CITY SURVEYING 133 



If Lo denotes the unsupported length of the tape, w the weight 

 of tape per unit of length, and P the pull, the shortening s due 

 to the sag is given by the formula 



24P* 



It should be observed that Lo is the length of the unsup- 

 ported part, which may not be the entire length of the tape. 



Since, when the tape sags, the distance between its two sup- 

 ports, as indicated by the nominal length of the tape, is 

 greater than the actual distance, or the length of the chord 

 subtended by the arc, the correction for the sag is negative, 

 and must be subtracted from the nominal length indicated 

 by the tape. If the length of a line, as measured, contains 

 n times the length Lo, and the sag is the same in all measure- 

 ments, the correction for sag is 



nw n -L<? 

 nS = ^R^ 



EXAMPLE. A line as measured with a 100-ft. tape weighing 

 .007 Ib. per ft., with a pull of 14 lb., is found to be 400 ft. 

 Determine the correction for sag. 



SOLUTION. Here, n = 4, w = .OQ7, Lo=100. and P=14. 

 Substituting these values in the formula, 

 4X.0072X100 3 



"15 042ft - 



If it is desired to pull the tape just enough to cause the stretch, 

 which is a positive error, to balance the sag, which is a negative 

 error, the proper pull P may be found by the following formula : 



24 



EXAMPLE. The weight of a 100-ft. tape is .008 lb. per ft., 

 and the sectional area is .002 sq. in. Taking E as 28,000,000 

 lb. per sq. in., determine the pull necessary fx> neutralize the sag. 



SOLUTION. In this example, w = .008, Lo= 100, A = .002, and 

 = 28,000,000. Substituting these values in the formula, 



o .0082 X 1002 X .002X28,000,000 



r* = A / = H-4 lb. 



24 



