Chapter 8, Problem 17P

To determine

Effectof cycle length when saturation flow rates at $10\%$ higher from given $\text{Problem 8-16}$.

Answer to Problem 17P

$C=133\sec$

Explanation of Solution

Given information:

Repeat $\text{Problem 8-16}$ using saturation flow rates at $10\%$ higher.

Shows a detailed layout of the phasing system and the intersection geometry used.

  

Calculation:

Equivalent hourly flow for north approach at left turn is given by,

${\text{E}}_{\text{q}}\text{=}\frac{{\text{v}}_{\text{p}}}{\text{PHF}}\mathrm{.....}(1)$

Equivalent hourly flow $\text{=}{\text{E}}_{\text{q}}$

Peak hour volume $\text{=}{\text{v}}_{\text{p}}\text{=133 veh/hr}$

Peak hour factor $\text{=PHF=0}\text{.95}$

Substitute the values in equation $(1)$ we get,

$\begin{array}{l}{\text{E}}_{\text{q}}\text{=}\frac{{\text{v}}_{\text{p}}}{\text{PHF}}\\ {\text{E}}_{\text{q}}\text{=}\frac{\text{133}}{0.95}\\ {\text{E}}_{\text{q}}\text{=}140\text{ veh/hr}\end{array}$

Calculate for other approaches like left turn, right turn and through movement in $\text{Table 1}$.

$\text{Table 1}$ :

Approach	North $\left(\text{56ft}\right)$	South $\left(\text{56ft}\right)$	East $\left(\text{68ft}\right)$	East $\left(\text{68ft}\right)$
Left turn	$140$	$77$	$177$	$141$
Through movement	$442$	$393$	$593$	$543$
Right turn	$147$	$142$	$178$	$187$

Calculate the critical volume for other approaches like left turn, right turn and through movement in $\text{Table 2}$,

Left to north approach is left turn is $140$.

Through movement is $147+442=589$.

$\text{Table 2}$ :

Approach	North $\left(\text{56ft}\right)$	South $\left(\text{56ft}\right)$	East $\left(\text{68ft}\right)$	East $\left(\text{68ft}\right)$
Left turn	$140$	$77$	$177$	$141$
Through movement	$589$	$535$	$771$	$730$

For the saturation flow rates at $10\%$ higher given below.

$\begin{array}{l}{s}_{11}=1600\times 1.10\\ s_{11}=1760\text{ veh/hr}\end{array}$

Similarly calculate the saturation flow rates for other approaches in $\text{Table 3}$,

Approach	saturation flow rates, $\text{veh/h}$
Through lanes	$1760$
Through right lanes	$1540$
Left lanes	$1100$
Left through lanes	$1320$
Left through right lanes	$1210$

Maximum value of the ratios of approach flow using formula,

${\text{Y}}_{\text{i}}\text{=}\frac{{\text{q}}_{\text{ij}}}{{\text{s}}_{\text{ij}}}\mathrm{.......}\left(2\right)$

Maximum value of ratios of approach flow to saturation flow to all lanes $\text{=}{\text{Y}}_{\text{i}}$

Flow on lane groups $\text{=}{\text{q}}_{\text{ij}}=177\text{ veh/hr}$

Saturation flow on lane group $\text{=}{\text{s}}_{\text{ij}}=1100\text{ veh/hr}$

Substitute the values in equation $(2)$ we get,

$\begin{array}{l}{\text{Y}}_{\text{i}}\text{=}\frac{{\text{q}}_{\text{ij}}}{{\text{s}}_{\text{ij}}}\\ {\text{Y}}_{\text{i}}\text{=}\frac{177}{1100}\\ {\text{Y}}_{\text{i}}\text{=}0.161\end{array}$

Calculate for other ${\text{Y}}_{\text{i}}$ approaches in $\text{Table 4}$.

$\text{Table 4}$ :

Approach	Phase $1$:E-W left	Phase $2$:E-W through	Phase $4$:N-S left	Phase $4$:N-S through
${\text{q}}_{\text{ij}}$	$177$	$771$	$140$	$589$
${\text{s}}_{\text{ij}}$	$1100$	$3300$	$1100$	$3300$
${\text{Y}}_{\text{i}}$	$0.161$	$0.234$	$0.127$	$0.178$

Sum of critical ratio is given by,

$\begin{array}{l}{\text{Y}}_{\text{sum}}\text{=0}\text{.161+0}\text{.234+0}\text{.127+0}\text{.178}\\ {\text{Y}}_{\text{sum}}\text{=0}\text{.700}\end{array}$

Total lost time is given by,

$L={\displaystyle \sum _{i=1}^{\varphi }{l}_i}+R\mathrm{.....}\left(3\right)$

Total lost time $L$.

Number of phases $\varphi =4\text{ phases}$

Lost time for phase $l_i=3.5\sec$

Total all red time $R=1.5\sec$

Substitute the values in equation $(3)$ we get,

$\begin{array}{l}L={\displaystyle \sum _{i=1}^{\varphi }{l}_i}+R\\ L={\displaystyle \sum _{i=1}^4\left(3.5+1.5\right)}\\ L=20\sec \end{array}$

Cycle length is given by,

$\begin{array}{l}{C}_o=\frac{\left(1.5\times 20\right)+5}{1-0.700}\\ C_o=116.67\sec \end{array}$

Total effective green time is given by,

$\begin{array}{l}{G}_{te}=C_o-L\\ G_{te}=120-20\\ G_{te}=100\sec \end{array}$

Allocated green time for phase $1$ is given by,

$\begin{array}{l}{\left(G+Y\right)}_1=\left(\frac{Y_1}{Y_{sum}}\times G_{te}\right)+l_1\\ {\left(G+Y\right)}_1=\left(\frac{0.161}{0.700}\times 100\right)+3.5\\ {\left(G+Y\right)}_1=26.5\sec \end{array}$

Allocated green time for other phases is given in the table $5$,

$\text{Table 5}$ :

Phase	Allocated green time in sec
${\left(G+Y\right)}_1$	$26.5$
${\left(G+Y\right)}_2$	$36.4$
${\left(G+Y\right)}_3$	$21.6$
${\left(G+Y\right)}_4$	$28.9$

Minimum green time for phase $1$ is given by,

${\text{G}}_{\text{P}}\text{=3}\text{.2+}\frac{\text{L}}{{\text{S}}_{\text{P}}}\text{+}\left(\text{0}\text{.27}{\text{N}}_{\text{Ped}}\right)\mathrm{.....}(4)$

Crosswalk length $\text{=L=56ft}$

Average speed of pedestrians $\text{=}{\text{S}}_{\text{P}}\text{=4ft/sec}$

Number of pedestrians crossing during an interval $\text{=}{\text{N}}_{\text{Ped}}\text{=}\left(\frac{1200}{3600}\right)\text{veh/sec}$

Substitute the values in equation $(4)$ we get,

$\begin{array}{l}{\text{G}}_{\text{P}}\text{=3}\text{.2+}\frac{\text{L}}{{\text{S}}_{\text{P}}}\text{+}\left(\text{0}\text{.27}{\text{N}}_{\text{Ped}}\right)\\ {\text{G}}_{\text{P}}\text{=3}\text{.2+}\frac{56}{4}\text{+}\left(\text{0}\text{.27}\left(\frac{1200}{3600}\right)120\right)\\ {\text{G}}_{\text{P}}\text{=}28\sec \end{array}$

Minimum Green time for other phases is given Table $6$,

$\text{Table 6}$ :

Phase	Minimum green time in sec
$G_{P1}$	$28$
$G_{P2}$	$28$
$G_{P3}$	$31$
$G_{P4}$	$31$

From the Table $5$ the $G_{P3}$ is greater than ${\left(G+Y\right)}_3$, $G_{P1}$ is greater than ${\left(G+Y\right)}_1$, $G_{P4}$ is greater than ${\left(G+Y\right)}_4$ is the allocated green and yellow time for phase $3$ is $\text{31sec }$ ,phase $1$ is $28\sec$ and phase $4$ is $31\sec$ from the table $34.1$ sec.

Sum of green and yellow time is given by,

$G_1=28\sec ,G_2=37\sec ,G_3=31\sec ,G_4=31\sec$

Total cycle length is given by,

$\begin{array}{l}C=\left(28+37+31+31\right)+\left(4\times 1.5\right)\\ C=133\sec \end{array}$

Conclusion:

Therefore by increasingsaturation flow rates at $10\%$ higher the required cycle length is reduced from $163\sec$ to 133 Sec.