Question

A crest vertical curve joining a +3 percent and a -4 percent grade is to be designed with a length of 2184 ft and the Station of BVC is ( 334 + 68) at an elevation of 217.24 ft. The distance from BVC at station (339 + :00) is

Answer #1

.A crest vertical curve joining a +3 percent and a-4 percent
grade is to be designed for 40 Km/h. If the tangents intersect at
station (K20+ 60.00) at an elevation of 200 m, determine the
stations and elevations of the BVC and EVC. Also, calculate the
elevations of intermediate points on the curve at the whole
stations.

An equal-tangent crest vertical curve is being designed for a
speed of 55 mph. The curve connects grades of 2% and -3.5% in the
direction of interest. The curve high point is at station 110+00
and has an elevation of 500 ft. What is the station and elevation
of the PVC and the PVI?

Problem #4: Vertical curve
A vertical curve is define by gradient lines. The gradient of
the back tangent is +4.2% and the
gradient of the forward tangent is -3.75%. The BVC station is at
45+00 and has an elevation of
215.00 feet. For a vertical curve with a length of 12 stations,
determine the following:
a. The station and elevation of the PVI.
b. The station and elevation of the EVC.
c. The elevation of the vertical curve at station...

A
-6.00% grade and a +2.00% grade intersect at station 20+00. The two
grades are to be connected by a vertical curve 400 ft long. The
elevation of the BVC is 348.52.
What is the...
a. Elevation on the curve of the low/high point?
b. Elevation of the curve at station 19+00
c. Elevation of the tangent at station 21+00
d. Elevation of the curve at station 21+50
e. Elevation of the curve at station 18+50

A vertical curve has an incoming grade of +4.4% and an exit
grade of -2.2%. The length of the curve is 450.00 ft. The
intersection of the grades is at 10+00 and the elevation at that
point is 440.00.
Compute the elevation of the +00 points on the curve. Then
determine the location and elevation of the highest point on the
curve. If there is an existing bridge at this location over the
proposed road, and the elevation of its...

a sag vertical curve joining -3.5%and a 1.5% grades the point of
curve PC station is 188 ft with an elevation of 175 ft assume K
value for this highway is 180 determine the minimum length of the
curve L determine the station of the endpoint of the curve PT
determine the elevation of the point of intersection PL

a vertical curve joining 2.5% and a -3.5% grades.the point of
curve (pc) station is (235.2)ft with an elevation of 170 ft assume
k value for this highway is 280 determine the minimum length of the
curve (L) determine the elevation of the point of intersection (pi)
determine the station of the point of intersection (pi)

A vertical curve is needed to connect an entry grade of ‐2.4% to
an exit grade of +1.5%. The PVI of the two grades is at station 33
+ 75 and elevation of 540 m above the datum level. The elevation at
station 34 + 40 must be 541 m above the datum level in order to
Accommodate a street, which crosses the road. Determine; (
According to TRH17)
i. The length of the curve. ii. The elevation and station...

) 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 vertical parabolic sag curve of Lapulapu underpass has grade
of -4% followed by a grade of +2% intersecting at station 12+150.80
at elevation 124.80 m. above sea level. The change of grade of the
sag curve is restricted to 6%.
Compute the length of curve.
Compute the elevation of the lowest point of the curve.
Compute the elevation at station 12+125.60

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 10 minutes ago

asked 19 minutes ago

asked 23 minutes ago

asked 45 minutes ago

asked 56 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago