Let H F K in Fig. 300 be the given ball, of which G is the center. Let D1 G2 C3 D3 E represent a plan of the octagon shaft which is required to Jit against the ball. Draw this plan in line with the center of the ball, as indicated by F E. From the angles of the plan project lines upward, cutting the circle and constituting the elevation of the shaft. From the point A or A1, where the side in profile cuts the circle, draw a line at right angles to the center line of the ball F E. cutting it in the point B, as shown. Through B, from the center G by which the circle of the hall was struck, describe an arc, cutting the two lines drawn from the inner angles C C3 of the plan, as shown at C and C1. Then MCC1 N will be the pattern of one side of an octagon shaft mitering against the given ball H F K. If it be desired to complete the elevation of the shaft meeting the ball, it may be done by carrying lines from C and C' horizontally until they meet the outer line of the shaft in the points D and D1. Connect C1 and D1, also C and D, by a curved line, the lowest point in which shall touch the horizontal line drawn through B. Then the broken line D C C1 D1 will be the miter line in elevation formed by the junction of the octagonal shaft with the ball.

Fig. 306.   The Pattern of an Octagon Shaft to fit Against a Ball.

Fig. 306. - The Pattern of an Octagon Shaft to fit Against a Ball.