Question 7 of 7: Alternate-System Signal Progression in a Grid

Question 7: Alternate-System Signal Progression in a Grid (20 marks: (a)–(d) 5 each)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A signalised CBD grid with two different signal spacings and one desired progression speed.

Find. (a) which alternate system suits each street family; (b) a single system-wide cycle length to the nearest 5 s; (c) the progression speeds that cycle actually delivers; and (d) the time–space diagram showing the through band and its width.

Approach. In an alternate system the cycle length is not free: it is fixed by the spacing and the desired speed. Compute the cycle each candidate system would demand, discard those that fall outside a practical range, choose one cycle for the whole grid so that the two street families can be coordinated together, then work backwards to the speeds and draw the time–space diagram to measure the band.

1. Part (a) — relate cycle length to spacing and speed. In a single-alternate system adjacent signals show opposite indications, so a platoon must cover one block in half a cycle; in a double-alternate system pairs of signals share an indication and the platoon covers two blocks per half cycle; in a triple-alternate system, three blocks. Writing \(n\) = 1, 2 or 3 for the group size, $$v=\frac{2nL}{C}\qquad\Longleftrightarrow\qquad C=\frac{2nL}{v}$$ so the required cycle rises in proportion to both the group size and the block length.
2. Evaluate the required cycle for every candidate. With \(v=50/3.6=13.889\) m/s: $$\begin{aligned} \text{150 m spacing:}\quad & C_{single}=\frac{2(150)}{13.889}=21.6\ \text{s},\quad C_{double}=43.2\ \text{s},\quad C_{triple}=64.8\ \text{s}\\ \text{200 m spacing:}\quad & C_{single}=\frac{2(200)}{13.889}=28.8\ \text{s},\quad C_{double}=57.6\ \text{s},\quad C_{triple}=86.4\ \text{s} \end{aligned}$$
3. Rule out the single-alternate system. A single-alternate system would need a cycle of 21.6 s on the north-south streets and 28.8 s on the east-west streets. Both are far below any workable CBD cycle: after allowing the minimum green a pedestrian needs to cross a two-lane street plus yellow and all-red clearance, a two-phase CBD signal cannot operate below roughly 40 s, and 40 to 60 s is normal. $$\boxed{\text{Single alternate is unsuitable — it demands a 22 to 29 s cycle}}$$ The triple-alternate system on the 200 m streets fails at the other end, needing 86.4 s, which lengthens pedestrian waits and delay unnecessarily. That leaves double alternate at 43.2 s or triple alternate at 64.8 s on the 150 m streets, and double alternate at 57.6 s on the 200 m streets.
4. Choose one system per street family so that a single cycle serves the whole grid. Every signal in a coordinated network must run the same cycle, so the requirement is one \(C\) that suits both families. The pair $$C_{triple}^{NS}=64.8\ \text{s}\qquad\text{and}\qquad C_{double}^{EW}=57.6\ \text{s}$$ bracket the same region, whereas no other combination comes close (the next-best pairing, double alternate on both, would need 43.2 s and 57.6 s — a 33 % mismatch). Hence: $$\boxed{\text{triple alternate on the north-south (150 m) streets, } \text{double alternate on the east-west (200 m) streets}}$$
5. Part (b) — settle the cycle length. A cycle between 57.6 s and 64.8 s satisfies both families as closely as possible; rounding to the nearest five seconds and balancing the two errors gives $$\boxed{C=60\ \text{s}}$$ This is a standard CBD cycle. Choosing 65 s instead would match the north-south streets almost exactly but leave the east-west progression 11 % slow, whereas 60 s splits the discrepancy: 8 % fast one way, 4 % slow the other.
6. Part (c) — compute the actual speeds of progression. Inverting \(v=2nL/C\) with \(C=60\) s: $$v_{NS}=\frac{6(150)}{60}=15.00\ \text{m/s} \qquad\Rightarrow\qquad \boxed{v_{NS}=54.0\ \text{km/h}}$$ $$v_{EW}=\frac{4(200)}{60}=13.33\ \text{m/s} \qquad\Rightarrow\qquad \boxed{v_{EW}=48.0\ \text{km/h}}$$ Both lie within 8 % of the 50 km/h target, which is well inside normal driver tolerance. The block travel times that go with these speeds are \(150/15.00=10.0\) s north-south and \(200/13.33=15.0\) s east-west.
7. Part (d) — construct the time–space diagram and read the band. Plot distance against time, mark each signal's green and red periods along its own horizontal line, and draw the two extreme vehicle trajectories at the progression speed that clear every signal. Their horizontal separation is the band width. For the north-south triple-alternate system with \(C=60\) s and 30 s of green, a platoon leaving the first signal between \(t=0\) and \(t=10\) s meets green at all seven signals, and one leaving even a second later is stopped at the third; hence $$\text{band width}_{NS}=10.0\ \text{s}=16.7\,\%\ \text{of the cycle}$$ For the east-west double-alternate system the corresponding window is $$\text{band width}_{EW}=15.0\ \text{s}=25.0\,\%\ \text{of the cycle}$$ In general, for a matched alternate system with a 50 % split the band width is the effective green divided by the group size, \(g/n\): 30 s for single alternate, 15 s for double, 10 s for triple. This is the price of the wider grouping — grouping signals lets a short block spacing tolerate a practical cycle, but each step from single to double to triple alternate cuts the through band.
8. Interpret the band width. A 10 s band on the north-south streets passes about 10 vehicles per lane per cycle at a 2 s saturation headway, or roughly 600 vehicles per lane per hour, without a stop; the 15 s east-west band passes about half again as many. Vehicles outside the band are stopped once and then join the following band. Because the bands are narrow, this grid would be a candidate for a computer-optimised offset plan (maximising bandwidth directly, rather than imposing an alternate pattern), which typically recovers a few more seconds of band on unequal spacings such as these.