Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 6 (Q.No. 15)
15.
The sag of 50 m tape weighing 4 kg under 5 kg tension is roughly
Discussion:
24 comments Page 3 of 3.
Sanku said:
5 years ago
Thanks @Johny Chauhan.
Vintage said:
3 years ago
50/ 24*5^2 = 0.0833.
(1)
K Andres said:
1 year ago
I think this is the formula;
To get w = 4kg - 5kg = -1kg
Then p = 5
L = 50.
(1)^2(50)/24(5)^2 = 0.08333.
To get w = 4kg - 5kg = -1kg
Then p = 5
L = 50.
(1)^2(50)/24(5)^2 = 0.08333.
Prazwol said:
10 months ago
(W^2×l)/24 P^2.
W = weight of tape
P = tension
So, sag correction = (4^2× 50)/ (24 × 5^20).
= 1.333.
W = weight of tape
P = tension
So, sag correction = (4^2× 50)/ (24 × 5^20).
= 1.333.
(2)
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