Two tangent of a vertical curve intersects at station 32+11.61 and elevation 64.
ID: 1714481 • Letter: T
Question
Two tangent of a vertical curve intersects at station 32+11.61 and elevation 64.18 ft. back tangent grade is gi--4% and forward tangent grade = +6%, and length L = 600 ft Calculate the station of PVC a. 28+11.61 b. 29+11.61 c. 30+11.61 d. 35+11.61 9. 10. Calculate the elevation of PVC in feet. a. 70.18 b. 76.18 c. 80.18 d. 82.18 11. Calculate the satiations of PVT in feet a. 28+11.61 b. 29+11.61 c. 30+11.61 d. 35+11.61 12. Calculate the elevation of PVT a. 70.18 b. 76.18 c. 80.18 d. 82.18 13. Calculate the station and elevation of low point in ft. a. 28.61.61: 40.38 b. 29.51.61: 68.38 c. 31+51.61:70.38 d. 31+51.61;72.38Explanation / Answer
VPI Station = 32+11.61
VPI Elevation = 64.18
g1 = -4.00%
g2 = +6.00%
Length = 600.00
BVC Station = 29+11.61
BVC Elevation = 76.18
Station x (Sta) g1*x r/2*x*x Elevation
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35+11.61 6.00 -24.00 30.00 82.18
35+00.00 5.88 -23.54 28.85 81.49
34+00.00 4.88 -19.54 19.88 76.52
33+00.00 3.88 -15.54 12.57 73.21
32+00.00 2.88 -11.54 6.93 71.58
31+00.00 1.88 -7.54 2.96 71.60
30+00.00 0.88 -3.54 0.65 73.30
29+11.61 0 0 0 76.18
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Minimum elevation = 71.38 @ station 31+51.61
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