Or multiply that half product by the natural sine of the said angle for the area. This rule is founded on proofs which will be found in Sect. III., on which it is unnecessary here to say more. Example. Required the area of a triangle whose sides are 30 and 40 feet respectively, and their contained angles 28° 57'. Rule 3. When the three sides are given, take half the sum of the three sides added together. Then subtract each side severally from such half sum, by which three remainders will be obtained. Multiply such half sum and the three remainders together, and extract the square root of the last product, which is the area of the triangle. This rule is founded on one of the theorems in Trigonometry to be found in the section relating to that subject. Example. Required the area of a triangle whose three sides are 20, 30, and 40. 20+30 +40=90, whose half sum is 45. 45-20=25, first remainder; 45-SO=15, second remainder; 45-40=5, third remainder. Then, 45 x 25 x 15 x 5=84375, whose root is 290-4737, area required. 1216. PROBLEM III. To find the area of a trapezoid. Add together the parallel sides, multiply their sum by the perpendicular breadth or distance between them, and half the product is the area. (See Geometry, 932.) Example 1. Required the area of a trapezoid whose parallel sides are 750 and 1225, and their vertical distance from each other 1540. 1217. PROBLEM IV. To find the area of any trapezium. Divide the trapezium into two triangles by a diagonal; then find the areas of the t triangles, and their sum is the area. Observation. If two perpendiculars be let fall on the diagonal from the other two oppos ingles, then add these two perpendiculars together, and multiply that sum by the diagon Half the product is the area of the trapezium. Example. Required the area o a trapezium whose diagonal is 42, and the two p pendiculars or it 16 and 18. Here, 16+18=34, whose half- 17; Then, 42 x 17-714, the area. 1218. PROBLEM V. To find the area of an irregular polygon. Draw diagonals dividing the proposed polygon into trapezia and triangles. Th having found the areas of all these separately, their sum will be the content quired of the whole polygon. Example. Required the content of the irregular figure ABCDEFGA (fig. 514.), wherein the given. AC=55, GC-44, Bn-18, Ep=8, 55 x 6.5 =357-5 44 x 11.5=506 B 1219. PROBLEM VI. Fig. 514. To find the area of a regular polygon. Rule 1. Multiply the perimeter of the polygon, or sum of its sides, by the perpendic lar drawn from its centre on one of its sides, and take half the product of the area; which is in fact resolving the poly- Example. Required the area of the regular pentagon ABCDE 25x5 2 Here =62.5= half the perimeter, and 62.5 x 17.2=1075 square feet area required. Fig. 515. Rule 2. Square the side of the polygon, and multiply the square by the tabular area or multiplier set against its name in the following table, and the product will be the area. This rule is founded on the property, that like polygons, being similar figures, are to o another as the squares of their like sides. Now the multipliers of the table the respective areas of the respective polygons to a side =1; whence the rule evident. In the table, the apothem of a regular polygon is the line O P, |