A 3% ascending grade meets a 5% descending grade at station 2+110 and elevation 100m. The tangents are connected by an unsymmetrical parabolic curve passing thru a point at station 2+150 having an elevation of 97.078m. If the length of the first curve is 60m, determine: A.) Length of the 2nd Curve = [L2] B.) Stationing of the Highest Point = [StaHP] C.) Elevation of the Highest Point = [ElevHP]

A 3% ascending grade meets a 5% descending grade at station 2+110 and elevation 100m. The tangents are connected by an unsymmetrical parabolic curve passing thru a point at station 2+150 having an elevation of 97.078m. If the length of the first curve is 60m, determine: A.) Length of the 2nd Curve = [L2] B.) Stationing of the Highest Point = [StaHP] C.) Elevation of the Highest Point = [ElevHP]

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 parabolic curve AB, 400 m. long is connected by tangents having an upgrade of +6.5% and a downgrade of -3% intersecting at sta. 20 + 800 at elevation 102.5 m. Find the distance from B to the highest point of the curve
A parabolic curve JS, 480 m long is connected by tangents having an upgrade of +5.9% and a downgrade of -2.8% intersecting at Sta 20+870 at elevation 460 m. Find the distance from S to the highest point of the curve.
A symmetrical parabolic curve passes through point A whose elevation is 23.23m at a distance of 54 meters from P.C. 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. Determine the stationing of the highest point.
A symmetrical vertical summit curve has tangents of + 4% and – 2%. The allowable rate of changeof grade is 0.3% per meter station. Stationing and elevation of P.T. is at 10 + 020 and 142.63m.,respectively. (Hint: 1 meter each station).1. Compute the length of the curve. 2. Compute the distance of the highest point of the curve from P.C. 3. Compute the elevation of the highest point of the curve.
A vertical parabolic curve passes thru point A whose elevation is 23.23 m at a distance of 54 meters from the point of curvature. The elevation of the point of curvature is 22.56 m at station 4 + 100m. The grade of the back tangent is +2% and the length of the curve is 120m. Determine the grade of the forward tangent.
An unsymmetrical parabolic curve has a forward tangent of -8% and backward tangent of +5%. The length of curve on the left side is 40m long while that of the right side is 60m long. PC is at Sta 6+780 and at elevation 110m. Determine the elevation at Sta 6+820. Determine the elevation at the summit. draw a free body diagram or grade diagram.
2. A +7% grade meets a -4% grade at the vertex V with elevation of 30m and stationing of 2+000. It is required to connect the two tangents with an asymmetrical vertical parabolic curve, 60m on the back tangent and 40m on the forward tangent. Compute the elevation at station 1+ 950m. Sketch the curve.
A parabolic sag curve has a length of 180m. The tangents are -2% and 4% respectively. Determine the vertical offset at the first quarter point.
1. A symmetrical parabolic curve passes through point A whose elevation is 23.23 m at a distance of 54 m from PC. The elevation of PC at Station 4+382.3 is 22.56 m. The grade of the back tangent is +(9÷3)% and the length of the curve is 120 m. a. Compute the grade of the forward tangent. b. Compute the stationing of the highest point of the curve. c. Compute the elevation of the highest point of the curve.
A 160m long parabolic curve connects a -8% grade to a +2% grade.These grades intersect at Sta. 1+240 with elevation of 14.87m. Find the elevation and station of the lowest/highest point on the curveand at the first quarter point of the curve. (DRAW CLEAR FIGURE)
A 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)
A symmetrical parabolic curve has an ascending grade of 3.2% and a descending grade of 2.8%. If the PC is at elevation 68.72m, what is the elevation at the PI, length of curve is 50m.