therefore (Art. 213. a.) the point in which a line drawn through the given point parallel to the axis of the cylinder will intersect the plane XOY, and from thence draw a line touching the base of the cylinder; this line will be the intersection of the required plane with xoY. Find next the intersection with zox of the line parallel to the axis; then a line joining that intersection with the point in which the tangent to the base meets ox, will be the intersection of the required plane with zox. For a cone: find (Art. 213.) the points in which a line drawn through the given point and the vertex of the cone, will cut the planes XOY, Zox; then a line drawn through the point in XOY, touching the base of the cone, will be the intersection of the required tangent plane with the same plane XOY, and a line drawn from the point in zox to that in which the tangent to the base meets ox will be the intersection of the required plane with zox. PROPOSITION XVIII. 230. To determine the position of a plane, which, passing through a given point on the surface of a sphere, may be a tangent to that surface. a E b Let ABC, on the plane XOY, be the projection of a great circle of the sphere parallel to that plane; let P be the centre of this circle, and, the line Ppq being perpendicular to ox, let pqa be the projection on Zox of half a great circle parallel to that plane. Let the given point on the surface of the sphere be designated M, and let m, m' be its projections on XOY, ZOX respectively; also, imagine a section parallel to xoY to pass through M; its projection on m N X XOY will be a circle whose circumference passes through m, and it has a line equal to Pm for its semi-diameter, while ba passing through m' perpendicularly to pq will be the projection of a quadrant on zox. Now a tangent to the section through M, parallel to XOY, will be a tangent to the sphere at that point; and its projection on XOY will be the line mn drawn perpendicular to Pm. But the required tangent plane is to pass through M, it will therefore pass through the tangent, whose projection is mn, mn. and its intersection with XOY will be in a line parallel to Draw ad touching the semicircle qab in a, and meeting ox in d; then if PD passing through m be made equal to cd, and DN be drawn parallel to mn, that line will be the intersection of the tangent plane with XOY. Next, since a tangent to the circular section passing through M, if conceived to be drawn from м till it meets zox, will be equal and parallel to mn, draw ne perpendicular to ox meeting ba produced in E; then the point E will be the intersection of the tangent line with zox, and a line drawn through N, E will be the intersection of the tangent plane with that coordinate plane. The position of a plane touching any solid of revolution at a point given on that surface may be determined in like manner, if the axis of revolution be perpendicular to XOY, and a section passing through it parallel to zox be projected on the latter plane. 231. Scholium. The projections on XOY, Zox of a normal, supposed to be drawn from a given point in a curve surface, are easily determined when the position of a plane touching the surface at that point has been found; since the projections are merely those of a line drawn from a given point in a plane perpendicularly to that plane. (See Art. 217.) PROPOSITION XIX. 232. To determine the projection, on one of the co-ordinate planes, of the lines in which two cylinders may intersect one another. Let the axes of the cylinders be at right angles to one another, and let them be in the plane on which the projection is to be made. For simplicity, let the bases of both cylinders be circles, but let their diameters be unequal. Let XOY be the plane of projection, and in that plane let PQ, RS be the directions of the axes of the cylinders; also, let Ꮓ 9 lines cd, ef, &c., parallel to ox; and through the greater, at X equal distances from OY, draw as many lines c'd', e'f', &c. perpendicular to ox. It is evident that these will be the projections, on zox and zoy, of planes parallel to xoY; and if on AQ straight lines be drawn, as in the figure, through c, e, &c., c',e', &c. parallel both to Ox and OY, these will be the projections on XOY of lines supposed to be drawn on the surfaces of the cylinders parallel to the two axes; the points in which they intersect one another respectively, being joined by a curve line, as in the figure, will be the required projection. A like curve must be understood to constitute the projection of the curve line in which the hemicylinders intersect one another on the side opposite to AB. It is evident that those opposite curves will be similar to one another; it is also evident that if the diameters of the two circular cylinders were equal to one another, or if the bases of the cylinders were any similar and equal curves, the projection, on XOY, of the curves in which the cylinders intersect one another, would be two straight lines constituting the diagonals of the square or parallelogram formed on that plane by the chords of the vertical sections agb, &c. If the axes of the two cylinders, though coinciding with the plane XOY, form oblique angles with one another and with the other co-ordinate planes, the problem would be solved in a similar manner by means of semicircles agb, a'c'g' described on diameters perpendicular to the axes, and by drawing, parallel to those axes, the lines which determine points in the curves of intersection. 233. In the annexed figure, ABCD represents on XOY the projection of a groined vault, formed by the intersection of two hemicylinders having equal semicircular bases, and having their axes in the directions PQ and RS at right angles to one another. The figure within es'b is a projection, on zox, of half the vault; the curve es'b is the semi-ellipse supposed to stand vertically over CB, which is here parallel to Ox, and the straight line As' is the projection of the elliptical quadrant over AS; the curves cpA, Arb are semiellipses which constitute the projections of the semicircles standing vertically over AC and AB. Since the intersecting hemicylinders have equal diameters, each of the heights As', ap, dr, is equal to AP or AR, the radius of the semicircle standing on either C m! m B of those lines; Ab and AC are each evidently equal to SB or SC. manner: Whatever be the nature of the curves which, standing on AB and AC, form the bases of the intersecting hemicylinders, their projections on zox may be determined in the following Take any number of points, as m on AB, for example, and from each of them draw a line, as mn' perpendicular to ox; then mn being an ordinate to the given section An SB of the hemicylinder, (that section being supposed to stand vertically over AB) make m'n' equal to mn. The point n' will be in the required projection of AnSB. like manner may any number of points in arb, and also in Apc and cs'b, be found. PROPOSITION XX. In 234. To determine the projections, on one of the co-ordinate planes, of the curve line in which two cones may intersect one another. For simplicity let the cones have circular bases; let their axes be at right angles to one another, and let them be in the plane of projection. H f B m el A h' a' K X D' a A' Let ABC, A'B'C' be the projections of the cones on the plane YOX, AD and A'D' being the axes. Let a'b' on ox be the projection of ab, in A'D', and on a'b' as a diameter describe a semicircle; this will be the projection on zox of a vertical section through the cone ABC on the diameter ab. On oz make OH equal to D'B' the semidiameter of the base of the cone; and on ox make OK equal to D'A'; then the triangle HOK will be the projection on zox of the half cone A'B'C'. Lines drawn perpendicular to ox from r and t, where HK cuts the semicircle on a'b', will cut A'D' in points and ť, through which the required curves are to pass. The points p and q are evidently those in which the projection of one of the curves of intersection will meet AB. F E h C' Ze YB Take any point, as E in B'C', and draw EA', cutting AB and AC in h and k; also, draw the ordinate EF to a semicircle on B'C' as a diameter. Make of equal to EF, and draw fK. On ox project hk and the point e in which it is bisected, in h', k', e', and describe a semi-ellipse on h'k' as a transverse axis, with a semi-conjugate axis én equal to the ordinate at e of a vertical section through the cone ABC, formed on a line drawn through e parallel to BC as a diameter; this curve will be intersected by fK in the points m and s, which, being projected on hk, will give two other points, m' and s', in the required curves. In like manner may any number of points be found. PROPOSITION XXI. 235. To determine the projections, on one of the co-ordinate planes, of the curve lines in which a cylinder may intersect a sphere. B' The cylinder is supposed to have a circular base, and its axis to be parallel to the plane on which the projection is to be made. Let AB be the projection of a great circle of the sphere on XOY, A'B' the projection of a great circle on Zox, C and c' their centres, being in a line perpendicular to ox; also, let the circle abcd be the projection of the base of the cylinder on the latter plane, ç" being its centre, and ab, cd diameters respectively perpendicular and parallel to ox. Parallel to ox let the lines ap, ch, bq, and any others, as em, be drawn ; these will represent the intersections on zox of the planes of circles of A A p C' B bi C C X the sphere parallel to XOY. Let them intersect the line cc', produced if necessary, in a', d', b', g, &c., and touch or cut the circle acbd in a, b, c, d, f, &c. Then, if lines perpendicular to ox be drawn through these last points, they will represent on Yox the projections of lines drawn on the surface of the cylinder parallel to its axis; and cutting the surface of the sphere on the circumferences of the small circles whose planes are projected in ap, ch, ef, &c. Therefore, with c as a centre, and a radius equal to a'p, describe arcs intersecting ab produced in a, and a; also with c as a centre, and a radius equal to b'q, describe arcs intersecting ab produced in b, and b,,. Next, |