RAILROAD CURVES 



175 



EXAMPLE. A spiral 600 ft. long connects a tangent with a 

 12 curve. Find the angle of deviation and deflection angle, 

 and angle NPC for a point 580 ft. from the P. Si. 



SOLUTION. The unit degree of curve, 



D. 12 



c = - = = 2 



L 6 

 and 1 = 5.8 stations; hence, d = JX2X5.8 2 = 33 38.4'. 



In determining the deflection angle 6, it is known that \ d 

 = 11 12.8'. Interpolating from the table, 



AT = 1.9' + J3X(2.4'-1.9')=2.0' 

 Therefore, = %d-N=U 12.8' -2.0' = 11 10.8'. 

 Coordinates of the Spiral. Let P, Fig. 2, be any point of a 

 spiral, and PR the perpendicular distance from this point to the 



FIG. 2 



original tangent. This perpendicular is represented by y, and 

 its value is given by the formula 



in which a and / have the same meanings as before. 





The value of M corresponding to any value of / may be taken 

 from the accompanying table. 

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