Civil Engineering - Highway Engineering - Discussion

Discussion Forum : Highway Engineering - Section 2 (Q.No. 4)
If x% is the gradient of an alignment and y% is the gradient after proper superelevation along a curved portion of a highway, the differential grade along the curve, is
(x + y)%
(x - y)%
(y - x)%
(x x y)%
(y + x)%
Answer: Option
No answer description is available. Let's discuss.
8 comments Page 1 of 1.

Hira said:   3 years ago
The grade is decreased in a curve, not increased. It is called grade compensation. Y-X % is differential grade. Although, y% is lesser or flatter than x%.

Anom said:   6 years ago
How the gradient is increased after attaining super elevation?

Please suggest.

Ankush said:   6 years ago
y-x is the right answer.

Garry said:   6 years ago
The Answer is correct because Y% is the gradient after attaining full superelevation.

Previously it has X%. So the difference will be {Y-X}%.

BB Bande said:   7 years ago
Please explain this.

Andal said:   7 years ago
The normal gradient ( slope of the ground ) is x% which is the same value at the point of Transition curve. The super elevation is starting from this Transition point and ending at the mid of curvature of the horizontal curve.

Hence, after super elevation attained, the difference between gradient is (y-x)%.

Ng reddy said:   7 years ago
That is called grade compensation.

At the time of horizontal curves, we reduce the gradient 75/r or (30 + r) /r.

Samm said:   8 years ago
Why not (x-y) %? Please explain.

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