Question

A horizontal curve is being designed for a new two-lane highway (12-ft lanes). The PI is at station 250 + 50, the design speed is 65 mi/h, and a maximum superelevation of 0.07 ft/ft is to be used. If the central angle of the curve is 38 degrees, design a curve for the highway by computing the radius and stationing of the PC and PT. Also, compute the Degree of curvature and Middle Ordinate.

Answer 1

given,

Speed of vehicle = 65 mi/hr

= 65 x 1.4667 = 95.33 ft/s

e = 0.07 ft/ft

f is the lateral friction, f = 0.11

central angle,Δ = 38°

The PI station is

PI = 250 + 50

= 25050 ft

using super elevation formula

`e + f = (v^2)/(rg)`

`0.07 + 0.11 =(95.33^2)/(r* 32.2)`

`r = (95.33^2)/(32.2* 0.18)`

r = 1568 ft

As the road is two lane with width 12 ft

R = 1568 + 12/2

R = 1574 ft

Length of the curve

`L = (\piR\Delta)/(180)`

`L = (\pi* 1574* 38)/(180)`

L = 1044 ft

Tangent of the curve calculation

`T = R tan((\Delta)/(2))`

`T = 1574 tan((38)/(2))`

T = 542 ft

The station PC and PT are

PC = PI - T

PC = 25050 - 542

= 24508 ft

= 245 + 8 ft

PT = PC + L

= 24508 + 1044

=25552

= 255 + 52 ft

the middle ordinate calculation

`MO = R(1-cos(\Delta)/(2))`

`MO = 1574* (1-cos(38)/(2))`

MO = 85.75 ft

degree of the curvature

`D = (5729.578)/(R)`

`D = (5729.578)/(1574)`

D = 3.64°