Example |
Ex12-1 A dumpy level was set up midway between two peg points 100m apart. The readings on the staff at the two pegs were 1.650m and 1.210m respectively. The instrument was then moved, by 10 m ahead of the second peg, in line with the two pegs. The respective staff readings were 1.405m and 0.935m. Calculate the staff readings on the two pegs to provide a horizontal line of sight. Solution: Let A and B be the two peg points at a distance of 100m and the instrument position be C when midway and D when 10m from B. While the instrument is at C, True difference in elevation between the peg points is obtained and is given by, d = 1.650-1.210 = 0.440m (B higher than A) Assuming there is no error in the line of collimation of the instrument, while the instrument is at D, If the staff reading at B i.e., 0.935 is correct. Then, the staff reading at A should be = (0.935 +0.440) m = 1.375 m (< observed reading i.e., 1.405m) This shows that the line of collimation of the instrument is not in adjustment and it is inclined upward. The amount of error = 1.405 - 1.375 =0.030m in 100m Thus, error in 10m i.e., the staff reading at B while the instrument is at D = 0.003m The correct staff reading at B while the instrument is at D should be = 0.935- 0.003 = 0.932m And that at A, 0.932+0.440=1.372m (Verification: The correct staff reading at B while the instrument is at D should be = 1.405 - 0.003 x 11 = 1.372 m) |