Question

.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.

Answer #1

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

For a vertical curve with the following data: L = 425.00 ft.;
g1= -2.50%; g2= +0.90%; VPI sta = 28+50.00 ft.; VPI elev = 609.35
ft., determine the: a. (1 pt.) BVC station b. (1 pt.) EVC station
c. (1 pt.) BVC elevation d. (1 pt.) EVC Elevation e. (3 pts.)
Elevations on the curve at full stations f. (3 pts.) Elevations on
the tangents at full stations g. (2 pts.) Tangent offsets at full
stations h. (2 pts.) High/low...

A 3.0% grade passing at station 20+00.00 at an elevation of
99.10 m meets a -2.0% grade passing at station 20+50.00 at an
elevation of 99.60 m. Please note that these two elevations and
stations DO NOT correspond to BVC and EVC.
a) Determine the station and elevation of the point of
intersection of the two grades.
b) If the highest point on the curve must lie at station
20+55.00, find the elevation of this point and the length of...

) 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 curve is joining -4% with -2% gradient, determine the
minimum length of the curve using all applicable criteria. Speed on
the curve will be 60mph. Also, find out elevations of intermediate
stations on the curve for layout. The PVI is at station 31+50 and
elevation 101ft. (using K-values to find the curve length)

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 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...

Problem
3. A
+2.35% grade meets a -2.20% grade at station 45+50.00 and elevation
695.00 feet (PIVC or VPI). An 800-foot (8 station) vertical curve
is planned. Draw a sketch and find the following:
a. The station and elevation of the
PCVC (BVC) (5 points)
b. The station and elevation of the
PTVC (EVC) (5 points)
c. The elevation (on the curve) of
station 44+00.00 (5 points)
d. The vertical offset for station
44+00.00 (5 points)
e. The station and...

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?

Use the following data to solve next 3 questions: A vertical
curve with the following parameters: L = 600', g1 = -4%, g2 = -1%,
STA_VPI = 25+00, ELEV_VPI = 672.05' The stations of the BVC and EVC
are:
a. 21+00 and 29+00 b. 19+00 and 31+00 c. 22+00 and 28+00 d.
20+00 and 27+00
The elevation of the BVC of the curve in the previous problem
is:
a.
669.05'
b.
660.05'
c.
684.05'
d.
676.05'
The elevation of the...

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