the innermost cehicle path is tentaively planned as 2500 ft

the innermost cehicle path) is tentaively planned as 2500 ft. What rate of superelevation is required for this curve?

Solution

rate of super elivation = V²/gR

given velocity V = 70 mi/hr = 31.29 m/sec

radious R = 2500ft = 762 m

g = 9.81 m/sec²

rate of super elivation = (31.29² / (9.81* 762)

= 0.131