Question

1. Two highways are planned, each running perpendicular to the other. The details of the lower highway are already fixed. The design process is continuing for the upper highway, which must maintain a 25 foot clearance over the highway below (as measured from highway centerlines). (a) Locate the minimum station where the new highway can be constructed. (b) Locate the maximum station where the new highway can be constructed. (c) What is the station of the lowest point on the curve? (d) What is the elevation of the lowest point on the curve? elev 770.0 82+ 00 25' minimum 94 +00 PIVC 88 00 elev 729

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Answer 1

theta_1 = 1/2 = 1200/2 = 600 ft station of PVC = station of PVT - D/100 ft = 90 + 00.00 - 600/100 = 89 + 94.00 PVC = 150 - (-4% ft/station times 6 station) = 182 ft D_2 = 1/2 = 600 ft Solution of PVT = 90 + 00.00 + 600 ft/100 ft = 90 + 06.00 Elevation of PVT = 150 + [+2% ft/station times 6 station] = 170 ft