Question:
Two points A and B are 1530 m apart across a wide river. The following reciprocal levels are taken with one level.
Level at | Reading on A | Reading on B |
A | 2.165 | 3.810 |
B | 0.910 | 2.355 |
The error in the collimation adjustments of the levels is -0.004 m in 100 m. Calculate the true difference of level between A and B and the refraction.
Solution for Two points A and B are 1530 m apart across a wide river
» The difference in level between A and B
= [(3.810 – 2.165) + (2.355 – 0.910)]/2
= 1.545 m
Error due to curvature = 0.07849 d^{2} m
= 0.07849 (1.53)^{2}
= 0.184 m
∴ When the level is at A, corrected staff reading on
B = 3.810 – (Cc – Cr) + C1
Where,
Cc = Correction due to curvature
= 0.184 m
Cr = Correction due to Refraction
C1 = Correction due to collimation
= (0.004/100) × 1530
= 0.0612 m
∴ Corrected staff reading on B
= 3.810 – (0.184 – Cr) + 0.0612
= 3.6872 + Cr
∴ True difference in level between A and B
= (3.6872 + Cr – 2.165)
= (1.5222 + Cr)
But it is quual to 1.545
∴ 1.5222 + Cr = 1.545
Cr = 1.545 – 1.5222
= 0.0228 ≃ 0.023 m