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

Answer

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

$e + f = \dfrac{v^2}{rg}$

$0.07 + 0.11 =\dfrac{95.33^2}{r\times 32.2}$

$r = \dfrac{95.33^2}{32.2\times 0.18}$

  r = 1568 ft

As the road is two lane with width 12 ft

R = 1568 + 12/2

R = 1574 ft

Length of the curve

$L = \dfrac{\piR\Delta}{180}$

$L = \dfrac{\pi\times 1574\times 38}{180}$

L = 1044 ft

Tangent of the curve calculation

  $T = R tan(\dfrac{\Delta}{2})$

  $T = 1574 tan(\dfrac{38}{2})$

      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

$MO = R(1-cos\dfrac{\Delta}{2})$

$MO = 1574\times (1-cos\dfrac{38}{2})$

     MO = 85.75 ft

degree of the curvature

$D = \dfrac{5729.578}{R}$

$D = \dfrac{5729.578}{1574}$

D = 3.64°