Let a b (Fig, 173) be the given breadth, 1 3/4 the given number of revolutions, and let the relative size of the regulating square to the eye be 1/3 of the diameter of the eye. Then, by the rule, 1 3/4 multipled by 4 gives 7, and 3, the number of times a side of the square is contained in the eye, being added, the sum is 10. Divide a b, therefore, into 10 equal parts, and set one from Hoc; bisect ac in e; then a e will be the length of the longest ordinate (1 d or 1 e). From a draw a d, from e draw e 1, and from b draw b f, all at right angles to a b; make e 1 equal to a, and through 1 draw 1 d parallel to a b; set b c from 1 to 2, and upon 1 2 complete the regu lating square; divide this square as at Fig. 172; then describe the arcs that compose the scroll, as follows: upon 1 describe d e, upon 2 describe e f, upon 3 describe f g, upon 4 describe g h, etc.; make dl equal to the width of the rail, and upon 1 describe l m, upon 2 describe m n, etc.; describe the eye upon 8, and the scroll is completed.
Bisect dl (Fig. 173) in 0, and make o v equal to 1/3 of the diameter of a baluster; make v w equal to the projection of the nosing, and e x equal to wl; upon 1 describe wy, and upon 2 describe y z; also, upon 2 describe x i, upon 3 describe ij, and so around to z; and the scroll for the step will be completed.
Bisect dl (Fig. 173) in 0, and upon 1, with 1 o for radius, describe the circle o ru; set the baluster at p fair with the face of the second riser, c2, and from p, with half the tread in the dividers, space off as at o, q, r, s, t, u, etc., as far as q2; upon 2, 3, 4, and 5 describe the centre-line of the rail around to the eye of the scroll; from the points of division in the circle o r u draw lines to the centre-line of the rail, tending to 8, the centre of the eye; then the intersection of these radiating lines with the centre-line of the rail will determine the position of the balusters, as shown in the figure.