Chapter 9, Problem 18P

To determine

The LOS, $\text{v}/\text{c}$, veh-mi in the peak $15\min$ and peak hr, and total travel time in the peak $15\min$.

Answer to Problem 18P

The level of service is $\text{E}$, volume to capacity ratio is $0.35$, the total number of vehicle miles during the peak $15\min$ period is $1276\text{veh}/\text{mi}$, the total number of vehicle miles during the peak period is $4950\text{veh}/\text{mi}$ and the total travel time in peak $15\min$ period is $33.5\text{veh}/\text{h}$.

Explanation of Solution

Given:

Base free flow speed is $55\text{mi}/\text{h}$.

Lane width is $11\text{ft}$.

Shoulder width is $3\text{ft}$.

Access points per mile is $15$.

Formula used:

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

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

Here, $P_{\text{T}}$ is decimal portion of trucks in the traffic stream, $P_{\text{R}}$ is 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.

Write the expression for peak $15\min$ hourly passenger car equivalent.

$v_{\text{p}}=\frac{V}{\left(PHF\right)\left(f_{\text{G}}\right)\left(f_{\text{HV}}\right)}$   ...... (II)

Here, $V$ is demand volume for the entire peak hour, $PHF$ is the peak hour factor, $f_{\text{G}}$ 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.

Write the expression to calculate the base percent time spent following.

$BPTSF=100\left(1-e^{-0.000879v_{\text{p}}}\right)$   ...... (III)

Here, $BPTSF$ is the base time percent spent following and $v_{\text{p}}$ is the peak $15\min$ hourly passenger car equivalent.

Write the expression to calculate the percent time following.

$PTSF=BPTSF+f_{\frac{\text{d}}{\text{np}}}$   ...... (IV)

Here, $f_{\frac{\text{d}}{\text{np}}}$ is the adjustment in $PTSF$ to account for the combined effect.

Write the expression to calculate the free flow speed.

$FFS=BFFS-f_{\text{LS}}-f_{\text{A}}$   ...... (V)

Here, $FFS$ is the free flow speed, $BFFS$ is the base free flow speed, $f_{\text{LN}}$ is adjustment for lane and shoulder width and adjustment for number of access point per mile.

Write the expression to calculate the average travel speed.

$ATS=FFS-0.00776v_{\text{p}}-f_{\text{np}}$   ...... (VI)

Here, $ATS$ is the average travel speed and $f_{\text{np}}$ is adjustment for the percentage of no passing zones.

Write the expression to calculate the volume to capacity ratio.

$\frac{v}{c}=\frac{v_{\text{p}}}{c}$   ...... (VII)

Here, $v$ is the volume and $c$ is the capacity.

Write the expression to calculate the total number of vehicle miles during $15\min$ peak.

$VMT=0.25\left(\frac{V}{PHF}\right)L_{\text{t}}$   ...... (VIII)

Here, $L_{\text{t}}$ is the total length of the analysis segment.

Write the expression to calculate the number of vehicle miles during the peak hour.

$VM{T}_{60}=V{L}_{\text{t}}$   ...... (IX)

Write the expression to calculate the total travel time in peak $15\min$ period.

$T{T}_{15}=\frac{VM{T}_{60}}{ATS}$   ...... (X)

Calculation:

Refer table 9.5, "Passenger car equivalents for trucks and RV's to determine the percent time spent following on two way and directional segments.

The value of $E_{\text{T}}$ for level terrain is $1.1$.

The value of $E_{\text{R}}$ for level terrain is $1.0$.

Substitute $10\%$ for $P_{\text{T}}$, $1.1$ for $E_{\text{T}}$, $7\%$ for $P_{\text{R}}$ and $1.0$ for $E_{\text{R}}$ in equation (I).

$\begin{array}{c}{f}_{\text{HV}}=\frac{1}{1+10\%\left(1.1-1\right)+7\%\left(1-1\right)}\\ =\frac{1}{1+0.10\left(0.1\right)}\\ =0.99\end{array}$

Substitute $1100\text{veh}/\text{h}$ for $V$, $0.97$ for $PHF$, $1$ for $f_{\text{G}}$ and $0.99$ for $f_{\text{HV}}$ in equation (II).

$\begin{array}{c}{v}_{\text{p}}=\frac{1100\text{veh}/\text{hr}}{\left(0.97\right)\left(1\right)\left(0.99\right)}\\ =1146\text{pc}/\text{h}\end{array}$

Substitute $1146\text{pc}/\text{h}$ for $v_{\text{p}}$ in equation (III).

$\begin{array}{c}BPTSF=100\left(1-e^{-0.00879\times 1146}\right)\\ =63.48\%\end{array}$

Refer table 9.3, "Adjustment for combined effect of directional distribution of traffic and percentage of no passing zones on percent time spent following on two way segments.

The value of $f_{\frac{\text{d}}{\text{np}}}$ is $7.47\%$.

Substitute $63.48\%$ for $PTSF$ and $7.47\%$ for $f_{\frac{\text{d}}{\text{np}}}$ in equation (IV).

$\begin{array}{c}PTSF=63.48\%+7.47\%\\ =70.95\%\end{array}$

Substitute $55\text{mi}/\text{h}$ for $BFFS$, $3\text{mi}/\text{h}$ for $f_{\text{LS}}$ and $3.75\text{mi}/\text{h}$ for $f_{\text{A}}$ in equation (V).

$\begin{array}{c}FFS=55\text{mi}/\text{h}-3\text{mi}/\text{h}-3.75\text{mi}/\text{h}\\ =48.25\text{mi}/\text{h}\end{array}$

Refer table 9.6, "Adjustment for effect of no passing zones on average travel speed on two way segments".

The value of $f_{\text{np}}$ is $1.284\text{mi}/\text{h}$.

Substitute $48.25\text{mi}/\text{h}$ for $FFS$, $1146\text{pc}/\text{h}$ for $v_{\text{p}}$ and $1.284\text{mi}/\text{h}$ for $f_{\text{np}}$ in equation (VI).

$\begin{array}{c}ATS=48.25\text{mi}/\text{h}-\left(0.00776\times 1146\text{pc}/\text{h}\right)-1.284\\ =38.07\text{mi}/\text{h}\end{array}$

Refer table 9.1, "Level of service criteria for two lane highways in class I.

The value of $PTSF$ is $70.95\%$ and $ATS$ IS $38.07\text{mi}/\text{h}$, therefore the level of service is $\text{E}$.

Substitute $1146\text{pc}/\text{h}$ for $v_{\text{p}}$ and $3200\text{pc}/\text{h}$ for $c$ in equation (VII).

$\begin{array}{c}\frac{v}{c}=\frac{1146\text{pc}/\text{h}}{3200\text{pc}/\text{h}}\\ =0.35\end{array}$

Substitute $1100\text{veh}/\text{h}$ for $V$, $0.45\text{mi}$ for $L_{\text{t}}$ and $0.97$ for $PHF$ in equation (VIII).

$\begin{array}{c}VMT=0.25\left(\frac{1100\text{veh}/\text{hr}}{0.97}\right)\times 4.5\text{mile}\\ =1276\text{veh}/\text{mi}\end{array}$

Substitute $1100\text{veh}/\text{h}$ for $V$ and $4.5\text{mile}$ for $L_{\text{t}}$ in equation (IX).

$\begin{array}{c}VM{T}_{60}=1100\text{veh}/\text{h}\times 4.5\text{mile}\\ =4950\text{veh}/\text{mile}\end{array}$

Substitute $1276\text{veh}/\text{mi}$ for $VMT$ and $38.07\text{mi}/\text{h}$ for $T{T}_{15}$ in equation (X).

$\begin{array}{c}T{T}_{15}=\frac{1276\text{veh}/\text{mi}}{38.07\text{veh}/\text{h}}\\ =33.5\text{veh}/\text{h}\end{array}$

Conclusion:

Therefore, the level of service is $\text{E}$, volume to capacity ratio is $0.35$, the total number of vehicle miles during the peak $15\min$ period is $1276\text{veh}/\text{mi}$, the total number of vehicle miles during the peak period is $4950\text{veh}/\text{mi}$ and the total travel time in peak $15\min$ period is $33.5\text{veh}/\text{h}$.