# K L G

## draw, shown, line, fig, hip and pattern

Page: 1 2 3

K L G. A line traced through the points thus obtained, as shown by X Y Z, will be the pattern for the heads.

Where a hip ridge is re quired to miter with the apron of a deck moulding, as shown in Fig. 329, in which B repre sents the apron of the deck cornice, A and A the hip ridges mitering at a and a, a slightly different process from that described in the preceding problem is used. In this case the part elevation of the mansard roof must first be drawn as shown in Fig. 330. Let A B C K represent the pact elevation of the mansard, the section of the deck moulding and apron being shown by D B E. Draw E X par allel to B C. E X then represents the line of the roof. In its proper position, at right angles to B C, draw a half-section of the hip mould, as shown by F G, which is an exact reproduction of B E of the deck mould. Through the corners of the hip mould at Y and G, draw lines parallel to B C, which intersect by lines drawn parallel to B .A from V, W, and E in the deck cornice. Draw the miter-line H I, which completes the part elevation of the mansard.

Before the patterns can be obtained, a developed surface of the mansard must be drawn. Therefore, from B (Fig. 330), drop a ver tical line, as B J, intersecting the line C K at J. Now take the dis tance of B C, and place it on a vertical line in Fig. 331, as shown by B Through these two points draw the horizontal lines B A and C K as shown. Take the projection J to C in Fig. 330, and place it as shown from to C in Fig. 331, and draw a line from C to B. Then will AB C K be the developed surface of AB C K in Fig. 330.

As both the profiles B V W E and F 1 G are similar, take a tracing of either, and place it as shown by D and respectively in Fig. 331. Divide both into the same number of equal spaces, as shown. Bisect the angle A B C by establishing a and b, and, using these as centers, by describing arcs intersecting at c; then draw d B, which represents the miter-line. Through the points in D and D', draw lines parallel to their respective moulds, as shown, intersecting the miter-line B d and the base-line C Cl.

For the pattern for the hip, draw any line, as E F, at right angles to B C, upon which place twice the stretchout of D, as shown by the divisions 6 to 1 to 6 on EF. Through these divisions draw lines at right angles to E F, intersecting similarly numbered lines drawn at right angles to B C from the divisions on B d and C C. Trace a line through the points thus obtained. Then will G H J L be the pattern for the hip ridge.

When bending this ridge in the machine, it is necessary to know at what angle the line 1 in the pattern will be bent. A true section must be obtained at right angles to the line of hip, for which proceed as shown in Fig. 330. Directly in line with the elevation, construct a part plan L M N 0, through which, at an angle of 45 degrees (because the angle L 0 N is a right angle), draw the hip line 0 M. Establish at pleasure any point, as PI on 0 M, from which erect the vertical line into the elevation crossing the base-line C K at P and the ridge-line C B at R. Parallel to 0 M in plan, draw 0' P, equal to 0 P', as shown. Extend as which make equal to PR in elevation.

Draw a line from R' to Then represents a true section on OP' in plan. Through any point, as a, at right angles to OM, draw bc, cutting L 0 and ON at b and c respectively. Extend b c until it intersects at d. From d, at right angles to RI, draw the line d e. With d as center, and de as radius, draw the arc e e', intersecting at e', from which point, at right angles to OM in plan, draw a line intersecting OM at e". Draw a ling from b to e" to c, which repre sents the true section of the hip after which the pattern shown in Fig. 331 is formed.

The pattern for the deck mould D B in Fig. 330 is obtained in the same way as the square miter shown in Fig. 277; while the pattern for the apron in Fig. 331 is the same as the one-half pattern of the hip ridge shown by n H 1 6.

Page: 1 2 3