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?

Expert Solution

Ally Wells answered 10 months ago

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?

A 1000-ft vertical curve connects a -1% grade to a -3.5% grade. If the vertical tangents...

A 1000-ft vertical curve connects a -1% grade to a -3.5% grade. If the vertical tangents intersect at sta. 38+60.24 and elevation 1,400.13 ft, develop an Excel spreadsheet to calculate the elevations of the grade line at each station from sta. 31+00 to 45+00. Include the elevations of the PVC and PVT.

A 500 feet equal tangent vertical curve connects two tangents that intersects at station 5+34.00 and...

A 500 feet equal tangent vertical curve connects two tangents that intersects at station 5+34.00 and elevation 1210 ft. The initial grade is +4% and the final grade is -2.5%. i. determine the station and elevation of the high point, PVC and PVT ii. What will you recommend as the maximum safe design speed for this road to provide adequate stopping sight distance?

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 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.

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.

A vertical summit curve has tangent grades of +2.5% and -1.5% intersecting at station 12+460.12 at...

A vertical summit curve has tangent grades of +2.5% and -1.5% intersecting at station 12+460.12 at an elevation of 150m above sea level. If the length of the curve is 182m: a. Compute the length of the passing sight distance. b. Compute the stationing of the highest point of the curve. c. Compute the elevation of the highest point of the curve.

Compute and tabulate full-station elevations for an unequal-tangent vertical curve to fit the requirements of grade...

Compute and tabulate full-station elevations for an unequal-tangent vertical curve to fit the requirements of grade g1 = +1.25%, g2 = +3.75%, VPI at station 62+00 and elevation 1053.95 ft, L1 = 500 ft and L2 = 600 ft.

Tabulate station elevations (stakeout at full stations) for an equal-tangent vertical curve for the following data...

Tabulate station elevations (stakeout at full stations) for an equal-tangent vertical curve for the following data given. (20 pts) • 500-ft curve • g1 = -3.00% • g2 = -1.25% • VPI at station 38 + 00 and elevation 560.00 ft

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