Question

A 200m equal tangent sag vertical curve has the PVC at station 3 + 700.000 and elevation
321m. The initial grade is -3.5% and the final grade is 0.5%. Determine the elevation and
stationing of the PVI, PVT, and lowest point on the curve.

Answer 1

Ans) We know,

Station at PVI = station at PVC + 1/2 Length of curve

=> Station of PVI = (3 + 700) + (0 + 100) = 3 + 800

Elevation of PVI = Elevation of PVC + (initial grade x length/2)

=> PVI elevation = 321 + (-0.035 x 200/2) = 317.5 m

Station of PVT = Station at PVC + Length of curve

=> Station of PVT = (3 + 700) + (0 + 200) = 3 + 900

  Elevation of PVT = Elevation of PVI + (final grade x length/2)

=> PVI elevation = 317.5 + (0.005 x 200/2) = 318 m

From the values initial and final grades it is clear that the lowest point on the vertical curve will occur when the first derivative of parabolic function is zero because the initial and final grades are opposite in sign. So ,

y = A ² + B + C ...........................(1)

=> dy/dx = 2 A + B = 0

Also, A = (G₂ - G₁) / 2 L = [0.005 - (-0.035)] / (2 x 200)

=> A = 0.0001   

  B = G₁ = - 0.035

Putting values,

2 (0.0001) + (-0.035) = 0

=> = 175 m

This gives the stationing of low point 3 + 875 (175 m from PVC) .For the elevation of lowest point on vertical curve , the value of A, B and C(elevation of PVC) and x are required to be substituted in equation 1,

Putting A = 0.0001 , B = -0.035 and C = PVC elevation = 321 m

=> y = 0.0001(175)²- 0.035(175) + 321

=> y = 3.0625 - 6.125 + 321

=> y = 317.9375 m 317.94 m