Question

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

Answer #1

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 sag vertical curve has the PVC at station 88+00 and PVT at
station 94+25. The back grade is -3.5% and the forward grade is
+2%. The length of the vertical curve is.................ft

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 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)

A 1,800 ft. long sag vertical curve (equal tangent) has PVC at
Station 145+10 and elevation 1,280 ft. The initial grade is -2.5%
and final grade is +4.8%. Determine the elevation and stationing of
PVI and PVT.

Please I need The right answer for this question as soon as
possible.
A sag vertical curve (equal tangent) has PVI at station 212+00
and elevation 540.75 ft. The initial grade is -2.5% and the final
grade is +4.5%. The length of the curve is 900 ft. Determine the
following,
1. Stationing of the low point, PVC, and PVT.
2. Elevation at station 213+00, PVC, low point, and PVT.

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

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 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...

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