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& Turreli

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 . 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.*

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. 1. In fig. 2, join ec, and draw al parallel to ec. 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 R'T 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 wnole 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.

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