a roadway is designed to have an entrance grade of 4.5%, an exit grade of 2.5%, PVI at station 200+00 with elevation of 72.50' and vertical curve length of 700' in profile view. What is the top of pavement elevation at station 199+00?
G1 = 4.5%
G2 = 2.5%
Station of PVI = 200+00
Elevation of PVI = 72.50 ft
Length of vertical curve = 700 ft
Station of PVC = Station of PVI - (L/2) = (200+00) - (3+50) = 196+50
Elevation of PVC = Elevation of PVI - (G1 * L/2)/100 = 72.5 - (4.5 * 350)/100 = 56.75 ft
Equation of curve:
y = ax2 + bx + c
b = 4.5
c = 56.75
Top of Pavement Elevation at Station 199+00:
x = (199+00) - (196+50) = (2+50) = 2.5
y = -0.00143 * 2.52 + (4.5 * 2.5) + 56.75 = -0.00894 + 11.25 + 56.75 = 67.99106 ft
a roadway is designed to have an entrance grade of 4.5%, an exit grade of 2.5%,...
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