On a two-lane road with 12-ft lanes, a horizontal curve is designed for 50 mi/h with a superelevation of 10%. The PI of the curve is at 220+48 and the PC is at station 216+74. Determine the station of the PT and the middle ordinate for stopping sight distance.
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- The design speed for a two-lane highway (lane width=12 ft.) is 60 mph. The localjurisdiction governs 8% superelevation. The central angel of the curve is 40 degrees. The PI station is at 452+50.What is the curve radius to the travel path (R)? What is the curve radius (R)? What is the curve length (L)? What isthe station of PC? What is the station of PT? What is the SSD? How many feet must be cleared from the lane'sshoulder edge to provide adequate stopping sight distance?A horizontal curve is being designed for a new two-lane highway (3.6-m lanes). The PI is at station 7 + 640, the design speed is 105 km/h, and a maximum superelevation of 0.08 m/m is to be used. If the central angle of the curve is 35 degrees, design a curve for the highway by computing the radius and stationing of the PC and PT.Q: A horizontal curve is being designed through mountainous terrain for a four- ane with lanes that are 3 m wide. The central angle is known to be 40 degrees, the tangent distance is 155m, and the stationing of the tangent intersection (PI) 'is 8 + 23. Under specified conditions and vehicle speed, the roadway surface is determined to have a coefficient of side friction of 0.08, and the curve's superelevation is 0.09. what is the stationing of the PC and PT and what is the safe vehicle speed?
- JA horizontal curve on a two-lane highway (12-ft lanes) has PC at station 124 + 00 and PT at station 130 + 00. The central angle is 30 degrees, and 25 ft is cleared (for sight distance) from the inside edge of the innermost lane. a. Determine the radius of this horizontal curve. b. Determine a maximum safe speed for this curve section.Q1) Answer the following A) A section of a two-lane highway (12-ft lanes) is designed for 75 mi/h. At one point a vertical curve connects a -2.5% and +1.5% grade. The vertical curve ends at station 36 + 50. It is known that a horizontal curve starts 294 ft before the starting of the vertical curve. If the superelevation of the horizontal curve is 8% and the central angle is 38 degrees, what is the station of the beginning of the horizontal curve?A horizontal curve on a four-lane (two lane in each direction) highway (11-ft lanes) has PC at station 124+30 and PT at station 130 + 15. The central angle is 40 degrees, the superelevation is 0.12, and 23.5 ft is cleared (for sight distance) from the inside edge of the innermost lane. Determine a maximum safe speed (assuming current design standards) to the nearest 5 mi/h.
- 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 section of a two-lane highway (12-ft lanes) is designed for 75 mph. At one point, a vertical curve connects a -2.5% and +1.5% grade. The VPT of this curve is at station 25+00. It is known that a horizontal curve starts (has PC) 292 ft before the vertical curve VPC. If the superelevation of the horizontal curve is 0.08 and the central angle is 38 degrees, what is the station of the PT?Traffic Engineering vertical curve problem. Solve the d,e,f part
- A section of a two-lane, two-way highway (2 @ 3.5 m) is to be designed with a circular curve. Sta. Pl at 200 + 100.000 Design speed = 60 kph Intersection angle = 40 degrees Rate of change of acceleration: 0.5 m/sec2A sag vertical curve connects a -1.5 percent grade with a +2.5 percent grade on a rural arterial highway. If the criterion selected for design is the minimum stopping sight distance, and the design speed of the highway is 70 mi/h, compute the elevation of the curve at 100-ft stations if the grades intersect at station (475 +00) at an elevation of 300 ft. Use L = KA to find minimum curve length for this problem.4. Horizontal Curve: P.C. 12+00.00 P.I T 4= 30° P.T. This highway has a 60 mph speed limit requiring a minimum radius for any curve to be 1500'. If the curve is to begin at Station 12+00.00 and is to have a radius of 1610.10', how long will the curve be? At what station will the PI and PT go? How long will E be? (Answers should all be in hundredths.)