Ex9-6 A 30 m tape weighing 0.50 kg and with a cross–sectional area of 0.025 cm2 was standardized and found to be 30.006 m at 18o C, with 5-kg tension and supported at the 0 and 30 m points. The tape was used to measure a distance of about 120 m over terrain of a uniform 3 percent slope. The average temperature during measurement was 25o C, the tape was fully supported throughout, and tension of 4 kg was applied to each tape length. The observed distances were 30.0 m. 30.0 m and 30 m. Calculate the horizontal distance between the points.
Given a = 1.16 x 10-5 / C.
Solution : Slope temperature and standardization are in direct proportion to the length and can be calculated for the total distance. On the other, the sag correction must be calculated for each segment.
Distance corrected for slope
H = (s 2 – h 2)1 / 2 = [ (120.0)2 – (0.03 x 120) 2 ]1 / 2
= [ (120)2 (1 – 0.0009) ]1 / 2
= 120 x 0.999549 = 119.946 m
