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A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the distance that must be cleared from the inside edge of the inside lane to provide sufficient sight distance. Also determine the superelevation rate for the curve.

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%

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