Question

An equal-tangent sag vertical curve is designed with a PVC as station 109+00 and elevation 950ft, the PVI has a station of 110+77 and elevation of 947.34ft, and the low point at station 110+50. Determine the design speed of the cure.

Answer 1

K = 96 and the design speed of the curve = 50mph

Step-by-step explanation:

109+00 = 10900

Elevation = 950ft

110+77 = 11077

Elevation = 947.34ft

Station of low point = 110+50 = 11050

To get grade of curve

Gi = 947.34-950/11077-10900

= -2.66/177

= -0.015x100

= -1.5%

Locate of low point (XL)

= 11050-10900

= 150

To get the value of K

XL = |GL| x K

When we substitute values

150 = 1.5 x K

150 = 1.5K

K = 150/1.5

K = 100

The suitable and most nearest value is K = 96

Then we use the standard chart to get the design speed for K = 96

On this chart, the design speed for the curve = 50mph

Therefore K = 96 and speed = 50mph