Question

Part A. The following information is for a proposed horizontal curve in a new subdivision: PI Station: 12+40.00 R-572.96 feet Deflection angle A-12.30 What is the Station of PT? PI PC PT Part B. An existing highway has a historic site located at the PT of an existing curve with a PI located at Station 80+00, an extemal angle of 65°, and a radius of 1,200 ft. The curve must be relocated, moving the PI to Station 78+00. The external angle and the PC will remain the same. What is the radius of the new curve? Not to Scale PI-STA 80+00 200 ft Δ-65° Historical Site PT PC Part C. A horizontal curve on a 4-lane highway (two lanes each direction and no median) has a superelevation of 4% and a central angle of 40 degrees. The PT of the curve is at station 322 +50 and the PI is a 320+08. The road has 12-ft lanes and 6-ft shoulders on both sides with high retaining walls going up immediately next to the shoulders. What is the highest safe speed of this curve (highest in 5 mi/hr increments) and what is the station of the PC?

Answer 1

Given PI @ 12 + 40.00 R = 572.96' and delta = 12.30 degree Degree of curvature, D = 360/2pi times 100/R Therefore D = 360/2pi times 100/572.96 = 10 degree length of curve, L = 100 delta/D = 123' Tangent length, T = R tan delta/2 = 572.96 tan(12.3/2) = 55.52 PC @ station PI - T (12 + 40) - 55.52 Therefore PC @ station 11 + 84.48 PT @ station Pc + L (11 + 84.48) + 123 Therefore PT @ station 13 + 0.743 Part B Given PI @ 80 + infinity: Radius = 1200 ft, delta = 65 degree Tangent length, T = R tan delta/2 Therefore T = 1200 tan (65/2) = 672.03 Therefore PC@station PI - T (80 + infinity) - 673.03' = 73.26.97 For New curve PI @ 78 + infinity: delta = 65 degree PC@73 + 26.97 Therefore Tangent length, L = (78 + infinity) - (73 + 26.97) R tan (65/2) = 473.03 Therefore R = 844.66'