Q6. To help improve the safety of at-grade intersection of two roads, the intersection is being redesigned so that the E-W road will pass underneath the N-S road. Currently the vertical alignment of the E-W road consists of a crest vertical curve joining a 4% upgrade to a 3% downgrade. The existing vertical curve is 138 m long, the PVC of this curve is at station 14+47, and the elevation of the PVC is 477 m. The centerline of the N-S road is at station 15+45. Your job is to find the shortest vertical curve that provides 6.0 m of clearance between the new E-W road and the N-S road.

A crest vertical curve joining a + 3 percent and – 4 percent grade is to be designed for 75 mph. If the tangent intersect at station (345 + 6000) at an elevation of 250 ft, determine the stations and elevations of the PVC and PVT.

An equal-tangent curve connects a +1.0% and a -0.5% grade. The PVC is at station 54+24 and the PVI is at station 56+92. Is this curve long enough to provide passing sight distance for a 60-mi/r design speed? Problem 2

1) An equal-tangent crest vertical curve has a 50 mi/h design speed. The initial grade is +3%. The high point is at station 33+37.43 and the PVT is at station 37+18.26. What is the elevation difference between the high point and the PVT?   2) A 1400-ft-long sag vertical curve (equal tangent) has a PVC at stateion 115+00 and elevation 750ft.  The intial grade is -3.5% and the final grade is +6.5%.  Determine the elevation and stationing of the low point, PVI, and PVT.   If you could show as many steps taken as possible that would be helpful in me actually understanding how to do the problem.  Thanks