A vertical curve has its PVI at station 14+75.00 and elevation
76.29 ft. The grade of...
A vertical curve has its PVI at station 14+75.00 and elevation
76.29 ft. The grade of the back tangent is 3.4%, the grade of the
forward tangent is -4.8%, and the curve length is 450 ft.
Determine the stations and elevations of the BVC and EVC.
Determine the location and elevation of the high point of the
curve.
Determine the curve elevation at station 14+00 and station
16+00.
A −2.60% grade meets a +1.95% grade at station 44+25 and
elevation 3386.96 ft, 315-ft curve, stakeout at half stations.
Part A
Tabulate station elevations for an equal-tangent parabolic curve
for the data given above.
Express your answers, separated by commas, to six significant
figures from lowest to highest elevation.
ftft, ftft, ftft, ftft, ftft, ftft, ftft, ftft, ftft, ftft
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Curve 2: A +7.00% grade meets a -6.00% grade at station 4+80 and
elevation 705.70 ft, 80-ft curve, staked 10ft apart. Get a rough
estimate in cubic yards of how much soil that must be brought in
(fill) and/or removed (cut) to create a 14-foot wide roadway that
starts at your PVC/BVC, ends at your PVT/EVC, and follows your
curve elevations. Make sure to state which curve your group was
assigned in the calculations. Trapezoidal Area Estimation is key.
ground...
Given the following vertical curve data: PVI at 19+ 00,L= 500
ft,g1= +2.5%,g2= -1%, and elevation of PVI = 723.86 ft.Compute the
elevationsof the curve summit (i.e. high
point)andfullstations.Assume that all symbols have their usual
meaning.
A 0% grade meets a 3% grade at station 24+00.00 and an elevation
of 372.15 ft. For a passing sight distance of 2,000 ft, find the
following:
a. Minimum length of the vertical curve that allows for passing
using the minimum curve length equations.
b. Calculate the elevation of the middle point of the curve.
c. Calculate the elevation of the road at station 23+00.00.
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 −2.60% grade meets a +1.70% grade at station 44+25 and
elevation 3386.96 ftft, 305-ftft curve, stakeout at half
stations.
Tabulate station elevations for an equal-tangent parabolic curve
for the data given above.
Express your answers, separated by commas, to six significant
figures from lowest to highest elevation.
The following information is given about a vertical curve:
Station of VPI = 150+00.00 feet
Elevation of VPI = 2500.00 feet
L = 1000.00 feet
g1 = -3.0%
g2 = 1.0%
(a) Find the stations and elevations of the BVC and EVC.
(b) Also, what is the station and elevation of the lowest point
of the curve?
(c) Calculate the first and second differences on the curve
elevations of the full stations
A vertical tangent at 4.7% intersects a tangent at -3.9% at a
PVI station of 310+11.23 with an elevation of 903.00 ft. Using a
vertical curve of length 493 ft., find the elevation of each full
station along the curve. What is the station of the turning point
on the curve? {If your calculations result in more than 6 stakes on
the curve, evaluate only the first 6}.
A vertical tangent at 4.7% intersects a tangent at -3.9% at a
PVI station of 310+11.23 with an elevation of 903.00 ft. Using a
vertical curve of length 493 ft., find the elevation of each full
station along the curve. What is the station of the turning point
on the curve? {If your calculations result in more than 6 stakes on
the curve, evaluate only the first 6}.
A vertical tangent at 4.7% intersects a tangent at -3.9% at a
PVI station of 310+11.23 with an elevation of 903.00 ft. Using a
vertical curve of length 493 ft., find the elevation of each full
station along the curve. What is the station of the turning point
on the curve? {If your calculations result in more than 6 stakes on
the curve, evaluate only the first 6}.