Traffic Control Devices for the FE Civil Exam: Signs, Markings and Signal Timing

NCEES FE Civil spec: Transportation Engineering, D. Traffic control devices. FE Reference Handbook: Civil Engineering chapter, Transportation section (Traffic Signal Timing).

What the FE Civil Spec Covers

Section D is traffic control devices: signs, pavement markings and traffic signals. The computational side is signal timing: yellow change interval, red clearance interval, minimum pedestrian green, lane group capacity and cycle length. The conceptual side is the MUTCD: device categories, colors and when a signal is justified.

Key Concepts

MUTCD sets national standards. Sign shapes and colors carry meaning: red for stop and prohibition, yellow for warning, white for regulatory, green for guidance, orange for work zones.
Signal warrants are criteria (such as eight-hour and four-hour vehicle volumes, pedestrian volume and crash experience) that should be met before installing a signal. Meeting a warrant does not by itself require a signal.
Yellow change interval gives a driver who cannot stop comfortably time to reach the stop line.
Red clearance (all-red) interval gives a vehicle that entered on yellow time to clear the intersection.
Saturation flow is the maximum rate a lane discharges during green, about 1,800 to 1,900 veh/h/ln, which corresponds to a saturation headway near 2 seconds.

Core Formulas

Yellow interval ( $v$ in ft/s, $a$ in ft/s², $G$ as a decimal, uphill positive):

y = t + \frac{v}{2 a + 64.4 G}

Red clearance interval ( $W$ = curb-to-curb width, $l$ = vehicle length):

r = \frac{W + l}{v}

Minimum pedestrian green ( $L$ = crosswalk length, $S_{p}$ = walking speed, default 3.5 ft/s):

G_{p} = 3.2 + \frac{L}{S _{p}} + 0.27 N_{p e d}

Lane group capacity and Webster's optimum cycle length ( $L$ = total lost time, $Y$ = sum of critical flow ratios):

c = s \frac{g}{C} C_{o} = \frac{1.5 L + 5}{1 - Y}

Signs, Markings and Signal Operations

Pavement marking color tells drivers which way traffic flows: yellow lines separate opposing directions and mark the left edge of a divided road, while white lines separate lanes moving the same way and mark the right edge. A broken line may be crossed; a solid line discourages or prohibits crossing; a double solid yellow centerline prohibits passing in both directions.

Signals can be pretimed (fixed cycle and splits, good for closely spaced downtown grids), actuated (detectors extend or skip phases, good for variable or low demand) or coordinated along a corridor. Left turns can be protected (green arrow, no conflict), permitted (yield to oncoming traffic on a green ball) or protected-permitted. Each cycle loses a few seconds to start-up and clearance; that lost time is why longer cycles raise capacity slightly but also raise delay.

Saturation flow and saturation headway are two views of the same thing: $s = 3600/ h_{s}$ , so a 1.9 s headway gives about 1,895 veh/h/ln. The critical lane group in each phase is the one with the highest flow ratio $v / s$ , and those critical ratios add up to $Y$ in the cycle length formula.

Where It Is in the FE Reference Handbook

The Transportation section of the Civil Engineering chapter opens with Traffic Signal Timing (yellow, red clearance and pedestrian green), right before stopping sight distance and peak hour factor. Webster's cycle length and the MUTCD sign and marking rules are not printed there, so know them coming in or expect the problem to supply them.

Worked Example: Signal Change Intervals

An approach has a speed of 40 mph, perception-reaction time 1.0 s and deceleration 10 ft/s². The intersection is 60 ft curb to curb and the design vehicle is 20 ft long. The crosswalk is 60 ft long and carries up to 10 pedestrians per interval. Find the yellow on a level approach and on a 3% downgrade, the red clearance and the minimum pedestrian green.

Step 1: convert speed.

v = 40 (1.467) = 58.7 ft/s

Step 2: yellow, level approach.

y = 1.0 + \frac{58.7}{2 ( 10 ) + 0} = 1.0 + 2.93 = 3.9 s

Step 3: yellow, 3% downgrade ( $G = - 0.03$ ).

y = 1.0 + \frac{58.7}{20 - 64.4 ( 0.03 )} = 1.0 + \frac{58.7}{18.07} = 4.2 s

Step 4: red clearance.

r = \frac{60 + 20}{58.7} = 1.4 s

Step 5: pedestrian green.

G_{p} = 3.2 + \frac{60}{3.5} + 0.27 (10) = 3.2 + 17.1 + 2.7 = 23.0 s

For capacity: a lane with saturation flow 1,900 veh/h and $g / C = 0.45$ serves $1900 (0.45) = 855$ veh/h. With 12 s of lost time and $Y = 0.60$ , Webster gives $C_{o} = (18 + 5) /0.40 = 57.5$ s.

Common Mistakes

Plugging speed in mph into the yellow or red clearance formula.
Getting the grade sign backward. Downhill is negative and lengthens the yellow.
Leaving out vehicle length in the red clearance interval.
Treating a satisfied signal warrant as a mandate to install a signal.
Using total cycle length instead of the effective green ratio when computing lane capacity.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

A signalized lane has saturation flow 1,900 vehicles/h/lane and effective green ratio g/C = 0.45. What is its capacity? See the worked solution or practice the Transportation deck
A signalised intersection has a total lost time of 12 seconds per cycle and a sum of critical flow ratios of 0.75. Using Webster's formula, the optimum cycle length is most nearly: See the worked solution or practice the Transportation deck
The average time headway in a lane is 2 s per vehicle. What is the approximate lane flow rate? See the worked solution or practice the Transportation deck
Traffic demand is 900 vehicles/h and capacity is 1,200 vehicles/h. What is the volume-to-capacity ratio? See the worked solution or practice the Transportation deck

Related Calculators and Examples

Frequently Asked Questions

What document governs traffic control devices in the United States?

The Manual on Uniform Traffic Control Devices (MUTCD), published by the Federal Highway Administration. It sets the standards for signs, pavement markings and signals, including warrants for installing a signal.

Should approach speed be in mph or ft/s in the yellow interval formula?

In ft/s, because deceleration is in ft/s squared. Multiply mph by 1.467 to get ft/s.

Does a downhill approach need a longer or shorter yellow?

Longer. Gravity works against braking on a downgrade, so the grade term reduces the denominator and increases the yellow time.

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