364 



THE CIVIL ENGINEER AND ARCHITECTS JOURNAL. 



[October, 



ON THE CONSTRUCTION OF OBLIQUE ARCHES. 



Sir — In compliance with the request contained in your letter of the 

 Kith, I forward you the metliod by which the formula and construction 

 given in your last number, for finding the angles between the joint lines 

 in the face and soffit of an oblique arch, were obtained. The approxi- 

 mations thus arrived at are as accurate as it is possible to work to, 

 the discrepance being much too small to be detected practically. As 

 the subject is somewhat complex, I must be excused if the explanation 

 here given is not strictly mathematical, the deductions however will 

 be found correct. 



Fig. 1. r=-; 



Let A B Cfig. 1.) be the elevation of an oblique arch, on a plane at 

 right angles to the axis of the cylinder on which it is formed, and let 

 c be the centre, and a the point at which any joint a m in the face of 

 the arch meets the intrado. 



Fig. 2. 



Let C D E (fig. •2,") be the plan of the same, (part of the arch being 

 supposed to be removed), and let o' be the position in plan of the 

 point a (fig. 1.) Suppose a vertical plane P P to pass through the 

 point a', intersecting the spiral surface c/ a' m' in the line a' V. From 

 /' let/ ' m' be drawn a tangent to the spiral line of the extrado meet- 

 ing D E in m'. And because the plane P F is perpendicular to the 

 axis of till' cylinder, the line of intersection /' a' will be a straight line 

 at right angles to the tangent I' m'. The triangle a' I' m' is therefore 



the projection of a right angled triangle; and in the short distance 

 /' m' the tangent very nearly coincides with the plane of the spiral 

 surface, hence the angle I' a' m' will be a very close approximation to 

 that formed between the chord of the curved joint in the face and the 

 line /' a' \ which is the angle represented by ii / o, fig. 3, and that 

 which it is required to find. 



FiS- 3. 



From o' where the axis intersects D E in plan draw o' u' parallel to 

 /' m', meeting P P in j.', produce r o' to c'. In fig. 1 let / a be the pro- 

 jection in elevation of the line /' a' ; join a c and produce it to u, mak- 

 ing I a : a u : : I' a' : a' u'. Through u draw a o at right angles to 

 a 0, meeting the vertical line B c produced, in o; and through 7 draw 

 i m at right angles to a I, meeting o a produced, in 7ji. 



From the nature of the spiral surface it follows that / a and a u will 

 be in the same straight line, and without going through the detail of 

 each part of the construction, it will be seen that the triangle repre- 

 sented by o' o' It' in plan, and a o u in elevation, is a right angled 

 triangle similar to, and in the same plane with the trangle a' m' I', 

 having the angle o' a' u' equal the angle I' a' ;«'. Also that the side 

 represented by a v and a' o' is the hypotheneuse, and is the line drawn 

 from tlie point where the joint intersects the intrado to the point of 

 convergence of the joints in the face; and that the side represented 

 by H and o' «', which is the perpendicular of the triangle a u o, is the 

 hypotheneuse of another triangle whose perpendicular is c' o', and the 

 angle c' o' ii! equal to the angle of extrado. In the construction given 

 in your last numhoi', it will be seen that these are the two sides made 

 use of to obtain the angle «' a' o' =il' a' hi', but in the formula it is 

 more convenient to work with the base of the triangle w u' o', which 



is the line a 

 have given 



z 

 z 



L 



and a c 

 then drawing a c 



u in fig. 1, and represented by a' ti in fig. 2, for when we 



C D E ^ 6 the angle of obliquity. 



c' o' u' = (j> the angle of extrado. 



o c B := A the angle from the vertical, 

 r the radius ; 



t right angles to B c fig. 1, 

 (Fig. 1> a c =: )■ sin. \, =; a' c' (fig. '2). 

 (Fig. 2) o' c' = )• sin. A, cot. e, 



(Fig. 2) o' u' = r sin ^, cot. 8, sec. <f = the perpendicular. 

 (Fig. 1) !< =; f sin. a, cot. 9, tan 9. 



(Fig. 1; u c:=r sin. A, cot. 0, tan. ip, cot. A =: »•, cos. A, cot. 9, tan. <f. 

 (Fig. \) a u:=r ■\- (r cos. A, cot. 9, tan. <p) = the base. 



_, , , , , '■ sin. A, cot. e, sec. a> 



Therefore tan Z u' a' z:z — — — 7 



r -\-(r COS. A, cot, e, fan. <f) 



Separating the constant from the variable quantities, and calling 



r cot. 6, sec. <p =: a, 



antl )■ cot. e, tan. (p =: 6, 



. , , „ , , " sin. A 



tan. Z li a := tan. Z I a m = --; — . 



(6 COS. A) + r 



Hence if the arch be a semicircle on the square section, the expression 



for the angle I' a' m' at the springing will be tan. Z /' a' m' = cot. fl, 

 sec. f. 



