The conditions given in this problem are essentially the same as those of Problem 193, but the following solution differs from that of the former problem in the method of finding the distances from points assumed in the base to those of the top, and is introduced as showing varieties of method: In Fig. 640, CGHD represents the side view of the article, of which G K H is a half profile of the top and C F D a half profile of the base. For convenience in obtaining the pattern the half profiles are so drawn that their center lines coincide with the upper and lower lines of the elevation.

Divide both of the half profiles into the same number of equal parts - in the present instance eight. From the points obtained in the half profiles drop perpendiculars cutting G H and C D. Connect the points secured in G H with those in C D, as a n, b m, etc. Also connect the points in G H with those in C D, as indicated by the dotted lines 1 n, o m, etc. Reference to Problem 193 will show that in order to obtain the correct lengths represented by the several solid and dotted lines drawn across the elevation complete sections upon those lines were constructed, as shown in Figs. 628 and 629. In the present case these distances will be derived from a series of triangles whose bases are the differences between the lengths of the lines drawn across the half profile of the top and those of the bottom.

Therefore, to obtain the triangles giving the true distances represented by the solid lines proceed as follows:

Fig. 640.   Elevation and Profiles of an Irregular Fluring Article. Showing Method of Triangulation

Fig. 640. - Elevation and Profiles of an Irregular Fluring Article. Showing Method of Triangulation

First set off from C D upon each line in the base C F D the length of the corresponding line in the top; thus make n o equal to a 2, m p equal to b 3, l q equal to c4, etc. For the bases of the triangles represented by the dotted lines set off from C D the length of corresponding lines in G K H, as shown by the small figures in C O D. Thus make m 2' equal to a 2, l 3' equal to b 3, k 4' equal to c 4, etc. The triangles represented by solid lines in the elevation, and shown in Fig. 641, are obtained as follows: Draw the line P Q, and from a convenient point erect the perpendicular R S. From R on R S set off the lengths of solid lines in C G H D. and from R on P Q set off the lengths of corresponding lines in C O D F, as indicated by the small letters in COD. To avoid a confusion of lines the lengths from C to O are set off to the left of R, and the lengths from O to D on R Q. Thus make R a' of diagram equal to n a of Fig. 640, R o' of diagram equal to o 17 of half profile of base, and connect a o'. Make R b' of diagram equal to m b of Fig. 640, R p' of diagram equal to p 16 of the base, and connect p' b', etc.

Fig. 641.   Diagram of Triangles Representing Solid Lines of the Elevation.

Fig. 641. - Diagram of Triangles Representing Solid Lines of the Elevation.

Fig. 642.   Diagram of Triangles Representing Dotted Lines of the Elevation.

Fig. 642. - Diagram of Triangles Representing Dotted Lines of the Elevation.

Fig. 643.   Pattern of Shape Shown. in Fig. 640.

Fig. 643. - Pattern of Shape Shown. in Fig. 640.

The triangles represented by dotted lines in C G H D are obtained in a similar manner. Draw the line T U in Fig. 642 and erect the perpendicular V W. Prom V, on V W, set off the lengths of dotted lines in C G H D of the elevation. Thus make V 1" equal to n 1 of Fig. 640, V a" equal to m, a, V b" equal to l b, etc. Upon V T or V U set off the lengths of the lines in C O D F of Fig. 640, as indicated by the small figures. Thus make V X" equal to n 17 of the base in Fig. 640, and draw X" 1". Make V 2" equal to 2' 16 of the base, and draw 2" a", etc.

To obtain the pattern first draw the line C G of Fig. 643, in length equal to C G of Fig. 640. From C, with radius equal to C 17 of the half profile of base, strike a small arc (o), which intersect with another arc struck from G as center, and 1" X" of Fig. 642 as radius, thus establishing point o in the curve of the pattern. From G of pattern as center, and G

2 of the half profile of top as radius, strike a small arc, which intersect with another arc struck from o of pattern as center, and a' o' of Fig. 641 as radius, thus establishing point a in the upper curve of the pattern. From o of pattern as center, and 17 16 in C F as radius, strike a small arc (p), which intersect with another arc struck from a of pattern as cenlet, and a" 2" of Fig. 642 as radios, thus locating point p of the pattern. From a of the pattern as center, and 2 3 in GK as radius, strike a small arc, which intersect with another arc struck from p of pattern as center, and p' b' of Fig. 641 as radius, thus locating point b of pattern. Proceed in this manner, using in the order named the spaces in the lower profile, the hypothenuses of triangles in Fig. 642, the spaces in the upper profile, and the hypothenuses of triangles in Fig. 641, The points thus obtained, as indicated by the letters in Fig. 643, are the points through which the pattern lines are to be traced. Then C G H D is the required pattern for one-half the article. The other half of pattern, as shown by C G H' D', can be obtained by any convenient method of duplication.