Civil Engineering - Highway Engineering - Discussion
Discussion Forum : Highway Engineering - Section 1 (Q.No. 5)
5.
The total length of a valley formed by two gradients - 3% and + 2% curve between the two tangent points to provide a rate of change of centrifugal acceleration 0.6 m/sec2, for a design speed 100 km ph, is
Discussion:
30 comments Page 3 of 3.
Dame abebe said:
8 years ago
Yes, L =84.60 so option "C" is the answer.
SANDI said:
8 years ago
L = 2X(NV3/C)^.5,
N = .05,
V = 27.78,
L = 84.498.
N = .05,
V = 27.78,
L = 84.498.
(1)
VAMSI said:
8 years ago
Answer is C.
(1)
David Bielstein said:
8 years ago
I thought the centrifugal acceleration at the beginning of a parabolic curve would be 0 m/sec^2, increasing to the maximum acceleration at the point directly above or below the PI, and then back to 0.
Is 0.6 m/sec^2 assumed to be the maximum instantaneous centrifugal acceleration?
Is 0.6 m/sec^2 assumed to be the maximum instantaneous centrifugal acceleration?
(2)
Shahbaz Bashir said:
7 years ago
N is the difference between raising gradient and falling gradient i.e N = n1 - n2 and also falling gradient = negative (-) similarly raising gradient =positive (+).
(1)
Shaguftasultana said:
6 years ago
N=|-3-2|=5,
L=2* {(Nv^3)/C},
=2* 5*(100/3.6)/0.6,
= 84.57.
L=2* {(Nv^3)/C},
=2* 5*(100/3.6)/0.6,
= 84.57.
(10)
Prince said:
4 years ago
Both the gradients are positive so it should be subtracted right?
(3)
Sangla bagang said:
3 years ago
It should be option D. 84.54m only satisfies the comfort criteria, and the actual length of the curve should satisfy min. So, the sight distance criteria i.e 210.21m.
(2)
Shubham said:
2 years ago
For c = 0.6
L=0.38√(NV^3)
Where v is in kmph.
So, here;
L = 0.38*(5*1000000)^0.5.
L=0.38√(NV^3)
Where v is in kmph.
So, here;
L = 0.38*(5*1000000)^0.5.
(12)
Farouq algradi said:
12 months ago
L = 2 * √(N * (V^3)/c)
N = |N2-N1| = |-0.03-0.02| = |-0.05|
= 0.05.
V = 100kmph = 100 * 1000/(60*60)
= 27.7m/s.
L = 2 * √(0.05*(27.7^3)/0.6),
= 84.52m.
N = |N2-N1| = |-0.03-0.02| = |-0.05|
= 0.05.
V = 100kmph = 100 * 1000/(60*60)
= 27.7m/s.
L = 2 * √(0.05*(27.7^3)/0.6),
= 84.52m.
(5)
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