Civil Engineering - Highway Engineering - Discussion

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

[A]. 32.44 km/hour
[B]. 42.44 kg/hour
[C]. 52.44 km/hour
[D]. 62.44 km/hour
[E]. 72.44 km/hour

Answer: Option C

Explanation:

No answer description available for this question.

Abhishek said: (Feb 27, 2015)  
E+f = v^2/(127R).

V^2 = (e+f) 127R = 52.4563.

Anoop Rastogi said: (Aug 8, 2015)  
e+f = (v^2)/127R.

V^2 = (0.066667+0.15)127*100 = 52.4563.

Roy said: (Aug 26, 2017)  
But by above formula, it comes 43.64. Am I right?

Udoetuk said: (Nov 9, 2017)  
ROY: You are wrong.

ANOOP RASTOGI is right.

Bhavesh said: (Dec 1, 2017)  
How 0.066667? Please explain.

Sumon Sarkar said: (Jan 3, 2018)  
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.

Nabin Majumder said: (Jan 4, 2018)  
e= 1 in 15 then 100/15=6.6667 but how Sumon has calculated the value is .066667.

Er Abrar Ansari said: (Jan 11, 2018)  
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.

Vijayan said: (Apr 9, 2018)  
6% - 1 in 15,
Then 6/100=0.06.

Virender Paul said: (Aug 8, 2018)  
How? I am not getting this. Please explain me.

Nikhil said: (Dec 10, 2018)  
How 127 comes?

Please explain.

Waleed said: (Dec 19, 2018)  
@Nikhil.

127 is constant.

Nitin M said: (Aug 20, 2019)  
1 in 15 = 1/15 = 0.0666.

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