Question

2. A 500 foot vertical curve connects a -4% grade to a +2 percent grade with a PVI at station 35+60.55. The PVI has an elevation of 438.24 ft. a. Calculation the location and elevation of the PVC, the PVT, and the high point or low point on the curve. b. What should be the maximum posted speed limit on this curve?
please show steps and do it clearly

Answer 1

G1 = G2 = GIVEN DATA Back Tangent or Initial Grade, Forward Tangent or Final Grade, Length of curve, Station of PVT Elevation of P VI 4% 2 % 500 ft 35 + 60.55 438.24 = = e CALCULATIONS a) Station of PVC 35 + 60.55 2 + 50.00 33 + 10.55 438.24 - 448.24 ft Elevation of PVC 1111111111 0.040 x 250 Station of PVT 35 + 60.55 2 + 50.00 38 + 10.55 438.24 + 443.24 ft Elevation of PVT 0.020 x 250 Rate of slope change, G2-G1 2L 0.000060000 0.04 448.24 Initial grade, Elevation of VPC = = = b C Thus, the equation of the curve can be written as: y = ax ? + bx + = 0.000060000 x ?. 0.04 x + 448.24 At the lowest/highest point, dy /dx = 0, = 20 x + b = 0.000120000 x Therefore, x = 333.33 ft - 0.04 Hence, station of lowest point, 33 + 10.55 3 + 33.33 36 + 43.88 Elevation of lowest point, x 333.33 ) - 0.04 x( 333.33 ) + 448.24 = 0.000060000 = 441.57 ft

Algebraic difference in grades, A = G1-G2 = 6 L = KXA 500 = 6 K Minimum length of curve, Therefore, K = 83.33 Hence, design speed, V = 50 mph