In Fig. 144, j'ge represents a pitch-board of the first flight, and d and i the pitch-board of the second flight of a platform stairs, the line e f being the top of the platform; and abc is the plan of a line passing through the centre of the rail around the cylinder. Through i and d draw i k, and through j and c drawy jk; from k draw k l parallel to fe; from b draw bm parallel to gd; from l draw lr parallel to kj; from n draw n t at right angles to jk; on the line ob make ot equal to nt; join c and t; on the line jc, Fig. 145, make ec equal to en at Fig. 144; from c draw c t at right angles to jc, and make ct equal to c t at Fig. 144; through t draw p l parallel to j c, and make t l equal to tl at Fig. 144; join l and c, and complete the parallelogram ecls; find the points 0, 0, 0, according to Art. 551; upon e, o, o, o, and l, successively, with a radius equal to half the width of the rail, describe the circles shown in the figure; then a curve traced on both sides of these circles, and just touching them, will give the proper form for the mould. The joint at l is drawn at right angles to c l.
This simple method for obtaining the face-moulds for the hand-rail of a platform stairs appeared first in the early editions of this work. It was invented by a Mr. Kells, an eminent stair-builder of this city. A comparison with Fig. 142 will explain the use of the few lines introduced. For a full comprehension of it reference is made to Fig. 146, in which the cylinder, for this purpose, is made rectangular instead of circular. The figure gives a perspective view of a part of the upper and of the lower flights, and a part of the platform about the cylinder. The heavy lines, im, wc, and cj, show the direction of the rail, and are supposed to pass through the centre of it. Assuming that the rake of the second flight is the same as that of the first, as is generally the case, the face-mould for the lower twist will, when reversed, do for the upper flight; that part of the rail, therefore, which passes from e to c, and from c to l, is all that will need explanation.
Suppose, then, that the parallelogram eaoc represent a plane lying perpendicularly over eabf, being inclined in the direction ec, and level in the direction co; suppose this plane eaoc be revolved on ec as an axis, in the manner indicated by the arcs o n and a x, until it coincides with the plane ertc, the line ao will then be represented by the line xn; then add the parallelogram xrtn, and the triangle ct/, deducting the triangle ers; then the edges of the plane eslc, inclined in the direction ec, and also in the direction cl, will lie perpendicularly over the plane eabf. From this we gather that the line co, being at right angles to to ec, must,in order to reach the point l, be lengthened the distance nt, and the right angle ec t be made obtuse by the addition to it of the angle tc l. By reference to Fig. 144, it will be seen that this lengthening is performed by forming the right-angled triangle cot, corresponding to the triangle cot in Fig. 146. The line ct is then transferred to Fig. 145, and placed at right angles toec; this angle ect is then increased by adding the angle t c l, corresponding to tcl, Fig. 146. Thus the point / is reached, and the proper position and length of the lines ec and cl obtained. To obtain the face-mould for a rail over a cylindrical well-hole, the same process is necessary to be followed until the length and position of these lines are found; then, by forming the parallelogram ecls, and describing a quarter of an ellipse therein, the proper form will be given.