Civil Engineering - Highway Engineering - Discussion
Discussion Forum : Highway Engineering - Section 1 (Q.No. 1)
1.
A district road with a bituminous pavement has a horizontal curve of 1000 m for a design speed of 75 km ph. The super-elevation is
Discussion:
95 comments Page 8 of 10.
Shailendra kumar said:
9 years ago
I think the formula of superelevation e= v^2/225R.
So we put all values as given in the question e = 75 * 75/225 * 1000.
e = 1/40.
So we put all values as given in the question e = 75 * 75/225 * 1000.
e = 1/40.
Jack said:
9 years ago
A District road having the mixed traffic condition so speed is reduced by 25%.
Super elevation e = V^2 / 225 R.
e = 75 * 75/225 * 1000
= 0.025.
so for 1m,
1/0.025 = 40.
=> 1 in 40.
Super elevation e = V^2 / 225 R.
e = 75 * 75/225 * 1000
= 0.025.
so for 1m,
1/0.025 = 40.
=> 1 in 40.
Jatav shurjeet singh said:
9 years ago
It is the right answer, it will use v^2/225r.
RAVINDRA said:
9 years ago
The given answer is true, don't confuse it.
Miki said:
9 years ago
Both the formula is correct, the only difference is in e = v^2/127R, uses 75% of speed, and in e = V^2/225R, we are directly used.
Osi said:
9 years ago
But we have studied it in the formula = v^2/126R. I am confusing which one is correct.
IES AKP said:
10 years ago
1000 m is radius, it can't be length because there is no relation to provide length.
Prashant said:
10 years ago
1000 m is length of curve here, not radius.
Mohamed Yaagob said:
10 years ago
F = 0 V = 0.75 V design speed.
E = (0.75 V)^2/(127 R).
E = (0.75 V)^2/(127 R).
Pratik said:
10 years ago
Radius is not mentioned.
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