PRIN.OF HIGHWAY ENGINEERING&TRAFFIC ANA.
7th Edition
ISBN: 9781119610526
Author: Mannering
Publisher: WILEY
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Chapter 3, Problem 11P
To determine
The design speed of the curve.
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A +3.00% grade intersects a -2.40% at station 46+70.00 and elevation 853.48 ft. A 400.00 ft curve will be used to connect the two grades. Compute Station and elevation for the curve's endpoints.
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.
1. If PVI at sta 5+650 has an elevation of 122.45 m what is the elevation of the grade tangent at station 5+684? Given grade is +3%.
2. If a 26-degree curve consists of a 14 degree simple curve, what is the angle of the transition curves?
Chapter 3 Solutions
PRIN.OF HIGHWAY ENGINEERING&TRAFFIC ANA.
Ch. 3 - Prob. 1PCh. 3 - Prob. 2PCh. 3 - Prob. 3PCh. 3 - Prob. 4PCh. 3 - Prob. 5PCh. 3 - Prob. 6PCh. 3 - Prob. 7PCh. 3 - Prob. 8PCh. 3 - Prob. 9PCh. 3 - Prob. 10P
Ch. 3 - Prob. 11PCh. 3 - Prob. 12PCh. 3 - Prob. 13PCh. 3 - Prob. 14PCh. 3 - Prob. 15PCh. 3 - Prob. 16PCh. 3 - Prob. 17PCh. 3 - Prob. 18PCh. 3 - Prob. 19PCh. 3 - Prob. 20PCh. 3 - Prob. 21PCh. 3 - Prob. 22PCh. 3 - Prob. 23PCh. 3 - Prob. 24PCh. 3 - Prob. 25PCh. 3 - Prob. 26PCh. 3 - Prob. 27PCh. 3 - Prob. 28PCh. 3 - Prob. 29PCh. 3 - Prob. 30PCh. 3 - Prob. 31PCh. 3 - Prob. 32PCh. 3 - Prob. 33PCh. 3 - Prob. 34PCh. 3 - Prob. 35PCh. 3 - Prob. 36PCh. 3 - Prob. 37PCh. 3 - Prob. 38PCh. 3 - Prob. 39PCh. 3 - Prob. 40PCh. 3 - Prob. 41PCh. 3 - Prob. 42PCh. 3 - Prob. 43PCh. 3 - Prob. 44PCh. 3 - Prob. 45PCh. 3 - Prob. 46PCh. 3 - Prob. 47PCh. 3 - Prob. 48PCh. 3 - Prob. 49PCh. 3 - Prob. 50PCh. 3 - Prob. 51PCh. 3 - Prob. 52PCh. 3 - Prob. 53PCh. 3 - Prob. 54PCh. 3 - Prob. 55PCh. 3 - Prob. 56PCh. 3 - Prob. 57PCh. 3 - Prob. 58PCh. 3 - Prob. 59PCh. 3 - Prob. 60PCh. 3 - Prob. 61PCh. 3 - Prob. 62PCh. 3 - Prob. 63PCh. 3 - Prob. 64PCh. 3 - Prob. 65PCh. 3 - Prob. 66PCh. 3 - Prob. 67P
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.Similar questions
- 1. An equal-tangent crest vertical curve is being designed for a speed of 45 mph. The curve connects grades of 1.3% and -2.4% in the direction of interest. The curve high point is at station 110+30 and has an elevation of 930 ft. What is the station and elevation of the PVC and the PVI? [Answer: PVI station: 110+50.3; PVI elevation: 930.6']arrow_forwardA parabolic curve has a descending grade of -0.8% and an ascending grade of +0.4% that intersect at station 10 + 020 with an elevation of 240.600 m. If the maximum allowable rate of change of grade for this curve is 0.15% per 20 m stations, determine the following: a. The maximum probable length of the parabolic curve b. The location and elevation of the lowest point of the curve c. The elevation at station 10 + 000arrow_forwardAn equal-tangent curve connects a +1.0% and a -0.5% grade. The PVC is at station 54+24 and the PVI is at station 56+92. Is this curve long enough to provide passing sight distance for a 60-mi/r design speed? Problem 2arrow_forward
- A +4% grade line intersects a -2 % grade line at station 6+800, where the elevation is 642.40 m. as shown in the figure. Determine the elevations of 100-m stations for this 1,000-m unequal tangent curve. Determine the station and curve elevation of PC, PT, D, E and CVC where in CVC is compound vertical curve.arrow_forward1. A vertical parabolic summit curve was designed in order to have a clear sight distance of 120m. The grade lines intersect a Station 9 + 000 at elevation 160.50m. The curve was so designed such that when the height of the driver's eye to 1.50m above the pavement would just see an object whose height is 0.10m above the pavement. Determine the max. speed that a car could travel along this curve. The grade lines has an upward grade of 5% and a downward grade of -3%.arrow_forwardA 3.5% grade passing at station 49+45.00 at an elevation of 174.83 ft meets a -5.5% grade passing at station 49+55.00 at an elevation of 174.73 ft. Determine the station and elevation of the point of intersection of the two grades as well as the length of the curve, L, if the highest point on the curve must lie at station 48+61.11arrow_forward
- An old equal-tangent crest vertical curve was designed with a PVI at station 117+45 and elevation of 960 ft., and a PVC at station 110 + 85 and elevation of 927 ft. If the highest point on the curve is at station 117 + 65, what is the maximum safe design speed for this curve? (explore and report your answer in 5 mph increment) [hint: Use design tables. But this time, find the speed based on K]arrow_forwardGiven an equal tangent vertical curve, in feet, with the following known information: STA A (Beginning of Curve) = 17+00 Elevation A = 95.00 feet G1 = +4.00% and G2 = -2.00% The vertical curve must begin at the fixed VPC (station A) and pass through station 21+00 at the elevation of 99.00 feet, the new calculated design length of the vertical curve, in feet, is:arrow_forwardA vertical crest curve has an approach grade +4% and a departure grade -3%. The Vertical Intersection Point level is 125m Australian Height Datum (AHD), with Chainage 3405m. It is proposed to use a curve 550m long. Calculate: Chainage and Level at TP1 Chainage and Level at TP2. Chainage and level at the apex point.arrow_forward
- Design the curve, of An equal tangent vertical curve is to be constructed between grades of -2.0% (initial) and 1.0% (final). The PVI is at station 11 + 000.000 and at elevation 420 m. Due to a street crossing the roadway, the elevation of the roadway at station 11 + 071.000 must be at 421.5 m. GIVEN DIMENSIONS ON THE ATTACHED PICTUREarrow_forwardAn equal-tangent vertical curve is to be constructed between grades of -2.0% (initial) and +1.0% (final). The PVI is at station 110 + 00 and at elevation 420 ft. Due to a street crossing the roadway, the elevation of the roadway at station 112 + 00 must be at 424.5 ft. Design the curve.arrow_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_forward
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