190 RAILROAD CURVES 



EXAMPLE. Two tangents that intersect at an angle of 

 80 20' are to be connected with a 6 circular curve by two 

 equal spirals, each 300 ft. long. The tangents intersect at 

 Sta. 36. Lay out the two spirals and the circular curve. 



D c 6 

 SOLUTION. The unit degree of spiral c = = = 2; the 



spiral offset F = .072709 aL* = .072709 X 2 X 3 = 3.93 ft. ; CV = i 

 length-/ cor. = 150- .000127 &L?>= 150-0.1 = 149.9. R = "^ 



5,730 80 20' 

 = = 955 ft. Cr=| length-/ cor. + (R+F) tan 



= 149.9 + (955+3.93) tan 40 10' = 959.3ft. 



Since T is at Station 36, the station number of the P. Si is 

 36- (9+59.3) = 26+40.7. 



It will be assumed that stakes are set 50 ft. apart on the spirals 

 and at the even stations on the circular curve. The spiral deflec- 

 tions are then figured as shown in example under the heading 

 Angle of Deviation and Angle of Deflection. They are: 

 to first stake, 5' 



to second stake, 20' 



to third stake, 45' 



to fourth stake, 1 20' 



to fifth stake, 2 5' 



(A) Angles to be 

 deflected from the 

 tangent. Vernier 

 set at 0'. 



to P. 82 at 29+40.7, 3 0'^ 



The deviation angle A = i cL 2 = $X2X3 2 = 9. Therefore, 

 the central angle of circular curve = /-2A = 8020'-2X9 

 = 62 20'. The length of AB is therefore 62 20' -f- 6= 10.389 

 Sta. and the station number of B is 29+40.7+ (10+38.9) 

 = 39+79.6. 



The angle between the chord CA and the tangent to the 

 circular curve at A is A - =9- 3 = 6. 



Transit at P. Sz. The deflection angles to the stakes on the 

 circular curve are as follows: 



to Sta. 30, .593X3= 147'; to Sta. 35, 1647' 

 to Sta. 31, 4 47'; to Sta. 36, 19 47' 



to Sta. 32, 7 47' ; to Sta. 37, 22 47' 



to Sta. 33, 10 47' ; to Sta. 38, 25 47' 



to Sta. 34, 1347';toSta.39,2847' 



to B, 3110' 



(B) Angles to 

 be deflected 

 from tangent 

 to circular 

 curve. Ver- 

 niersetat60'. 



