. The Civil engineer and architect's journal, scientific and railway gazette. Architecture; Civil engineering; Science. Fig. ii m Let L S I and M s ni be two square sections at right angles to the axis, and passing through the points S and s respectively. b K=6 which is called the obliquity of the bridge. a= axil length (see Buck on oblique arches). *â R=^E F, E G radii of the semicylinder respectively. w=H N, H=3-14, &c. V « V m=X and V S V Z=X'. Let X, y, z, be the rectangular co-ordinates of the point s, origin at H, Z measured vertically, and x on the line H K. First for this poin
. The Civil engineer and architect's journal, scientific and railway gazette. Architecture; Civil engineering; Science. Fig. ii m Let L S I and M s ni be two square sections at right angles to the axis, and passing through the points S and s respectively. b K=6 which is called the obliquity of the bridge. a= axil length (see Buck on oblique arches). *â R=^E F, E G radii of the semicylinder respectively. w=H N, H=3-14, &c. V « V m=X and V S V Z=X'. Let X, y, z, be the rectangular co-ordinates of the point s, origin at H, Z measured vertically, and x on the line H K. First for this point s we have the following equations: a: = rco3. X, y = n-\-°'-^, Z=r Sin. X (V) 11 ^ ^ These equations are too obvious to need any further explanation. Next, to find the equation of the elliptical face a F 6, let x' y' z' be the rectangular co-ordinates of any point iu this face, referred to the same axis and origin as before, Y the angle made by a section passing through this point parallel to the section L S I (1 + COS. Y) 6 . y' = rcos. X,* y'= â , z' = r sin, X (2) * This equation is obtained m the following manner :â S'' = tan. z E 0 H X (,⢠+ r los. Y) but tan. .d E o H = .,t y = _(l+cos. V)& as stated above. At the point of intersection of these two lines, namely, the spiral and the ellipse, which is formed in the skew elevation, we must have the following obvious conditions:â x = x', y=y', z=^z' .â¢.X = Y ,, ., a X (1 + cos. Y) h Consequently w+ â= By reduction we have xâ-âcos. X = (6â2w)â (3) 2 o 2 a the equation for the intrado. The equation to the exfrado is a similar one to that of the intrado, viz.â X'-^ COS. X' = (6'-2?b') â (4) 2 a ^ 2a where 6' = 6 + 2 K K'and 7v' = w + Â¥.'K .â¢.6'-2j»' = 6-2j!; + 2K K'-2 KK' = 6-2 w by similar triangles 2r : (R-r) : : 6 : K K' R6-br ,. 6R .2KK' 6'=" r r Substituting these values in equation (4) we shall have for the equation of the extradoâ X , - COS. X' = (5-2 7») â (5) r ^ 2 o Now it
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