# Civil Engineering - Highway Engineering - Discussion

### Discussion :: 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

 [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

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.