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.
Compute and tabulate full-station
elevations for an unequal-tangent vertical curve to fit the
requirements in Problems 2 through 5.
2. A +3.50% grade meets a −2.25% grade
at station 60+00 and elevation 1310.00 ft. Length of first curve
600 ft, second curve 400 ft.
Compute and tabulate full-station elevations for an
unequal-tangent vertical curve to fit the requirements in Problems
2 through 5.
4. Grades g1 of +5.00% and g2 of -2.00%, meet at the VPI at
station 4+300 and elevation 154.960 m. Lengths of curves are 200
and 350 m. (Use 40-m stationing)
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 –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 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 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.
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}.
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?