PROBLEM 2. Section of a cylinder by a plane perpendicular to its generatrixes. 48. Let abc be the horizontal trace of the cylinder, and let the outlines of the cylinder be determined according to the construction to Prob. 1, p. 53. The cutting plane being perpendicular to the generatrixes, its traces must be per pendicular to the projections of these lines (Prob. 14, p. 41); let qp and gp' be these traces. The first step is to find the intersections of the generatrixes with the plane pop: let a vertical plane be conceived to pass through one of these, ae a' é' for example. The horizontal projection of the intersection of this plane with pap' is ae, and one point in the vertical projection is obtained by drawing hk' perpendicular to xy. To determine a second point, draw a parallel, zo, zo' to pl, through any point z,z' of the trace qp'; the point o, in which zo meets ae, is the horizontal projection of a point common to the two planes, the vertical projection o' of this point is deduced, and subsequently h'o', which is that of the line in which the two planes cut each other; the point e' in which aé' meets l'o' will be the vertical projection of one point of the section sought, e therefore is the horizontal projection. The vertical planes passing through the different generatrises will cut the sectional plane in parallel lines, the vertical projections of which will therefore be parallel to h'o'. The intersections of these parallels with the vertical projections of the generatrixes, will determine any requisite number of points in the same manner as e, é' was determined, through which points the section of the cylinder by the plane is to be drawn. To draw. a tangent to this curve at any point m, m', draw kr, a tangent to the horizontal trace of the cylinder, through that of the generatrix which contains the proposed point. This line kr will be the horizontal trace of a plane, tangential to the cylinder. Now the tangent to the section will lie both in this plane, and in the plane of the section, it is therefore their common intersection: hence the point r, in which kr meets pq, is a point in the tangent sought; mr is consequently its horizontal, and m’r its vertical, projection. The true figure and magnitude of the section will be found by turning the plane pap' round on its trace pq, into the horizontal plane; but the constructions requisite will be readily comprehended both from what was explained on this subject in Prob. 1, and from the figure. When the cylinder is developed, the sectional curve made by the plane qap', being perpendicular to the generatrixes, will develop into a straight line, perpendicular to the development of the generatrixes. · Draw a straight line, and set off on it the lengths of the arcs EF FG... of the section*, from e to 0, $ to y, &c.; then make the perpendiculars ea, B... drawn through the points €, 6, 7, &c., equal to the real length of the segments of the corresponding generatrixes, intersected between the sectional and the horizontal planes. By this proceeding points in the developed base of the solid may be obtained. The true lengths of these generatrixes are to be deduced by Prob. 5, p. 34. Let the ordinate ku in the development be that which corresponds to mk, m'k' on the cylinder, passing through the point of contact of the tangent; krm, k'r'm' are the projections of a right-angled triangle, of which kr, k'r is the hypotheneuse, and r m, j k are respectively equal to its two sides; by making up equal to mr, the line joining the points kp will be the tangent at K. * The generatrixes which pass through the extremities of the equal arcs, ab, bc, cd .... of the base of the cylinder, will not intercept equal arcs of the elliptic section, made by a plane perpendicular to those generatrixes, as will be perceived from the figure E F G... hence the segments of the straight line ed, dy... which are these arcs developed, are not equal, as those in the corresponding construction of the last Problem were. Scholium. When the cutting plane is oblique to the generatrixes, the projections of the curve and of the tangent are found by constructions similar to the foregoing. But to turn the curve down on the horizontal plane, it is necessary to trace the projections and construct the true magnitudes of the perpendiculars, drawn from the various points of this curve to the horizontal trace of the cutting plane. To draw the development of the cylinder, it is necessary previously to obtain the section of the solid by a plane perpendicular to the generatrixes, as in the two previous problems. PROBLEM 3. Section of a cone by a plane perpendicular to the certical plane of projection. 49. 1°. It is evident that the vertical projection of the section is identical with the vertical trace of the cutting-plane pap'. Let om, oʻm" be the projections of any generatrix of the cone; it is easy to find the projections n, n" of its intersection with the plane pop'; and by repeating the construction for as many generatrixes as may be deemed necessary, the data for drawing the horizontal projection efg of the conic section may be obtained. The cone in the figure is a right one with a circular base; its axis is vertical, and the two generatrixes parallel to the vertical plane are oa, oʻa', and od, o'd'. The points of the curve efg may also be obtained by cutting the cone by horizontal planes: let h'i' be the vertical trace of such a plane, which will obviously cut the cone in a circle, the horizontal projection of which will be an equal circle, described on the centre o, with a radius equal to half l'i'; and it cuts the plane pap' in a line perpendicular to the vertical plane, the horizontal projection nn of which per E pendicular is readily determined. The points n,n, in which this perpendicular cuts the circle, are common to the horizontal projection of the elliptic section, other points in which may be found by a repetition of the construction made with other horizontal planes. By means of this construction, the points situated on the line o'o, perpendicular to xy, may be found, which by the first construction could not readily be done. 2°. To draw a tangent to the curve efg, at the point n, for example, draw the tangent mr to the circular base at the point in which the generator, passing through the proposed point n, cuts that base. This tangent mr will be the horizontal trace of a plane touching the cone in the gene |