ANGULAR SURVEYING 63 



and on a latitude range, it is 



* _5/ 



The altered length of the course is then 



In these formulas, S g , S/. and Si have the same significance 

 as in the formula for e- I is the length of the corresponding 

 course; and /i and gi are the corrected latitude range and 

 longitude range, respectively. 



EXAMPLE. The accompanying table contains the bearings 

 and lengths of the courses of a compass survey. The lengths 

 as measured, and the ranges, as calculated from the measured 

 lengths and bearings, are printed horizontally opposite the 

 letters denoting the corresponding corners. Above these num- 

 bers are placed in parentheses the corrected values of the 

 lengths and ranges. Verify these corrected values and deter- 

 mine the relative error of closure. 



SOLUTION. First, determine the corrected latitude ranges. 

 Here the sum of the courses, or S/, is 29.55. The sum of the 

 northings is 10.00, and that of the southings is 10.10. There- 

 fore, the algebraic sum of the latitude ranges is S*= 10.00 

 + ( 10.10) = .10. Applying the above formula 



. 



S/ 29.55 2,955 

 Therefore, 



c t for AB=10.63X -.003= -.03 

 c,forBC= 4.10X-.003=-.01 

 c t for CD = 7.69 X - .003 = - .02 (See below) 

 c t for DA = 7.13 X - .003 = - .02 



The sum of these corrections should be equal to St, or .10, 

 but it is only .08. A correction of .01 therefore must be 

 applied to two of the ranges. As the lengths of the third and 

 fourth courses are nearly equal, 1 li. will be added arithmetically 

 to the correction for CD and that for DA, writing ct for CD 

 = .03, and c t for DA = .03. Subtracting algebraically the 

 corrections just found from the corresponding latitude ranges, 

 the corrected ranges are found to be 



