ENGINEERING SURVEY I $ II
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Problem solved: 2.1 The readings obtained..
The readings obtained from a two-peg test carried.. 2.1 The readings obtained from a two-peg test carried out on an automatic level with a staff placed on two pegs A and B 50 m apart are: With the level midway between A and B Staff reading at A = 1.283 m Staff reading at B = 0.860 m With the level positioned 5 m from peg B on line AB extended Staff reading at A = 1.612 m Staff reading at B = 1.219 m Calculate the collimation error of the level per 50 m sighting distance and the horizontal reading that should be observed on the staff at A with the level in position 5 m from B. Expert Answer ♥♥♥ S1 = 0.860 m S2 = 1.283 m S3 = 1.219 m S4 = 1.612 m From Equation of collimation error e = (S1-S2) – (S3-S4) per L m e = (0.860 – 1.283) – (1.219 – 1.612) per 50m = (-0.423-(-0.393)) = -0.030m per 50m For the instrument in position 5 m from peg B, the horizontal reading that should have been obtained with the staff at A is S’4 = S4 – [Collimation error per meter x sighting distance DA] = 1.612 – 55 = 1.645m This is checked by computing S’3, where The readings obtained..
Problem Solved: 2.12, Collimation Error, The observations..
Collimation Error 2.12 The observations listed below were recorded when testing the adjustment of an automatic level. Use them to calculate the collimation error in the level. Level Position Staff Reading Sighting distance Staff Reading L1 A 25m 1.225m L2 B 25m 1.090m L3 A 55m 1.314m L4 B 5m 1.155m Expert Answer ♥♥♥ ‘ Problem 2.12 Instrument at I1, Instrument at I2, True difference in level between A and B Apparent level difference between A and B = 1.225 – 1.090 = 1.314 – 1.155 = 0.135m A is lower = 0.159m A is lower Therefore, connecting reading at A = Reading at B + 0.135 = 1.155 + 0.135 = 1.290m Observed reading at A = 1.314m 1.314m > 1.290, the line of collimation is upward elevated, Collimation Error C’ error in 50m, = 1.314 – 1.290 = 0.024m C’ error in 55m, = 0.024 x = 0.0264m