Gl 



trades), and in which o=OK, oa=OA, and 5=AB, are the same 

 as in the primitive arch fig. 1 . 



Fig- 1. 



Fig. 2. 



Solution. On any vertical plane passing through BK, and not 

 coinciding with the plane of fig. 1, draw cod of the given length and 

 inclination, intersecting COD in O. Join Ce, Dd, and project the 

 whole of fig. 1 on the new plane by lines parallel to Cc, Dd. The 

 projection so obtained will he the figure of the arch and abutments 

 required. Moreover, if the lines B, B, fig. 1, represent in length, 

 direction, and position, the resultants of the pressures of the abut- 

 ments on their foundations in the original arch, then will r, r, fig. 2, 

 the projections of R, R, represent the corresponding resultants in 

 the new arch ; and in like manner, the thrust at a is the projection 

 of the thrust at A. 



W. J. MACQUORN RANKINE. 



Glasgow, 18th February, 1856. 



Note. The horizontal foundation courses in fig. 2 do not form part 

 of the projection of fig. 1, but are supposed to be added after the 

 completion of the projection. 



