Question

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.

Answer #1

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.

An
existing equal-tangent sag vertical curve is designed for 60 mi/h.
The initial grade is -3% and the elevation of the PVT is 754 ft.
The PVC of the curve is at station 134 + 16 and the PVI is at 137 +
32. An overpass is being constructed directly above the PVI. The
highway is for cars only (AASHTO minimum and recommended structure
clearances do not apply) and the overpass design assumes the
driver's eye height is set conservatively...

An existing equal-tangent sag vertical curve is designed for 60
mi/h. The initial grade is -3% and the elevation of the PVT is 754
ft. The PVC of the curve is at station 134 + 16 and the PVI is at
137 + 32. An overpass is being constructed directly above the PVI.
The highway is for cars only (AASHTO minimum and recommended
structure clearances do not apply) and the overpass design assumes
the driver's eye height is set conservatively...

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

An equal tangent sag vertical curve is 600 ft long (i.e. 6
stations). The elevation of PVC is 1000 ft. The initial grade is
-0.035 ft/ft (-3.5%) and the final grade is +0.005 ft/ft (+0.5%).
What is the elevation of a point that is 350 ft (3.5 stations) from
PVC?
A) 1005.42 ft
B)991.83 ft
C) 991.01 ft
D) 996.83 ft

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?

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...

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.

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?

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