Vertical Alignment

A highway design includes the intersection of a +5.8% grade with a –2.9% grade at station 1052+75 at elevation 100.50 feet above sea level. Calculate the center line elevation along this highway for every 100-ft station on a parabolic vertical curve of 600-ft length. You’ll need to use concepts introduced in the Geometric Design Chapter.

[Solution shown below]

Solution

The VPC is at station 1052+75 – 3+00 = 1049+75

The elevation of the VPC equals the elevation of the VPI minus ½ of the curve length times the initial grade = 100.50-(300*0.058)= 83.10 feet.

The VPT is at station 1052+75 + 3+00 = 1055+75

The elevation of the VPT equals the elevation of the VPI minus ½ of the curve length times the second grade = 100.50-(300*0.029)= 91.80 feet.

The equation for a parabolic vertical curve is y = (r/2)*x2 + g1*x + (elevation of VPC) where y = station elevation, r = rate of change of the grade of the curve [(g2-g1)/(length of curve in stations)] and x = stations beyond the VPC.

First, r = (-2.9-5.8)/6 = -1.45% per station.

The computations are shown in the following table.

 Station x x2 (r/2)*x2 g1*x Elevation VPC Elevation Curve 1049+75 0 0 0 0.00 83.10 83.10 1050+75 1 1 -0.73 5.80 83.10 88.18 1051+75 2 4 -2.90 11.60 83.10 91.80 1052+75 3 9 -6.53 17.40 83.10 93.98 1053+75 4 16 -11.60 23.20 83.10 94.70 1054+75 5 25 -18.13 29.00 83.10 93.98 1055+75 6 36 -26.10 34.80 83.10 91.80