TRANSITION CURVE. 



151 



[I] DF 



For, by similar triangles 



BK 



Hence BK 



BD 



BD' 



Again, BR ^ AB sin 



Hence 



BK 



BR 



4 



BD 



2 



"rT 



1 c 



8R^ 



% 



^A 



BK 



;BR 



(12) 



81.. 



lcv3c 



IcOc 



(approximately) 



and hence 



DF 



BR 



(J) From (11) and (12) it is evident that EF = 



DF 



or that E is the mid-point of DF. 



With the foregoing characteristics in mind the laying out 

 of the curve becomes a simple matter. The problem as usu- 

 ally met with will be the following : Given the intersection 

 angle I, the radius Re of the circular curve and the length Ic 

 of the transition curve, to locate the complete curve, that is, 

 the central circular curve and the transition curves at each 

 end. Both for simplicity and also to obtain more accurate 

 results it is preferable to run the first transition from A to B and 

 then the second transition from A' back to B' and then run in 

 the circular curve between these points with transit at B or B'. 



The first step, therefore, is to calculate the total tangent 

 AX. This is made up of three parts, as follows : 



