An equal-tangent crest vertical curve is designed for 100 km/h. The initial grade is +3.4% and...

An equal-tangent crest vertical curve is designed for 100 km/h. The initial grade is +3.4% and the final grade is negative. Draw the curve. What is the elevation difference between the PVC and the high point of the curve?

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Ally Wells answered 1 year ago

An equal-tangent crest vertical curve connects an initial grade of +2.5% and a final grade of...

An equal-tangent crest vertical curve connects an initial grade of +2.5% and a final grade of –0.5%. The curve is designed for 70 mi/h and the station of the PVT is 132+62 and is at elevation 833 ft. What is the station and elevation of the curve's high point?

A vertical curve is designed for 90 km/h and has an initial grade of + 2.5%...

A vertical curve is designed for 90 km/h and has an initial grade of + 2.5% and a final grade of -1.0%. The PVT is at station 3 + 480. It is known that a point on the curve at station 3 + 440 is at elevation 75 m. What is the stationing and elevation of the PVC? What is the stationing and elevation of the high point on the curve?

A vertical curve is designed for 75 km/h and has an initial grade of + 3.5%...

A vertical curve is designed for 75 km/h and has an initial grade of + 3.5% and a final grade of -2.0%. The PVT is at station 3 + 560. It is known that a point on the curve at station 3 + 520 is at elevation 85 m. What is the stationing and elevation of the PVC? What is the stationing and elevation of the high point on the curve?

A vertical curve is designed for 55 mi/h and has an initial grade of +2.5% and...

A vertical curve is designed for 55 mi/h and has an initial grade of +2.5% and a final grade of −1.0%. The PVT is at station 114 + 50. It is known that a point on the curve at station 112 + 35 is at elevation 245 ft. What is the stationing and elevation of the PVC? What is the stationing and elevation of the high point on the curve?

Vertical Curves: Problem 1 Given an equal tangent parabolic curve, a +0.50% grade meets a -2.50%...

Vertical Curves: Problem 1 Given an equal tangent parabolic curve, a +0.50% grade meets a -2.50% grade at Station 12+35.00 with an elevation of 305.12’, with L = 6 sta. Calculate the high point station. Provide the elevations at the PVC, PVT, High point, and at all full stations.

A vertical curve of 1000 feet is designed to connect a grade of +4% to a...

A vertical curve of 1000 feet is designed to connect a grade of +4% to a grade of -5%. The PVI is located at station 1500+55 and has a know elevation of 500 feet. Find the following: The station and elevation of the PVC The station and elevation of the PVT The elevation of points along the vertical curve at 100-feet intervals The location and elevation of the high point on the curve

Use the following data to calculate the layout of an equal tangent vertical curve by offsets...

Use the following data to calculate the layout of an equal tangent vertical curve by offsets for every full station. g1 = -1.20% g2 = +3.50% X PVI = 49+70 Y PVI = 5,893.48’ L = 1500’ , find the station and elevation of the high (or low) point of the curve.

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An equal-tangent crest vertical curve is designed for 100 km/h. The initial grade is +3.4% and...

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