Fig. 416. - Diagonal Section.

Fig. 417.   True Face of Octagonal Side and Part of Pattern.

Fig. 417. - True Face of Octagonal Side and Part of Pattern.

Patterns for a Hip Molding Mitering Against the Bed Molding of a Deck Cornice on a Mansard Roof which is Square at the Base and

Octagonal at the Top.

At right angles to the line of the molding in the true face lay off a stretchout equal to that portion of the profile thus used, as shown by P N, through the points in which draw measuring lines in the usual manner. Place the T-square at right angles to the lines of the molding in the true face, and, bringing it against the several points in the miter line between the hip and bed molding at E2,. cut corresponding measuring lines drawn through the stretchout. Then a line traced through these points, as shown by S T, will be the miter line for that portion of the pattern corresponding to the part of the profile thus used.

For the other half of the hip molding, being that portion which lies on the face of the transition piece, another operation must be gone through. Construct a section of the roof corresponding to the line F G in the plan. At any convenient point lay off F1 C in Fig. 415, equal in length to F G. From the point C1 erect a perpendicular, C1 B3, in length equal to C B of the section. Connect F1 and B3. Then F1 B3 is the length of the transition side of the roof through that portion corresponding to F G of the plan. It will be well to add to this at B3 a section of the bed mold as it appears in the section below, thus establishing the true relation between it and the transition side of the roof. By means of this section and the plan, construct a true face of one-half the transition side of the roof, by means of which to obtain the miter of the remaining portion of the roll. To do this first redraw the section B3 F1, placing the line of the roof in a vertical position, as shown by B4 F2, Fig. 417, from the points in which project horizontal lines, as shown to the right, upon each of which set off from an assumed vertical line the width of the roof as given in the plan. Thus make G1 E4 equal to G E of the plan, and F3 D4 equal to F D. Connect D4 and E4. Then G1 E4 D4 F3 is the true face of that portion of the roof represented by G E D F in the plan.

In connection with the vertical section just described place so much of the stay as was not used for the pattern already delineated, and in the elevation of the transitional face of the roof place a corresponding portion of the profile, as shown, each of which divide into the same number of spaces. From the points thus obtained carry lines parallel to the lines of the respective views of the part, those in the vertical section cutting the bed molding, and those in the elevation being produced indefinitely. From the points in the bed molding of the vertical section carry lines horizontally, intersecting those drawn from the profile in the elevation, thus establishing the miter line, as indicated at E4. At right angles to the line D4 E4 set off a stretchout of the profile, as shown by R P2 through the points in which draw the usual measuring lines. With the T-square placed parallel to this stretchout line, or, what is the same, at right angles to the line D4 E4, and being brought successively against the points in the miter line at E4, cut corresponding measuring lines, as shown. Points also are to be carried across, in the same manner as described, corresponding to the bottom of the apron or fascia strip in connection with the bed molding. Then a line traced through these points, as indicated by the line drawn from U to T1, will be the pattern of the other half of the hip molding. By joining the two patterns thus obtained upon the dividing line of the stay, corresponding to P T of the first piece or P2 T1 of the second piece, the pattern will be contained in one piece.

Fig. 415.   Plan, Elevation, True Face and Part of Pattern.

Fig. 415. - Plan, Elevation, True Face and Part of Pattern.

Fig. 418.   Diagonal Section.

Fig. 418. - Diagonal Section.

Fig. 417.   True Face of Octagonal Side and Part of Pattern.

Fig. 417. - True Face of Octagonal Side and Part of Pattern.

Patterns for a Hip Molding Mitering Against the Bed Molding of a Deck Cornice on a Mansard Roof which is Square at the Base and correspond to the point 8. Locate the point 8 on the first section of the hip obtained near O, as shown, and use the remainder of profile 8 to 14 for another operation. Lay off a stretchout of the entire prorfile of the hip molding, as shown by W V, through the points in which draw the usual measuring lines. With the T-square placed at right angles to the lines of the hip, as shown in the true face, and brought against the points in the miter line S T U, cut so many of the measuring lines drawn through the stretchout W V as correspond to those points. By this means that portion of the pattern shown by S1 T1 U1 will be obtained. For the portion of the pattern corresponding to the part of the profile which miters against the other hip mold it will be necessary first to construct a true face of the octagon side of the roof. To do this, obtain a diagonal section of the roof corresponding to the line D E in the plan, viz.: Lay off D2 E1 equal to D E of the plan, and from E1 erect a perpendicular, E1 A2, equal to C A of the section in Fig. 419. Connect A1 and D2. Then A2 D2 is the length of the diagonal face of the roof measured on the line D E of the plan. Upon any convenient straight line lay off D4 A4 in Fig. 420, in length equal to D' A2, and from A4 set off, at right angles to it, A4 C3, in length equal to E C of the plan. Then D4 A4 C3shows in the flat one-half of the diagonal face of the roof, or what is represented by DEC in the plan. At right angles to D4C3draw the remaining portion of the stay not used in connection with the true face, placing it in such a manner that the point O3 corresponding to O of the hip section, shall fall upon the line D4 C, which represents the angle of the hip. Through the point 8 of the section L5,M7 corresponding to 8 of the section L2 M2 of Fig. 418, draw a line parallel to D4 C3, as shown by S2 Y1.Then S2 Y1 corresponds to S Y of the true face in Fig. 418.