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 1 of 3.
Alik Biswas said:
1 decade ago
We Know, L= 2*(N*v^3/c)^0.5 (Length of valley curve).
Here, N = 2%-(-3%) (Algebraic difference in grades).
= 5% Or 0.05.
v = 100 km/hr or 27.778 m/sec (Design Speed).
c = 0.6 m/sec3 (Centrifugal acceleration).
Therefore, L= 2*(0.05*27.778^3/0.6)^0.5 = 84.53 m.
Say,84.60 m ([C] is the correct answer.
Here, N = 2%-(-3%) (Algebraic difference in grades).
= 5% Or 0.05.
v = 100 km/hr or 27.778 m/sec (Design Speed).
c = 0.6 m/sec3 (Centrifugal acceleration).
Therefore, L= 2*(0.05*27.778^3/0.6)^0.5 = 84.53 m.
Say,84.60 m ([C] is the correct answer.
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)
Elsadig Ahmed Ali said:
9 years ago
V = 1000 km/hr = 27.8 m/s.
N = 5%.
C = .6.
L = 2 * ((N * V^3) / C)^.5,
L = 2 * ((.05 * 27.8^3) / .6)^.5,
L = 2 * ((.05 * 21484.952) / .6)^.5,
L = 2 * (1074.2 / .6)^.5,
L = 2 * (1790.3)^.5,
L = 2 * 42.3.
L = 84.6.
N = 5%.
C = .6.
L = 2 * ((N * V^3) / C)^.5,
L = 2 * ((.05 * 27.8^3) / .6)^.5,
L = 2 * ((.05 * 21484.952) / .6)^.5,
L = 2 * (1074.2 / .6)^.5,
L = 2 * (1790.3)^.5,
L = 2 * 42.3.
L = 84.6.
(1)
Ayebare Junior jeezy said:
9 years ago
If V is the speed =100km/hr.
Difference in grades = A = (m - n).
= (-3 - 2) = 5% of acceleration
= 0.6.
But V =100 x 1000/60 x 60 = 27.8m/s.
L = 2*(aV^3)^0.5.
=((2)*0.05 x 27.8^3/0.6)^0.5).
= 84.53m.
Difference in grades = A = (m - n).
= (-3 - 2) = 5% of acceleration
= 0.6.
But V =100 x 1000/60 x 60 = 27.8m/s.
L = 2*(aV^3)^0.5.
=((2)*0.05 x 27.8^3/0.6)^0.5).
= 84.53m.
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)
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)
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)
Elsadig Ahmed Ali said:
9 years ago
@Samm.
N = |n2 - n1| = |-3 - 2| =|-3 + -2| = |-5| = 5.
@Ravindra.
C = centrifugal acceleration = 0.6 m/s^2.
N = |n2 - n1| = |-3 - 2| =|-3 + -2| = |-5| = 5.
@Ravindra.
C = centrifugal acceleration = 0.6 m/s^2.
Rishabh said:
10 years ago
L is the length of valley curve.
And this formula is applicable in this case only 2(Nv^3/c)^0.5.
And this formula is applicable in this case only 2(Nv^3/c)^0.5.
Samm said:
9 years ago
But N is not a difference in grade, it is deviation angle in radians. Please clarify this.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers