Principles of Highway Engineering and Traffic Analysi (NEW!!)
6th Edition
ISBN: 9781119305026
Author: Fred L. Mannering, Scott S. Washburn
Publisher: WILEY
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Chapter 3, Problem 43P
To determine
The station of the PT.
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A horizontal curve on a two-lane highway has its PC at station 123+70 and its PI at station 130+90. The curve has a super elevation of 0.06ft/ft and is designed for 70mi/h. SOLVE FOR horizontal alignment
A 200-m symmetrical parabolic curve is to be placed between grades of g1=1.25% and g2=-2.75% intersecting at station 18+000, which has an elevation of 270.190m above mean sea level. a)Calculate elevations for every 20-m station on the vertical curve b)Determine the station and elevation of the highest point on the curve. c)Determine the rate of change of the vertical curve.
Given two grades intersecting at station 1 + 250m, one grade at +6%, the other at -3%. Use a vertical curve of 183m composed of two sections, L1 of 121.95m and L2 of 61.05m. Find the elevation on the curve at station 1 + 220 and at station 1 + 280.
Chapter 3 Solutions
Principles of Highway Engineering and Traffic Analysi (NEW!!)
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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- A horizontal curve has a maximum speed of 50 mph. If e = 0.10, and f = 0.16, find the degree of the curve. a. 4.84 deg b. 5.84 deg c. 3.84 deg d. 6.24 degarrow_forwardQ3/ A vertical curve below has a lower point (A) which exists at station (53+50) with elevation (1271.2 m). the back grade of (-4%) meet the forward grade of (+3.8%) at (PVI) station (52+00) with elevation (1261.5 m). determine the length of the curve with the stations of (PVC) and (PVT)?arrow_forward25. A parabolic curve has a descending grade of -0.8% which meets an ascending grade of 0.4% at sta. 10 + 020. The max. Allowable change of grade per 20 m. station is 0.15%. Elevation at station 10 + 020 is 240.60 m. Compute the lowest elevation point of the curve.arrow_forward
- Please show illustration if possible. A compound curve has a length of the chord of the first curve (passing thru PC to PT) of 470m. And a length of the chord (from PCC to PT) of 230m. If the angle that the long chord makes with the chord from PC to PCC and from PT to the PCC is 6 degrees and 9 degrees respectively, find the length of the long chord.arrow_forwardAn equal-tangent crest vertical curve is designed with a PVI at station 110 + 00 (elevation 927.2 ft) and a PVC at station 107 + 43.3 (elevation 921.55 ft). If the high point is at station 110 + 75.5, what is the design speed of the curve?arrow_forwardA parabolic curve has a descending grade of -0.8% which meets an ascending grade of 0.4% at sta. 10 + 020. The max. Allowable change of grade per 20 m. station is 0.15%. Elevation at station 10 + 020 is 240.60 m. Compute the lowest elevation point of the curve. CORRECT ANSWER: 240.81arrow_forward
- A vertical parabolic happy curve located in Malabon underpass has a grade of -4% followed by a grade of 2% intersecting at Sta. 12+150.60 at elevation of 124.80 m above sea level. The change of grade of the sag curve is restricted to 0.6%. b. Compute the elev. of the lowest point of the curve.arrow_forwardA section of a two-lane highway (3.60 meter lanes) is designed for 120km/h. At one point a vertical curve connects a 2.5% and +1.5 % grade. The PVT of this curve is at station 100+000. It is known that a horizontal curve starts (has PC) 89.6 meters before the vertical curve's PVC. If the super elevation of the horizontal curve is 0.08 and the central angle is 38 degrees, what is the station of the PT?arrow_forwardA symmetrical parabolic curve has a descending grade of -0.80% which meets ascending rate of +0.4% at station 10+020 having an elevation of 240.62m. The maximum allowable change in grade per 20m station is 0.15%. Determine the stationing of the lowest point of the curve. (Answer: 10+046.67) Determine the elevation of the lowest point of the curve. (Answer: 240.83)arrow_forward
- A 6.2% grade is followed by -2.8% grade; the grades intersecting at sta. 10+800 of elevation 1060 m the symmetrical parabolic curve is 300m. a. Locate the highest point on the curve. b. Locate Sta. C on the curve that lies on the left of the summit and is at elec. 1054 m.arrow_forwardThe ascending of a vertical curve is 1/45 and descending gradient is 1/55. If the design speed is 60 km/hr and rate of change of acceleration of 0.4 m/s3, calculate the minimum length of the curve.arrow_forwarda 400-ft equal tangent sag vertical curve has its PC t station 100 + 00 and elevation 500ft. The initial grade is -4.0% and the final grade is +2.5%. Determine the stationing of the lowest point.(take note that 1 station consists of 100 ft.)arrow_forward
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