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.
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)
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)
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.
Ram kumar said:
1 decade ago
Length is 84.16 means, then what s the answer? option A or option C.
Prince said:
4 years ago
Both the gradients are positive so it should be subtracted right?
(3)
Manikanth said:
1 decade ago
Lenth of valley curve for comfort condition is 2(Nv^3/c)^0.5.
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)
Selvi said:
1 decade ago
L = 0.378(NV3)^1/2.
= 0.378(.05*100*100*100)^.5.
= 86.4 m.
= 0.378(.05*100*100*100)^.5.
= 86.4 m.
(1)
Doha said:
8 years ago
L = (A*V^2)/1300a,
L = (5 * 100^2)/1300 * 0.6,
L = 64.1m.
L = (5 * 100^2)/1300 * 0.6,
L = 64.1m.
(2)
Twizere kagunga elie said:
1 decade ago
I don't understand well if N is load or if is median.
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