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Surveying - Engineering

Q1:

A level when set up 25 m from peg A and 50 m from peg B reads 2.847 on a staff held on A and 3.462 on a staff held on B, keeping bubble at its centre while reading. If the reduced levels of A and B are 283.665 m and 284.295 m respectively, the collimation error per 100 m is

A 0.015 m

B 0.030 m

C 0.045 m

D 0.060 m

ANS:D - 0.060 m

hi = 283.665+2.847 = 286.512m.
Rl of point B = 286.512-3.462 = 283.048m.
Collimaton error = 284.295-283.048 = 1.247m in 25m.
Collimation error for 100m = 1.247*4 = 4.988m. Try to use it as a check and see after applying this correction you wouldn't get (.63) as δh rather use 4.980 as error per 100 meters you will get the correct answer. The distance between A and B is 75 possible then why taking only 25 by subtracting? Please explain in detail. Actual height difference between A and B = 284.295 - 283.665 = 0.63.

Height difference between A and B based on staff readings = 3.462 - 2.847 = 0.615.

Distance between A and B = 50-25 = 25 m.

Therefore collimation error per 100 m.

= (0.63 - 0.615)*(100/25).

= 0.060 m.