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.

Solutions

Expert Solution

Ally Wells answered 1 year ago

Next > < Previous

Related Solutions

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

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?

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 276 m equal-tangent sag vertical curve has the PVC at station 3 + 700.000 and...

A 276 m equal-tangent sag vertical curve has the PVC at station 3 + 700.000 and elevation 321 m. The initial grade is -3.5% and the final grade is +0.5%. Determine the stationing and elevation of the PVI, the PVT, and the lowest point on the curve.

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?

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 280m long equal tangent sag vertical curve has its PVC at chainage 2950.00m and elevation...

A 280m long equal tangent sag vertical curve has its PVC at chainage 2950.00m and elevation 76. 520m.The initial grade is -3.89% and the final grade is 1.02%. (20) a. Make a neat free hand drawing to illustrate a long section of this curve. b. Calculate the chainage and elevation of PVI c. Calculate the chainage and elevation of PVT d. Calculate the chainage and elevation of the lowest point on the curve.

For the curve r=1+sin(theta), find: Location of horizontal tangent lines: Location of vertical tangent lines:

Solve the problem 43) Find equations for the horizontal and vertical tangent lines to the curve...

Solve the problem 43) Find equations for the horizontal and vertical tangent lines to the curve r = 1 - sinθ, 0 ≤ θ < 2π. Please check if your answer is correct with the following: Horizontal: y = 1/4 at (1/2, π/6), y = 1/4 at (1/2, 5π/6), y = -2 at (2, 3π/2) Vertical: x = 0 at (0, π/2), x = -3sqrt(3)/4 at (3/2, 7π/6), x = 3sqrt(3)/4 at (3/2, 11π/6)

You have a symmetrical sag vertical curve of Length = 6 stations. The entering tangent from...

You have a symmetrical sag vertical curve of Length = 6 stations. The entering tangent from the left has a gradient of –6% and the exiting tangent has a gradient of +4%. The VPC Station has an elevation of 250.00 feet (which will be your C-Value) and its station distance is measured to be at 40+00 stations. You will now do the following: Find the elevation and station of the VPI. Find the elevation and station of the VPT Find...

Subjects