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 3 of 10.
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.
RAVINDRA said:
9 years ago
The given answer is true, don't confuse it.
Jatav shurjeet singh said:
9 years ago
It is the right answer, it will use v^2/225r.
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.
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.
Kruti nayak said:
9 years ago
v^2/225R is correct.
v^2/127R when f = 0.
v^2/127R when f = 0.
Joel said:
9 years ago
The provided answer 1 in 40 is correct.
Happy said:
9 years ago
But, here v^2/225 is the highest unit of super-elevation.
Mukeah senani said:
9 years ago
E = V^2/225R -> this method is Right.
Jayrajsinh said:
9 years ago
Practically super elevation e = v^2/225R.
So answer 1 in 40 is correct.
So answer 1 in 40 is correct.
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