Question

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?

Answer #1

An unequal tangent vertical curve has the following elements:
g1=-3.25%, g2=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’ What is the elevation of the PCC?
276.98’
277.63’
285.63’
265.12’

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...

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...

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?

Vertical Curve Information :
VPISTA = 15+00
VPIELEV = 320.50 L = 800.00
g1=-4.00% g2=+2.50%
What is the elevation of the low point on the curve?

A sag vertical curve is to join a grade of g1=−3.25% to a grade
of g2=−1.00%. What is the minimum length of the curve for
acceptable passing sight distance at night, with headlight
illumination?

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.

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 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.

The PVI of a vertical curve is at station 110 + 00 and elevation
1100.00 ft. The vertical curve is equal tangent, 600ft long, and
connects an initial grade of +1.30% and a final grade of –1.10%.
The vertical curve crosses a 3ft diameter pipe at right angles. The
pipe is located at station 112 + 00 and its centerline is at
elevation 1093.00 ft.
a) Using offsets, determine the depth, below the surface of the
curve, to the top...

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