41.5.2. Lane capacity

The ratio of effective green time to the cycle length (g_i) is defined as green ratio. We know that saturation flow rate is the number of vehicles that can be moved in one lane in one hour assuming the signal to be green always. Then the capacity of the lane can be computed as,

c_i = s_i (41.6)

where c is the capacity of lane in vehicle per hour, s is the saturation flow rate in vehicle per hour per lane, C is the cycle time in seconds.

This chunk introduces the concept of lane capacity, which pertains to how many vehicles can effectively pass through a lane at an intersection during a given timeframe. The 'green ratio' helps measure how much of the signal's cycle time is actually available for vehicular movement (effective green time) versus how long the entire signal cycle lasts (cycle length). The formula provided indicates that the capacity of the lane can be calculated by multiplying the saturation flow rate (how many vehicles can pass when the signal is green) with the effective green time, normalized over the cycle length.

Consider a busy intersection with a traffic light. When the light is green, vehicles can pass through. If the green time is shorter than the total cycle (which includes yellow and red lights), then not all vehicles queued can move. Think of it as a runway at an airport: if it is not used effectively (like if planes wait on the ground for a longer time than they fly), then fewer planes can take off in an hour. Hence, optimizing the green time relative to the cycle helps maximize traffic flow.

where c is the capacity of lane in vehicle per hour, s is the saturation flow rate in vehicle per hour per lane, C is the cycle time in seconds.

The chunk repeats the formula, which emphasizes the necessity to understand that 'capacity' (c) indicates how many vehicles can pass through the lane per hour under ideal conditions. The saturation flow rate (s) represents the maximum flow of vehicles if the signal were constantly green. The cycle time (C) is key because it tells us how often vehicles are given the opportunity to enter the intersection. Thus, knowing these components allows traffic engineers to design signals that maximize vehicle movement.

Imagine that for every hour of green light, 150 cars can theoretically go through an intersection if traffic was perfectly efficient. However, if the signal only stays green for 30 seconds of a 60-second cycle, only a portion of those 150 will actually get to move. You can think of it like filling a bathtub; if the drain (representing signal cycles) is only open part of the time, you won't fill it as fast as if it were open constantly.

For a movement at a signalised intersection green time is given as 27 seconds, 3 seconds for yellow plus all red out of a 60 seconds cycle. If the saturation headway is 2.4 seconds/vehicle, the start-up lost time is 2 seconds/phase and the clearance lost time is 1 second/phase, find the capacity of the movement per lane?

In the example, we are tasked with calculating the lane capacity. First, we find the total losses due to start-up and clearance times, which add up to 3 seconds per cycle. Next, we compute the effective green time: 27 seconds of green time plus 3 seconds of yellow minus the 3 seconds of lost time gives us an effective green time of 27 seconds. Using the saturation flow rate formula, we find the saturation flow rate based on a saturation headway of 2.4 seconds, which translates to 1500 vehicles per hour. Finally, we apply the capacity formula to find that the lane can effectively manage 675 vehicles per hour based on the given conditions.

Consider this like a fast-food drive-thru: if the window is open for 27 seconds (green light), but it takes time to process (lost time) orders or if a lot of cars need to be served in that 27 seconds, the total number of cars that can get served in that period will be lower than if no order taking was involved. Just like in traffic, the efficiency of the service and the timings impact how many cars can go through.

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