RAILROAD LOCATION 



233 



notation as before, the correction in grade at the point V will 



be vu- 



The correction in grade at any point a will be given by the 

 preceding formula for am, as before, but this correction is nov? 

 to be added to the old elevation of grade at a to obtain the 

 corrected elevation. 



EXAMPLE. The grade of CV, Fig. 4, is 1.2%, that of 

 VD is +.6%, and the elevation of V is +49.2 ft. Find the 



FIG. 4 



corrections in grade and the corrected elevations at stakes 

 100 ft. apart, if the length of the vertical curve is 600 ft. 

 SOLUTION. The uncorrected grade elevations are as follows: 



Along CV 



At first stake 



At second stake 



At third stake 



At fourth stake, V. 



52.8 

 51.6 

 50.4 

 49.2 



Along VD 



At fifth stake 49.8 



At sixth stake 50.4 



At seventh stake, D 51.0 



Therefore, } (+') = * (52.8+51.0) = 51.9; and, by the pre- 

 ceding formula, 



VM= } (51.9-49.2) = 1.35 ft. 

 The formula for am may now be applied. 

 Correction in grade at second stake, 100 ft. from C, is 

 2 1 



X 1.35 = -X 1.35 = . 15 = correction at sixth stake. 



/HKA 

 \300/ 



Correction at third 

 'correction at fifth stake. 



stake ' '55 



X1.35 = -X 1.35 =.60 



