Chapter 9, Problem 8P

To determine

The number of additional lanes required in each direction.

Answer to Problem 8P

The number of additional lanes required to maintain the level of service B is four.

Explanation of Solution

Given:

Length of the segment is $2\text{mi}$.

Sustained grade is $4\%$.

Design volume is $3000\text{veh}/\text{h}$.

Percentage of trucks is $10\%$.

Percentage of buses is $2\%$.

Percentage of RV's is $3\%$.

Percent hour factor is $0.95$.

Free flow speed is $70\text{mi}/\text{h}$.

Right side lateral obstruction is $5\text{ft}$.

Design LOS is $\text{B}$.

Formula used:

Write the expression to calculate the $f_{\text{HV}}$.

$f_{\text{HV}}=\frac{1}{1+P_{\text{T}}\left(E_{\text{T}}-1\right)+P_{\text{R}}\left(E_{\text{R}}-1\right)}$   ...... (I)

Here, $P_{\text{T}}$ is decimal portion of trucks in the traffic stream, $P_{\text{R}}$ is the decimal portion of RV's in the traffic stream, $E_{\text{T}}$ is passenger car equivalent for trucks, and $E_{\text{R}}$ is passenger car equivalent for RV's.

Find the number of lanes in each direction using the following relation.

$v_{\text{p}}=\frac{V}{PHF\times N\times f_{\text{p}}\times f_{\text{HV}}}$ ...... (II)

Here, $V$ is demand volume for the entire peak hour, $PHF$ is peal hour factor, $f_{\text{c}}$ is grade adjustment factor for level or rolling terrain and $f_{\text{HV}}$ is adjustment factor to account for heavy vehicles in the traffic stream.

Calculate the free flow speed by using the formula.

$FFS=BFFS-f_{\text{LW}}-f_{\text{LC}}-f_{\text{N}}-f_{\text{HD}}$   ...... (III)

Here, $BFFS$ is the basic free flow speed, $f_{\text{LW}}$ is the adjustment for lane width, $f_{\text{LC}}$ is the adjustment for right shoulder lateral clearance, $f_{\text{N}}$ is the adjustment for number of lanes and $f_{\text{HD}}$ is the adjustment for interchange density.

Write the expression to calculate the speed.

$S=FFS-\left[\frac{1}{9}\left(7FFS-340\right){\left(\frac{v_{\text{p}}+30FFS-3400}{40FFS-1700}\right)}^{2.6}\right]$   ...... (IV)

Here, $S$ is the speed.

Write the expression to calculate the density using the formula.

$D=\frac{v_{\text{p}}}{S}$   ...... (V)

Here, $D$ is the density, $v_{\text{p}}$ is the passenger car equivalent and $S$ is the speed.

Calculation:

Refer table 9.26, "PCE's for trucks and buses on upgrades, multilane highways and basic freeway sections".

The value of passenger car equivalent for trucks $\left(E_{\text{T}}\right)$ is $3.0$.

The value of passenger car equivalent for recreational vehicles is $\left(E_{\text{R}}\right)$ is $2.75$.

Substitute $10\%$ for $P_{\text{T}}$, $3.0$ for $E_{\text{T}}$, $3\%$ for $P_{\text{R}}$ and $2.75$ for $E_{\text{R}}$ in equation (I).

$\begin{array}{c}{F}_{\text{HV}}=\frac{1}{1+10\%\left(3-1\right)+3\%\left(2.75-1\right)}\\ =\frac{1}{1+0.10\left(3-1\right)+0.03\left(2.75-1\right)}\\ =0.80\end{array}$

Consider the driver population factor as $1.0$.

Consider $2$ lanes:

Substitute $3000\text{veh}/\text{h}$ for $V$, $0.95$ for $PHF$, $2$ for $N$, $1$ for $f_{\text{p}}$ and $0.80$ for $f_{\text{HV}}$ in equation (II).

$\begin{array}{c}{v}_{\text{p}}=\frac{3000\text{veh}/\text{h}}{0.95\times 2\times 1\times 0.80}\\ =1974\text{pc}/\text{h}/\text{in}\end{array}$

Consider $3$ lanes:

Substitute $3000\text{veh}/\text{h}$ for $V$, $0.95$ for $PHF$, $3$ for $N$, $1$ for $f_{\text{p}}$ and $0.80$ for $f_{\text{HV}}$ in equation (II).

$\begin{array}{c}{v}_{\text{p}}=\frac{3000\text{veh}/\text{h}}{0.95\times 3\times 1\times 0.80}\\ =1316\text{pc}/\text{h}/\text{in}\end{array}$

Consider $4$ lanes:

Substitute $3000\text{veh}/\text{h}$ for $V$, $0.95$ for $PHF$, $4$ for $N$, $1$ for $f_{\text{p}}$ and $0.80$ for $f_{\text{HV}}$ in equation (II).

$\begin{array}{c}{v}_{\text{p}}=\frac{3000\text{veh}/\text{h}}{0.95\times 4\times 1\times 0.80}\\ =987\text{pc}/\text{h}/\text{in}\end{array}$

Refer table 9.29, "Adjustment for lane width".

The value of $f_{\text{LW}}$ is $0\text{mi}/\text{h}$.

Refer table 9.30, "Adjustment for right shoulder lateral clearance".

At the right side lateral clearance, the value of reduction in free flow speed $f_{\text{LC}}$ is,

For two lanes the value is $0.6\text{mi}/\text{h}$.

For three lanes the value is $0.4\text{mi}/\text{h}$.

For four lanes the value is $0.2\text{mi}/\text{h}$.

Refer table 9.31, "Adjustment for number of lanes" for the value of $f_{\text{N}}$.

For two lanes the value is $4.5\text{mi}/\text{h}$.

For three lanes the value is $3\text{mi}/\text{h}$.

For four lanes the value is $1.5\text{mi}/\text{h}$.

Refer table 9.32, "Adjustment for interchange density" for the value of $f_{\text{HD}}$.

The value is $2.5\text{mi}/\text{h}$.

Consider $2$ lanes.

Substitute $70\text{mi}/\text{h}$ for $BFFS$, $0$ for $f_{\text{LW}}$, $0.6\text{mi}/\text{h}$ for $f_{\text{LC}}$, $4.5\text{mi}/\text{h}$ for $f_{\text{N}}$ and $2.5\text{mi}/\text{h}$ for $f_{\text{HD}}$ in equation (III).

$\begin{array}{c}FFS=70\text{mi}/\text{h}-0-0.6\text{mi}/\text{h}-4.5\text{mi}/\text{h}-2.5\text{mi}/\text{h}\\ =62.4\text{mi}/\text{h}\end{array}$

Substitute $62.4\text{mi}/\text{h}$ for $FFS$ and $1974\text{pc}/\text{h}/\text{in}$ for $v_{\text{p}}$ in equation (IV).

$\begin{array}{c}S=62.4\text{mi}/\text{h}-\left[\frac{1}{9}\left(7\times 62.4\text{mi}/\text{h}-340\right){\left(\frac{1974\text{pc}/\text{h}/\text{in}+\left(30\times 62.4\text{mi}/\text{h}\right)-3400}{40\times 62.4\text{mi}/\text{h}-1700}\right)}^{2.6}\right]\\ =60.01\text{mi}/\text{h}\end{array}$

Substitute $1974\text{pc}/\text{h}/\text{in}$ for $v_{\text{p}}$ and $60.01\text{mi}/\text{h}$ for speed in equation (V).

$\begin{array}{c}D=\frac{1974\text{pc}/\text{h}/\text{in}}{60.01\text{mi}/\text{h}}\\ =32.89\text{pc}/\text{mi}/\text{lane}\end{array}$

Refer table 9.33, "Level of service criteria for multilane highways".

The level of service is D.

Consider $3$ lanes.

Substitute $70\text{mi}/\text{h}$ for $BFFS$, $0$ for $f_{\text{LW}}$, $0.4\text{mi}/\text{h}$ for $f_{\text{LC}}$, $3\text{mi}/\text{h}$ for $f_{\text{N}}$ and $2.5\text{mi}/\text{h}$ for $f_{\text{HD}}$ in equation (III).

$\begin{array}{c}FFS=70\text{mi}/\text{h}-0-0.4\text{mi}/\text{h}-3\text{mi}/\text{h}-2.5\text{mi}/\text{h}\\ =64.1\text{mi}/\text{h}\end{array}$

Substitute $64.1\text{mi}/\text{h}$ for $FFS$ and $1316\text{pc}/\text{h}/\text{in}$ for $v_{\text{p}}$ in equation (IV).

$\begin{array}{c}S=64.1\text{mi}/\text{h}-\left[\frac{1}{9}\left(7\times 64.1\text{mi}/\text{h}-340\right){\left(\frac{1316\text{pc}/\text{h}/\text{in}+\left(30\times 64.1\text{mi}/\text{h}\right)-3400}{40\times 64.1\text{mi}/\text{h}-1700}\right)}^{2.6}\right]\\ =64.25\text{mi}/\text{h}\end{array}$

Substitute $1316\text{pc}/\text{h}/\text{in}$ for $v_{\text{p}}$ and $60.01\text{mi}/\text{h}$ for speed in equation (V).

$\begin{array}{c}D=\frac{1316\text{pc}/\text{h}/\text{in}}{64.25\text{mi}/\text{h}}\\ =20.48\text{pc}/\text{mi}/\text{lane}\end{array}$

Refer table 9.33, "Level of service criteria for multilane highways".

The level of service is C.

Consider $4$ lanes.

Substitute $70\text{mi}/\text{h}$ for $BFFS$, $0$ for $f_{\text{LW}}$, $0.2\text{mi}/\text{h}$ for $f_{\text{LC}}$, $1.5\text{mi}/\text{h}$ for $f_{\text{N}}$ and $2.5\text{mi}/\text{h}$ for $f_{\text{HD}}$ in equation (III).

$\begin{array}{c}FFS=70\text{mi}/\text{h}-0-0.2\text{mi}/\text{h}-1.5\text{mi}/\text{h}-2.5\text{mi}/\text{h}\\ =65.8\text{mi}/\text{h}\end{array}$

Substitute $65.8\text{mi}/\text{h}$ for $FFS$ and $987\text{pc}/\text{h}/\text{in}$ for $v_{\text{p}}$ in equation (IV).

$\begin{array}{c}S=65.8-\left[\frac{1}{9}\left(7\times 65.8-340\right){\left(\frac{987\text{pc}/\text{h}/\text{in}+\left(30\times 65.8\right)-3400}{40\times 65.8-1700}\right)}^{2.6}\right]\\ =65.18\text{mi}/\text{h}\end{array}$

Substitute $987\text{pc}/\text{h}/\text{in}$ for $v_{\text{p}}$ and $65.18\text{mi}/\text{h}$ for speed.

$\begin{array}{c}D=\frac{987\text{pc}/\text{h}/\text{in}}{65.18\text{mi}/\text{h}}\\ =15.14\text{pc}/\text{mi}/\text{lane}\end{array}$

Refer table 9.33, "Level of service criteria for multilane highways".

The level of service is B.

Conclusion:

Therefore, the number of additional lanes required to maintain the level of service B is four.