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 2 of 3.
VAMSI said:
8 years ago
Answer is C.
(1)
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)
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)
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.
L = 2*((0.05 * 27.78 * 27.78 * 27.78)/0.6)"0.5.
L = 84.53m.
(1)
Ravindra said:
9 years ago
@Alik Biswas, Well. Here, 'c' means what?
Dame abebe said:
8 years ago
Yes, L =84.60 so option "C" is the answer.
Adikeri said:
9 years ago
Not understanding. Explain in detail.
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.
VENKATESH said:
1 decade ago
Here N = 0.03+0.02 = 0.05.
L = 0.38(NV^3)^0.5.
L = 84.76MTS APPROX.
L = 0.38(NV^3)^0.5.
L = 84.76MTS APPROX.
Samm said:
9 years ago
But N is not a difference in grade, it is deviation angle in radians. Please clarify this.
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