Let DS be the axis of the cone, A' L'B' the circle of its base, and the triangle A BS its projection on the vertical plane; and let C, C', be the projections of the centre, and the circles E' K' F' and E G F those of the circumferences of the sphere.

This problem, like most others similar to it, can be solved only by the aid of imaginary intersecting planes. Let a b (fig. 3) represent the projection of a horizontal plane; it will cut the sphere in a circle whose diameter is a b, and which is to be drawn from the centre C' in the plan. Its intersection with the cone is also a circle described from the centre S' with the diameter cd; the points e' and f', where these two circles cut each other, are the horizontal projections of two points in the lower curve, which is evidently entirely hidden by the sphere. The points referred to are projected vertically upon the line ab at e and f. The upper curve, which is seen in both projections, is obtained by a similar process; but it is to be observed that the horizontal cutting planes must be taken in such positions as to pass through both solids in circles which shall intersect each other. For our guidance in this respect it will be necessary, first, to determine the vertices m and n of the curves of penetration.

For this purpose, conceive a vertical plane passing through the axis of the cone and the centre of the sphere; its horizontal projection will be the straight line C' L' joining the centres of the two bodies. Let us also make the supposition that this plane is turned upon the line C C' as on an axis, until it becomes parallel to the vertical plane; the points S' and L' will now have assumed the positions S2 and L2, and consequently the axis of the cone will be projected vertically in the line D' S3, and its side in S3 L3, cutting the sphere at the points p and r. Conceive the solids to have resumed their original relative positions, it is clear that the vertices or adjacent limiting points of the curves of penetration must be in the horizontal lines p o and r q, drawn through the points determined as above; their exact positions on these lines may be ascertained by projecting vertically the points m' and n', where the arcs described by the points p and r, in restoring the cone to its first position, intersect the line S L.

It is of importance further, to ascertain the points at which the curves of penetration meet the outlines AS and S B of the cone. The plane which passes through these lines being projected horizontally in A' B', will cut the sphere in a circle whose diameter is j'; this circle, described in the elevation from the centre C, will cut the sides A S and S B in four points at which the curves of penetration are tangents to the outlines of the cone. Figs. 5 and 6.-To find the lines of penetration of a cylinder and a cylindrical ring or torus.

Let the circles A' E' B', F' G' K', represent the horizontal, and the figure A CBD the vertical projection of the torus, and let the circle H'f' L', and the rectangle H I M L be the analogous projections of the cylinder, which passes perpendicularly through it. Conceive, as before, a plane a b (fig. 5), to pass horizontally through both solids; it will obviously cut the cylinder in a circle which will be projected in the base H'ƒ' L' itself, and the ring in two other circles, of which one only, part of which is represented by the are ƒ' b3 b', will intersect the cylinder at the points f' and b3, which being projected vertically to fig. 5, will give two points ƒ and b2 in the upper curve of penetration.

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Another horizontal plane, taken at the same distance below the centre line A B as that marked a b is above it, will evidently cut the ring in circles coinciding with those already obtained; consequently the points ƒ' and b3 indicate points in the lower as well as in the upper curves of penetration, and are projected vertically at d and e. Thus, by laying down two planes at equal distances on each side of A B, by one operation four points in the curves required are determined.

To determine the vertices m and n, following the method explained in the preceding problem, draw a plane O n', passing through the axis of the cylinder and the centre of the ring, and conceive this plane to be moved round the point O as on a hinge, until it has assumed the position O B', parallel to the vertical plane; the point n', representing the extreme outline of the cylinder in plan, will now be at r', and being projected vertically, that outline will cut the ring in two points p and r, which would be the limits of the curves of penetration in the supposed relative position of the two solids; and by drawing the two horizontal lines rn and p m, and projecting the point n' vertically, the intersections of these lines, the two points m and n, are the vertices of the curves in the actual position of the penetrating bodies.

The points at which the curves are tangents to the outlines HI and LM of the cylinder, may readily be found by describing arcs of circles from the centre O through the points H' and L', which represent these lines in the plan, and then proceeding, as above, to project the points thus obtained upon the elevation. Lastly, to determine the points, as j, z, &c., where the curves are tangents to the horizontal outlines of the ring, draw a circle P's' j' with a radius equal to that of the centre line of the ring, namely, PD; the points of intersection z' and j' are the horizontal projections of the points sought.

Required to represent the sections which would be made in the ring now before us, by two planes, one of which, N' T', is parallel to the vertical

plane, while the other, T' E', is perpendicular to both planes of projec

tion.

The section made by the last-named plane must obviously have its vertical projection in the line C D, which indicates the position of the plane; but the former will be represented in its actual form and dimensions in the elevation. To determine its outlines, let two horizontal planes g q and i k, equidistant from the centre line A B, be supposed to cut the ring; their lines of intersection with it will have their horizontal projections in the two circles g'o' and h' q' which cut the given plane N' T' in o' and q'. These points being projected vertically to o, q, k, &c., give four points in the curve required. The line N' T' cutting the circle A' E' B' at N', the projection N of this point is the extreme limit of the curve.

The circle P's'', the centre line of the rim of the torus, is cut by the planes N' T' at the point s', which being projected vertically upon the lines D P and C 7, determines s and 7, the points of contact of the curve with the horizontal outlines of the ring. Finally, the points t and u are obtained by drawing from the centre O a circle T'v' tangent to the given plane, and projecting the point of intersection v' to the points v and x, which are then to be replaced upon C D by drawing the horizontals v t and x u.

PENETRATIONS OF CYLINDERS, PRISMS, SPHERES, AND CONES.

Plate VI., figs. 1 and 2.-Required to delineate the lines of penetration of a sphere and a regular hexagonal prism whose axis passes through the centre of the sphere.

The centres of the circles forming the two projections of the sphere are, according to the terms of the problem, upon the axis C C' of the upright prism, which is projected horizontally in the regular hexagon D' E' F' G′ H'I'. Hence it follows, that as all the lateral faces of the prism are equidistant from the centre of the sphere, their lines of intersection with it will necessarily be circles of equal diameters. Now, the perpendicular face represented by the line E' F'in the plan, will meet the surface of the sphere in two circular arcs E F and L M (fig. 1), described from the centre C, with a radius equal to 'b' or a'c'. And the intersections of the two oblique faces D' E' and F'G' will obviously be each projected in two arcs of an ellipse whose major axis dg is equal to the diameter of the circle a c b, and the minor axis is the vertical projection of that diameter, as represented at e'ƒ' (fig. 2). But as it is necessary to draw small portions only of these curves, the following method may be employed.

Draw D G; through the points E, F, divide the portions E F and F G respectively into the same number of equal parts, and drawing perpendiculars through the points of division, set off from F G the distances from the corresponding points in E F to the circular arc E C F, as points in the elliptical arc required. The remaining elliptical arcs should be traced by the same method.

Figs. 3 and 4.-Required to draw the lines of penetration of a cylinder and a sphere, the centre of the sphere being without the axis of the cylinder. Let the circle D' E' L' be the projection of the base of the given cylinder, the elevation of which is shown at fig. 3, and let A B be the diameter of the given sphere.. If a plane, as c' d', be drawn parallel to the vertical plane, it will evidently cut the cylinder in two straight lines G Gʻ, HH', parallel to the axis, and projected vertically from the points G' and H'. This plane will also cut the sphere in a circle whose diameter is equal to c' d', and which is to be described from the centre C with a radius of half that line; its intersection with the lines G G' and H H' will give so many points in the curves sought, viz., G, H, I, K.

The planes a' b' and e' f', which are tangents to the cylinder, furnish only two points respectively in the curves; of these points E and F alone are visible, the other two, L and M, being concealed by the solid; therefore, the planes drawn for the construction of the curves must be all taken between a' b' and e' f'. The plane which passes through the axis of the cylinder cuts the sphere in a circle whose projection upon the vertical plane will meet at the points D, N, and g, h, the outlines of the cylinder, to which the curves of penetration are tangents.

Figs. 5 and 6.—To find the lines of penetration of a truncated cone and a prism.

The straight line C D is the axis of a truncated cone, which is represented in the plan by two circles described from the centre C'; and the horizontal lines M N and M' N' are the projections of the axis of a prism of which the base is square, and the faces respectively parallel and perpendicular to the planes of projection.

In laying down the plan of this solid, it is supposed to be inverted, in order that the smaller end of the cone, and the lines of intersection of the lower surface F G of the prism may be exhibited. According to this arrangement, the letters A' and B' (fig. 6) ought, strictly speaking, to be marked at the points I' and H', and conversely; but as it is quite obvious that the part above M' N' is exactly symmetrical with that below it, the distribution of the letters of reference adopted in our figures can lead to no confusion.