The design hour traffic and saturation headway at a four-legged intersection are: Approach North (N) South (S) East (E) West (W) --------------- Design hour flow 450 430 350 315 Saturation headway (sec) 2.25 2.33 2.5 2.75 Design a two phase signal for the intersection based on Webster method and draw the phase diagram. Use amber time of 2 sec on each phase for clearance and start-up loss time phase per phase of 1.5 sec.

Phases: Phase 1 = North-South, Phase 2 = East-West. Saturation flow from saturation headway . Step 1: Flow ratios Flow ratio for each approach; the larger value in a phase is the critical one. Phase Approach q s y = q/s --------------- Phase 1 North 450 1600 0.281 (critical) Phase 1 South 430 1545 0.278 Phase 2 East 350 1440 0.243 (critical) Phase 2 West 315 1309 0.241 Step 2: Lost time Lost time per phase = start-up loss 1.5 s + amber 2 s = 3.5 s. Step 3: Optimum cycle time (Webster) Adopt C = 35 s. Step 4: Green times Total effective green s, divided in proportion to : Displayed green…

What is Y for the intersection?

0.281 + 0.278 + 0.243 + 0.241 = 1.04

2079 Bhadra · 8 marks

Webster two-phase signal with saturation headways

The design hour traffic and saturation headway at a four-legged intersection are:

Approach	North (N)	South (S)	East (E)	West (W)
Design hour flow	450	430	350	315
Saturation headway (sec)	2.25	2.33	2.5	2.75

Design a two phase signal for the intersection based on Webster method and draw the phase diagram. Use amber time of 2 sec on each phase for clearance and start-up loss time phase per phase of 1.5 sec.

Phases: Phase 1 = North-South, Phase 2 = East-West. Saturation flow from saturation headway $s = 3600/h$ .

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	North	450	1600	0.281 (critical)
Phase 1	South	430	1545	0.278
Phase 2	East	350	1440	0.243 (critical)
Phase 2	West	315	1309	0.241

Y = 0.281 + 0.243 = 0.524

Step 2: Lost time

Lost time per phase = start-up loss 1.5 s + amber 2 s = 3.5 s.

L = (1.5+2.0) + (1.5+2.0) = 7.0\ \text{s}

Step 3: Optimum cycle time (Webster)

C_0 = \frac{1.5L + 5}{1 - Y} = \frac{1.5(7.0) + 5}{1 - 0.524} = 32.6\ \text{s}

Adopt C = 35 s.

Step 4: Green times

Total effective green $= C - L = 35 - 7.0 = 28.0$ s, divided in proportion to $y$ :

g_{1} = \frac{y}{Y}(C-L) = \frac{0.281}{0.524}(28.0) = 15.0\ \text{s}

g_{2} = \frac{y}{Y}(C-L) = \frac{0.243}{0.524}(28.0) = 13.0\ \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	15.0	1.5	17	2
Phase 2	13.0	1.5	14	2

Check: 17 + 2 + 14 + 2 = 35 s.

Phase diagram

|      P1 green 17      |  A2 |    P2 green 14    |  A2 |

✓ Answer checked by Claude Opus 5.5 (high effort)

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