1- A 1000 feet vertical curve connects a 2.5% approach grade and a-35% departure grade. The PVI is at 95+65 with elevation 545.22 a) b) Calculate the station and elevation for EVC. (7.5+7.5 points) Calculate the station and elevation of the low point or high point. (15+10 points) Show your calculations) Solution 1) L = 1000ft Station at EVC = PVI + (L/2) = (95+65)+ (10+.00)/2 = (95+65)+ (5+00) Station at EVC = 100+65 Elevation at EVC = Elevation at PVI + (g2 x L/2) Elevation at EVC = 545.22 + (-3.5% x 1000/2) Elevation at EVC = 527.72ft SO elevation of BVC = Elevation at PVI - (g1 x L/2) BVC = 545.220 - (2.5%x 1000/2) = 532.72ft Highest point elevation x = G1 xL/ (G1 - G2) x = 0.025x1000/(0.025+0.035) at x = 416.67ft So Y =x2 [(G2 - G1)/2L] + g1 x + BVC Y = 416.672 [(-0.035-0.025)/2000] + 0.025 x416.67 + 532.72 Y = -5.208 + 10.416 + 532.72 Y = 537.928ft 2) elevation of BVC = Elevation at PVI - (g1 x L/2) elevation of BVC = 800 - (0.03 x 1000/2) = 785ft Y =x2 [(G2 - G1)/2L] + g1 x + BVC where we PVI - L/2 = 3050-500 = 2550 = 25+50 So L = 10stations X = 32+50 - 25+50 = 7stations Y = 72 [(-5-3)/2x10] + 3 x 7+ 785 Y = -19.6+21 + 785 Y = 786.14ft.