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

The given Speed is 70 mil/hr converting it into ft.sec

v = 70 mi/hr = 369600 ft/hr = 102.67 ft/sec = 112.654 km/ hr

and given radius of curvature (r) = 2500 ft = 762 meters

g = 32.15 ft/sec^{2 and}

now

superelevation ratio in terms of rise over run = v² / (g*r) = (102.67)² / (32.15*2500) = 0.131

now, convert the superelevation ratio into an angle

tan () = 0.131

= tan^- (0.131) = 7.46⁰

Superelevation = 7.46⁰

As we know the

v = 70 mi/hr = 112.654 km/ hr

and given radius of curvature (r) = 2500 ft = 762 meters

Superelevation rate (e_max) = 7.6 % 0r 0.076

and minimum length of rub off (L₁) = 72 meter