Respuesta :
Answer:
distance = 22.57 ft
superelevation rate = 2%
Explanation:
given data
radius = 2,300-ft
lanes width = 12-ft
no of lane = 2
design speed = 65-mph
solution
we get here sufficient sight distance SSD that is express as
SSD = 1.47 ut + [tex]\frac{u^2}{30(\frac{a}{g}\pm G)}[/tex] ..............1
here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here
so put here value and we get
SSD = 1.47 × 65 ×2.5 + [tex]\frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}[/tex]
solve it we get
SSD = 644 ft
so here minimum distance clear from the inside edge of the inside lane is
Ms = Rv ( 1 - [tex]cos (\frac{28.65 SSD}{Rv})[/tex] ) .....................2
here Rv is = R - one lane width
Rv = 2300 - 6 = 2294 ft
put value in equation 2 we get
Ms = 2294 ( 1 - [tex]cos (\frac{28.65 \times 664}{2294})[/tex] )
solve it we get
Ms = 22.57 ft
and
superelevation rate for the curve will be here as
R = [tex]\frac{u^2}{15(e+f)}[/tex] ..................3
here f is coefficient of friction that is 0.10
put here value and we get e
2300 = [tex]\frac{65^2}{15(e+0.10)}[/tex]
solve it we get
e = 2%
