--- title: "Signal timing: Road 1 14 m, Road 2 7 m" subject: "Transportation Engineering II" chapter: 1 key: "semester_7/transportation_engineering_ii:signal-webster-road1-14m-road2-7m" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/signal-webster-road1-14m-road2-7m site: BCT Engineering --- # Signal timing: Road 1 14 m, Road 2 7 m Transportation Engineering II · Chapter 1: Traffic Engineering - Asked in: 2081 Bhadra, 8 marks - Video: [transportation engineering 1 module 5 part 2 signal design by webster's method](https://www.youtube.com/watch?v=H7-pS6ONqik) — Domicile Designs - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/signal-webster-road1-14m-road2-7m - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1 ## Question At a right-angled intersection of two roads, Road 1 has four lanes with a total width of 14m and Road 2 has two lanes with a total width of 7m. The volume of traffic approaching the intersection during the design hour are 850 and 750 PCU/hr on the two approaches of Road 1 and 200 and 250 PCU/hr on the two approaches of Road 2. Take saturation flow of each direction of road 1 and 2 as 3675 PCU/hr and 1890 PCU/hr respectively. Amber time 3 sec, lost time 2 sec per phase for both the roads. ## Short answer Using Webster's method, critical flow ratios 0.231 (Road1) and0.132 (Road2), sum Y=0.364, total lost time10 s. Optimum cycle35 s; displayed green: Road1=18 s, Road2=11 s, amber3 s each. ## Answer Road 1 and Road 2 each run in one phase, so this is a two-phase signal. The design hour flows of the two approaches of each road are given and the saturation flow is the same for both directions of a road (3675 and 1890 PCU/h). #### Step 1: Flow ratios Flow ratio $y = q/s$ for each approach; the larger value in a phase is the critical one. | Phase | Approach | q | s | y = q/s | |---|---|---|---|---| | Phase 1 | Road 1 (850) | 850 | 3675 | 0.231 (critical) | | Phase 1 | Road 1 (750) | 750 | 3675 | 0.204 | | Phase 2 | Road 2 (200) | 200 | 1890 | 0.106 | | Phase 2 | Road 2 (250) | 250 | 1890 | 0.132 (critical) | $$ Y = 0.231 + 0.132 = 0.364 $$ #### Step 2: Lost time Lost time per phase = start-up lost time 2 s + amber 3 s = 5 s; no all-red time is given. $$ L = (2.0+3.0) + (2.0+3.0) = 10.0\ \text{s} $$ #### Step 3: Optimum cycle time (Webster) $$ C_0 = \frac{1.5L + 5}{1 - Y} = \frac{1.5(10.0) + 5}{1 - 0.364} = 31.4\ \text{s} $$ Adopt **C = 35 s**. #### Step 4: Green times Total effective green $= C - L = 35 - 10.0 = 25.0$ s, divided in proportion to $y$: $$ g_{1} = \frac{y}{Y}(C-L) = \frac{0.231}{0.364}(25.0) = 15.9\ \text{s} $$ $$ g_{2} = \frac{y}{Y}(C-L) = \frac{0.132}{0.364}(25.0) = 9.1\ \text{s} $$ Displayed green $G = g + $ start-up loss (so that $\sum G + \sum I + R = C$): | Phase | Effective g (s) | Start-up loss (s) | Green G (s) | Amber (s) | |---|---|---|---|---| | Phase 1 | 15.9 | 2.0 | 18 | 3 | | Phase 2 | 9.1 | 2.0 | 11 | 3 | Check: 18 + 3 + 11 + 3 = 35 s. #### Phase diagram ```text | P1 green 18 | A3 | P2 green 11 | A3 | ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which flow ratios are summed to get Y? A. All four approach ratios B. The largest ratio in each phase: 0.231 + 0.132 C. Only road 1 ratios D. The smallest in each phase Answer: B. Only the critical (largest) y in each phase counts, so Y = 0.231 + 0.132 = 0.364. (Key point: Critical flow ratio) 2. How is the total lost time L = 10 s obtained? A. 2 s start-up + 3 s amber per phase, for two phases B. 3 s per phase only C. 2 s total D. Cycle time minus green Answer: A. Each phase loses 2 + 3 = 5 s, so L = 10 s. (Key point: Lost time) 3. Which is Webster's optimum cycle formula? A. C0 = (1.5L + 5)/(1 - Y) B. C0 = (1.5L + 5)(1 - Y) C. C0 = L/(1 - Y) D. C0 = 1.5L - Y Answer: A. C0 = (1.5 x 10 + 5)/(1 - 0.364) = 31.4 s, adopted as 35 s. (Key point: Webster cycle time) 4. How is the effective green divided between phases? A. Equally B. In proportion to critical y: g = (y/Y)(C - L) C. Proportional to road width D. In proportion to amber Answer: B. g1 = (0.231/0.364)(25) = 15.9 s and g2 = 9.1 s; adding the 2 s start-up loss gives 18 s and 11 s. 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