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 2 of 2.
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)
Bhavesh said:
8 years ago
How 0.066667? Please explain.
(1)
UDOETUK said:
8 years ago
ROY: You are wrong.
ANOOP RASTOGI is right.
ANOOP RASTOGI is right.
Roy said:
8 years ago
But by above formula, it comes 43.64. Am I right?
(1)
Anoop rastogi said:
1 decade ago
e+f = (v^2)/127R.
V^2 = (0.066667+0.15)127*100 = 52.4563.
V^2 = (0.066667+0.15)127*100 = 52.4563.
Abhishek said:
1 decade ago
E+f = v^2/(127R).
V^2 = (e+f) 127R = 52.4563.
V^2 = (e+f) 127R = 52.4563.
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