A prism is shown in perspective at Fig. 139, cut by an oblique plane. The points abed are the angles of the horizontal base, and abg, bcf, cdef, and adeg are the vertical sides; while efbg is the top, the form of which is to be shown.

267. - Form of Top of Prism - In Fig. 139 the form of the top of the prism is shown as it appears in perspective..

not in its real shape; this is now to be developed. In Fig. 140, let the sauare a b c d represent by scale the actual form and size of the base, abcd, of the prism shown in Fig. 139. Make c c' and dd' respectively equal to the actual heights at cf and de, Fig. 139; the lines dd, and cc, being set up perpendicular to the line dc. Extend the lines dc and d'c, until they meet in h; join b and h. Now this line b h is the intersection of two planes: one, the base, or horizontal plane upon which the prism stands; the other, the cutting plane, or the plane which, passing obliquely through the prism, cuts it so as to produce, by intersecting the vertical sides of the prism, the form b feg, Fig. 139.

266 A Prism Cut By An Oblique Plane 175

Fig. 139.

266 A Prism Cut By An Oblique Plane 176

Fig. 140.

Illustration By Planes

To show that b k is the line of intersection of these two planes, let the paper on which the triangle dhd' is drawn (designated by the letter B) be lifted by the point d' and revolved on the line dh until d' stands vertically over d, and c' over c; then B will be a plane standing on the line dh, vertical to the base-plane A. The point h being in the line cd extended, and the line cd being in the base-plane A, therefore h is in the base-plane A. Now the line d'c' represents the line ef of Fig. 139, and is therefore in the cutting plane; consequently the point h, being also in the line d' c' extended, is also in the cutting plane. By reference to Fig. 139 it will be seen that the point b is in both the cutting and base planes; we must therefore conclude that, since the two points b and h are in both the cutting and base planes, a line joining these two points must be the intersection of these two planes. The determination of the line of intersection of the base and cutting planes is very important, as it is a controlling line; as will be seen in defining the lines upon which the form of the face-mould depends. Care should therefore be taken that the method of obtaining it be clearly understood.

It will be observed that the intersecting line bh, being in the horizontal plane A, is therefore a horizontal line. Also, that this horizontal line b h being a line in the cutting plane, therefore all lines upon the cutting plane which are drawn parallel to b h must also be horizontal lines. The importance of this will shortly be seen. Through a, perpendicular to bh, draw the line b" dv' and parallel with this line draw dd""; on d as centre describe the arc d' d""; draw d"" dv parallel with dd", and extend the latter to d"' on d" as centre describe the arc dv d"'; join b" and d". We now have three vertical planes which are to be brought into position around the base-plane A, as follows: Revolve B

upon dh, E upon dd" and C upon b" d" each until it stands perpendicular to the plane A, Then the points d' and d"" will coincide and be vertically over d; the points d"' and dv will coincide and stand vertically over dn; and c' will cover c. These vertical planes will enclose a wedge-shaped figure, lying with one face, b"d"dh, horizontal and coincident with the base-plane A, and three vertical faces, b" d" d"', dd" dv d"", and hdd' By drawing the figure upon a piece of stout paper, cutting it out at the outer edges, making creases in the lines hd, dd"d"b"' then folding the three planes B, E, and C at right angles to A, the relation of the lines will be readily seen. Now, to obtain the form of the top or cover to the wedge-shaped figure, perpendicular to b" d"' draw b"h' and d"'e; on b" as centre describe the arc hh'; make d"'e equal to d" d; join e and h' Now the form of the top of the wedge-shaped figure is shown within the bounds d"' b"h" c. By revolving this plane D on the line b" d"' until it is at a right angle to the plane C, and this while the latter is supposed to be vertical to the plane A, it will be perceived that this movement will place the plane D on top of the wedge-shaped figure, and in such a manner as that the point e will coincide with d"" d., and the point h' will fall upon and be coincident with the point h, and the lines of the cover will coincide with the corresponding lines of the top edges of the sides of the figure; for example, the line b"d"' is common to the top and the side C; the line d"'e equals d" d, which equals dvd"" therefore, the line d"'e will coincide with dvd"" of the side E; the line eh' will coincide with d'h of the side B; and the line b"h' will coincide with the line b"h. Thus the figure D bounded by b"d"'ch' will exactly fit as a cover to the wedge-shaped figure. Upon this cover we may now develop the form of the top of the prism.

Preliminary thereto, however, it will be observed, as was before remarked, that lines upon the cutting plane which are parallel to the intersecting line b" h, are horizontal; and each, therefore, must be of the same length as the line in the base-plane A vertically beneath it. For example, the line d"' t' is a line in the cutting plane D, parallel with the line b" h' in the same plane, and this line b"h will (when the cutting plane D is revolved into its proper position) be coincident with the intersecting line b" h; therefore, the line d"'e is a line in the cutting plane D, drawn parallel with the intersecting line b" h. Now this line d"' e, when in position, will be coincident with the line d"'dv, which lies vertically over the line d"d of the base-plane A; its length, therefore, is equal to that of the latter. In like manner it may be shown that the length of any line on the plane D parallel to b" h" is equal in length to the corresponding line upon the plane A vertically beneath it.

Therefore, to obtain the form of the top of the prism, we proceed as follows: Perpendicular to b" dv draw c c"' and aa"'; perpendicular to b"d"' draw c"'f and equal to c"c; on b" as centre describe the arc b b'; join b' a"' b'f, and a"' e. Now we have here in plane D the form of the top of the prism, as shown in the figure bounded by the lines a"'b'fe. This will be readily seen when the plane D is revolved into position. Then the point a"' will be vertically over a; the point e coincident with d' d"" and vertically over d; the point f coincident with c' and vertically over c; while b' will coincide with b of the base-plane A.

The figure a"' b'fe, therefore, represents correctly both in form and size the top of the prism as it is shown in perspective at b f eg, Fig. 139. The line ef, Fig. 140, is equal to the line d' c', and so of the other lines bounding the edges of the figure.

The cutting plane b f eg, Fig. 139, may be taken to represent the surface of the plank from which the wreath of hand-railing is to be cut; the wreath curving around from b to e, as shown in Fig. 141, the lines b g and g e being tangent to the curve in the cutting plane; while ab and ad are tangents to the curve on the base plane, or plane of the cylinder. The location of the cutting plane, however, is usually not at the upper surface of the plank, but midway between the upper and under surfaces. The tangents in the plane are found to be more conveniently located here for determining the position of the butt-joints. For a moulded rail two curved lines, each with a pair of tangents, are required upon the cutting plane, one for the outer edge of the rail, and the other for the inner edge; but for a round rail only one curve with its tangents is required, as that from b to e in Fig. 141, which is taken to represent the curved line running through the centre of the cross-section of the rail. As an easy application of the principles regarding the prism, just developed, an example will now be given.

Illustration By Planes 177

Fig. 141.