Chapter 3, Problem 1P

To determine

The elevation and station of the high point, $PVC$, and $PVT$.

Answer to Problem 1P

The elevation of $PVC$ is $1,324.90\text{ft}$ ,and station of $PVC$ is $\text{33,740}\text{ft}$, elevation of $PVT$ is $1,325.06\text{ft}$, and $\text{Station}\text{of}PVT$ is $342+60\text{Stations}$, and the station of high point is $\text{340}+\text{60}\text{stations}$, the elevation of high point is $1,331\text{ft}$.

Explanation of Solution

Given:

The distance of vertical curve is $520\text{ft}$.

The vertical curve intersects at station $340+00$ and the elevation $1325\text{ft}$.

The initial grade is $+4.0\%$.

The final grade is $-2.5\%$.

Formula used:

The formula to calculate the distance when the curve is an equal-tangent vertical curve is given by

  $D_1=\frac{L}{2}$ ...... (I)

Here, $D_1$ is thedistance between the point of vertical intersection $PVI$ and initial point of vertical curve $PVC$, $L$ is the distance of the vertical curve.

The formula to calculate the station of the initial point of vertical curve is given by

  $\text{Station}\text{of}PVC=\text{Station}\text{of}PVI-\frac{D_1}{100\text{ft}}$ ...... (II)

Here, $\text{station}\text{of}PVC$ is the station of the initial point of vertical curve and $\text{Station}\text{of}PVI$ is the station of the initial point of vertical intersection of the vertical tangents.

The formula to calculate the station of the final point of vertical curve is given by

  $\text{Station}\text{of}PVT=\text{Station}\text{of}PVI+\frac{D_1}{100\text{ft}}$ ...... (III)

Here, $\text{Station}\text{of}PVT$ is the station point of the vertical tangent.

The formula to calculate the elevation of the $PVC$ is given by

  $\text{Elevation}\text{of}PVC=\text{Elevation}\text{of}PVI-\left(G_1\times N_1\right)$ ...... (IV)

Here, $G_1$ is the initial grade and $N_1$ is the number of stations between $PVC$, and $PVI$.

The formula to calculate the distance when the curve is an equal-tangent vertical curve is given by

  $D_2=\frac{L}{2}$ ...... (V)

Here, $D_2$ is thedistance between the final point of vertical tangent $PVT$ and initial point of vertical intersection $PVI$.

The formula to calculate the elevation of the $PVT$ is given by

  $\text{Elevation}\text{of}PVT=\text{Elevation}\text{of}PVI-\left(G_2\times N_2\right)$ ...... (VI)

Here, $G_2$ is the final grade and $N_2$ is the number of stations between $PVT$, and $PVI$.

The equation of the vertical curve is given by

  $y=a{x}^2+bx+c$ ...... (VII)

Here, $y$ is the roadway elevation at distance $x$ from the beginning of the vertical curve $PVC$ in stations, $x$ is the distance from the beginning of the vertical curve station, $a\text{and}b$ are the coefficients and $c$ is the elevation of the $PVC$.

Differentiate equation (V) with respect to $x$ ,to find the station of the high point on the vertical curve.

  $\begin{array}{c}\frac{dy}{dx}=\frac{d\left(a{x}^2+bx+c\right)}{dx}\\ \frac{dy}{dx}=2ax+b\end{array}$ ...... (VIII)

The relation between $b$ and $G_1$ is given by

  $b=G_1$

The formula to calculate the coefficient $a$ by using the above relation is given by

  $a=\frac{G_2-G_1}{2L}$ ...... (IX)

The formula to calculate the station of the high point from the $PVC$ ,

  $\text{Station}\text{of}\text{High}\text{Point}=\text{Station}\text{of}PVC+x$ ...... (X)

Calculation:

Substitute $520\text{ft}$ for $L$ in equation (I).

  $\begin{array}{c}{D}_1=\frac{520\text{ft}}{2}\\ =260\text{ft}\end{array}$

The above calculation represents that the point $PVT$, final point of the vertical curve is exactly $260\text{ft}$ away from the intersection point $PVI$ .It can be represent as $\left(2\text{stations}+60\text{ft}\right)$.

The point $PVT$, final point of the vertical curve is exactly $520\text{ft}$ away from the initial point of the curve $PVC$ .It can be represent as $\left(5\text{stations}+20\text{ft}\right)$.

Substitute $340$ for $\text{Station}\text{of}PVI$ and $260\text{ft}$ for $D_1$ in equation (II).

  $\begin{array}{c}\text{Station}\text{of}PVC=340-\frac{260\text{ft}}{100\text{ft}}\\ =340\times 100\text{ft}-260\text{ft}\\ =\text{33,740}\text{ft}\end{array}$

Divide the length of above calculation in $\text{ft}$ by $100$ and put the remainder in section as follows:

  $\text{Station}\text{of}PVC=337+40\text{Stations}$.

Hence, the stationing of $PVC$ is at $337+40$.

Substitute $340$ for $\text{Station}\text{of}PVI$ and $260\text{ft}$ for $D_1$ in equation (III).

  $\begin{array}{c}\text{Station}\text{of}PVT=340+\frac{260\text{ft}}{100\text{ft}}\\ =340\times 100\text{ft}+260\text{ft}\\ =34,260\text{ft}\end{array}$

Divide the length of above calculation in $\text{ft}$ by $100$ and put the remainder in section as follows:

  $\text{Station}\text{of}PVT=342+60\text{Stations}$.

Hence, the stationing of $PVT$ is at $342+60$.

The number of stations between $PVC$ and $PVI$ is $2.6\text{stations}$ or $\left(2\text{stations}+60\text{ft}\right)$.

Substitute $1,325$ for $\text{Elevation}\text{of}PVI$, $+4.0\%\text{ft}/\text{station}$ for $G_1$ and $2.6\text{stations}$ for $N_1$ in equation (IV).

  $\begin{array}{c}\text{Elevation}\text{of}PVC=1,325\text{ft}-\left(+4.0\%\frac{\text{ft}}{\text{station}}\times 2.6\text{stations}\right)\left(\frac{1}{100}\right)\\ =1,324.90\text{ft}\end{array}$

Substitute $520\text{ft}$ for $L$ in equation (V).

  $\begin{array}{c}{D}_2=\frac{520\text{ft}}{2}\\ =260\text{ft}\end{array}$

The number of stations between $PVT$ and $PVI$ is $2.6\text{stations}$ or $\left(2\text{stations}+60\text{ft}\right)$.

Substitute $1,325$ for $\text{Elevation}\text{of}PVI$, $-2.5\%\text{ft}/\text{station}$ for $G_2$ and $2.6\text{stations}$ for $N_2$ in equation (VI).

  $\begin{array}{c}\text{Elevation}\text{of}PVT=1,325\text{ft}-\left(-2.5\%\frac{\text{ft}}{\text{station}}\times 2.6\text{stations}\right)\left(\frac{1}{100}\right)\\ =1,325.06\text{ft}\end{array}$

The $L$ is $5.20$ or $\left(5\text{stations}+20\text{ft}\right)$.

Substitute $+4.0$ for $G_1$ and $-2.5$ for $G_2$ in equation (IX).

  $\begin{array}{c}a=\frac{\left(-2.5\right)-\left(+4.0\right)}{2\left(5.20\right)}\\ =-0.625\end{array}$

Substitute $-0.625$ for $a$ and $4.0$ for $b$ in equation (VIII).

  $\begin{array}{c}\frac{dy}{dx}=2\left(-0.625\right)x+\left(4.0\right)\\ =-1.25x+4.0\end{array}$

Substitute $0$ for $\frac{dy}{dx}$ in above calculation to find $x$.

  $\begin{array}{c}0=-1.25x+4.0\\ 1.25x=4.0\\ x=\frac{4.0}{1.25}\\ x=3.2\text{stations}\end{array}$

Substitute $337,40\text{ft}$ for $\text{Station}\text{of}PVC$ and $3.2\times 100\text{ft}$ for $x$ in equation (X).

  $\begin{array}{c}\text{Station}\text{of}\text{High}\text{Point}=337,40\text{ft}+3.2\times 100\text{ft}\\ =\text{34,060}\text{ft}\\ =\text{340}+\text{60}\text{stations}\end{array}$

Substitute $3.2\text{stations}$ for $x$, $-0.625$ for $a$ and $4.0$ for $b$, and $1,324.90\text{ft}$ for $c$ in equation (VII).

  $\begin{array}{c}y=-0.625{\left(3.2\right)}^2+\left(4\times 3.2\right)+1,324.90\text{ft}\\ =1,331\text{ft}\end{array}$

Conclusion:

Thus, the elevation of $PVC$ is $1,324.90\text{ft}$ and station of $PVC$ is $\text{33,740}\text{ft}$, elevation of $PVT$ is $1,325.06\text{ft}$ and $\text{Station}\text{of}PVT$ is $342+60\text{Stations}$ and the station of high point is $\text{340}+\text{60}\text{stations}$, the elevation of high point is $1,331\text{ft}$.