Assignment #$4$ Important Note to expert : Please dont forget to add the graph for Part C and dont forget to compare the answer with part A as asked in Part $2$B$.$

Field data from a four$-$legged intersection included the following:

$(1)$ Vehicle headways starting from a standing queue: $4\mathrm{.}4,3\mathrm{.}8,3\mathrm{.}3,2\mathrm{.}8,2\mathrm{.}5,2\mathrm{.}1,2\mathrm{.}1,2\mathrm{.}1$dots

$(2)$ Change$-$of$-$phase delay is approximately the same as the all$-$red time from field

observations

$(3)$ Yellow time is $3sec$ and all$-$red is $2sec$ in each phase

$(4)$ Two lanes in the N$-$S approaches and three lanes in the E$-$W approaches

$(5)$ Phase split for the N$-$S and E$-$W directions is $40$:$60,$ respectively, for A and B$.$

$(6)$ PHF is $0\mathrm{.}90$

Requirements:

A$.$ Assume the cycle time is $90$ seconds, estimate the capacity of the intersection if a

two$-$phase operation $($without left turns$)$ is implemented. Show your work clearly

step by step.

B$.$ If a two$-$phase operation is considered without a given cycle time, estimate and

recommend the cycle time for each of the following two cases:

$(1)$ the combined demand in the N$-$S plus E$-$W direction is $4800vph($hint: since the

number of lanes is different for N$-$S and E$-$W$,$ use the equation in lecture notes

with different number of lanes for N$-$S from $E-W$ to solve for cycle time$)$;

$(2)$ the demand in the N$-$S direction is $1800vph$ and in the $E-W$ is $3000vph($hint:

break the equation into a $N-S$ direction term and an $E-W$ direction term by using

the corresponding volume and the allocated phase split $(40\%$ or $60\%.$ Solve

each term for the needed cycle time and compare$/$select which one to use for the

intersection$).$ Why is the cycle time different from the result in $(1)$ even though

the total intersection volume is still $4800vph?$

C$.$ A four$-$phase plan is to be implemented after intersection expansion, where the

left turns $($for N$-$S and E$-$W directions$)$ use one extra left$-$turn lane and $15\%$ of the

cycle in every approach. The rest $70\%$ of the cycle time is equally split for the

through movements in the N$-$S and E$-$W directions. Assume a cycle time to

estimate the capacity of the entire intersection. Repeat the process with different

assumed cycle times and plot the cycle time vs$.$ capacity curve in the range of

$45150$ seconds. $($Hint: add two similar terms for the left$-$turn phases to the

original equation, making sure that all the phase split percentages add up to the

entire cycle time.$)$