Chapter 15, Problem 4P

To determine

(a)

Using the vertical curvature determine minimum length of vertical curve.

Answer to Problem 4P

$579ft$

Explanation of Solution

Given information:

Station $\left(535+24.25\right)$

Elevation $\text{300ft}$

Grades $G_1=+2\%$

Grades $G_2=-1\%$

Design speed of $\text{65mi/h}$.

Calculation:

Length of curve is given by formula,

$\text{L=KA}\mathrm{......}\left(1\right)$

A is the grade difference between $G_1=+2\%$ and $G_2=-1\%$

$\begin{array}{l}\text{A=}{\text{G}}_{\text{1}}\text{-}{\text{G}}_{\text{2}}\\ \text{A=2-}\left(\text{-1}\right)\\ \text{A=3\%}\end{array}$

Rate of vertical curvature from textbook Table $15.4$ which is based on stopping sight distance $\text{K=65mph=193}$

  

Substitute in equation $\left(1\right)$ we get,

$\begin{array}{l}\text{L=KA}\\ \text{L=193}\times \text{3}\\ \text{L=579ft}\end{array}$

Minimum length of vertical curve is $579ft$.

To determine

(b)

Stations and elevations of BVC and EVC.

Answer to Problem 4P

Station at BVC is $\left(532+34.75\right)$ and station at EVC is $\left(532+13.75\right)$.

Tangent elevation at BVC is $294.21ft$ and at EVC is $297.1ft$.

Explanation of Solution

Given information:

Station $\left(535+24.25\right)$

Elevation $\text{300ft}$

Grades $G_1=+2\%$

Grades $G_2=-1\%$

Design speed of $\text{65mi/h}$.

Calculation:

$\begin{array}{l}\text{BVC station=300ft elevation station-}\frac{L}{2}\\ \text{BVC station=}\left(532+24.25\right)-\frac{579}{2}\\ \text{BVC station=}\left(532+34.75\right)\end{array}$

$\begin{array}{l}\text{EVC station=300ft elevation station+}\frac{L}{2}\\ \text{EVC station=}\left(532+24.25\right)+\frac{579}{2}\\ \text{EVC station=}\left(532+13.75\right)\end{array}$

Therefore station at BVC is $\left(532+34.75\right)$ and station at EVC is $\left(532+13.75\right)$.

$\begin{array}{l}\text{Tangent elevation at BVC=Y-}\frac{{\text{G}}_{\text{1}}\text{x}}{\text{200}}\\ \text{Y is elevation point and x is elevation at length of curve}\text{.}\\ \text{Tangent elevation at BVC=300-}\frac{\text{2×579}}{\text{200}}\\ \text{Tangent elevation at BVC=294}\text{.21ft}\end{array}$

$\begin{array}{l}\text{Tangent elevation at EVC=Y+}\frac{{\text{G}}_2\text{x}}{\text{200}}\\ \text{Y is elevation point and x is elevation at length of curve}\text{.}\\ \text{Tangent elevation at EVC=300+}\frac{\left(-1\right)\text{×579}}{\text{200}}\\ \text{Tangent elevation at EVC=297}\text{.1ft}\end{array}$

Therefore tangent elevation at BVC is $294.21ft$ and at EVC is $297.1ft$.

To determine

(c)

Elevation at each $\text{100ft}$ station.

Answer to Problem 4P

Station	BVC distance $\text{ft}\left(\text{x}\right)$	Tangent elevation	offset	Curve elevation(Tangent elevation)
$532+34.75$	$0$	$294.21$	$0$	$294.21$
$533+00$	$65.25$	$295.52$	$0.12$	$295.4$
$534+00$	$165.25$	$297.52$	$0.71$	$296.81$
$535+00$	$265.25$	$299.52$	$0.83$	$297.69$
$536+00$	$365.25$	$301.52$	$3.46$	$298.06$
$537+00$	$465.25$	$303.52$	$5.61$	$297.91$
$538+00$	$565.25$	$305.52$	$8.28$	$297.24$
$538+13.75$	$579$	$305.79$	$8.69$	$297.1$

Explanation of Solution

Given information:

Station $\left(535+24.25\right)$

Elevation $\text{300ft}$

Grades $G_1=+2\%$

Grades $G_2=-1\%$

Design speed of $\text{65mi/h}$.

Calculation:

Determine the offset,

$\begin{array}{l}\text{Y=}\frac{\text{A}{\text{x}}^{\text{2}}}{\text{200L}}\\ \text{x=distance from BVC elevation=532+24}\text{.75-533=65}\text{.25}\\ \text{Y=}\frac{\text{3×65}\text{.2}{\text{5}}^{\text{2}}}{\text{200×579}}\\ \text{Y=0}\text{.12ft}\end{array}$

$\begin{array}{l}\text{Curve elevation=Elevation at tangent-offset}\\ \text{Curve elevation=}\left(294.21+\frac{65.25}{100}\times 2\right)-0.12\\ \text{Curve elevation=295}\text{.52-0}\text{.12}\\ \text{Curve elevation=295}\text{.40ft}\end{array}$

For every $\text{100ft}$ station the values are tabulated below,

Station	BVC distance $\text{ft}\left(\text{x}\right)$	Tangent elevation	offset	Curve elevation(Tangent elevation)
$532+34.75$	$0$	$294.21$	$0$	$294.21$
$533+00$	$65.25$	$295.52$	$0.12$	$295.4$
$534+00$	$165.25$	$297.52$	$0.71$	$296.81$
$535+00$	$265.25$	$299.52$	$0.83$	$297.69$
$536+00$	$365.25$	$301.52$	$3.46$	$298.06$
$537+00$	$465.25$	$303.52$	$5.61$	$297.91$
$538+00$	$565.25$	$305.52$	$8.28$	$297.24$
$538+13.75$	$579$	$305.79$	$8.69$	$297.1$

To determine

(d)

Station and elevation of high point.

Answer to Problem 4P

$\text{Station of high point=}536+20.75$

$\text{Elevation of high point=298}\text{.07ft}$

Explanation of Solution

Given information:

Station $\left(535+24.25\right)$

Elevation $\text{300ft}$

Grades $G_1=+2\%$

Grades $G_2=-1\%$

Design speed of $\text{65mi/h}$.

Calculation:

Distance of the high point of BVC is given by,

$\begin{array}{l}{\text{X}}_{\text{high}}\text{=}\frac{\text{L}{\text{G}}_{\text{1}}}{\left({\text{G}}_{\text{1}}\text{-}{\text{G}}_{\text{2}}\right)}\\ {\text{X}}_{\text{high}}\text{=}\frac{\text{579×2}}{\left(\text{2-}\left(\text{-1}\right)\right)}\\ {\text{X}}_{\text{high}}\text{=386ft}\end{array}$

$\begin{array}{l}\text{Station of high point=station of BVC+}{\text{X}}_{high}\\ \text{Station of high point=}\left(532+34.75\right)+386\\ \text{Station of high point=}\left(532+34.75\right)+3+86\\ \text{Station of high point=}536+20.75\end{array}$

Difference beteween both elevation of BVC and highpoint.

$\begin{array}{l}\text{Elevation of high point=Tangent elevation of BVC+}{\text{Y}}_{\text{high}}\\ \text{Elevation of high point=294}\text{.21+3}\text{.86}\\ \text{Elevation of high point=298}\text{.07ft}\end{array}$

Conclusion:

Therefore elevation of high point is $\text{298}\text{.07ft}$ and $\text{Station of high point is }536+20.75$