Question

7.4 A pretimed four-timing-stage signal has critical lane group flow rates for the first three timing stages of 200, 187, and 210 veh/h (saturation flow rates are 1800 veh/h/ln for all timing stages). The lost time is known to be 4 seconds for each timing stage. If the cycle length is 60 seconds, what is the estimated effective green time of the fourth timing stage?

Answer 1

Answer:

16 seconds

Explanation:

Given:

C = 60

L = 4 seconds each = 4*4 =16

In this problem, the first 3 timing stages are given as:

200, 187, and 210 veh/h.

We are to find the estimated effective green time of the fourth timing stage. The formula for the estimated effective green time is:

$g = (\frac{v}{s}) (\frac{C}{X})$

Let's first find the fourth stage critical lane group ratio $\frac{v}{s}$ , using the formula:

$C = \frac{1.5L +5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}$

$60 = \frac{1.5*16 + 5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}$

$60 = \frac{24+5}{1 - (0.332 + ( \frac{v}{s}))}$

Solving for $(\frac{v}{s})$, we have:

$(\frac{v}{s}) = 0.185$

Let's also calculate the volume capacity ratio X,

$X = (\frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800} + 0.185)(\frac{60}{60-16}$

X = 0.704

For the the estimated effective green time of the fourth timing stage, we have:

$g_4 = (\frac{v}{s}) (\frac{C}{X})$

Substituting figures in the equation, we now have:

$g_4 = (0.185) (\frac{60}{0.704})$

$g_4 = 15.78 seconds$

15.78 ≈ 16 seconds

The estimated effective green time of the fourth timing stage is 16 seconds