Respuesta :
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
[tex]e + f = \dfrac{v^2}{rg}[/tex]
[tex] 0.07 + 0.11 =\dfrac{95.33^2}{r\times 32.2}[/tex]
[tex]r = \dfrac{95.33^2}{32.2\times 0.18}[/tex]
r = 1568 ft
As the road is two lane with width 12 ft
R = 1568 + 12/2
R = 1574 ft
Length of the curve
[tex]L = \dfrac{\piR\Delta}{180}[/tex]
[tex]L = \dfrac{\pi\times 1574\times 38}{180}[/tex]
L = 1044 ft
Tangent of the curve calculation
[tex]T = R tan(\dfrac{\Delta}{2})[/tex]
[tex]T = 1574 tan(\dfrac{38}{2})[/tex]
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
[tex]MO = R(1-cos\dfrac{\Delta}{2})[/tex]
[tex]MO = 1574\times (1-cos\dfrac{38}{2})[/tex]
MO = 85.75 ft
degree of the curvature
[tex]D = \dfrac{5729.578}{R}[/tex]
[tex]D = \dfrac{5729.578}{1574}[/tex]
D = 3.64°
