A large grain elevator is located 40 feet from the centerline of a two-lane highway, which has 12-foot wide lanes. The elevator is situated on the inside of a horizontal curve with a radius of 500 feet. Assuming that the elevator is the only sight restriction on the curve, what is the minimum sight distance along the curve?

[Solution Shown Below]

Solution

The first thing that we need to do is calculate the distance from the edge of the grain elevator to the center of the nearest lane. This turns out to be 40 ft - 6f. = 34 ft. Next, we need to calculate the degree of the curve using the equation R = 5730/D. The degree of the curve turns out to be 11.46°. The last step involves solving for the sight distance using the equation below.

M = (5730/D)*(1 - cos(SD/200))

Where:
M = Distance from the center of the inside lane to the obstruction, 34 ft
D = Degree of the curve, 11.46°.

Where R = 5730/D

S = Sight distance (ft)
R = Radius of the curve, 500 ft