Fundamentals of Transportation/Vertical Curves/Solution

Example

Solution:

A.) What is the curve length?

The curves are equal tangent, so we know that: PVI - PVC = PVT - PVI = L/2

From this, we know:

• PVC of First Curve: 48+24
• PVI of Both Curves (this is a constant): (48+24.00 + (450/2) = 50+49)
• $x_{rr}$: Distance between PVI and railroad tracks: (51+50 - 50+49 = 101 feet)
• PVI Elevation: 1591 + (0.04*225) = 1600 feet

Use the offset formula for the first curve to find the vertical distance between the tangent and the curve:

$Y=\frac{A{x}^2}{200L}$

Where:

• $A$ = 7 (Change in Grade)
• $L$ = 450 feet
• $x$ = 51+50 - 48+24 = 326 feet

The offset, Y, at the railroad tracks's station is computed to be 8.27 feet.

The road is to be lowered an additional 20 feet. Therefore, the new offset at that sight would become 28.27 feet. The equation for the second curve becomes:

$Y=28.27=\frac{7x^2}{200L}$

Neither L (length of curve) or x (distance of offset from PVC) are known. However, we know this is an equal tangent curve, meaning the distance from PVC to PVI is L/2 for the curve in question. Also, the distance between PVI and the railroad tracks is $x_{rr}$, which is 101 feet. Therefore, (L/2) + $x_{rr}$ equals the distance from PVC to the railroad tracks, which is what we want for x. Thus, we are left with one unknown and one equation.

$Y=28.27=\frac{7(L+101{)}^2}{200L}$

The new curve length is found to be 2812 feet.

B.) How deep is the cut at the PVI?

To find this, the offset formula can be used again, using the length L/2 as the distance from the PVC on any curve.

For the old curve, the offset is:

$Y=\frac{A(L/2{)}^2}{200L}=\frac{7(450/2{)}^2}{200(450)}=3.935feet$

For the new curve, using data found from before, the offset is:

$Y=\frac{A(L/2{)}^2}{200L}=\frac{7(2812/2{)}^2}{200(2812)}=24.605feet$

The difference between the two offsets is 20.67 feet. This is the depth of the cut at the PVI.

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