Civil Engineering - Highway Engineering - Discussion
Discussion Forum : Highway Engineering - Section 5 (Q.No. 27)
27.
If the width of carriage way is 12.5 metres, outer edge 50 cm higher than the inner edge, the required super elevation is
Discussion:
8 comments Page 1 of 1.
Harsh Shukla said:
7 years ago
Elevations that we generally encounter here in highway engineering are written in the form of 1/n which means for 1 unit vertical distance we move 'n' unit of horizontal distance so tan(theta)=1/n.
Here instead of 1 unit vertical distance, it is given 50cm (0.5m) vertical distance and 'n' is given as 12.5m.
So here tan(θ) = 0.5/12.5 = 1 in 25.
Here instead of 1 unit vertical distance, it is given 50cm (0.5m) vertical distance and 'n' is given as 12.5m.
So here tan(θ) = 0.5/12.5 = 1 in 25.
(2)
Sachin said:
7 years ago
e= tanθ = (y/w) = (50cm/1250cm) = 0.04 (1in 25).
Sebastian said:
4 years ago
Tan = P/B.
= 0.5/12.5,
= 0.04,
= 4% = 1:25.
= 0.5/12.5,
= 0.04,
= 4% = 1:25.
(1)
Nitin mukare said:
6 years ago
E = B*e.
0.5 = 12.5*e.
e = 0.04 =1 in 25.
0.5 = 12.5*e.
e = 0.04 =1 in 25.
(1)
Himangkar said:
7 years ago
(50 cm/1250 CM) = 0.04 (1 in 25).
RAihan said:
1 decade ago
E = tan e = 50/1250 = 1 in 25.
Dinesh said:
7 years ago
How come 1250 here?
(1)
Sushil THaKuR said:
7 years ago
@Dinesh.
1m=100cm.
1m=100cm.
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