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the parabola ; and from C" to G, m, i on one side, and on the oth or to G, m, i, for tl byperbola. Each of these is represented in figs. 506. and 507.
DEVELOPEMENT or BODIES OR SOLIDS WHOSE SURFACES HAVE A DOUBLE CURVATURL 1184. The developement of the sphere and other bodies whose surface has a doub flexure would be impossible, unless we considered them as consisting of a great number plane faces or of simple curvatures, as the cylinder and the cone. Thus a sphere or sphero may be considered, — I. As a polyhedron of a great number of plane faces formed t truncated pyramids whose base is a polygon, as fig. 508. II. By truncated cones, formir 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 i order to obtain the summits P, q, r, D of the truncated pyramids which form the semi-poly hedron ADB; then from the points P, q, r, with the radii PA, P1, 91, 92, r2, 13 and D: describe the indefinite arcs AB', 16', 16", 2f", 2f", 3g', and Sg, upon which, after havin transferred the divisions of the demi-polygons AEB, 1e6'", 2e'5'", 3e", 4", from all th transferred points, as A, 4, 5', 6', 7', 8', '9', B , for each truncated pyramid draw lines to th summits Per D, and other lines which will form inscribed polygons in each of the arcs AB 16', 16", &c. These lines will represent for each band or zone the faces of the truncate pyramids whereof they are part.
1188. We may arrive at the same developement by raising upon the middle of each sid of the polygon A E B indefinite perpendiculars, upon which must be laid the height of th faces of the elevation in 1, 2, 3, 4; through which points draw parallels to the base, upo which transfer the widths of each of the faces taken on the plan, whereby trapezia will b formed, and triangles similar to those found in the first developement, but ranged in anothe manner. This last developement, which is called in gores, is more suitable for geographica globes; the other method, for the formation of the centres, moulds, and the like, of spherica vaults.
1189. The developement of the sphere by conic zones ( fig. 509.) is obtained by tl same process as that by truncated pyramids, the only difference being, that the develop ment of the arrisses A B', 16', 28", 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 gore (fig. 510.) is conducted in the second manner, but instead of joining with lines the point 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 ol engles, 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, wliereof we are about to speak, are those formed by the meeting of two surfaces joining in an edge.
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.
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 equs to GH, and consequently the angle IFK so much smaller than GFH as it is more distant.
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 B to E, determined by an arc made from the point D with a radius equal to BA. The angle sought will be found to be 108 degrees.
1199. For the icosahedron (fig. 482.), draw the parallels Aa, Bb, Cc, and after having made bc 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, PCD are 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
from the point C with a radius equal to the diagonal AD (of the square representi the square of the base) describe arcs intersecting each other. The angle FDG will 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'o' the right pyramid mpo, whose developement is shown in fig. 490., by the portion of t polygon a, b", c', e", d", d'F.
1203. By means of this base and the part developed, proceeding as we have already e plained for the right pyramid, we shall find the angles formed by the meeting of the face and they will differ but little from those of the little polygon oqmo'o'.
1204. In respect to the angles formed by the faces inclined to the base, that of the fa answering to the side De of the base is expressed by the angle ADP of the vertical proje tion, fig. 489.
1205. As to the other faces, for instance, that which corresponds to the side AE of t1 base (fig. 490.), through any point g draw of perpendicular to it, meeting the line AF, show ing the projection of one of the sides of the inclined face; upon the developement of th face, expressed by A"EʻF, raise at the same distance from the point E' another perpe dicular g'm', which will give the prolongation of the line shown on the base by Af. If transfer A"m of the developement upon Am, which expresses the inclination of the arr represented by this line, we shall have the perpendicular height mf of the point m above th base, which, being transferred from fm" upon a perpendicular to gf, we shall have the tw sides of a triangle whose hypothenuse gm' will give m"gf, the angle sought.
1206. In the oblique prism (fig. 491.), the angles of the faces are indicated by the plar 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 an 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 CcDd an CcAb, draw the perpendiculars csbt, whose projection in plan are indicated by s'd' and b' then upon fc, drawn aside, raise a perpendicular c"c"' equal to cs of the profile, fig. 49 Through the point c" draw a line parallel to fc, upon which, having transferred c's' of th projection in plan (fig. 492.), draw the hypothenuse s"c", and it will give the angle s'é formed by the face Ce Dd with the inferior base.
1209. To obtain the angles of the face CcAb, raise upon Fl", drawn on one side, a per pendicular 6"t", equal to bt (fig. 492.), and drawing as before a parallel to 6" through th point t", transfer bt of fig. 492. to t"'t"; and drawing t'b", the angle t""F is tha required.
1210. As the bases of this prism are parallel, these faces necessarily form the same angle with the superior base.
1211. An acquaintance with the angles of planes is of the greatest utility in the prepare tion of stone, as will be seen in chap. iii., and a thorough acquaintance with it will well repa the architectural student for the labour he may bestow on the subject.
1212. The area of a plane figure is the measure of its surface or of the space containe within its extremities or boundaries, without regard to thickness. This area, or the conten of the plane figure, is estimated by the number of small squares it contains, the sides o each whereof may be an inch, a foot, a yard, or any other fixed quantity. Hence the are 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 squar
1214. PROBLEM I. To find the area of a parallelogram, whether it
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 14 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. 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 194-25
= 21:584 yards. 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'. By natural numbers :
First, 1 x 40 x 30=600.
Then, 1 : 600:: -484046 (sin. 28° 57') : 290:4276.
Sin. 28° 57'=9.684887
2.463038 = 290-4276, as above. Rule 3. When the three sides are given, take half the sum of the three sides added to
gether. 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.
20 + 30+40=90, whose half sum is 45.
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 trapezrid. Add together the parallel sides, multiply their sum by the perpendicular breadth or dis
tance 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.
1225 + 750 x 770=1520750, the area. Example 2. Required the area of any quadrangular figure (fig. 513.) wherein AP is 110 feet,
AQ 745 feet,
DQ 595 feet.
For PCDQ, 595 +952 x 635+2=300672.5
365 x 595
Area=428620-0 feet. 1217. PROBLEM IV. To find the area of any trapezium. Divide the trapezium into two triangles by a diagonal; then find the arcas of the
triangles, and their sum is the area. Observation. If two perpendiculars be let fall on the diagonal from the other two oppe angles, then add these two perpendiculars together, and multiply that sum by the diago Half the product is the area of the trapezium. Example. Required the area o a trapezium whose diagonal is 42, and the two
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. I
having found the areas of all these separately, their sum will be the conten
quired of the whole polygon.
figure ABCDEFGA (fig. 514.), wherein the
AC=55, GC=44, Bn=18, Ep=8,
55 x 6.5 =357-5
lar drawn from its centre on one of its sides, and take half
gon into so many triangles.
(fig. 515. ), whose side AB or BC, &c. is 25 ft., and
perpendicular OP 17.2 ft.
square feet area required.
Length of Length Length of
of Rad. Radius.
Rad = 1 Apt. = 1 Side = 1
30 120 •5000 1.732051 2.0090 5773503 0-43301 45
90 •7071 1.414214 1:4142 7071068 1.00000 54
72 •8090 1.175570 1-2960 8506508 1.72047 60
60 .8660 1.000000 1.1547 1.00000001 2.59807 6451.25.9010 867767 1.1095 1.1523824 3.63391 673 45 -9239 -765367 1.0823 1.3065628 4.82842 70
40 -9397 684040 1•0642 1.4619029 6.181824 72
36 -9511 618034 1.0515 1.6180340 7694205 7:11 32:43 1 9595 •563465 1.0422 1.7747324 9-365635 75
30 9659 •517638 1.03521.9318517 11.196155