Timing one approach

Signal timing · Webster's method on one approach

An intersection runs a 90-second cycle and loses 12 seconds of it to change intervals. The critical flow ratios across all phases sum to 0.72, of which this approach's movement contributes 0.24. Its saturation flow is 1,800 veh/h, and the count board says demand is running 450 veh/h. Split the green by Webster's rule, find the capacity that green creates, the degree of saturation the demand produces, and the uniform delay a driver on this approach averages.

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given

C = 90 s — Cycle length
L = 12 s — Lost time per cycle
y_i = 0.24 — — This phase's critical flow ratio
Y = 0.72 — — Sum of critical flow ratios
s = 1,800 veh/h — Saturation flow
q = 450 veh/h — Demand

Determine

(a)the effective green this phase gets
(b)the capacity that green creates
(c)the degree of saturation
(d)the uniform delay per vehicle

Step 1 of 4(a) · solve for Effective green for this phase

Webster's split is proportional representation for traffic: after the 12 lost seconds are paid, the 78 that remain divide by demand pressure. A third of the critical load gets a third of the usable cycle — 26 seconds of effective green.

Rearranged for g_i

g_i = \frac{y_i}{Y}\,(C - L)

y_i

Y

C

L

Your values, in your units

g_i = \frac{\left( 0.24 \right)}{\left( 0.72 \right)}\left(\left( 90\ \text{s} \right) - \left( 12\ \text{s} \right)\right)

Answer

g_i = 26\ \text{s}

in

Carried onward at full precision, not this rounded figure.

Open the Green Split by Critical Flow Ratio solver →

Step 2 of 4(b) · solve for Approach capacity

Capacity is saturation flow rationed by green time: 1,800 veh/h flowing 26 seconds out of every 90 is 520 veh/h. The movement's real ceiling is not the lane's — it is the lane's times the fraction of the cycle it is allowed to move.

Rearranged for c

c = s\,\frac{g}{C}

s

g26 scarried from step 1

C

Your values, in your units

c = \left( 1{,}800\ \text{1/h} \right) \cdot \frac{\left( 26\ \text{s} \right)}{\left( 90\ \text{s} \right)}

Answer

c = 144.44\ \text{mHz}

in

Carried onward at full precision, not this rounded figure.

Open the Capacity of a Signalised Approach solver →

Step 3 of 4(c) · solve for Degree of saturation

Demand over capacity: 450 against 520 is x = 0.87 — working, but hot. Signal folk treat 0.85 as the comfort line and 0.9 as the alarm, because delay grows gently below it and viciously above.

Rearranged for x

x = \frac{q}{c}

q

c144.44 mHzcarried from step 2

Your values, in your units

x = \frac{\left( 450\ \text{1/h} \right)}{\left( 0.144444\ \text{Hz} \right)}

Converted to base units

x = \frac{\left( 450\ \text{1/h} \right)}{\left( 520\ \text{1/h} \right)}

Answer

x = 865.38\ \text{per mille}

in

Carried onward at full precision, not this rounded figure.

Open the Degree of Saturation solver →

Step 4 of 4(d) · solve for Uniform delay per vehicle

Webster's first term prices the stop-and-go itself: with a green ratio of 26/90 and x = 0.87, the average driver pays 30.3 seconds of uniform delay every cycle. This is the delay of a PERFECTLY regular arrival pattern — randomness and platoons only add to it.

Rearranged for d₁

d_1 = \frac{0.5\,C\,(1 - \lambda)^{2}}{1 - \lambda x}

C

λ

x865.38 per millecarried from step 3

Your values, in your units

d_1 = \frac{0.5 \cdot \left( 90\ \text{s} \right) \left(1 - \left( 0.288889 \right)\right)^{2}}{1 - \left( 0.288889 \right) \cdot \left( 0.865385 \right)}

Answer

d_1 = 30.341\ \text{s}

in

Carried onward at full precision, not this rounded figure.

Open the Webster Uniform Delay per Vehicle solver →

Why this order

The chain is the skeleton of every signal-timing calculation since Webster: allocate green in proportion to demand pressure, convert green to capacity, compare capacity to demand, and only then talk about delay, because delay is the OUTPUT of the first three, not a knob of its own. The quantity doing the quiet work is the lost time — 12 of 90 seconds here, 13 % of the intersection's whole existence, spent on ambers and all-reds that serve nobody. Short cycles lose that 13 % more often; long cycles lose less of it but make everyone wait longer per stop, and the optimum cycle length is exactly the truce between those two facts.

The trap is reading x = 0.87 as “13 % spare”. Delay is convex in x: the last ten percent of capacity costs more delay than the first fifty, and past x = 1 the uniform-delay model simply breaks — queues grow cycle over cycle and no formula with a steady answer applies. An approach at 0.87 at 4 pm is an approach failing at 5.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.