Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 5 (Q.No. 49)
49.
A sewer is laid from a manhole A to a manhole B, 250 m away along a gradient of 1 in 125. If the reduced level of the invert at A is 205.75 m and the height of the boning rod is 3 m, the reduced level of the sight rail at B, is
Discussion:
16 comments Page 2 of 2.
Sophia said:
6 years ago
Please can someone explain why you're dividing it by 125?
(1)
Xavi said:
5 years ago
@Sophia.
The distance between A and B is 250m and B is at a lower level, so in layman terms, for every 125m distance there is a 1m decrease in RL, thus for 250m distance there will be 2m decrease in RL. In mathematical terms, they are simply dividing 250/125 to obtain 2.
The distance between A and B is 250m and B is at a lower level, so in layman terms, for every 125m distance there is a 1m decrease in RL, thus for 250m distance there will be 2m decrease in RL. In mathematical terms, they are simply dividing 250/125 to obtain 2.
(1)
Ajay Kumar Yadav said:
5 years ago
205.75 - (250/125) + 3 =206.75.
(3)
Dawood khan said:
5 years ago
Thankyou for the explanation @Xavi.
Nabin said:
4 years ago
Invert is the lowest point of sewer and crown is the highest point, therefore, 205.75-2=203.75 is the invert of sewer B. Boning rod is T shaped rod with a vertical portion at invert and a horizontal portion at the crown. So when the boning rod reading is 3 it gives the crown reading of sewer B.
Therefore RL of B =invert RL of B + boning rod height =(203.75+3) = 206.75.
Therefore RL of B =invert RL of B + boning rod height =(203.75+3) = 206.75.
(8)
Ranjith said:
3 years ago
Very useful, Thanks all for explaining the answer.
(2)
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