Traffic Signals

7% of exam

Signal timing kinematics: yellow change and all-red clearance intervals, pedestrian crossing time, cycle length and green splits, lost time and capacity, left-turn phasing, and signal warrants.

6 concepts

Exam areas

A. Traffic signal timing

Yellow & All-Red Clearance Intervals

Time the ITE yellow change and all-red clearance intervals, then locate the dilemma zone, keeping the grade sign and the mph-to-ft/s conversion the way examinees keep dropping them.

The change-and-clearance intervals look like pure plug-and-chug, which is exactly why they bleed points on the PE: the speed never gets converted from

\text{mph}

\text{ft/s}

, the grade term carries the wrong sign, or the all-red width is read off the wrong reference. Yet these two short formulas decide whether a driver caught at the onset of yellow can legally clear the intersection or is trapped in a dilemma zone where they can neither stop nor go. The PE Civil Reference Handbook gives you the ITE kinematic forms in §5.4.3 Interval Timing and the dilemma-zone geometry in §5.4.1 Dilemma Zones; this concept builds the fluency to apply them without the bookkeeping slips that NCEES distractors are built from.

The yellow change interval from the ITE kinematic model

A driver who is too close to stop comfortably must have enough yellow to reach the stop line legally. The ITE model adds a perception-reaction time

t

(use

1.0\,\text{s}

if none is given) to the braking distance

v^2/2a

divided by the approach speed

v

(half the full stopping time) at a comfortable deceleration

a

(use

10\,\text{ft/s}^2

if none is given), corrected for the approach grade

G

. The speed

v

in this form is in

\text{ft/s}

, so an approach posted in

\text{mph}

must first be multiplied by

1.47

. Grade enters as a signed decimal: a downgrade (

G<0

) shrinks the denominator and lengthens the required yellow, because gravity now fights the brakes.

Y = t + \dfrac{v}{2a + 2gG}

The all-red clearance interval

The yellow lets a committed vehicle enter the intersection legally; the all-red then gives that vehicle time to physically clear the conflict before cross traffic receives green. ITE sizes it from the distance the vehicle must travel — the intersection width

w

measured from the near-side stop line to the far edge of the conflict, plus the vehicle length

L

(use

20\,\text{ft}

if none is given) — divided by the approach speed in

\text{ft/s}

. Keep

v

\text{ft/s}

here too; mixing

\text{mph}

into

w/v

is a classic factor-of-1.47 blunder.

R = \dfrac{w + L}{v}

Where the dilemma zone comes from

Picture a vehicle a distance

X

upstream of the stop line when the indication turns yellow. There is a minimum distance

X_c

inside which it cannot stop in time, and a maximum distance

X_0

beyond which it cannot reach and clear the intersection during the available yellow. When

X_c > X_0

, the band between them is the dilemma zone: a driver located there can neither stop safely nor legally clear, regardless of choice. The handbook §5.4.1 defines the two boundaries kinematically —

X_c

from perception plus braking,

X_0

from the distance covered during the yellow.

The cannot-stop boundary

The minimum safe stopping distance is the perception-reaction distance traveled at constant speed plus the braking distance. With

V

\text{mph}

, multiply by

1.47

to convert the approach speed to

\text{ft/s}

before squaring it in the braking term — squaring an un-converted

\text{mph}

value is one of the most punishing errors on this topic because the result is off by

1.47^2 \approx 2.16

X_c = 1.47\,V\,t_{stop} + \dfrac{(1.47\,V)^2}{2a_2}

The cannot-clear boundary

A vehicle that decides to go must cover the distance to the stop line plus the intersection-plus-vehicle dimension

W

within the yellow

Y

. Traveling at constant approach speed, the farthest it can start and still clear is

X_0 = 1.47\,V\,Y - W

. The handbook's full form lets the driver react for

t_{passing}

and then accelerate at

a_1

, which pushes

X_0

outward; in the constant-speed limit (

a_1 = 0

) it collapses to the expression below. Setting

X_0 = X_c

and solving for

Y

gives the yellow that just eliminates the dilemma zone — a favorite exam twist.

X_0 = 1.47\,V\,Y - W

Exam strategy

Convert the approach speed to

\text{ft/s}

first and write it down — almost every wrong answer here is a

1.47

error. For the yellow, read the grade sign carefully: downgrade is negative and lengthens

Y

. For the all-red, confirm whether the problem wants you to include the vehicle length

L

and which width definition (stop line to far edge) applies. For dilemma-zone problems, compute

X_c

and

X_0

separately, compare, and remember that a longer yellow shrinks the zone by pushing

X_0

outward. Carry one extra digit through the division and round only the final interval to

0.1\,\text{s}

, the way signal timing sheets are specified.

Key equations

ITE yellow change interval

Y = t + \dfrac{v}{2a + 2gG}

Yellow time in s. $t$ = reaction time (1.0 s default), $v$ = approach speed in ft/s, $a$ = deceleration (10 ft/s² default), $g = 32.2\,\text{ft/s}^2$ , $G$ = grade as signed decimal (downgrade negative).

All-red clearance interval

R = \dfrac{w + L}{v}

Red clearance in s. $w$ = width from near stop line to far edge of conflict (ft), $L$ = vehicle length (20 ft default), $v$ = approach speed in ft/s.

mph to ft/s conversion

v\,[\text{ft/s}] = 1.47\,V\,[\text{mph}]

Exact factor 5280/3600 = 1.4667 ≈ 1.47. Apply before any kinematic term.

Cannot-stop distance

X_c = 1.47\,V\,t_{stop} + \dfrac{(1.47\,V)^2}{2a_2}

Minimum safe stopping distance (ft). $V$ in mph, $t_{stop}$ = stopping perception time (s), $a_2$ = stopping deceleration (ft/s²).

Cannot-clear distance

X_0 = 1.47\,V\,Y - W

Max distance from which a vehicle at constant speed clears during the yellow (ft). $W$ = intersection width plus vehicle length (ft), $Y$ = yellow (s).

Dilemma-zone existence

X_c > X_0 \;\Rightarrow\; \text{zone width} = X_c - X_0

A dilemma zone forms only when the cannot-stop boundary lies upstream of the cannot-clear boundary.

Yellow that eliminates the zone

Y_{min} = \dfrac{X_c + W}{1.47\,V}

Set the constant-speed X₀ (a₁ = 0) equal to X_c and solve; the smallest yellow that closes the dilemma zone at speed V (mph). With a₁ > 0, solve the handbook's full X₀ instead.

Total change-plus-clearance

CP = Y + R

Sum of yellow and all-red. Only its unused tail (the clearance lost time) is lost; effective green g = G + Y + R − t_L.

Worked examples

Yellow and all-red on a downgrade approach

Problem. An approach is posted at $45\,\text{mph}$ on a $-3\%$ grade. The intersection is $60\,\text{ft}$ wide from the near stop line to the far edge of the conflict, and the design vehicle is $20\,\text{ft}$ long. Using ITE defaults ( $t = 1.0\,\text{s}$ , $a = 10\,\text{ft/s}^2$ ), find the yellow change and all-red clearance intervals.

Solution. Convert speed:

v = 1.47(45) = 66.15\,\text{ft/s}

. Denominator:

2a + 2gG = 2(10) + 2(32.2)(-0.03) = 20 - 1.932 = 18.07\,\text{ft/s}^2

Y = 1.0 + \dfrac{66.15}{18.07} = 1.0 + 3.66 = 4.66\,\text{s}

. All-red:

R = \dfrac{w+L}{v} = \dfrac{60 + 20}{66.15} = 1.21\,\text{s}

. Answer:

Y \approx 4.7\,\text{s}

R \approx 1.2\,\text{s}

. Sanity check: the downgrade lengthened the denominator subtraction, so

Y

is longer than the

4.3\,\text{s}

a level approach (

G=0

) would give — consistent with gravity reducing braking effectiveness.

Y = 1.0 + \dfrac{66.15}{18.07} = 4.66\,\text{s}

Locating and closing a dilemma zone

Problem. A $50\,\text{mph}$ approach has a yellow of $4.0\,\text{s}$ . Use $t_{stop} = 1.0\,\text{s}$ , $a_2 = 10\,\text{ft/s}^2$ , and $W = 70\,\text{ft}$ (intersection width plus vehicle length). Determine whether a dilemma zone exists, its width, and the yellow that would eliminate it.

Solution. Convert:

1.47(50) = 73.5\,\text{ft/s}

. Cannot-stop:

X_c = 73.5(1.0) + \dfrac{(73.5)^2}{2(10)} = 73.5 + 270.1 = 343.6\,\text{ft}

. Cannot-clear:

X_0 = 73.5(4.0) - 70 = 294 - 70 = 224\,\text{ft}

. Since

X_c = 344\,\text{ft} > X_0 = 224\,\text{ft}

, a dilemma zone exists; its width is

343.6 - 224 = 120\,\text{ft}

. To close it set

X_0 = X_c

Y_{min} = \dfrac{343.6 + 70}{73.5} = 5.63\,\text{s}

. Answer: dilemma zone

\approx 120\,\text{ft}

; a yellow of about

5.6\,\text{s}

removes it. Sanity check: the required yellow exceeds the provided

4.0\,\text{s}

, which is precisely why the zone exists.

X_c - X_0 = 343.6 - 224 = 120\,\text{ft}

Common pitfalls

•Leaving the approach speed in mph inside the kinematic terms. Multiply by 1.47 first; the braking term squares the speed, so an un-converted value is off by about 2.16×.
•Sign error on the grade: a downgrade is G < 0 and lengthens the yellow. Plugging +0.03 for a downgrade shortens Y and underestimates the safe interval.
•Confusing the two perception times: t_stop (for stopping, in X_c) versus the reaction time t in the yellow formula. They are not always equal in a problem.
•Forgetting the vehicle length in the all-red: w is the width to the far edge, but the clearing distance is w + L. Dropping L underestimates R.
•Treating all of yellow plus all-red as lost time (or all of it as green). Drivers use the early yellow; only the unused tail is clearance lost time, so effective green g = G + Y + R − t_L.
•Assuming a dilemma zone always exists. It forms only when X_c > X₀; if the yellow is generous, X₀ exceeds X_c and drivers have an option zone, not a dilemma.

References

NCEES PE Civil Reference Handbook — §5.4.3 Interval Timing
NCEES PE Civil Reference Handbook — §5.4.1 Dilemma Zones
ITE Traffic Engineering Handbook, 6th ed. — Yellow Change and Red Clearance Intervals

Cycle Length, Splits & Pedestrian Timing

Find Webster's optimum cycle from total lost time and critical flow ratios, allocate green in proportion to demand, and size the pedestrian walk-plus-clearance interval.