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CIVL2700 Chap.10 Signal Timing I: Phasing, Clearance & Lane Groups

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Chapter 10 of 12 · CIVL2700

Signal Timing I: Phasing, Clearance & Lane Groups

This Week-10 topic of CIVL2700 Transport Systems at the University of Sydney turns a signal plan into numbers.

You build the phases, size the between-phase clearance with the ITE yellow y = τ + v/(2a) and all-red AR = (w + l)/v, test for a dilemma zone, and read the saturation flow to split each approach into lane groups.

Taking the critical (maximum v/s) lane group per phase gives the sum of critical flow ratios Yc and the total lost time L — the two inputs the Webster cycle uses in the next chapter.

In this chapter

What this chapter covers

  • 01

    Phasing recap: a phase serves non-conflicting movements; 2-5 phases; protected turn phase from the cross-product warrant

  • 02

    Yellow (amber) time y = τ + v/(2a + 2Gg); level grade gives y = τ + v/(2a); τ ≈ 1.0 s, a ≈ 3.05 m/s²

  • 03

    All-red time AR = (w + l)/v with w = width to clear and l = vehicle length (≈ 6 m); intergreen I = y + AR

  • 04

    Speed converted to m/s (km/h ÷ 3.6); both y and AR rounded UP to the nearest 0.5 s

  • 05

    Grade sign: a downgrade (G negative) lengthens the yellow; an upgrade shortens it; grade does not enter AR

  • 06

    Dilemma zone: exists when the amber is shorter than ψmin = τ + v/(2a) + (W + L)/v, so a driver can neither stop nor clear

  • 07

    Saturation flow s = 3600/h from the settled saturation headway h; higher for through than turning lanes

  • 08

    Effective green g = G + I − l, effective red r = R + l, cycle C = g + r; movement capacity c = s·g/C

  • 09

    Lane groups: same-lane movements together, exclusive turn lanes separate; flow ratio v/s = arrival flow / saturation flow

  • 10

    Critical lane group = maximum v/s in a phase; Yc = Σ(v/s)ci and L = Σ(tL)ci, one per phase, need Yc < 1

Worked example · free

Yellow & all-red at a 50 km/h level approach (ITE)

Q [4 marks]. A signal phase serves a level approach with a 50 km/h speed limit. Take perception-reaction tau = 1.0 s, deceleration a = 3.05 m/s^2, vehicle length l = 6 m, and a distance to clear of w = 24 m. Find the yellow time, the all-red time and the intergreen, rounding each clearance up to the nearest 0.5 s.
Displayed vs effective green.Displayed vs effective green. The driver sees green G, then yellow y, then all-red AR; the yellow + all-red is the intergreen I that clears the intersection. The saturation flow does not start the instant green shows (a start loss ) and runs a little into the yellow (an end gain ), so the effective green is g = G + I − l, where l is the lost time for the phase.DISPLAYED GREEN GyARsignal shown to the driverintergreen I = y + AREFFECTIVE GREEN g = G + I - lstartlossend gaintime the saturation flow actually occurstime
Displayed vs effective green. The driver sees green G, then yellow y, then all-red AR; the yellow + all-red is the intergreen I that clears the intersection. The saturation flow does not start the instant green shows (a start loss) and runs a little into the yellow (an end gain), so the effective green is g = G + I − l, where l is the lost time for the phase.
  • +1Convert the speed to m/s: v = 50/3.6 = 13.89 m/s (the ITE clearance formulae use m/s).
  • +1Yellow (level grade, so the grade term is 0): y = tau + v/(2a) = 1.0 + 13.89/(2 x 3.05) = 1.0 + 2.28 = 3.28 s, round up to 0.5 s -> y = 3.5 s.
  • +1All-red: AR = (w + l)/v = (24 + 6)/13.89 = 30/13.89 = 2.16 s, round up to 0.5 s -> AR = 2.5 s.
  • +1Intergreen: I = y + AR = 3.5 + 2.5 = 6.0 s of clearance between this phase and the conflicting one.
v = 13.89 m/s; yellow y = 3.5 s, all-red AR = 2.5 s, intergreen I = 6.0 s. Each clearance is rounded up to 0.5 s before adding.
Sia tip — Always convert km/h to m/s first, and round each of y and AR UP to the nearest 0.5 s before summing them - rounding down or rounding the total can give a different, unsafe answer. On a downgrade the yellow gets longer, not shorter.
Glossary

Key terms

Phase
The part of a signal cycle during which a chosen set of non-conflicting movements has right of way. A cycle is the full sequence of phases, typically 2 to 5.
Intergreen (clearance interval)
The yellow plus all-red between one phase ending and the conflicting phase starting: I = y + AR. It lets committed vehicles get clear of the intersection.
Yellow (amber) time
The ITE clearance y = tau + v/(2a + 2Gg); on the level y = tau + v/(2a). Rounded up to the nearest 0.5 s; a downgrade (G negative) lengthens it.
All-red time
The interval AR = (w + l)/v when every approach shows red, sized so a vehicle just entering can cross width w. Rounded up to 0.5 s; grade does not affect it.
Dilemma zone
A stretch of road, when the amber is too short, where a driver can neither stop safely before the line nor clear the junction in the amber. Avoided when amber >= tau + v/(2a) + (W + L)/v.
Saturation flow (s) & effective green (g)
s = 3600/h is the maximum discharge a lane sustains during a fully saturated green (settled headway h); the effective green g = G + I - l is the time that flow is assumed to occur, giving movement capacity c = s*g/C.
Lane group
A set of lanes analysed together: same-lane movements form one group and an exclusive turn lane is its own group. Each has a flow ratio v/s.
Critical lane group / flow ratio
In a phase, the lane group with the highest flow ratio v/s (arrival flow / saturation flow); it sizes that phase's green. Yc = sum of the critical v/s, one per phase.
FAQ

Signal Timing I: Phasing, Clearance & Lane Groups FAQ

Why are the yellow and all-red rounded UP to 0.5 s and never down?

Because they are safety clearances: the yellow gives a driver time to stop or commit, and the all-red flushes committed vehicles before the conflicting green. Rounding down would shorten that protection and could create a dilemma zone, so both y and AR are rounded up to the nearest 0.5 s - and each is rounded before you add them for the intergreen I = y + AR.

What is the difference between displayed green and effective green?

Displayed green G is what the driver sees; effective green g = G + I - l is the time the saturation flow is actually assumed to occur. Discharge does not start the instant green shows (a start loss) and runs a little into the yellow (an end gain), so capacity uses the effective green, c = s*g/C, not the displayed green.

Can AI help me with signal timing, clearance and lane groups in CIVL2700?

Yes - Sia can explain the ITE yellow and all-red formulae step by step, check that you converted speed to m/s and rounded up to 0.5 s, and walk through how the critical lane group's v/s builds Yc and the total lost time on your own practice numbers. It is a study aid that explains the method; it will not sit your assessment or guarantee a mark, so always confirm final answers against your Canvas materials.

Study strategy

Exam move

Drill the mechanical order every time: convert the approach speed to m/s (km/h divided by 3.6), apply the ITE yellow y = tau + v/(2a) (add the grade term only if the approach is not level) and all-red AR = (w + l)/v, round each up to 0.5 s, then add for the intergreen. Be able to explain a dilemma zone and test whether the amber beats the minimum clearance psi_min.

For the lane-group thread, write the saturation flow s = 3600/h, form each lane group's v/s, take only the critical (maximum) one per phase, and sum to Yc with the total lost time L = one start-up plus one clearance per critical group - showing the working, since a bare Yc rarely earns full marks.

The final exam is a 2.5-hour, 40% hurdle task, so budget time in proportion to the marks on each part and confirm the exam's open- or closed-book status and exact timing in the June 2027 Semester-1 exam period on Canvas.

Working through Signal Timing I: Phasing, Clearance & Lane Groups in CIVL2700? Sia is AskSia’s AI Engineering tutor — ask any CIVL2700 Signal Timing I: Phasing, Clearance & Lane Groups question and get a clear, step-by-step explanation grounded in how CIVL2700 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.

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