parallel to the base A B C. Vide Fig. 21, Plate 2, HEIGHTS, DISTANCES, and PRACTICAL GEOMETRY. A Frustrum, or Trunk, is the part ABCDEF, that remains at the bottom after the segment is cut off. A Cone is a round pyramid, of which the base is a circle. The Axis of a Solid is a line from the vertex (or point) to the centre of the base, or through the centres of the two ends. When the axis is perpendicular to the base, it is a Right prism, pyramid or cone; otherwise it is Oblique. A Sphere is a solid contained under one convex surface, and is described by the revolution of a semicircle about its diameter, which remains fixed. The Centre of the Sphere is such a point within the solid, as is every where equally distant from the convex surface, or circumference of it. The Diameter (or Axis) of a Sphere is a straight line, which passes through the centre, and is terminated by the convex surface. A Segment of a Sphere is a part cut off by a plane, the section of which is always a circle, called the Base of the Segment. A Sector of a Sphere is that which is composed of a segment, less than an hemisphere, and of a cone. A Prism is a solid, the sides of which are parallelograms, and its ends equal, and similar plane figures. Prisms are named according to the number of angles in the base. A Cylinder is a solid, the two ends of which are circular; and it is described, or formed, by the revolution of a right angled parallelogram about one of its sides, which remains fixed. To find the Superficies of a Prism, or Cylinder. Multiply the perimeter of one end of the prism by the length, or height of the solid, and the product will be the surface of all its sides. To which add also the area of the two ends of the prism, when required. Or, compute the Areas of all the sides and ends separately, and add them altogether. Example.-Required the surface of a Cube, whose sides are each 5 inches. 5 +5 +5 +5 = 20 perimeter of one end. 20 X 5100 surface of sides. 5X5 25 area of one end. 100+25+25 150 S. Inches. Surface of Cube. To find the Surface of a Pyramid, or Cone. Multiply the perimeter of the base by the slant height, or length of the side, and half the product will be the Surface of the sides; to which add the area of the base when required. Example.-Required the upright surface of a Triangular Pyramid, the slant height being 20 feet, and each side of the base 3 feet. 3 +3 +39 perimeter of base. To find the Surface of the Frustrum of a Pyramid, or Cone. Add together the perimeters of the two ends, and multiply their sum by the slant height, and take half the product. Example.-How many square feet are in the surface of the Frustrum of a Square Pyramid, whose slant height is 10 feet, each side of the base 3 feet, and each side of the less end 2 feet. 3 +3 +3 +3 = 12 perimeter of base. 12+ 8 X 10 2 100 Feet. Surface required. To find the solid Content of a Prism, or Cylinder. Find the area of the base, or end, aud multiply it by the length of the Prism, or Cylinder. For a Cube, multiply its side twice by itself; and for a Parallelopipedon, multiply the length, breadth, and depth together for the Content. H H Example.-Required the solid content of a Cube, whose side is 24 inches. 24 X 24 X 24 13824 S. Inches. Content required. To find the Content of the solid part of a hollow Cylinder. From the content of the whole cylinder considered as a solid, subtract the content of the hollow part also considered as a solid, and the difference will be the Solidity required. Example.-Required the content of the solid part of the hollow Cylinder whose exterior diameter is 12 inches, the interior diameter 8 inches, and height 20 inches. 12 X 12 X 7854113.0976 Area of base of cylinder. The Solidity required. To find the Solidity of the Frustrum of a Cylinder. Multiply the area of the base by half the greatest, and the least lengths, and the product will be the solidity. Example.-Required the solidity of a Frustrum, whose diameter is 24 inches, the greatest length 36 inches, and the least length 20 inches. 24 X 24576. Square of the diameter. 36+ 20 452.3904 X 2 Area of the base. =12666-9312 Cubic inches. Solidity required. To find the Content of a Pyramid, or Cone. Find the area of the base, and multiply that area by the perpendicular height, and take of the product. Example.-Required the solidity of a Square Pyramid, each side of its base being 30, and its perpendicular height 25. 30 X 30 900 Area of base. To find the Solidity of the Frustrum of a Cone, or Pyramid. Add into one sum, the areas of the two ends, and the mean proportional between them: take of that sum for the mean area, which multiply by the perpendicular height, or length of the frustrum. Note.-To find a mean Proportional. As one of the sides of the base is to the homologous, or corresponding side of the other end; so is the area of the base, to the mean proportional required. Example.-Required the number of solid feet in a piece of Timber, whose bases are squares, each side of the greater end being 15 inches, and each side of the less end 6 inches; also the length of the perpendicular altitude 24 feet. 15 X 15225 Area of the base. 6 X 6 36 Area of the top. As 15: 6:225: 90 mean proportional. 24 feet 288 inches. 2253690 X 288 33696 Cubic inches. 3 19 Cubic feet. To find the Surface of a Sphere, or any Segment. Multiply the circumference of the Sphere by its diameter, which will give the whole Surface. Or, Square the diameter, and multiply by 3.1416. Or, Square the circumference, and multiply by 3183, or divide by 3.1416. Note. For the surface of the Segment, or Frustrum, multiply the whole circumference of the Sphere by the height of the part required. Example. - Required the superficies of a Globe, whose diameter is 24 inches. 24 X 24 X 3.14161809.5616 S. Inches. To find the Solidity of a Sphere, or Globe. 1.-Multiply the surface by the diameter, and take of the product. Or multiply the square of the diameter by the circumference, and take of the product. 2. Cube the diameter and multiply by 5236. 3. Cube the circumference, and multiply by 01688. Example.-Required the content of a Sphere, whose axis is 12. 12 X 12 X 12 X 5236 904·7808. Content required. To find the Solidity of a Hemisphere. Find the solidity of the sphere, and half the content will be that of the Hemisphere. Note 1.-Any Sphere or Globe twice the diameter of another contains four times the Superficies, or Area, of the other; and eight times the Solid Content. Hence the superficies of Spheres are as the squares, and the solidity as the cubes of their diameters. Note 2.-The cube of the diameter of a Sphere in inches multiplied by 00188 will give the number of imperial Gallons it will contain. To find the Solid Content of a Spherical Segment. 1. From three times the diameter of the sphere take double the height of the segment; then multiply the remainder by the square of the height, and this product by 5236. 2. Or, To three times the square of the radius of the segment's base add the square of its height; then multiply the sum by the height, and the product by 5236. Example.-Required the content of a Spherical Segment, 2 feet in height, cut from a Sphere of 8 feet diameter. (3 X 8)-(2 X 2) = 20 20 X 22 X 523641.888 Cubic Feet. Content required. To find the Diameter of a Sphere, its solidity being given. Divide the solidity by 5236, and take the cube root of the quotient. Example.-The solidity of a Sphere being 113·0976 solid inches, what will be its diameter. 113.0976 5236 216, the Cube root of which is 6 inches, the Diameter required. |