Civil Engineering - Highway Engineering - Discussion
Discussion Forum : Highway Engineering - Section 3 (Q.No. 28)
28.
If the coefficient of friction on the road surface is 0.15 and a maximum super-elevation 1 in 15 is provided, the maximum speed of the vehicles on a curve of 100 metre radius, is
Discussion:
16 comments Page 1 of 2.
Er Abrar Ansari said:
8 years ago
Here is given that means 1 horizontal and 15 vertical it is max. So we know that e+f=v^2/127R.
Then find V = 127R*(e+f)=Ans.
Then find V = 127R*(e+f)=Ans.
(5)
SUMON SARKAR said:
8 years ago
Superelevation 1 in 15 = 0.066667 (1/15 = 0.066667)
e+f = V^2/127R,
V^2 = (e+f)/127R.
V^2= (0.066667+.15)/127*100,
V^2=2751.6,
V=\/'''''''2751.6.
V=52.45.
e+f = V^2/127R,
V^2 = (e+f)/127R.
V^2= (0.066667+.15)/127*100,
V^2=2751.6,
V=\/'''''''2751.6.
V=52.45.
(3)
Nabin majumder said:
8 years ago
e= 1 in 15 then 100/15=6.6667 but how Sumon has calculated the value is .066667.
(3)
Nitin m said:
6 years ago
1 in 15 = 1/15 = 0.0666.
(2)
Prakash said:
5 years ago
The Correct answer is 52.4563.
(2)
Roy said:
8 years ago
But by above formula, it comes 43.64. Am I right?
(1)
Bhavesh said:
8 years ago
How 0.066667? Please explain.
(1)
Nikhil said:
7 years ago
How 127 comes?
Please explain.
Please explain.
(1)
Raju said:
5 years ago
Can we derive using formula SSD vt + v^2/2gf.
(1)
Paramjit said:
5 years ago
Simple formula use v^2/127R to get the answer in km/hr.
(1)
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