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 6 of 6.
Swethakondisetty said:
8 years ago
Slope correction =h^2/2l.
Juhi said:
8 years ago
Thank you all for your explanations.
Sathya said:
8 years ago
30 m length, h = 1.5.
Given gradient is 1 in 20 i.e 1/20.
The equation of slope correction is,
= h^2/2l.
= 1.5^2 / 2*30 m,
= 2.25 / 60m,
= 0.0375 m or 3.75 cm.
Given gradient is 1 in 20 i.e 1/20.
The equation of slope correction is,
= h^2/2l.
= 1.5^2 / 2*30 m,
= 2.25 / 60m,
= 0.0375 m or 3.75 cm.
Vijay bhaskar said:
8 years ago
How h= 1.5?
Avijit Patra said:
8 years ago
Here, l= 30 m.
Gradient= 1 in 20, i.e h= 30/20= 1.5
We know, slope correction (Ch)= - h^2/2l
= - (1.5)^2/(2*30) m.
= -3.75 cm.
Gradient= 1 in 20, i.e h= 30/20= 1.5
We know, slope correction (Ch)= - h^2/2l
= - (1.5)^2/(2*30) m.
= -3.75 cm.
Asad Ali Chandia said:
8 years ago
Slop correction Formula = h^2/2*L.
h = 30*(1/20) = 1.5
Hence, correction = 3.375 cm.
h = 30*(1/20) = 1.5
Hence, correction = 3.375 cm.
Abhinandan kumar said:
7 years ago
1 in 20 means 0.05,
So in 30 means 0.05*30=1.5(which is h),
And we know the formula for a lope correction =h^2/2L,
So 1.5^2/2*30=0.0375m and in cm.
We get 3.75 m.
So in 30 means 0.05*30=1.5(which is h),
And we know the formula for a lope correction =h^2/2L,
So 1.5^2/2*30=0.0375m and in cm.
We get 3.75 m.
Rasheedrash79 said:
7 years ago
How are they getting 1.5*2=2.25? Please explain.
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