Traffic and Highway Engineering
5th Edition
ISBN: 9781305156241
Author: Garber, Nicholas J.
Publisher: Cengage Learning
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Question
Chapter 15, Problem 21P
To determine
(a)
Minimum length of crest vertical curves
To determine
(b)
The minimum length of sag vertical curve.
To determine
(c)
Maximum super elevation.
To determine
(d)
Maximum degree of curve.
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Design a horizontal curve for the highway by computing the radius and stationing
of PC and PT using the following data,
# of lanes = 2
Width per lane = 12 ft
Pl is at station 120+20
Design speed is 60 mi/h
Superelevation = 0.06 ft/ft
Central angle of the curve = 37 degrees
(b)
For the safe road design and smooth traffic flow, combination of transition and
circular curve will be constructed to combine two tangent line that have bearing of
N 70° E (entry tangent) and S 80° E (exit tangent) respectively. With the aid of
sketch, design the minimum length of transition curve and circular curve that has
the following characteristics
• Design speed = 110 km/hr
coefficient of side friction, f= 0.15
Superelevation, e = 5 %
• Radial acceleration, c = 0.9 m/s³
PI sta = 18+00, PI elev = 300.00, L = 20 stations, incoming grade = +2%, outgoing grade = -3% The high point of the curve is at station ?
Note: I need right solution.. Don't copy from other expert solution.
Chapter 15 Solutions
Traffic and Highway Engineering
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Similar questions
- The project speed is 100 km/h on a divided state road with a vertical curve. The driver's transition reaction time is 2.0 seconds; the friction coefficient is 0.4. Accordingly, calculate the following: a) The slopes of g, and g2. b) The vertical curve length (L). c) Calculate the grade line elevation of the points T1 and T2 as well every 30 meters and present them in a table format. Note: use the Stopping sight distance to solve thiaarrow_forwardB- Unequal-Tangent Parabolic (Vertical) Curve: Example: A grade gl 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. Stations 00 00 00 00 EVC (752.84) BVC (751.24) A CVC (748.04) (747.24) (747.56) -2.00% 400 +1.60% V(743.24) >と じ。 Solution: The CVC is defined as a point of Compound Vertical Curvature. We can determine the station and elevation of points A and B by reducing this unequal tangent problem to two equal tangent problems. Point A is located 200' from the BVC and Point B is located 300' from the EVC. Knowing this we can compute the elevation of points A and B. Once A and B are known we can compute the grade from A to B thus allowing us to solve this problem as two equal tangent curves. Point A STA 85+00: Elev. = 743.24+2 (2) = 747.24'…arrow_forwardA compound curve has the following data: intersection angle of the first curve I sub 1 = 28 degrees, intersection angle of the second curve I sub 2 = 31 degrees, D sub 1 = 3 degrees, and D sub 2 = 4 degrees. Find the stationing of P.C.C. Use P.I. equals 30 + 120.5.arrow_forward
- Example: 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_forwardIt is desired to replace the compound curve with a simple curve that will be tangent to the three tangent lines, and at the same time forming a reversed curve with parallel tangents and equal radii, solve for the ff: a. Common radius of the reversed curve Distance between the parallel tangents c. Stationing of the new PT b.arrow_forward5. (a) Using a well annotated diagram, present the five point permissible approximations or assumptions used for the computations related to vertical curve design and setting out. (b) A 100 metre vertical curve is to connect a downgrade of 1.75% to an upgrade of 1.25%. If the level of the intersection point of the two grades is 150m, calculate the curve levels at 20m intervals, showing the second difference check on the computation.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 degree of curve of a sample curve is 5°. Compute the desired super elevation required if the design speed of a car passing thru the curve is 80kph and the skid resistance is equal to 0.12.arrow_forwardDetermine the P.C. (Point of Curve) and P.T. (Point of Tangent) for the horizontal curve using the following information: • P.I. (Point of Intersection) = 56 + 77.21 • Intersection Angle, I = 16° 41’ • Degree of Curvature, D = 5° 40’arrow_forward
- A simple horizontal curve has 35 degrees of curvature and subtends an angle of 110 degrees. If the PC is at station 3 + 65.00 ft, what is the intermediate chord length (ft) (i.e. the straight line distance between two sequential full stations)? Response Feedback: 8 C =2Rsin- D 2arrow_forwardUse the simplified version of the basic curve equation to calculate the minimum radius (ft) for a horizontal curve given a design speed of 60 mph, a maximum superelevation of 6.0% and a maximum friction factor of 0.12. Response Feedback: R = min 15(0.01e V² max + f maxarrow_forward2. Vertical Curve Given the following vertical curve data, compute the elevations of the curve, including low/high point, and even 20m stations. Use either Tangent Offset or General Equation method. PVI @ 2+200 g1=-1% L=100m PVI El.=560.000 g2=1.5%arrow_forward
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