Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 1 (Q.No. 42)
42.
The slope correction for a length of 30 m along a gradient of 1 in 20, is
Discussion:
58 comments Page 2 of 6.
Adarsh Krishna said:
5 years ago
h^2/2l = 1.5*1.5/(2*30) = .0375m its change cm. = 3.75cm
(1)
Ankit said:
6 years ago
Slope correction is negative or positive? Give details about it.
Charan Singh Majhi said:
6 years ago
Slope correction = h^2/2l = 1.5^2/2*30 = 0.0375mtr = 3.75cm.
Karn chaurasiya said:
6 years ago
150/ n^2 because 30 metre chain used and now slope current is 150/ 20^2 = 150 ÷ 400 = 0.375 m or 3. 75 cm.
Sathya said:
6 years ago
@Sindhu.
It is 0.075.
It is 0.075.
Sindhu said:
6 years ago
If 30m replace 60m then h=?
Ritesh Kumar singh said:
6 years ago
Thanks for the explanation @Shubhajit Chakrabarty.
Don chandu said:
6 years ago
For 20 m vertical 1m.
For 30m h = 30/20 = 1.5.
Slope correction = h^2/2l = 1.5^2/(2 * 30) = 0.0375m = 3.75cm.
For 30m h = 30/20 = 1.5.
Slope correction = h^2/2l = 1.5^2/(2 * 30) = 0.0375m = 3.75cm.
(1)
Omgir said:
7 years ago
Helpful explanation. Thank you all.
Manoj said:
7 years ago
By using of similarity of the triangle.
H = 30÷401 and putting into slope correction formula.
H = 30÷401 and putting into slope correction formula.
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