Lane Capacity from Saturation Flow

c = s \frac{g}{C}, X = \frac{v}{c}

s ≈ 1900 veh/h green per lane × fHV × fG × … (HCM). g/C is green split. X=v/c; Webster delay uses X. Cycle C from Webster 1.5L+5/(1−ΣY) already in-app.

$s$ (veh/h): Saturation flow rate
$g$ (s): Effective green
$C$ (s): Cycle length

Step-by-step solved example

s=1800 vphg, g=30 s, C=90 s. c? If v=500 vph, X?

Green split g/C along the intersection timeline; c = s g/C.

1. c
1800×30/90=600 vph.
2. X
500/600=0.83.

Answer: c=600 vph, X=0.83

10 practice questions

0/10 correct

1.If g/C=0.5 and s=1900, c=

2.X>1 means

3.Lost time L per phase is

4.Heavy vehicles lower s through

5.v=800, c=1000: X=

6.Protected lefts use their own

7.Effective green g =

8.A longer cycle at the same g/C

9.s=1200, g=40, C=100: c=

10.Critical v/s (Y) of a phase is