Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 5 (Q.No. 10)
10.
Correction per chain length of 100 links along a slope having a rise of 1 unit in n horizontal units, is
100 n2
.Discussion:
19 comments Page 1 of 2.
Hemanth said:
3 years ago
Given: gradient 1 in N.
Wkt gradient = H/L.
Chain length = 100links=20m chain
Gradient = H/L = 1/N.
H = L/N.
Correction slag = H^2/2L.
CS = (L/N)^2/(2*L).
Substitute L =20m.
We will get 100/n^2.
Wkt gradient = H/L.
Chain length = 100links=20m chain
Gradient = H/L = 1/N.
H = L/N.
Correction slag = H^2/2L.
CS = (L/N)^2/(2*L).
Substitute L =20m.
We will get 100/n^2.
(3)
Kcube said:
6 years ago
Correction per chain length of 100 links =100*α^2, here α is Radians. α=1/n(rise of 1 unit in n horizontal units).
Therefore the correction = 100/n^2.
Therefore the correction = 100/n^2.
(1)
Prasanta said:
6 years ago
We know 100 links =20 m chain.
Son horizontal =1 unit
20 m horizontal =20/n
i.e. h=20/n
L=20m.
Slope correction = h^2/2l
So (20/n)^2/2*20
= 100/n^2(ans)
Son horizontal =1 unit
20 m horizontal =20/n
i.e. h=20/n
L=20m.
Slope correction = h^2/2l
So (20/n)^2/2*20
= 100/n^2(ans)
(2)
Roy said:
7 years ago
n horizontal=1 vertical, so 100 horizontal=100/n vertical.
So, here h=100/n &l=100. Now, the formula is h^2/2l becomes 50/n^2.
So, here h=100/n &l=100. Now, the formula is h^2/2l becomes 50/n^2.
(1)
Mithun Chitra said:
6 years ago
Correct Answer should be = -50\n2.
So, here the answer should be none of the above.
So, here the answer should be none of the above.
(2)
Suresh said:
7 years ago
100 Links along the slope not a horizontal length.
Era said:
6 years ago
@Prasant it comes 10/n^2 how you got 100/n^2.
(4)
Vaibhav singh said:
6 years ago
Prasnta 1 link=20 cm not 20 m.
(2)
Prem said:
8 years ago
Hypotenuse allowance= 50/n^2.
SRIMANTA said:
8 years ago
The correct answer is 50/n^2.
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