In Fig. 734 is shown the plan and elevation of an arch in a curved wall, such as might be used as the head of a window or door, the jambs and head to have the same splay. In the plan, C E D represents the inner curve of wall and A F B the outer. J G M in elevation represents the inner curve and K H L the outer. In order to arrive at a system of triangles by means of which to measure the splayed surface, first divide J G of the elevation into any convenient number of parts, in this case five, as indicated by the small figures. From the points in the curve thus established drop lines parallel to the center line G F, cutting the inner curve of the plan C E, as shown. Next carry lines from points in J G in the direction of the center X, intersecting the outer curve and establishing the points 7, 8, 9, etc., in it. From these points drop lines to the outer curve in plan A F, establishing the points 7, 8, 9, etc. Connect opposite points in J G with those in K H, as 1 and 7, 2 and 8, 3 and 9, 4 and 10, 5 and 11, and 6 and 12. Likewise connect 1 and 8, 2 and 9, 3 and 10, etc., as shown by the dotted lines. In the same manner connect corresponding points in the inner and outer lines of plan, as shown. The solid and dotted lines of the plan will form the bases of a series of right angled triangles whose altitudes can be found in the elevation, and whose hypothenuses will give correct measurements across the face of the splayed surface.

Fig. 734.   Plan, Elevation and Extended Sections of Splayed Arch.

Fig. 734. - Plan, Elevation and Extended Sections of Splayed Arch.

To determine the hights of these triangles extend to the left the horizontal lines drawn through K H of the elevation until they intersect lines dropped from J G, as shown by the points b, d. f and h. Then b 2. d 3, etc., will be the required hights. To construct the triangles which represent the solid lines in plan and elevation, proceed as shown in Fig. 735; For the first triangle, G H 7. draw the right angle G H 7, making the altitude equal to G H of elevation and the base equal to E F, 1 7. of plan, and draw the hypothenuse, which represents the actual distance between the points

1 and 7 of the levation or plan. In like manner the hypothenuse of the second triangle 2 b 8 shows the actual distance between the points 2 and 8 of the elevation; 2 b in Fig. 735 is made equal to 2 b of the elevation, while b 8 equals 2 8 of the plan.

The remaining triangles in Fig. 735 are constructed in a similar manner, each of the triangles representing a vertical section through the head on the lines of corresponding numbers in the plan.

The triangles shown in Pig. 786 correspond to similar sections taken on the dotted lines in plan. The base a 8 of the first triangle is equal to 1 8 of the plan, and the hight 1 a to 1 a of elevation, the point a of the elevation being on a level with 8, as shown by the dotted line 8 a. The hypothenuse 1 8 is then drawn, which gives the distance between the points 1 and 8 in plan or elevation. The bases of the remaining triangles are derived from the dotted lines in plan, and the hights from the distances 2 c, 3e, 4 g and 5 k of elevation.

Fig. 735.   Diagram of Triangles Based Upon the Solid Lines of the Plan.

Fig. 735. - Diagram of Triangles Based Upon the Solid Lines of the Plan.

Before the correct measurements or stretchouts for the inner and outer lines of the pattern can be obtained it will be neces-sary to develop extended sections of the inner and outer curves of the arch in elevation, as shown at the left and right of that view. For the development of the outer curve, as shown to the right by A' H' F', proceed as follows: On X M extended lay off a stretchout equal to A F of the plan. transferring it point by point. From the points thus established in A' F' carry lines vertically, extending them indefinitely, as shown, and then from the points in the outer curve K H of the elevation carry lines horizontally to the right, intersecting the corresponding lines just drawn from A' F', and through the points thus established trace the curved line A' IF. The development of inner curve, as shown to the left, is accomplished in a similar manner. On X J extended lav off a stretchout of C E of plan, and from the points thus established carry lines vertically. From the points in the inner curve J G carry lines horizontally to the left, intersecting lines of similar number, and through the points thus established trace the curve C' G'.

To describe the pattern shown in Fig. 737, first draw the line g h, or 1 7, in length equal to the hypothenuse 1 7 of the first triangle in Fig. 735. From 7 as center, with 7 8 of the development of the outer curve as radius, strike a small arc, as shown at 8 in the pattern. From 1 as center, with 1 8 of the first triangle in Fig.

Fig. 736.   Diagram of Triangles Based Upon the Dotted Lines of the Plan.

Fig. 736. - Diagram of Triangles Based Upon the Dotted Lines of the Plan.

Fig. 737.   Pattern for Splayed Arch

Fig. 737. - Pattern for Splayed Arch.

736 as radius, intersect the arc last struck, thus establishing the point 8. From 1 as center, with radius equal to 1 2 of the development of inner curve, strike a small arc, as shown at 2. Then from 8 as center, with 8 2 of the second triangle in Fig. 735 as radius, intersect the arc at 2 already drawn, thus definitely establishing the point 2 in the upper line of the pattern. Continue in this way, using the hypothenuses of the several triangles, as shown, for measurements across the pattern, and the spaces of the inner and outer curves as developed in Fig. 734 for the distances along the edges of the pattern, establishing the several points, as shown. Lines traced through the points from c to g and from a to h will complete the pattern for one-half the arch. The complete pattern is shown in Fig. 737.