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 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.
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?
3.48 A horizontal curve is being designed for a new
two-lane highway (12-ft lanes). The PI is at station
250 + 50, the design speed is 65 mi/h, and a
maximum superelevation of 0.07 ft/ft is to be used. If
the central angle of the curve is 38 degrees, design a
curve for the highway by computing the radius and
stationing of the PC and PT.
3.49 You are nal
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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- 3.35 A horizontal curve on a two-lane highway (10-ft lanes) is designed for 50 mi/h with a 6% superelevation. The central angle of the curve is 35 degrees and the PI is at station 482 + 72. What is the station of the PT and how many feet have to be cleared from the lane's shoulder edge to provide adequate stopping sight distance?arrow_forwardA horizontal curve is being designed for a new two-lane highway (12-ft lanes). The PC is at station 150+50, design speed is 55 mi/h, and a superelevation of 0.08 fuft is to be used. If the central angle of the curve is 40 degrees, design a curve for the highway by computing the minimum radius and stationing of the PI and PT??arrow_forwardThe angle between a 5-m full station of a certain curve is 10°. The curve has 7 intermediate stations between P.C. (10 + 300m) and P.T. After laying out the curve, it was found out that the P.T. falls in a property lot and it should be moved in a way that the P.T. will be free from crossing the lot. Compute all the necessary elements of the old curve and new relocated curve.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_forwardA 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 alignmentarrow_forwardA horizontal Curve is designed for a two lane mountainous terrain. The following data are known. intersection angle: 40 degrees tangent length: 436.76 ft Station on PI: 2700+10.65 fs=0.12 e=0.08 Find the following Design speed Station of the PC Station of the PT Deflection Angle and chord length to the first even 100 ft stationarrow_forward
- A reverse curve with diverging tangent is to be designed for a proposed highway as illustrated below. The reverse curve is composed of one compound curve, from point A to C, and one simple curve, from point C to D. Point A is located at the Point of Curvature having Station 12+808.35, point B is located at the Point of Compound Curve, point C is located at the Point of Tangency. The back tangent of the reverse curve is having an located at the Point of Reverse Curve, and point D azimuth of 388.5 degrees and the forward tangent is having an azimuth of 748.8 degrees. The common tangent for the reverse curve has an azimuth of 313.1 degrees from south. It has been found out that the long chord of curve A to C is thrice the value of the chord in curve C to D. The common tangent of the curve A to C is having an azimuth of 254.1 degrees from south and the chord of curve C to D is 161.3m long.arrow_forward2. A 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_forwardQ6. To help improve the safety of at-grade intersection of two roads, the intersection is being redesigned so that the E-W road will pass underneath the N-S road. Currently the vertical alignment of the E-W road consists of a crest vertical curve joining a 4% upgrade to a 3% downgrade. The existing vertical curve is 138 m long, the PVC of this curve is at station 14+47, and the elevation of the PVC is 477 m. The centerline of the N-S road is at station 15+45. Your job is to find the shortest vertical curve that provides 6.0 m of clearance between the new E-W road and the N-S road.arrow_forward
- 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. Reversed curve with diverging tangents is to pass through three lines to form a center line of a proposed road. the first line AB has a bearing of N 88° E. and a distance of 125 m., BC has a bearing of N 60° E and a distance of 350 m., while on that of CD has a bearing of S 40° E and a distance of 305 m. If the first tangent has a distance of only % of the common tangent measured from the point of intersection of the first curve determine the radius of the curves and station of PT if PC is 12 + 350arrow_forward(TRAFFIC AND HIGHWAY ENGINEERING) A highway segment has an equal-tangent sag curve and a horizontal curve with the same design speed. It is known that the sag curve has an initial grade of –1.8% and a positive final grade that is not known. The station of the PVC is 232 + 70, and the station of the low point of the sag vertical curve is 235 + 14.8. The horizontal curve has a superelevation of 8%, central angle of 30 degrees, and a PT at Station 302 + 20. The road has two 11-ft lanes (one in each direction). What is the station of the PI?arrow_forward
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