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A horizontal curve is being designed for a new four-lane roadway with 11-ft lanes. The PT is located at station 1050+20, the design speed is 45 mph and maximum superelevation of 4%. If the central angle of the curve is 30 degrees, what is the radius of the curve and location of the PC and PI?

A horizontal curve is being designed for a new four-lane roadway with 11-ft lanes. The PT is located at station 1050+20, the design speed is 45 mph and maximum superelevation of 4%. If the central angle of the curve is 30 degrees, what is the radius of the curve and location of the PC and PI?

Traffic and Highway Engineering
Traffic and Highway Engineering
5th Edition
ISBN:9781305156241
Author:Garber, Nicholas J.
Publisher:Garber, Nicholas J.
Chapter15: Geometric Design Of Highway Facilities
Section: Chapter Questions
Problem 8P
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Example Prob Below: A horizontal curve is being designed for a new three-lane roadway with 12-ft lanes. The PT is located at station 35+70, the design speed is 60 mph and maximum superelevation of 6%. If the central angle of the curve is 25 degrees, what is the radius of the curve and location of the PC and PI? 3-lane roadway: 12ft lanes; PT=35+70; V = 60mph=88fps; emax 0.06; A= 25° Find: Solution: From Table 3.5, for V = 60-f₂ = 0.12 Ry Ry=- R R, PC, PI V² 88² g(e+f) 32.2(0.06 +0.12) = =1336.1ft R = R₁ +12= 1336.1 + 12 = 1348.1ft L = RA= (1348.1) (25) 588.2ft 180 180 PC = PT-L = (35+70) - (5+88.2) -29+81.8 Δ T =R x tan=1348.1 x tan PI = PC + T = (29+81.8)+(2+98.8)=32 +80.6 (²³) = 298.8ft
Transcribed Image Text:Example Prob Below: A horizontal curve is being designed for a new three-lane roadway with 12-ft lanes. The PT is located at station 35+70, the design speed is 60 mph and maximum superelevation of 6%. If the central angle of the curve is 25 degrees, what is the radius of the curve and location of the PC and PI? 3-lane roadway: 12ft lanes; PT=35+70; V = 60mph=88fps; emax 0.06; A= 25° Find: Solution: From Table 3.5, for V = 60-f₂ = 0.12 Ry Ry=- R R, PC, PI V² 88² g(e+f) 32.2(0.06 +0.12) = =1336.1ft R = R₁ +12= 1336.1 + 12 = 1348.1ft L = RA= (1348.1) (25) 588.2ft 180 180 PC = PT-L = (35+70) - (5+88.2) -29+81.8 Δ T =R x tan=1348.1 x tan PI = PC + T = (29+81.8)+(2+98.8)=32 +80.6 (²³) = 298.8ft
A horizontal curve is being designed for a new four-lane roadway with 11-ft lanes. The PT is located at station 1050+20, the design speed is 45 mph and maximum superelevation of 4%. If the central angle of the curve is 30 degrees, what is the radius of the curve and location of the PC and PI?
Transcribed Image Text:A horizontal curve is being designed for a new four-lane roadway with 11-ft lanes. The PT is located at station 1050+20, the design speed is 45 mph and maximum superelevation of 4%. If the central angle of the curve is 30 degrees, what is the radius of the curve and location of the PC and PI?
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