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 63P
To determine
The elevation station
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A horizontal curve is being designed with tangent length of 850 ft and central angle of 0.7315 radians. If the Pl is at station 215+00, determine the station of
PT.
A.
222+74
B.
223+50
C. 274+64
D. 271+95
A horizontal curve is being designed around a pond with a tangent length of 800 ft and central angle of 0.6109 radians. If the PI is at station 85 + 00, determine the station of PT.
A.
97 + 35
B.
92 + 50
C.
89 + 75
D.
101 + 45
Simple Curve
A railway curve is to connect intersecting tangents with the following design data:
Back tangent = N 20° 25' E
Forward tangent = S 49°35' E
Degree of curve = 5⁰
Design speed = 70 Kph
Station PI = 5+555.55
1.Determine the length of line from PC to Pl.
a) 234.67m
b) 244.67m
2.Compute the straight distance from PC to PT.
a) 402.34m
b) 375.60m
c) 245.89m
d) 327.42m
c) 389.78m
d) 423.67m
3. If the width of the land intended for the railway is 10m, determine the area of land from
PC to PT.
a) 4,456.56 sqm
b) 3,456,43 sqm
c) 5,112.34 sqm d) 4,399.92 sqm
4.How long would a 300m long train able to leave the curve completely.
a) 25.67 sec
b) 38.06m sec
c) 40.34 sec
d) 29.67 sec
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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- 2. In the figure shown AB = 103.20m with A at station 10 + 158.93. The angle VAC is 15'21' and angle VBC is 18°31', where Cis the point to which a simple curve is to be constructed tangent to the line AB. a) Determine the radius of the curve b) Determine the stationing of point C. c) Determine the stationing of P.T. 10+158.93 A 521'C 18 31B P.C.arrow_forwardWhat is the station of the PT of a horizontal curve with the following data: A=40°50', R=600 ft and sta of PI = 20+00? O 20+00.00 O 22+04.27 O 22+23.13 O 24+27.30 O 25+04.30arrow_forwardTwo roads AB and BC intersect at a chainage 184+65. These are intersected by a line PQ and the angles APQ and PQC are 112° and 102° respectively. The radius of the first arc of the curve is 330m and that of the second arc is 540m. Calculate the chainages of the first and the second tangent points P.C R₂ 2 P.T 185+87.15, 172+18.67 O 170+88.19, 182+14.97 O 176+48.69, 172+18.67 O 185+87.19, 152+14.97 O 190+84.32, 177+12.34 Oarrow_forward
- The perpendicular distance between the parallel tangents of a reversed curve is 35 m. The azimuth from north of the back tangent is 270° while the azimuth from north of the common tangent is 300°. If the radius of the back curve is 150 m and the station PRC is 10 + 400. Find the stationing of P.C. 10+242.92 10+321.46 10+322.35 10+359.81 Find the stationing of P.T. 10+458.25 10+398.23 10+557.68 10+447.87 Length Second Radius 119.33 123.45 108.69 111.24arrow_forwardGiven a simple circular curve, in feet, with the following known information: o | = 25°10'45" Da = 8°15'00" o STA P.I. = 30+89.45 What is the stationing of the C.T.? O 32+39.55 32+39.68 36+97.40 O 33+94.65arrow_forwardA parabolic curve having a forward tangent of -3% intersects a back tangent of 6.5% at sta 10=800 whose elevation is 5260m. If the length of the curve is 400m, Find the stationing of the highest point of curve. a.10+877.5 b. 10+875.6 c. 10+871.1 d. 10+873.7arrow_forward
- For the PI Inaccessible problem: Sta. X = 75+50.00, XY =800.00 ft, angle x = 35°00′00”, angle y = 30°00'00", and R for the simple curve = 1,000.00 ft. The station of A (T.C.) is: 74+82.51 73+54.28 None of these. 71+08.65 74+19.23arrow_forwardPlease help me asap ???? The intersection of two proposed tangent AB and CD are inaccessible. Therefore, from point E which is at Sta 10+266 on Tangent AB. The line EF is run to the point F onn the Tangent CD . The angle between the prolongation of AB and the line EF as measured by the transit is found to be 37 degrees. The line EF is measured and found to be 134.50 m and the angle between EF and the prolongation of CD is 24 degrees. If Tangent AB and CD are to be connected by 5 degrees, Find : a. Sta PC b. Distance from PT to Farrow_forwardQ#03: Two tangents AB & BC are intersected by a line having angles 40o & 35o respectively. The radius of 1st arc is 1000m and of 2nd arc is 800m. Carry out all the necessary calculation to find the chainage of tangent points points and the point of compound curvature given that the chainage of intersection point is 5000 marrow_forward
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