Plot the curve and show solution. A vertical summit curve is having tangents which intersects at point Pl at Station 12+609.84 and 139.82 meters elevation. The grade for the forward tangent is -6.04%. Station PC is at station 12+500.65 and having an elevation of 115.52 meters. Determine the elevation of the highest point in meters

Plot the curve and show solution. A vertical summit curve is having tangents which intersects at point Pl at Station 12+609.84 and 139.82 meters elevation. The grade for the forward tangent is -6.04%. Station PC is at station 12+500.65 and having an elevation of 115.52 meters. Determine the elevation of the highest point in meters

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 16P

See similar textbooks

Similar questions

NEED ASAP A summit curve has tangents intersects at Pl at station 14+750 with elevation of 76.30m. The grade of the back tangent is 3.4% and the forward tangent is -4.8%. What is the elevation of the highest point if the length of the curve is 300 meters.
a vertical summit curve is having tangents which intersects at point Pl at station 12+618.33 and 136.44 m elevation. The grade for the forward tangent is -7.34% station PC is at station 12+527.72 and having an elevation of 122.14 m. Determine the elevation of the highest point in m
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.
A symmetrical parabolic curve passes through point A whose elevation is 23.23m at a distance of 54m from the PC. The elevation of the PC at station 4+100 is 22.56m. The grade of the back tangent is +2% and the length of curve is 120m. a) Compute the grade of the forward tangent. b) Compute the stationing of the highest point of the curve from the PC. c) Compute the elevation of the highest point of the curve.
NEED ASAP A vertical summit curve is having tangents which intersects at point Pl at station 12+583 and 128.0 meters elevation. The grade for the forward tangent is -5.54%. Station PC is at station 12+530 and having an elevation of 123.744 meters. 1. Determine the value of the maximum offset in meters. 2. Determine the elevation of the highest point.
NEED ASAP A vertical summit curve is having tangents which intersects at point Pl at station 12+583 and 128.0 meters elevation. The grade for the forward tangent is -5.54%. Station PC is at station 12+530 and having an elevation of 123.744 meters. Determine the value of the maximum offset in meters. Determine grade of back tangent. Determine horizontal length of curve in meters.
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]
ASAP!!!!!! PLEASE HELP!!!! A vertical summit curve is having tangents which intersects at point P) at Station 12+604.36 and 138.47 meters elevation. The grade for the forward tangent is -5.78%. Station PC is at station 12+522.29 and having an elevation of 116.75 meters. What is the elevation of point A along the curve with Station 12+547.78?
The change in grade of a vertical parabolic curve is 0.60%. The grade of the said curve is -4% and +2% respectively. Both tangents intersects at station 12+150.60 at elevation 124.80m above sea level. Determine the elevation of the lowest point of the curve above sea level.
A vertical curve crosses a 1-m diameter pipe at right angles. The pipe is located at station 11 + 025.000 and its centerline is at elevation 1091.60 m. The PVI of the vertical curve is at station 11 + 000.000 and elevation 1095.20 m. The vertical curve is equal tangent, 150 m long, and connects an initial grade of +1.20% and a final grade of -1.08%. Using offsets, determine the depth, below the surface of the curve, to the top of the pipe, and determine the station of the highest point on the curve.
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 summit vertical curve has its PVI at station 14+750 with elevation of 76.30m. The grade of the back tangent is +3.4% and forward tangent of -4.8%. If the length of the curve is 300m. Compute the location of the vertical turning point from the PVI.Compute the elevation of the vertical curve turning point.