RAILROAD CURVES 177 



This correction must be subtracted from the half length of 

 spiral, expressed in feet, to obtain the distance CV, in feet. 



The values of F and t are given in the fifth and eighth 

 columns of the tables for transition spirals, which follow. 

 The value of I in the first column, corresponding to which is 

 found F and the t correction, is to be taken as the whole length 

 of the spiral. 



EXAMPLE. Find the distances EV and CV for a spiral 

 400 ft. long that connects with a 2 curve. 



D 2 

 SOLUTION. Here, a = ~ = = \ . 



The whole length of spiral is 4 sta. Therefore, substituting in 

 the formula, F = .072709 X * X 4 = .072709 X X 64 = 2.33 ft. 



By the formula for the t correction, t = .000127 X 2 2 X 4 s = .033 ft. 

 Therefore, CF=|X400 ft. -.033 ft. = 199.97 ft. 



FIG. 3 



The Middle Point of the Spiral Offset. If M', Fig. 2, is the 

 middle point of the spiral, and M'K is the offset from the < 

 original tangent, M'K is almost exactly equal to one-half the 

 spiral offset VE. The distance CK from the P. Si to the foot 

 of M'K is almost exactly equal to the distance CV from the 



