TRN-05

Webster uniform delay

d = 0.5 C (1−λ)² /(1−λ x) for an undersaturated approach.

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SignalsWebster 1958HCM

Governing equation

d=\dfrac{C(1-\lambda)^2}{2(1-\lambda x)},\quad \lambda=g/C,\ x=q/s

$g \in \left[5, 120\right]$

where

C: Cycle (s)
g: Effective green (s)
q: Arrival rate (veh/s)
s: Saturation flow (veh/s)
\lambda: Green ratio (—)
x: Degree of saturation (—)
d: Uniform delay (s)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

d=\dfrac{C(1-\lambda)^2}{2(1-\lambda x)},\quad \lambda=g/C,\ x=q/s

$g \in \left[5, 120\right]$

$g = 5$ $\lambda = 41.05$

Along the curve

②

Domain of the variable — blue

g_{\min} = 5,\quad g_{\max} = 120

$g \in \left[5, 120\right]$

$g = 5$ $g = 5$

Along the curve

③

Response curve — green

d=\dfrac{C(1-\lambda)^2}{2(1-\lambda x)},\quad \lambda=g/C,\ x=q/s

$g \in \left[5, 120\right]$

$g = 5$ $\lambda = 41.05$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}\lambda}{\mathrm{d}g}

$g \in \left[5, 120\right]$

$g = 5.723$ $\lambda^{\prime} = -0.776$

Along the curve

Lecture brief

Historical brief

Highway and traffic theory crystallized with Greenshields’ 1935 speed–density line, then AASHTO stopping sight and equivalent axles. The lab is that operational arithmetic of flow, braking and pavement load. This sheet (TRN-05 — Webster uniform delay) is the form associated with Webster 1958 · HCM. Working symbols: $C$ , $g$ , $q$ , $s$ $\rightarrow$ $\lambda$ , $x$ , $d$ . Webster integrated the triangular queue of a uniform arrival pattern over a signal cycle; a random term is omitted here.

Purpose

Purpose: compute

\lambda

x

d

from

C

g

q

s

in Transport via

d=\dfrac{C(1-\lambda)^2}{2(1-\lambda x)},\quad \lambda=g/C,\ x=q/s

d = 0.5 C (1−λ)² /(1−λ x) for an undersaturated approach. Use it when a real transport question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

C = 90.000\,\mathrm{s}

g = 30.000\,\mathrm{s}

q = 0.200\,\mathrm{veh/s}

s = 0.500\,\mathrm{veh/s}

, the governing relation

d=\dfrac{C(1-\lambda)^2}{2(1-\lambda x)},\quad \lambda=g/C,\ x=q/s

yields

\lambda = 0.333\,\mathrm{—}

x = 0.400\,\mathrm{—}

d = 23.077\,\mathrm{s}

. Fixed-time signal, x < 1, no overflow. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Cycle C90 sEffective green g30 sArrival rate q0.2 veh/sSaturation flow s0.5 veh/s

Outputs

Green ratio \lambda0.333 —
Degree of saturation x0.400 —
Uniform delay d23.077 s

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Narration of this film

Fixed-time signal, x < 1, no overflow.

Webster integrated the triangular queue of a uniform arrival pattern over a signal cycle; a random term is omitted here.

Reading speed

Watch on YouTube