Figure G.—Given K L M N, the plan of an oblong dome, and the rib A a; to find the hip and the rib parallel to the longi tudinal side, also the covering upon the longitudinal and transverse sides.
Divide the curve A a into any number of equal parts, at the points of division 1, 2, 3, and draw lines 1 k, 2 i, 3 parallel to N K, the longitudinal side cutting the seat of the hip x,o : from the points of intersection in K G draw lines parallel to K L, the breadth of the dome, to the points n, o ; draw c E parallel to N K, and produce it to F; also produce A C to D ; take the parts of the given rib A B, and extend them on c D, front c to 1, from 1 to 2, from 2 to 3, and from 3 to D : make 1 in, 2 n, 3 0, on each side of c D, respectively equal to the parallel distances from A K. comprehended be tween the lines A 0, and K G ; from the points d, e, cut by the lines parallel to K L, make e b, and f c, respectively equal to the several heights of the given rib A a, and trace a curve through the points E, a, 6, c A: upon the straight line E F, extend the parts E a, a b, b e A, of the arc E from E to a, from a to 6, from b to c, and from e to F ; through the points a, b, e, in E F, draw k k, i h h, parallel to K L ; make the parts a k, b e h, respectively equal to the parallels of E x, comprehended between E G and K ; then K F L, is the form of the boarding for each end, and L D m, that of the sides.
The angle-rib is found the same as in the square dome.
Figure 7, No. 1.—To find the covering of an oblong poly gonal donne.
Given the plan ABCDEFGHT, and the axal section through its breadth, a setni-circle. Take any straight line, N Q, No. 2; in No. 1, draw lines from the middle point f, perpendicular to the sides o H, n A, A a, of the polygon, and let these perpendiculars be I K, I H, and I L : on the straight line m o, No. 2, make m o equal to I m r equal to n, and N Q equal to i L, No. 1. In No. 2, draw m N perpen dicular to m o ; from the centre m, with the radius id o, describe the quadrant o N: divide the arc o N into any num ber of equal parts, say four, w z, z y, y x, x at the points z, y, x; from the points x, y, z, No 2, draw lines x y z w, cutting m o at a,r,w ; transfer the distances m at, It v, v2v, w o, to the straight line I K, No. 1. Through the points of divi
sion, draw lines parallel to it G, to cut the diagonals a i, and H I ; from the points of section in H r, draw lines parallel to H A, to cut the next diagonal A I: from the points of section in A I, draw lines parallel to A a, cutting a r: take the parts o z y, y x, x N, No. 2, and extend them to x a, No. 1, and draw lines through the points of section, parallel to n ; make the two parts of the lines so drawn on each side of x respectively equal to the lines drawn parallel to H o, termi nated by K 1, and H I. With the greater semi-axis m P, and lesser semi-axis m x, describe the quadrant of an ellipsis N P m : through the points z, y, x, draw lines parallel to at r, to cut the elliptic curve : extend the elliptic curve so cut, upon the straight line n s, and through the points of section, draw lines parallel to n ; transfer each of the lines contained between II i and A r, respectively, to each of the parallels on the other side of nI A ; or draw lines through the several points of sec tion in the diagonal A n. paral!el to t s, to cut the lines perpen dicular to Ii s, and the points of intersection will form the direction of the line for the edge of the covering n S A. In like manner, by describing the quadrant of the ellipsis m Q N, and by drawing lines through the points z, y, x, to cut the curve N Q, by proceeding as before, we shall obtain the cover ing A n T. The three coverings c a n, It S A, A T B, cover more than one-quarter of the whole, by the half coverings c a a, and L T B, of the side and end. The covering of each side so found answers to the opposite side, by turning the back for the front. Each covering, except that upon the side, and that upon the end, will cover four different sides : the covering upon the sides and ends only answer in two oppo site places.
The angle-ribs of this dome are found as usual. If the cir eutnference of each quadrant or rib perpendicular to the side, were ascertained by calculation, then the boarding could easily be laid out without the use of a plan, upon the same prin ciple as in Figure 5, as is obvious from what has already been said.