Question

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?

Answer 1

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