GEOMETRY OF SOLIDS. DEFINITIONS OF SOLIDS. 214. A RIGHT CYLINDER is that which is formed by the revolution of a rectangle about one of its sides; the line round which the rectangle revolves is called the axis (plural axes); and the circles generated by the two opposite sides of the rectangle, perpendicular to the axes, are termed the ends or bases. The surface of the cylinder, generated by the line parallel to the axis, is termed the curved surface, which is either straight or convex, according as a straight edge is applied, parallel to the axis, or in any other direction. 215. A RIGHT CONE is that which is formed by supposing a right-angled triangle to revolve about one of its legs or perpendicular sides; the fixed leg, or line, is called the axis; the surface generated by the other leg is called the base; and the surface formed by the hypothenuse, or side opposite the right angle, is denominated the curved surface, which is either straight or convex, according as a straight edge is applied upon the surface from the vertex, or in any other direction. 216. A SPHERE or GLOBE is that which is formed by supposing a semi-circle to revolve upon its diameter; the diameter upon which the semi-circle revolves is called the axis, and the surface formed by the arc of the semi-circle is called the curved surface, which is convex, in whatever way it may be tried by a straight edge. 217. An ELLIPSOID is formed or generated by supposing a semi-ellipse to revolve upon one its axes; the axis thus fixed is called the axis of the ellipsoid, and the surface generated by the curve is termed the curved surface. 218. To describe a conic section, from the cone, through a line given in position, in the section passing through the axis. Let ABC, (figures 1, 2, 3, pl. VI,) be the section of a right cone, and let DE be the line of section. Through the apex or top of the cone, C, draw CF, parallel to the base AB of the section, and produce ED to meet AB in D, as in figures 2 and 3, or AB produced in G, as in fig. 1, as also to meet CF in F. On AB describe a semi-circle, which will be equal to half the base of the cone. In the semi-circle take any number of points, a, b, c, &c. Draw Dd, in figures 2 and 3, and Gd in fig. 1, perpendicular to AB and Gd', in fig. 1, perpendicular to GF; as, also, Dd', figures 2 and 3, perpendicular to DF. From the points a, b, c, &c. draw lines, ae, bf, cg, &c., cutting Gd (figure 1) and Dd (figures 2 and 3) in the points e, f, g, &c. In figure 1, make in Ge', Gf', Ggʻ, &c. equal to Ge, Gf, Gg, &c.; and in Dd', (figures 2 and 3,) make De', Df', Dg', &c. equal to De, Df, Dg, &c. Through the points e', f', g', &c. draw lines to F. Through the points a, b, c, &c. draw lines perpendicular to AB. From the points of section, in AB, draw lines to the vertex C of the cone, cutting the sectional line, DE, in l, m, n, &c. Through the points of section, l, m, n, &c., draw lh, mi, nk, &c. perpendicular to DE. Through the points D, h, i, k, &c. in fig. 1, or d', h, i, k, &c. figures 2 and 3, draw a curve, which will be the conic section required. OBSERVATIONS. 219. In the first of these figures, the line of section cuts both sides of the section of the cone; in this case, the curve Dhik and eE is an Ellipse. In fig. 2, the line of section DE is parallel to the side AC of the section of the cone; in this case, the curve d' hi, &c. E, is a Parabola. In fig. 3, the line of section, DE, is not parallel to any side of the cone; it must, therefore, when produced with the sides of the section through the axis, meet each of these two sides in different points: in this case, the section d', h, i, &c., E, is either an Ellipse or Hyperbola; but the case is determined to be an hyperbola by the line of section meeting the opposite side BC at AC, where it cuts above the vertex at the point B'. Here we may observe, that the line of section, DE, is the same as that which has before been called the abscissa, the part EB produced, contained between the two sides of the section, is called the axis major; and the line Dd, perpendicular to DE, an ordinate. Hence the same section may be found by the method already shown in the problem; viz. by drawing any straight line, deb, fig. 4: make de equal to DE, fig. 3, and eb equal to EB, fig. 3. Through d draw the straight line DD at right angles to db': make dD equal to Dd', fig. 3; then, with the axis major, b'e, the abscissa ed, and the ordinate dD, on each side of the abscissa describe the curve of the hyperbola, which will be of the same species as that shown in fig. 3. PROBLEM 38. 220. To describe a cylindric section, through a line given in position, upon the section passing through its axis (fig. 4, pl. VI). This is no more than a particular case of the last problem. For a cylinder may be considered as a cone, having its apex at an infinite distance from its base; or, practically, at a vast distance from its base. In this case all the lines, for a short distance, would differ insensibly from parallel lines; and this is the construction shown at fig. 5, which is therefore evident. But as the section of a cylinder so frequently occurs, I shall here give a more practical description of it. Thus Let ABHI be a section of a right cylinder, passing through its axis, AB being the side which passes through the base, and let DE be the line of section. On AB describe a semi-circle; and, in the arc, take any number of points, a, b, c, &c. from which draw lines perpendicular to the diameter, AB, cutting it in Q, R, S, &c: perpendicular to AB, or parallel to AI or BH, draw the lines Qq, Rr, Ss, &c. cutting the line of section, DE, in the points, q, r, s, &c. from the points of section, q, r, s, &c. draw the lines qi, rk, sl, &c. per pendicular to the line of section, DE. Make the ordinates qi, rk, sl, &c. each respectively equal to the ordinates Qa, Rb, Sc, &c.; and through the points D, i, k, l, &c. to E, draw a curve, which will evidently be the section of the cylinder, as required. The same may be done in this manner, viz.-Bisect the line of section DE in the point t. Draw tm perpendicular to DE. Make tm equal to the radius of the circle which forms the end of the cylinder; then, with the axis major, DE, and the semi-axis minor, tm, describe a semi-ellipse, which will be the section of the cylinder required. A DEFINITION. 221. A CUNEOID is a solid ending in a straight line, in which, if any point be taken, a perpendicular from that point may be made to coincide with the surface: the end of the cuneoid may be of any form whatever. The cuneoid, which occurs in architecture, has a semi-circular or a semielliptical end, parallel to the straight line to which the perpendicular is applied. PROBLEM 39. 222. To find the section of a cuneoid, with a semi-circular base, the given data being a section through the axis, perpendicular to the vertex, or sharp end, and the line of section upon that end. Let ABC, (fig. 6, pl. VI,) be the section through the axis, perpendicular to the sharp edge, and let DE be the line of section. This construction is similar to that of finding the section of a cylinder, excepting that, instead of drawing parallel lines from the base, AB, they are, in this figure, drawn from the points of section in AB to the point C, which is the vertex of the cuneoid: the ordinates, Qa, Rb, Sc, &c., being transferred, respectively, to qi, rk, sl, &c.; and the curve D, i, k, l, &c. to E, being drawn through the points, D, i, k, l, &c., by hand. PROBLEM. 40. 223. Given the position of the seats of three points, in the circumference of the base of a cylinder, and the lengths of the perpendiculars intercepted between the points and their seats, to find the section of the cylinder passing through these three points. Through the three points, A, B, C, (fig. 7, pl. VI,) describe the circumference of a circle. Join the two remote points, A and B, and draw AD, CF, and BE, perpendicular to AB. Make AD equal to the height upon A, BE equal to the height upon B, and CF equal to the height upon C. Produce BA and ED to meet each other in H: draw CG parallel to BH, and FG parallel to EH. Join GH. In GH take any point, G, and draw GK perpendicular to CG, cutting BH in K: from the point K draw KI, perpendicular to EH, cutting EH in L. From H, with the radius HG, describe an arc, cutting KI at I. Join HI. In the circumference, ACB, take any number of points a, b, c, &c., at pleasure, and draw ae, bf, cg, &c., parallel to GH, cutting AB at e, f, g, &c. Through the points e, f, g, &c., draw lines ei, fk, gl, &c., parallel to GK, or AD, or BE, cutting DE at i, k, l, &c.; from the points of section, i, k, l, &c., draw the lines in, ko, lp, &c., parallel to HI. Transfer the ordinates, ea, fb, gc, &c., to in, ko, lp, &c.; then, through the points D, n, o, p, &c. draw the curve Dnop, &c. to E, and it will be the section cut by the plane, as required. PROBLEM 41. 224. Given the great circle of a sphere, and the line of position of a section at right angles to that great circle, to find the form of the section. Let ABC, (fig. 8, pl. VI,) be the great circle, and AB the line of section. On AB, as a diameter, describe a semi-circle, which will be the section required: since all the sections of a sphere, or globe, are circles. PROBLEM 42. 225. Given the section of an ellipsoid, passing through the fixed axis, and the line of position of another section, at right angles to the first section, to find the form of the section through that line. Let ABCD, (fig. 9, pl. VI,) be the section through the fixed axis, and EF the line of position. Through the centre of the ellipsoid draw AC parallel to EF. Bisect EF in H, and draw HG perpendicular to EF. Find HG a fourth |