Module 2 : Traffic Measurement Procedures

Lecture 09 : Intrusive Technologies

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	Dual-loop Detectors Dual-loop detectors are also called speed traps, T loops, or double loop detectors. In a dual-loop system, two consecutive single inductance loops, called ``M loop'' and ``S loop'', are embedded a few distance apart as shown in Fig. 1. With such a design, when one of them detects a vehicle, timer is automatically started in the dual-loop system and runs until the same vehicle is detected by other loop.Thus, in addition to outputs of vehicle count and occupancy data, individual vehicle speeds can be trapped through the dividend of the distance between those two single loops $l_{dist}$ by the elapsed time. Speed trap is defined as the measurement of the time that a vehicle requires to travel between two detection points. Spot speed is measured by following Eqn. 1. $$Speed = \frac{l_{dist}}{t_2 – t_1}$$ (1) where, $l_{dist}$ = Distance between two loops in meters $t_1$ = Vehicle entry time at first loop in sec $t_2$ = Vehicle entry time at second loop in sec Dual-loop detectors can also be used to measure vehicle lengths with extra data extracted from controllers’ records. The length of vehicle is measured by following Eqn. 2: $$L_{vehicle} = \frac{Speed\vert ot_2 + 0t_1\vert}{2}$$ (2) where, $L_{vehicle}$ = Length of vehicle in meters. $ot_i$ = on-time for loop detector i; Speed in m/sec Figure 1: Schematic diagram of dual loop detectors Example-1 If the vehicle entering the freeway in loop M at time 8:32:22:00 am and leaving loop N at time 8:32:22:15 am, the distance between two loops will be 3.66 m. Find the spot speed of the vehicle. Also find the length of the vehicle if time occupancy for M - loop is 0.25sec and 0.29 for N - loop. Solution: Step 1 Spot Speed calculated from the equation 1, where given that the distance between two loops are 3.66m and entry, exit times are 8:32:22:00 and 8:32:22:15 substitute in Eqn. 1. $SpotSpeed = (3.66)/(15 - 0)/100 = 24.4~\mathrm{m/sec}$. Step 2 The vehicle length can obtained by the spot speed of the vehicle, so substitute the occupancy times at exit and entry in the Eqn. 2. $$L_{vehicle} = \frac{(52.7/3.6)\vert.25 + 0.29\vert}{2} = 3.95~\mathrm{m}.$$ (3)