Question

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.

Answer #1

A vertical summit parabolic curve has vertical offset of 0.375
m. from the curve to the grade tangent at Station 10 + 050. The
curve has a slope of +4% and -2% grades intersecting at the P.I.
The offset distance of the curve at P.I. is equal to 1.5m. If the
stationing of the P.C. is at 10 + 000.
Compute the
(a) required length of curve (b) horizontal distance of the
vertical curve turning point from the point of...

A vertical parabolic curve is to connect a back tangent of -3%
and a forward tangent of +4%. The change of grade is 0.60% per 20 m
station. The stationing of PC is 17+428 with an elevation of 200 m.
Compute the:
a) length of the parabolic curve;
b) stationing of PT (format: 00+000.00);
c) elevation of PT;
d) elevation of the lowest point of the curve;
e) elevation at Station 17+544.67.

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 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

An equal-tangent crest vertical curve is 1919.200m long and
connects a +2.5%and –1.5%grade. If the design speed of the roadway
is 88.5 kph, does this curve have adequate passing-sight
distance?

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.

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...

You are being asked to obtain the length of a vertical curve for
the following situation: The location having the vertical curve is
aligned under a crossroads at station 199 + 20. The bottom
elevation was measured as 158 m above sea level. The PI station is
located at 200 + 00 with road centerline elevation of 150 m. The
vertical curve is located at the intersection of the tangents with
grades -2.5% and 1.5%. The clear distance between the...

An unequal tangent vertical curve has the following elements:
g1=-3.25%, g2=1.75%, total length = 500.00’, length of curve 1 is
150’, length of curve 2 is 350’, PVI station is 12+00, PVI
elevation is 275.00’
Using the vertical curve, what is the station and elevation of
the high/low point of the curve?

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?

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