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Vertical Curve Elevation Calculation for Crest Curve Connecting Grades

This document discusses the calculations for a crest vertical curve with a length of 400 ft, connecting grades of +1.0% and -1.75%. It includes the location of the vertical point of intersection (VPI) at station 35+00 with an elevation of 549.2 ft. The calculations proceed to determine the elevations and stations for the vertical point of curvature (VPC) and the vertical point of tangency (VPT). The provided formulas illustrate how to derive these values, resulting in elevations of 547.20 ft for VPC and 545.70 ft for VPT, along with their corresponding stations.

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Vertical Curve Elevation Calculation for Crest Curve Connecting Grades

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  1. Example #5: Locating Vertical Curve Elevations A crest vertical curve with a length of 400 ft connects grades +1.0% and -1.75%. The VPI station 35+00 and elevation 549.2 ft. What are the elevations and stations of VPC & the VPT? VPI Elevation = VPC elev +G1*L/2 VPI Station = VPC station + L/2 VPC=549.20 - 0.01*400/2 VPC elevation = 547.20 ft VPC station = 3500–400/2 VPC station = 33+00 VPI Elevation = VPT elev - G2*L/2 VPT Station = VPC station + L VPT=549.20 + (-0.0175)*400/2 VPT elevation = 545.70 ft VPT station = 3500+400/2 VPT station = 37+00

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