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Example #6: Locating Vertical Curve Elevations

Example #6: 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 +G 1 *L/2

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Example #6: Locating Vertical Curve Elevations

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  1. Example #6: 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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