Draw Le parallel to JI, and Te perpendicular to JI. Draw Mh parallel to Le, and make Mh equal to the breadth of a step. In Le make Lg and ge each equal to the breadth of a step; and join Nh and he. Join also hf and fG. Produce hf to w, and draw wQ parallel to ƒ G. In the lines wh and wQ make wu and wv each equal to the hypothenuse of a step. Draw the curve uv to touch the straight lines wh and wQ, in u and v. The same being done below, where the two straight lines join at Q, the crooked line NhuvSZ will be the under edge of the falling-mould. From the under edge of the falling-mould draw a line, at the distance of the depth of the rail above it, and the falling-mould will be complete. 198. To find the Face-Mould, (fig. 2.)-Here nape is the convex side of the rail, being one quadrant, and pc, a tangent at p, being a portion of the straight part. Through C, fig. 1, draw AC perpendicular to dC, and make CA equal to the stretch-out or developement of the curve-line cpan, fig. 2. Let dC intersect the under edge of the fallingmould in X. Bisect AC in B, and draw BP and AO parallel to CX, meeting the under edge of the falling-mould in the points P, O:* 1. Having completed the inner line of the plan, fig. 2, draw the chord-line eg. Draw the lines ef, ab, cd, perpendicular to eg; ab being drawn through the centre q. Make ef, fig. 2, equal to CX, fig. 1; ab, fig. 2, equal to BP, fig. 1; cd, fig. 2, equal to AO, fig. In fig. 2, join ec, and draw al parallel to e c. Join fd: and produce fd to meet ec in m. Draw bl parallel to fm, and join lm, which produce to g. In lg take any point, k, and draw ki perpendicular to eg, meeting eg in i. Join fg; and draw ik perpendicular to fg. From g, with the radius gk, describe an arc intersecting ih in h; and join gh. In eg take any point, r. Draw rt parallel to gl, intersecting the inner curve of the plan of the rail in s, and the outer curve in t. Draw rR parallel to ef, meeting fg in R, and RT parallel to gh. In RT make RS equal to rs, and RT equal to rt; then will S be a point in the concave curve of the face-mould, and T a point in the convex curve of the face-mould. In the same manner we may find as many points as are necessary, and by this means complete the whole face-mould. In the same manner, by means of the three lines DX, ER, FS, fig. 1, we may construct the face-mould, fig. 3, in every respect similar to that in fig. 2. Then the face-mould, fig. 3, applies to the lower half where there are winders, and fig. 2, to the upper half. TO FIND THe moulds FOR A SEMI-CIRCULAR STAIR WITH A LEVEL LANDING. 199. To construct the Falling-Mould, (fig. 1, pl. LXXXI.)-Draw the straight line MN and IL perpendicular to MN, intersecting MN in K. From K, with a radius equal to that of the convex side of the rail, describe the semi-circle MIN. Make KL equal to the radius, together with three-quarters of it. Join LM, and produce LM to B; and join LN, which produce to C: then BC is equal to the developement of the semi-circumference of the winders. (See art. 46, Carpentry,) Proceed and complete the falling-mould as in the former cases. In this, FHG is the section of the lower flyer, DAS the section of the upper flyer, the whole height being three steps; the middle part of the falling-mould between AD and HF is level. This method gives the resting-point a, nearly, but not accurately; to find it correctly, draw a line parallel to ec, fig. 2, to touch the curve; or, from the centre of the circle, draw a line perpendicular to ec, and it will cut the curve in the restingpoint. Hence, instead of bisecting AC, fig. 1, in B, the point ought to be found by developing the exact place of the resting-point. 200. To describe the Face-Moulds.-Figures 2 and 3 exhibit the methods of tracing out the face-mould. In fig. 2 produce the straight line a5 to d; and, through the centre Y, draw Ye parallel to ad. Draw ac and YQ perpendicular to ad and Ye. Make the angles acd and YQe each equal to the angle ASD or HGF, fig. 1. Join de. Then, to find any point in the face-mould. In Ye take any point, m, and draw mu parallel to YQ, meeting Qe in u. Find the line ef, as in the description of plate LXXX, art. 198; draw uE parallel to eF, and m5 parallel to ed, intersecting the concave side of the plan in l, and the convex side in 5. Make uM equal to ml, u E equal to m5; then M is a point in the inside curve of the face-mould, and E a point in the outside curve: a sufficient number of points being thus found, the whole facemould may be completed; and the other part in the same manner. 201. Another Method. The preceding mode of obtaining the face-mould requires more thickness of plank than is necessary; hence, we propose to show how to do it in a more economical manner. Let ABCEF, fig. 3, be the plan of one quarter, with a straight portion of the rail. Join AC; for, on a little consideration, it will be evident, that all the resting-points of a plane on the squared rail would be on its convex side. From the centre, c, draw Be perpendicular to AC; then, B is the other resting-point. Let b be the place of the point B when put in the plan, fig. 1; and through b, from the point L, draw a line to BC, which gives the place of b on the developement. From Z, fig. 1, draw ZE, parallel to BC. From C and B, fig. 3, draw lines parallel to one another, as CU, and BD; and make CU equal to EU, in fig. 1, and BD equal to da, in fig. 1. Join UD, fig. 3, and produce the line till it meets a line drawn through the points BC; but, if it would require a greater extent of paper for them to meet, as in the engraving, then draw a line Aa, and another db, parallel to Aa; and make the line bd in the same proportion to Aa, as bm is to an, and draw a line through the points dA. Produce this line, and draw Ce perpendicular to it; also, make hg equal to CU, and from C, through the point g, draw the line Cf. Then, eCf is the angle which the plane of section makes with the base or plan. Therefore, to find any point in the mould, draw a line, from the point in the plan, perpendicular to eC, to meet the line Cf, thus let it be from the point F on the plan; then draw FH perpendicular to e C, and from the point H draw HI perpendicular to Cf, and equal to GF, by which the point I is determined. To find the joint-lines, produce AF to e, and make eƒ perpendicular to eC; then from ƒ, through T, draw a line, which is the line of the joint. Also, from c the centre, draw chg perpendicular to e C, and gi perpendicular to ƒC; make gi equal to hc, and join ¿C, which is the direction of the joint. 202. Plate LXXXII shows the falling-moulds and face-moulds for a rail, where the opening of the well-hole is only three inches; and, though the operation of finding the moulds for executing the rail is exactly the same as has been shown, their forms are entirely different. Fig. 1 is the plan, fig. 2 the falling-mould, and figures 3 and 4, the face-moulds; all of which are found by the method described in article 194 and 195. Application of the Moulds to the Plank. 203. THE face-mould, though last found, must be first applied to the plank, in the following First, bevel the edge of the plank, according to the angle Am I, shown in fig. 1, plate LXXVII, of the sections of a cylinder. Near the upper end of the bevelled edge apply manner. |