Question

a. A crest vertical curve 150 m in length connects an upward gradient of 1 in 100 to a downward gradient of 1 in 50. If the tangent point 71 between the first gradient and the curve is taken as datum, calculate the levels of points at intervals of 25 m along the curve until it meets the second gradient at 72 and the level and site of the highest point. If the Sight distance = 130 m ( SL) and the height of the driver height above the road surface = 1.05 calculate also an object height take the length of crest curve represent the minimum length curve for design.

Answer 1

Dezive Equation of cove x 9,- - QY. y= an²+bxtc 9,-1%. > BVC - at No y=0 (=150 y=ototo c=0 ; dyt du 2 a (0) + b +o gi => b=91 -> b- 001) loo 92-9, → a- E 2 L bod loo 2x 150 -1x10 4 y = (*1x104) * x² + 4 ore - se y (Elevation) Stapt o 25m 0.1875m 50m 0-25 m 75 0-1875 m 10 125 -0.3125 End 150 -0.75

③ we have relation > A=192-9,1 = 1 100 L = A Cole 2 (Jh. a 5h): = 0.03 -> his om => 3 = 130 m => L=isom =)150 = 0.03(130)? 2x (Tos +Wha)? (Trieste)? 1.69 > h2=? Sha=0.2753 =>/h2=0.07579m object height

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