Click to Select
Glossary
Exit this Topic
Roadway Design: Example Problems

 
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 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 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