This problem, like many others pertaining toman-Bard roofs, may reach the pattern cutter in drawings either more or less accurate. and in different stages of completion. Certain facts, however - viz., the profiles of the moldings, the pitch of the roof and the angle in plan - must be known before the work can be accomplished; but with these given the pattern cutter will have no difficulty in drawing such elevations as are necessary to produce the required patterns.

In Fig. 410, let A B C D be the given section of the mansard trimming shown, A C the profile of the bed molding and apron, and B D E the pitch of the roof. According to the statement of the problem above the angle of the plan is octagonal; it might be a special angle either greater or less than that of an octagon, but the principle involved and the operation of cutting the patterns would be the same. As in all other problems connected with mansard trimmings, the first requisite is an elevation of the "true face," in order to obtain the correct angle between the bed mold ing and the hip molding. A normal elevation, such as is likely to be met with in the architect's drawings, is shown in the engraving at the left of the section, merely for purposes of design. In obtaining the true face, shown below, it is best to use the section and plan only. Therefore, redraw the section as shown immediately below it, placing the line of the roof in a vertical position, all as shown. From all the points of this section lines may now be projected horizontally to the left, as the first step in developing the required true face. Immediately above the space allotted to the elevation draw a plan of the horizontal angle, as shown by I E1 K. As it will be impracticable to include the entire profile of the roof in the drawings, some point must be assumed at a convenient distance below the bed mold, as D, from which to measure hight and projection, which locate also in the section below, as shown at D1, making B1 D1 equal to B D. From A draw a line at right angles to the line of the roof, meeting it at B, which point may be assumed for convenience as the upper limit of that part of the roof under consider-ation. Now, from the point B drop a vertical line, which intersect with one drawn horizontally from D, as shown at E; then D E will represent the projection.

From the lines I E1 and E1 K. upon lines at right angles to each, set off the projection D E. as shown at I F and K H; through the points F and H draw lines parallel to the first lines, meeting in G; then a line from G to H will represent the plan of the angle or hip of that portion of the roof of which B D is the profile. Now, to complete the true face of that part of the roof drop a line from the point E1 intersecting the line from B1 at L, and one from G intersecting the one from D1 at M; then the angle B1 L M will be the correct angle of the miter between the bed mold and the hip mold.

As in Problems 101 and 102 preceding, it will next be necessary to obtain a correct section of the hip mold on a line at right angles to the line of the hip. To avoid confusion of lines, this operation is shown in Fig. 411, in which E1 G1, the base line, is made equal to E1 G of the plan in the previous figure. At the point F2 erect a perpendicular, making it equal in hight to B E of the sectional view. Connect B2 with G1, which will give the correct angle of the hip of the roof. As a means of constructing a correct section at right angles to this line, assume any two points on the original plan, as N and 0, equidistant from G and connect them by a straight line, cutting the angle or hip line in P. Set off from G1 on the line G1 E2 of Fig. 411 a distance equal to G P of the plan, as shown at P1, from which draw a line parallel to the hip G1 B2. Next intersect these two lines by another at right angles at any convenient point, as shown by P2 Q. From the point P2 set off the distances P2 O1 and P2 N1, making them equal to P O and PN. Connect the points 0' and N' with R, which is the intersection of P2 Q with the hip line; then the angle O1 R N1 will be a correct section of the roof upon the line P2 Q or upon any line cutting the hip at right angles, upon which the finished profile of the hip mold may now be constructed as follows: Setoff the projections of the fascia and fillet as given in the sectional view, Fig. 410, from the lines R O1 and R N1, continuing their lines to the center line P2 Q. From the intersection S as a center, with a radius of the bed mold, describe the roll.

As stipulated in the statement of this problem, the profiles of the bed mold and hip mold are to correspond. By this it is understood that the curves of their molded surfaces are alike and struck with the same radius, and so placed as to member or miter.

Fig. 411. - Diagonal Section of Hip,

Fig. 410.-The Pattern of a Hip Molding Upon an Octagon Angle, Mitering Against a Bed Molding of Corresponding Profile.

As the curve of the bed mold is only a quarter circle, while that of the hip mold is nearly three-quarters of a circle, it will be seen that the quarter circles in each half of the hip mold next adjacent to the fascias and fillets will miter with the arms of the bed mold on either side of the miter, and that a small space in the middle of the roll will remain between them, which must be mitered against the planceer, and the object of the operation shown in Fig. 411 is to determine exactly what this space is. The dotted lines (nun S to the points 10 drawn at right angles to R O1 and R N1 show the limit of the quarter circles or the parts that must miter with the bed mold, while the space between them (10 to 10) shows the part that must miter against the planceer. It might be supposed that the angle between the fascias of the hip mold, to fit over the angle of a mansard which is octagonal in plan, would be octagonal, but the demonstration shows that while the angle N GO of the plan, Fig. 410, is that of an octagon, the angle N1RO1, Fig. 411, is greater, because the distance N1 O1 is equal to N O, while the distance R P2 is less than G P, R P2 being at right angles to the line of the hip and G1 P1 being oblique to it.

The true face, Fig. 410, may now be completed, as follows: Upon any line, as S1 T, drawn at right angles to L M, representing the face of the roof, draw a duplicate of one-half the profile of the hip mold obtained in Fig. 411, placing the point S upon the line L M, as shown. Lines drawn through the angles of this profile parallel to L M will intersect with lines from corresponding points from the profile A1 C1, previously drawn, giving the miter line J X and completing the elevation of the true face.

Upon any line at right angles to the line L M, as U V, lay off a stretchout of the complete hip mold as obtained from the half profile S1 T, through which draw measuring lines as usual. Drop lines from the points from 1 to 10 of the profile, parallel to L M, cut-ting the miter line; then, with the T-square placed at right angles to L M and brought successively against the points in J X, intersect them with lines of corresponding number in the stretchout; then lines traced through the points of intersection, shown by d b and a c, will give the pattern for that part of the profile from 1 up to the point 10. The pattern of that portion of the roll which miters against the planceer muse be obtained from the diagonal section of the hip. From points 10, 11 and 12 in Fig. 411 carry lines parallel to G1 B2 intersecting the line of the planceer, as shown at W. It is only necessary to ascertain how much shorter the lines 11 and 12 are than the line 10, and then to transfer these distances to the pattern. This can be done by dropping lines from the intersection of points 11 and 12 with the planceer, in Fig. 411, at right angles to G1 B2, cutting line 10. These distances can then be transferred to line 10 of the pattern, Fig. 410, measuring down from the point 10 of pattern already obtained, after which they may be carried parallel to U V into the measuring lines 11 and 12, thus completing the pattern.

This portion of the work is necessarily very minute in the drawing, but it will be easily seen, in applying the principle to other similar cases, that if the angle of the plan I E1 K were less than that shown, for instance, if it were a right or an acute angle, a greater distance or more points would occur between the points 10 and 10, and further, that if the angle of the roof were less steep a greater curve or dip would occur between those points (a to b) of the pattern.