Traffic and Highway Engineering - With Mindtap
5th Edition
ISBN: 9781305360990
Author: Garber
Publisher: CENGAGE L
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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 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.
A +2.5% grade intersects with a –1.5% grade at station (53+524.25) at an elevation of 90 m.
If the design speed is 90 km/h, use AASHTO (2011) criteria to determine:
The minimum length of vertical curve using the rate of vertical curvature.
The stations and elevations of the BVC and EVC.
The elevation of each 20-m station.
The station and elevation of the highpoint.
Two roads with grades of +4.5% and -3% meet at station 15 + 100 at an elevation
of 72.344 m. If the required rate of change of grade for the parabolic curve to be
fitted between those two roads is 0.4% per 20 m stations, determine the following:
a. The required length of the curve
b. Stationing and elevation at PC and PT
c. Location and elevation of the apex of the curve
d. Elevation at a point Q that has a distance of 15 m from PI
Chapter 15 Solutions
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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?arrow_forwardHighway 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_forwardPROBLEM 7 A +2 percent grade on an arterial highway intersects with a -1 percent grade at station 535 + 24.25 at an elevation of 300 ft. If the design speed of the highway is 65 mi/h, determine the stations and elevations of the initial point of vertical curve (PVC), the final point of vertical curve (PVT), the high point, and the elevation of each 100-ft station.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_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
- 12. 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_forwardThe minimum simple curve radius you recommended for a highway with design speed (96) Km / hr, super elevation (5.50 percent and f = 0.11, is: O 493.799 m O 439.799 m O 339.970 m O 440.700 m 430.900 ft / sec.arrow_forwardProblem 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_forward
- Two roads with grades of +4.5% and -3% meet at station 15 + 100 at an elevation of 72.344 m. If the required rate of change of grade for the parabolic curve to be fitted between those two roads is 0.4% per 20 m stations, determine the following: a . Elevation at a point Q that has a distance of 15 m from PIarrow_forwardA 5% grade intersects a -3.4% grade at station 41+990 of elevation 42.30 m. Design a vertical summit parabolic curve connecting the two tangent grades to conform with the following safe stopping sight distance specifications.Design Velocity = 60 kphHeight of driver's eye from the road pavement = 1.37 mHeight of an object over the pavement ahead = 100 mm Perception-reaction time = 3/4 sCoefficient of friction between the road and the tires = 0.15arrow_forwardA +8 percent grade intersects with a -2 percent grade at station (535 + 24.25) at an elevation of 300 ft. (a) If the design speed is 45 mi/h, determine the minimum length (in ft) of vertical curve using the rate of vertical curvature. (Assume the stopping sight distance is less than the length of the curve.) 610 ft (b) Using the length found in part (a), find the stations and elevations (in ft) of the BVC and EVC and the elevation (in ft) of each 100 ft station. (In the table below, the first row corresponds to the BVC and the last row corresponds to the EVC. Round your elevations to at least one decimal place.) Station Elevation (ft) 532 + 19.25 275.6 533 + 00 306.1 534 + 00 535 + 00 299.7 536 + 00 537 + 00 538 + 00 538 + 29.25 293.9 (c) Using the length found in part (a), find the station and elevation (in ft) of the highpoint. station elevation ftarrow_forward
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