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
16.0 m
42.3 m
84.6 m
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
30 comments Page 2 of 3.

Nikhil said:   10 years ago
What is mean by V?

Santhi said:   10 years ago
Here v is the design speed of the vehicle.

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.

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.
(1)

Samm said:   9 years ago
But N is not a difference in grade, it is deviation angle in radians. Please clarify this.

Ravindra said:   9 years ago
@Alik Biswas, Well. Here, 'c' means what?

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.

Sarang Mote said:   9 years ago
L = 2x(NV3/c)"0.5.
L = 2*((0.05 * 27.78 * 27.78 * 27.78)/0.6)"0.5.
L = 84.53m.
(1)

Adikeri said:   9 years ago
Not understanding. Explain in detail.

Doha said:   8 years ago
L = (A*V^2)/1300a,
L = (5 * 100^2)/1300 * 0.6,
L = 64.1m.
(2)


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