Traffic And Highway Engineering
5th Edition
ISBN: 9781133605157
Author: Garber, Nicholas J., Hoel, Lester A.
Publisher: Cengage Learning,
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Question
Chapter 15, Problem 8P
To determine
The station and elevation of the low point using the sag vertical curve for a given rural arterial highway.
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A +2.5% grade intersects with a –1.5% grade at station (53+524.25) at an elevation of 90 m.
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Chapter 15 Solutions
Traffic And Highway Engineering
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Similar questions
- Please correct answerarrow_forwardExample: A grade g1 of -2% intersects g2 of +1.6% at a vertex whose station and elevation are 87+00 and 743.24, respectively. A 400' vertical curve is to be extended back from the vertex, and a 600’ vertical curve forward to closely fit ground conditions. Compute and tabulate the curve for stakeout at full stations.arrow_forwardDesign a vertical curve (i.e., Determine length of the curve as well as station and elevation of the vertical curve's PVC and PVT) with the given PVI (station and elevation given the figure below) to go through a future intersection location at the point at which the vertical curve is flat. Report the vertical curve's design speed to the nearest 5 miles per hour. PVC Future Intersection Station = 100+00 G1 = -2.6% PVT PVI G2 = 1.0% Station = 99+00 Elevation = 228 ft %3|arrow_forward
- Highway Engineering: You are designing a highway to AASHTO guidelines on rolling terrain where the design speed will be 65 mi/h. At one section, a +1.25% grade and a -2.25% grade must be connected with an equal-tangent vertical curve. Determine the SSD given the reaction time of 2.5 sec and deceleration of 3.4 m/s^2. Determine also the minimum length of curve.arrow_forward12. Find the minimum length of curve for the following scenarios. Entry Grade Exit Grade Design Speed Reaction Time A 3% 8% 45 mi/hr 2.5 s В -4% 2% 65 mi/hr 2.5 s 0 % -3% 70 mi/hr 2.5 sarrow_forwardA crest vertical curve joining a + 3 percent and – 4 percent grade is to be designed for 75 mph. If the tangent intersect at station (345 + 6000) at an elevation of 250 ft, determine the stations and elevations of the PVC and PVT.arrow_forward
- An equal-tangent sag vertical curve is designed with the PVC at station 109 + 00 and elevation 950 ft, the PVI at station 110 + 77 and elevation 947.34 ft, and the low point at station 110 + 50. Determine the design speed of the curve.arrow_forwardA vertical curve is to be designed to connect a -4% grade to a +1% grade on a facility with a design speed of 70 mi/h. For economic reasons, a minimum-length curve will be provided. A driver-reaction time of 2.5 seconds may be used in sight distance determinations. The V.P.I. of the curve is at station 5,100 + 22 and has an elevation of 1,285 ft. find the station and elevation of the V.P.C. and V.P.T., the high point of the curve, and at 100-ft intervals along the curve.arrow_forwardNote: Don't copy from other websites.arrow_forward
- Problem 2. This is a four-part problem. A +2.5% grade intersects with a –1.5% grade at station (735 + 30.75) at an elevation of 475 ft. Part A. If the design speed is 65 mi/h, determine the minimum length, in ft, of vertical curve. Part B. If the design speed is 65 mi/h, determine the elevation, in ft, of the point of vertical tangency also known as end of vertical curve. Part C. If the design speed is 65 mi/h, determine the elevation, in ft, of the highpoint of the curve. Part D. If the design speed is 65 mi/h, determine the distance, in ft, from the point of vertical curvature (also known as the beginning of curve) to the highpoint of the curve.arrow_forwardTraffic Engineering vertical curve problem. Solve the d,e,f partarrow_forwardA vertical parbolic curve was design in order to have a clear sight distance of 120 m . The grade lines intersect at Sta 9+000 at elev 160.50 . The curve was design such that when the height of the drivers eye is 1.50 m above the payment it would just see an object whose height is 0.10 m above the pavement . Determine the max speed that a car could travel grade of 5% and a downgrade of -3%.arrow_forward
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