Question

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.

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...

Given: L=200 ft equal tangent parabolic curve with g1
= -1.25%, g2 = +1.25%, VPI station = 146+00, and VPI
elevation = 1,895.00 ft.
a. Compute stationing of BVC.
b. Compute stationing of EVC.
c. Compute rate of change, r.
d. Compute elevation of BVC.
e. Compute elevation at Station 145+50.

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 parabolic curve has a descending grade of -0.8% which meets an
ascending grade of 0.6% at station 10+020. The maximum allowable
change of grade per 20 m station is 0.20%. Elevation at station
10+020 is 240.60 m.
a) Compute the length of the curve;
b) Compute the horizontal distance of the lowest point from
station 10+020;
c) Compute for the vertical distance from PI to the curve;
d) Compute the elevation of the lowest point of the curve;
e)...

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...

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.

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 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.

A 182.880m equal tangent sag curve has the PVC at station
5+181.600 and elevation 304.800m. The initial grade is -3.5% and
the final grade is 0.5%. Determine the stationing and elevation of
the PVT, PVI and the lowest point on the curve.

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