232. RAILROAD LOCATION 



and c' along DV are equal to those for the corresponding points 

 along CV. That is, if Ca = Da', then am = a'm'; if Cb = Db', 

 then bm b'm', etc. 



EXAMPLE. A +.4% grade meets a .5% grade at Sta. 190, 

 the elevation of which is 161.3 ft. If a vertical curve 400 ft. 

 long is inserted, what is the correction in grade and. the cor- 

 rected grade elevation at each station and half station? 



SOLUTION. In this example, VC= VD = 2QO ft. The ele- 

 vation of C is 161.3 -2X. 4 = 160.5 ft., = ; that of D is 161.3 

 -2X.5=160.3 ft, = E'; that of K is $ ('+) = J X (1G0.5 

 + 160.3) = 160.4 ft.; and that of V is H= 161.3 ft. Substi- 

 tuting these values in the formula for VM, 



VM= iX (161.3-160.4) = .45 ft. 



Since, for the first stake, Co = 50 ft. and CF = 200 ft., the 

 formula for am gives 



am= I J XV M = X.45=.03 ft. = a'm' 



Similarly, 



*I^I "-"- 4 



_JL _ 



The original and corrected grade elevations are as follows: 



188 160.50 .00 160.50 

 + 50 160.70 .03 160.67 



189 160.90 .11 160.79 

 +50 161.10 .25 160.85 



190 161.30 .45 160.85 

 +50 161.05 .25 160.80 



191 160.80 .11 160.69 

 +50 160.55 .03 160.62 



192 160.30 .00 160.30 



Vertical Curve at a Sag. If two grade lines, AV and VB, 

 Fig. 4, meet so as to form a sag, the vertical curve will evi- 

 dently be 'wholly above both grade lines. Using the same 



