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
3.75 cm
0.375 cm
37.5 cm
2.75 cm.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
58 comments Page 4 of 6.

HITESH KUMAR said:   1 decade ago
Slope correction= h/2*2l.

Priya said:   1 decade ago
Can any one explain this properly?

Dash said:   1 decade ago
What is Gradient?

Ahishek said:   1 decade ago
How h = 1.5?

Anil soni said:   1 decade ago
For 20 m, h = 1 m. So for 30 m, h = 1.5 m.

Pranab said:   1 decade ago
h = 1.5 m & l = 30 m.

So that = h2/2l = (1.5)2/2*30 = 0.0375 cm.

Prashant said:   1 decade ago
(H2/2L)+(H4/8L3) as gradient is 1 in 20. i.e For 20 m, h = 1 m.

So for 30 m, h = 1.5 m which is less than 3 m.

Therefore second term of equation should be neglected.

Answer is (1.5)2/2x3000 = 3.75 cm.

Engr Attaullah Khan Tareen said:   1 decade ago
For 20 m, h = 1 m. So for 30 m, h = 1.5 m.

So that = h^2/2l = (1.5) 2/2*30 = 0.0375 m.

1 m = 100 cm so 0.0375*100 = 3.75 cm.

Maheshkumar said:   1 decade ago
In 1 in 20 means, 20m = 1m.

So total 30m length = 1.5m.

In Gradient 1 in 20 Means.

Imagine Triangle, length-30m & Height-1.5m.

Use Pythagoras theorem.

X2 = 302+1.52.
X = 30.0375m.

Change length = 30.0375-30.

= 0.0375m.
= 3.75cm.

Kanhu said:   1 decade ago
Slope correction for a sloping ground = h^2/2l.

Where h = height.

l = length of the ground.


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