5. A compound curve, having Station PCC as 2+320, and Station PC as 2+020, has the Central Angle of the First Curve as 37 Degrees. The Chord Distance from PCC to Station PT is 160 m, and the Degree of Curve of the Second Curve is 7 Degrees. a) What is the Radius of the First Curve? 554.4753 m 464.5604 m 440.7368 m 520.8707 m 491.1067 m None of the Choices 419.2374 m b) What is the Central Angle of the Compound Curve? 93.5093 Degrees 91.5093 Degrees 99.5093 Degrees 89.5093 Degrees None of the Choices 97.5093 Degrees 95.5093 Degrees c) What is the Length of the Common Tangent? 245.9841 m 247.7682 m 248.4416 m None of the Choices 245.4647 m 246.5405 m 247.1349 m

5. A compound curve, having Station PCC as 2+320, and Station PC as 2+020, has the Central Angle of the First Curve as 37 Degrees. The Chord Distance from PCC to Station PT is 160 m, and the Degree of Curve of the Second Curve is 7 Degrees. a) What is the Radius of the First Curve? 554.4753 m 464.5604 m 440.7368 m 520.8707 m 491.1067 m None of the Choices 419.2374 m b) What is the Central Angle of the Compound Curve? 93.5093 Degrees 91.5093 Degrees 99.5093 Degrees 89.5093 Degrees None of the Choices 97.5093 Degrees 95.5093 Degrees c) What is the Length of the Common Tangent? 245.9841 m 247.7682 m 248.4416 m None of the Choices 245.4647 m 246.5405 m 247.1349 m

Traffic and Highway Engineering

5th Edition

ISBN:9781305156241

Author:Garber, Nicholas J.

Publisher:Garber, Nicholas J.

Chapter15: Geometric Design Of Highway Facilities

Section: Chapter Questions

Problem 7P

See similar textbooks

Similar questions

A simple curve connects two tangents AB and BC bearings N 85°30' E and S 68°30' E respectively. Point D along line AB has a coordinate of 20100 N and 20100 E while point E along line BC has coordinates of 20086.55 N and 20184.27 E 1. Find the distance of line BD. 2. Solve for the degree of simple curve that shall be tangent to the three AB, DE and BC 3. If point D at station 1 + 052.87 determine the stationing of PT.
A parabolic sag curve has tangent grades of –2% and +3%. The stationing of PT is at 10+300 and elevation of 100 m. The total allowable change of grades per 20 m station is 0.002 1. Compute the total length of the curve. a. 534 m b. 500 m c. 360 m d. 527 m 2. Find the stationing of PC. a. 7+589 b. 10+089 c. 9+800 d. 8+206 3. Find the elevation of PC. a. 98.5 m b. 102.8 m c. 97.5 m d. 82.3 m
58. The length of a parabolic curve is 130 m. The forward tangent has a slope of 2.6% back tangent has a slope of -2.5%. Compute the distance of the lowest point of the curve from the P.C.
A parabolic curve having a forward tangent of -5% intersects a back tangent of 7% at station 10 + 750 whose elevation is 435.8 m. The curve passes through point A whose elevation is 432.15 at sta. 10 + 767.30. determine the length of the curve. a. 240 m b. 250 m c. 260 m d. 270 m
Two tangents intersect at station 2 + 550. A compound curve laid on their tangents has the following data. I1 = 310, I2 = 360, D1 = 30, D2 = 50 a) Compute the stationing of the P.C., P.C.C. of the curve. b) If the P.T. is moved 15 m out, compute the station of the P.T. with the P.C.C. on the same point.
Prob. 6 The common tangent AB of a compound curve is 76. 42 m. with an azimuth of 268030’. The vertex V being in accessible. The azimuth of the tangents AV & VB are 247050’ & 282050’ respectively. If the stationing of A is 43 + 010.46 and the degree of curve for the first curve was fixed at 40 based on a 20 m. chord. Using chord basis, determine the stationings of PC,PCC, & PT.
A symmetrical parabolic curve passes through point A whose elevation is 23 23m at a distance of 54 meters from PC The elevation of the P.C. at station 4+100 is 22.56m. The grade of the back tangent is +2% and the length of the curve is 120m. Determine the grade of the forward tangent a. -2.3% b. -1.4% c. -3.3% d. 2.5%
A horizontal curve with P.C is stationed at 3+025.42 and having a radius of 120 m. A property line with an azimuth of 300° intersects the curve at point B. Then a random line was run from the P.C of the simple curve towards the property line which intersects at point A, with an azimuth and distance of 190° and 42.50m respectively. Determine the stationing of B, knowing that the back tangent of the curve has an azimuth of 220°. Show a Clear and Organized solution:
SITUATION 1: AB and BC are tangents of a simple curve. AB is due north and BC is N50degreeE. Degree of simple curve is 4degree and PT is at station 20 + 130. 2. Determine the external distance.a. 29.62 m c. 14.66 mb. 30.23 m d. 10.4 m 3. Determine the length of the line from point D on the simple curve to PI that makes an angle of 54degree with the tangent passing thru the PC.a. 29.62 m c. 14.66 mb. 30.23 m d. 10.4 m 4. Determine the stationing of point D.a. 20 + 249.3 c. 20 + 010.70b. 19 + 999.25 d. 18 + 888.25
A certain horizontal curve has a central angle of 42.2°. If STA PC is 18+363 and STA PT is at 18+708, determine the station of PI.
the three courses AB, BC and CD have bearing N56°30'E, N80°30'E and S32°17'E, respectively. if BC= 57.80m, determine the radius of the curve that shall be tangent to three courses
A vertical sag parabolic curve CD, 620 meters long is connected by tangents having a downgrade of 6.2% and an upgrade of 4.2% intersecting at Station 18+254.4 and elevation 54.25 m. Find the distance from D to the lowest point of the curve.