A symmetrical parabolic curve passes through point A whose elevation is 23.23 m at a distance of 54 m from the PVC. The elevation of the PVC at Station 4 + 100 is 22.56 m. The grade of the back tangent is +2% and the length of the curve is 120 m. 1. Compute the grade of the forward tangent. 2. Compute the stationing of the highest point of the curve. 3. Compute the elevation of the highest point of the curve.

Q: 23.15 A +4% grade intersects a -3% grade at station 3+ 000m at elevation 566.98 m. Compute the…

A: PVCSTATION=PVISTA-L2PVCsta=3000-6002=2+700PVCELV=PVIELV-g1L2PVCelv=566.98-0.04×6002PVCelv=554.98m

Q: Vertical Curve: If L=1320 ft, the BVC is at 5+00, What is the Station and Elevation for PVI ? G1 is…

A: Vertical Curve: A vertical curve is a transition element in the geometric design of a highway. A…

Q: How are passengers and their belongings safeguarded during vehicle transfers?

A: To explain the how the passengers and their belongings are safeguarded during vehicle transfers.

Q: Problem 2 A symmetrical vertical summit curve has tangents of +5% and -3%. The allowable rate of…

A:

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