Question

On a section of a proposed road, a falling gradient of 3.2% meets a rising gradient of 2.4% which in turn meets a falling gradient of 2.8%. The through chainages of the first and second intersection points are 3452.79 m and 3568.41 m, respectively, and the reduced level of the first intersection point is 229.86 m. The gradients are to be joined by two equal tangent length vertical curves consisting of a sag curve, AB, and a crest curve, BC, of the same length. Point B defines the end of the first curve and the start of the second. Calculate

(i) The through chainages of the tangent points of the two vertical curves (ii) The reduced levels of the tangent points of the two vertical curves (iii) The RLs of points on the sag curve AB at exact 25 m multiples of through chainage (iv) The through chainage and reduced level of the highest point on the crest curve BC​

Answer 1

Answer:

Explanation: ab cuvered