Transportation engineering

This chapter covers the following topics:

Vertical curves
Horizontal curves
Traffic signal timing
Stopping sight distance
Peak hour factor
Basic freeway segment highway capacity
Traffic flow relationships
Traffic safety equations
Highway pavement design

Vertical curves

A vertical curve is typically modeled with a parabolic equation. The equations below are commonly used to describe the curve shape, the grade change, and key points along the curve.

Equations:

$y = a x^{2}$

$A = ∣ g_{2} - g_{1} ∣ \times 100$

$a = \frac{g _{2} - g _{1}}{2 L}$

$E = a (\frac{L}{2})^{2}$

$r = \frac{g _{2} - g _{1}}{L}$

$K = \frac{L}{A}$

$x_{a} = - \frac{g _{1}}{2 a} = \frac{g _{1} L}{g _{1} - g _{2}}$

Tangent elevation:

$Y = Y_{P V C} + g_{1} x = Y_{P V I} + g_{2} (x - L /2)$

Curve elevation:

$Y = Y_{P V C} + g_{1} x + a x^{2} = Y_{P V C} + g_{1} x + [\frac{( g _{2} - g _{1} )}{( 2 L )}] x^{2}$

Where:

$P V C$ = point of vertical curvature, or beginning of curve
$P V I$ = point of vertical intersection, or vertex
$P V T$ = point of vertical tangency, or end of curve
$A$ = algebraic difference in grades (percentage)
$a$ = parabola constant
$E$ = tangent offset at PVI
$g_{1}$ = percent grade of back tangent divided by 100
$g_{2}$ = percent grade of forward tangent divided by 100
$h_{1}$ = height of driver’s eyes above the roadway surface ( $f t$ )
$h_{2}$ = height of object above the roadway surface ( $f t$ )
$K$ = rate of vertical curvature
$L$ = length of curve ( $f t$ )
$r$ = rate of change of grade
$S$ = sight distance ( $f t$ )
$x$ = horizontal distance from PVC to point on curve ( $f t$ )
$x_{a}$ = horizontal distance to $min / ma x$ elevation on curve ( $f t$ )
$y$ = tangent offset
$v$ = design speed ( $m p h$ )
$Y$ = elevation ( $f t$ )

Vertical curve: sight distance related to curve length

These equations relate sight distance $S$ and vertical curve length $L$ . The correct form depends on whether the required sight distance fits entirely on the curve, so check which case applies before picking a formula.

Case 1: When $S \leq L$

Crest vertical curve (standard eye height $h_{1} = 3.5$ ft, object height $h_{2} = 2.0$ ft):

$L = \frac{A S ^{2}}{100 ( 2 h _{1} + 2 h _{2} ) ^{2}}$

Under these standard values, this reduces to:

$L = \frac{A S ^{2}}{2158}$

Sag vertical curve (based on standard headlight criteria):

$L = \frac{A S ^{2}}{400 + 3.5 S}$

Case 2: When $S > L$

Crest vertical curve:

$L = 2 S - \frac{200 ( h _{1} + h _{2} ) ^{2}}{A}$

Under standard values:

$L = 2 S - \frac{2158}{A}$

Sag vertical curve:

$L = 2 S - \frac{400 + 3.5 S}{A}$

Where:

$A$ = absolute value of the difference in grades ( $%$ )
$S$ = sight distance
$h_{1}$ , $h_{2}$ = eye and object heights

Watch out: Always check whether $S \leq L$ or $S > L$ before picking a curve-length formula - using a Case 1 equation on a Case 2 problem gives a noticeably wrong length, and this check is a commonly tested trap.

Example: Crest vertical curve length

A crest vertical curve connects a $+ 2%$ grade to a $- 3%$ grade. The required stopping sight distance is $S = 500$ ft, using standard eye and object heights. Find the minimum curve length.

$A = ∣ g_{2} - g_{1} ∣ \times 100 = ∣ - 0.03 - 0.02∣ \times 100 = 5$

Assume Case 1 ( $S \leq L$ ): $L = \frac{A S ^{2}}{2158} = \frac{5 \times 50 0 ^{2}}{2158} = 579.2$ ft

Check: $L = 579.2$ ft $> S = 500$ ft, so $S \leq L$ holds and Case 1 was the right choice.

Answer: $L \approx 579.2$ ft

Horizontal curves

A horizontal curve is defined by its radius, central angle, and related geometric elements (tangent length, chord length, external distance, and middle ordinate).

Equations:

$R = \frac{5729.58}{D}$

$R = \frac{L C}{2 sin ( I /2 )}$

$T = R tan (I /2) = \frac{L C}{2 cos ( I /2 )}$

$L = R (\frac{I}{180} π)$

$M = R [1 - cos (I /2)]$

$\frac{R}{E + R} = cos (I /2)$

$\frac{R - M}{R} = cos (I /2)$

$c = 2 R sin (d /2)$

$l = R d (\frac{π}{180})$

$E = R (\frac{1}{cos ( I /2 )} - 1)$

Where:

$c$ = length of sub-chord
$d$ = angle of sub-chord (the sub-chord formulas for $c$ and $l$ use $d$ ; every whole-curve formula uses $I$ )
$D$ = degree of curve (arc definition)
$e$ = superelevation ( $%$ )
$E$ = external distance
$f$ = side friction factor
$I$ = intersection angle (also $Δ$ )
$L$ = length of curve, from PC to PT
$L C$ = length of long chord
$M$ = length of middle ordinate
$PC$ = point of curve
$P I$ = point of intersection
$PT$ = point of tangent
$R$ = radius
$S$ = sight distance ( $f t$ )
$T$ = tangent distance
$V$ = design speed ( $m p h$ )

Example: Horizontal curve station

A horizontal curve has radius $R = 1, 000$ ft and central angle $Δ = 40°$ . The PC is at station $20 + 00$ . Find the station of the PT.

Curve length: $L = R (\frac{Δ}{180} π) = 1, 000 \times (\frac{40}{180}) π = 698.1$ ft

Station of PT $=$ station of PC $+ L = (20 + 00) + 698.1$ ft

Answer: Station $26 + 98.1$

Additional horizontal curve equations

Side friction factor (based on superelevation):

$0.01 e + f = \frac{V ^{2}}{15 R}$

Spiral transition length:

$L = \frac{3.15 V ^{3}}{RC}$

Where:

$C$ = rate of increase of lateral acceleration

Sight distance (to see around obstruction):

$H SO = R [1 - cos (\frac{28.65 S}{R})]$

Where:

$H SO$ = Horizontal sight line offset, measured from the center of the inside lane ( $f t$ )
$R$ = radius to the center of the inside lane ( $f t$ )
$S$ = stopping sight distance ( $f t$ ); the angle $28.65 S / R$ is in degrees

Traffic signal timing

These equations are used to estimate key signal timing intervals (yellow, red clearance, and pedestrian minimum green).

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

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

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

Where:

$t$ = driver reaction time ( $sec$ )
$v$ = vehicle approach speed ( $f t / sec$ ; convert mph using 1 mph = 1.47 ft/sec)
$W$ = width of intersection, curb-to-curb ( $f t$ )
$l$ = length of vehicle ( $f t$ )
$y$ = length of yellow interval to nearest 0.1 sec ( $sec$ )
$r$ = length of red clearance interval to nearest 0.1 sec ( $sec$ )
$G_{p}$ = minimum green time for pedestrians ( $sec$ )
$L$ = crosswalk length ( $f t$ )
$S_{p}$ = pedestrian speed ( $f t / sec$ ), default 3.5 $f t / sec$
$N_{p e d}$ = number of pedestrian in interval
$a$ = deceleration ( $f t / se c^{2}$ )
$\pm G$ = percent grade divided by 100 (uphill grade “+”)

Example: Yellow interval timing

An intersection approach has a design speed of $v = 44$ ft/sec (30 mph), driver reaction time $t = 1.0$ sec, and deceleration $a = 10$ ft/sec $^{2}$ on level ground ( $G = 0$ ). Find the yellow interval $y$ .

$y = t + \frac{v}{2 a + 64.4 G} = 1.0 + \frac{44}{2 ( 10 ) + 64.4 ( 0 )} = 1.0 + \frac{44}{20} = 1.0 + 2.2$

Answer: $y \approx 3.2$ sec

Stopping sight distance

Stopping sight distance combines perception-reaction distance and braking distance. Intersection sight distance is based on the time gap needed to enter or cross the major road.

$SS D = 1.47 V t + \frac{V ^{2}}{30 ( \frac{a}{32.2} \pm G )}$

$I S D = 1.47 V_{maj or} t_{g}$

Where:

$a$ = deceleration ( $f t / se c^{2}$ )
$\pm G$ = percent grade divided by 100 (uphill grade “+”)
$SS D$ = stopping sight distance ( $f t$ )
$I S D$ = intersection sight distance ( $f t$ )
$t$ = driver reaction time ( $sec$ )
$t_{g}$ = time gap for vehicle entering roadway ( $sec$ )
$V$ = design speed ( $m p h$ )
$V_{maj or}$ = design speed of major road ( $m p h$ )

Watch out: a downhill grade enters as a negative $G$ , which shrinks the braking-term denominator $\frac{a}{32.2} \pm G$ and increases the required SSD; an uphill grade (positive $G$ ) decreases it. Entering a downhill grade as positive is a common error.

Peak hour factor

Peak hour factor (PHF) compares the hourly volume to the peak 15-minute flow rate within that hour.

$P H F = \frac{Hourly volume}{4 \times V _{15}}$

Where:

$P H F$ = peak hour factor
$V$ = hourly volume (veh/hr)
$V_{15}$ = peak 15-min. volume (veh/15 min)

Basic freeway segment highway capacity

Basic freeway segment analysis converts a demand volume into an equivalent passenger-car flow rate, then uses a speed-flow model to find the mean travel speed and, from that, the density used for level of service. The base segment capacity is $c = 2, 200 + 10 (FFS - 50)$ pc/h/ln (capped at $2, 400$ , for $55 \leq FFS \leq 75$ ), and the adjusted values used in the speed-flow equations below scale this by a speed adjustment factor ( $S A F$ ) and capacity adjustment factor ( $C A F$ ): $FF S_{a d j} = FFS \times S A F$ and $c_{a d j} = c \times C A F$ (both factors equal $1.00$ under base conditions). The breakpoint flow rate is $B P_{a d j} = [1, 000 + 40 (75 - FF S_{a d j})] \times C A F^{2}$ pc/h/ln, and density at capacity is $D_{c} = 45$ pc/mi/ln. The full parameter and adjustment tables are in the FE Handbook.

Equations

For $v_{p} \leq B P_{a d j}$

$S = FF S_{a d j}$

For $B P_{a d j} < v_{p} \leq c_{a d j}$

$S = FF S_{a d j} - [\frac{FF S _{a d j} - \frac{c _{a d j}}{D _{c}}}{( c _{a d j} - B P _{a d j} ) ^{a}}] (v_{p} - B P_{a d j})^{a}$

Free-flow speed (FFS) equation

This equation predicts $FFS$ from a base value and subtracts adjustments for geometric and operational conditions.

$FFS = BFFS - f_{L W} - f_{R L C} - 3.22 \cdot TR D^{0.84}$

Where

$FFS$ = free flow speed of basic freeway segment ( $m p h$ )
$BFFS$ = base free flow speed of basic freeway segment (default: 75.4 mph)
$f_{L W}$ = adjustment for lane width ( $m p h$ )
$f_{R L C}$ = adjustment for right-side lateral clearance ( $m p h$ )
$TR D$ = total ramp density ( $r am p s / mi$ )

The lane-width adjustment $f_{L W}$ and the right-side lateral clearance adjustment $f_{R L C}$ are both looked up from tables in the FE Handbook rather than calculated. Both are deductions that grow as conditions worsen: $f_{L W}$ is zero for 12-ft lanes and increases as average lane width narrows, and $f_{R L C}$ increases as right-side lateral clearance narrows.

Demand flow rate equation

Demand flow rate converts the observed demand volume into an equivalent passenger-car flow rate under base conditions.

$v_{p} = \frac{V}{P H F \times N \times f _{H V}}$

Where:

$v_{p}$ = demand flow rate under equivalent base conditions ( $p c / h / l n$ )
$V$ = demand volume under prevailing conditions ( $v e h / h$ )
$P H F$ = peak-hour factor
$N$ = number of lanes in analysis direction
$f_{H V}$ = adjustment factor for presence of heavy vehicles in traffic stream

Heavy vehicle adjustment factor

$f_{H V} = \frac{1}{1 + P _{T} ( E _{T} - 1 )}$

Where:

$P_{T}$ = proportion of single-unit trucks and tractor-trailers in the traffic stream
$E_{T}$ = passenger-car equivalent (PCE) of single unit truck or tractor-trailer in traffic stream, typically $2.0$ on level terrain and $3.0$ on rolling terrain

Density equation

Density is flow per lane divided by mean speed.

$D = \frac{v _{p}}{S}$

Where:

$D$ = density ( $p c / mi / l n$ )
$v_{p}$ = demand flow rate ( $p c / h / l n$ )
$S$ = mean speed of traffic stream under base conditions ( $m p h$ )

Example: PHF, demand flow rate, and speed

A freeway segment carries hourly volume $V = 3, 600$ veh/hr with peak 15-min volume $V_{15} = 1, 000$ veh, over $N = 3$ lanes. Trucks are $P_{T} = 10%$ of the stream on level terrain ( $E_{T} = 2.0$ ). $FF S_{a d j} = 70$ mph, $c_{a d j} = 2, 300$ pc/h/ln, $B P_{a d j} = 1, 300$ pc/h/ln, $D_{c} = 45$ pc/mi/ln, $a = 2.0$ . Find $P H F$ , $v_{p}$ , and $S$ .

$P H F = \frac{V}{4 \times V _{15}} = \frac{3 , 600}{4 \times 1 , 000} = 0.90$

$f_{H V} = \frac{1}{1 + 0.10 ( 2.0 - 1 )} = 0.909$

$v_{p} = \frac{V}{P H F \times N \times f _{H V}} = \frac{3 , 600}{0.90 \times 3 \times 0.909} = 1, 467$ pc/h/ln

Since $B P_{a d j} = 1, 300 < v_{p} = 1, 467 \leq c_{a d j} = 2, 300$ : $S = 70 - [\frac{70 - \frac{2 , 300}{45}}{( 2 , 300 - 1 , 300 ) ^{2}}] (1, 467 - 1, 300)^{2} = 70 - 18.89 (0.167)^{2} \approx 69.5$ mph

Answer: $P H F = 0.90$ , $v_{p} = 1, 467$ pc/h/ln, $S \approx 69.5$ mph

Traffic flow relationships

Greenshields model

The Greenshields model assumes a linear relationship between speed and density, which leads to a parabolic flow-density relationship.

$S = S_{y} (1 - \frac{D}{D _{J}})$

$V = S \cdot D = S_{y} D (1 - \frac{D}{D _{J}})$

$V_{ma x} = \frac{S _{y} D _{J}}{4}$

$D_{o} = \frac{D _{J}}{2}$

Where:

$D$ = density ( $v e h / mi / l n$ )
$S$ = speed ( $m p h$ )
$V$ = flow ( $v e h / h r / l n$ )
$V_{ma x}$ = maximum flow ( $v e h / h r / l n$ )
$D_{o}$ = optimum density ( $v e h / mi / l n$ )
$D_{J}$ = jam density ( $v e h / mi / l n$ )
$S_{y}$ = theoretical speed ( $m p h$ )

Gravity model

The gravity model estimates trips between zones using productions, attractions, and impedance (via friction factors), with optional socioeconomic adjustments.

$T_{ij} = \frac{P _{i} A _{j} F _{ij} K _{ij}}{\sum A _{j} F _{ij} K _{ij}}$

Where:

$T_{ij}$ = number of trips from Zone i to Zone j
$P_{i}$ = trips produced in Zone i
$A_{j}$ = trips attracted to Zone j
$F_{ij}$ = friction factor (inverse of travel time between i and j)
$K_{ij}$ = socioeconomic adjustment factor

Logit models

Logit models use a utility value for each alternative and convert those utilities into probabilities.

Utility function:

$U_{i} = \sum β_{k} x_{k}^{i}$

Probability (2 modes):

$P (A) = \frac{e ^{U_{A}}}{e ^{U_{A}} + e ^{U_{T}}}$

Probability (n modes):

$P (x) = \frac{e ^{U_{x}}}{\sum e ^{U_{i}}}$

Traffic safety equations

Crash rates at intersections

$RME V = \frac{A}{A D T \times 365} \times 1, 000, 000$

Where:

$RME V$ = crash rate per million entering vehicles
$A$ = average number of crashes per year
$A D T$ = average daily traffic entering the intersection

Crash rates for roadway segments

$R V M V = \frac{A}{A D T \times T} \times 1, 000, 000$

Where:

$R V M V$ = crash rate per million vehicle miles
$A$ = number of crashes during the study period
$A D T$ = average daily traffic
$T$ = time (days in study period) × length (miles)

Crashes prevented

$Crashes prevented = N \times CR (\frac{A D T _{after}}{A D T _{before}})$

Composite reduction factor

$CR = C R_{1} + (1 - C R_{1}) C R_{2} + (1 - C R_{1}) (1 - C R_{2}) C R_{3} + \dots + (1 - C R_{1}) (1 - C R_{2}) \dots (1 - C R_{m - 1}) C R_{m}$

Highway pavement design

AASHTO structural number equation

The structural number $SN$ is a weighted sum of layer thicknesses, adjusted by drainage coefficients for unbound layers.

$SN = a_{1} D_{1} + a_{2} D_{2} m_{2} + a_{3} D_{3} m_{3} + \dots + a_{m} D_{m} m_{m}$

Where:

$SN$ = structural number for pavement
$a$ = layer coefficient
$D$ = thickness of layer (inches)
$m$ = drainage coefficient

Example: Pavement structural number

A flexible pavement has an asphalt surface ( $a_{1} = 0.44$ , $D_{1} = 4$ in), a base course ( $a_{2} = 0.14$ , $D_{2} = 8$ in, $m_{2} = 1.0$ ), and a subbase ( $a_{3} = 0.11$ , $D_{3} = 6$ in, $m_{3} = 0.9$ ). Find $SN$ .

$SN = a_{1} D_{1} + a_{2} D_{2} m_{2} + a_{3} D_{3} m_{3} = (0.44 \times 4) + (0.14 \times 8 \times 1.0) + (0.11 \times 6 \times 0.9)$

$SN = 1.76 + 1.12 + 0.594$

Answer: $SN \approx 3.47$

Load equivalency factor (LEF)

Pavement design also accounts for the fact that heavier axle loads cause disproportionately more damage than lighter ones. The load equivalency factor (LEF) converts a given axle’s passes into an equivalent number of passes of a standard 18,000-lb single axle, and the equivalent single axle load (ESAL) total sums these equivalent passes across all axle types and traffic over the design period. LEF values for specific axle configurations are read from AASHTO tables (reproduced in the FE Handbook) rather than calculated directly.

Example: Applying a load equivalency factor

A pavement section carries $200$ passes per day of a tandem axle with $L EF = 0.5$ relative to the standard 18,000-lb single axle. Find the number of equivalent single axle loads (ESALs) contributed per day.

$ES A L = passes \times L EF = 200 \times 0.5 = 100$

Answer: $100$ ESALs/day

This chapter covers the following topics:

Vertical curves
Horizontal curves
Traffic signal timing
Stopping sight distance
Peak hour factor
Basic freeway segment highway capacity
Traffic flow relationships
Traffic safety equations
Highway pavement design

Vertical curves

A vertical curve is typically modeled with a parabolic equation. The equations below are commonly used to describe the curve shape, the grade change, and key points along the curve.

Equations:

$y = a x^{2}$

$A = ∣ g_{2} - g_{1} ∣ \times 100$

$a = \frac{g _{2} - g _{1}}{2 L}$

$E = a (\frac{L}{2})^{2}$

$r = \frac{g _{2} - g _{1}}{L}$

$K = \frac{L}{A}$

$x_{a} = - \frac{g _{1}}{2 a} = \frac{g _{1} L}{g _{1} - g _{2}}$

Tangent elevation:

$Y = Y_{P V C} + g_{1} x = Y_{P V I} + g_{2} (x - L /2)$

Curve elevation:

$Y = Y_{P V C} + g_{1} x + a x^{2} = Y_{P V C} + g_{1} x + [\frac{( g _{2} - g _{1} )}{( 2 L )}] x^{2}$

Where:

$P V C$ = point of vertical curvature, or beginning of curve
$P V I$ = point of vertical intersection, or vertex
$P V T$ = point of vertical tangency, or end of curve
$A$ = algebraic difference in grades (percentage)
$a$ = parabola constant
$E$ = tangent offset at PVI
$g_{1}$ = percent grade of back tangent divided by 100
$g_{2}$ = percent grade of forward tangent divided by 100
$h_{1}$ = height of driver’s eyes above the roadway surface ( $f t$ )
$h_{2}$ = height of object above the roadway surface ( $f t$ )
$K$ = rate of vertical curvature
$L$ = length of curve ( $f t$ )
$r$ = rate of change of grade
$S$ = sight distance ( $f t$ )
$x$ = horizontal distance from PVC to point on curve ( $f t$ )
$x_{a}$ = horizontal distance to $min / ma x$ elevation on curve ( $f t$ )
$y$ = tangent offset
$v$ = design speed ( $m p h$ )
$Y$ = elevation ( $f t$ )

Vertical curve: sight distance related to curve length

Case 1: When $S \leq L$

Crest vertical curve (standard eye height $h_{1} = 3.5$ ft, object height $h_{2} = 2.0$ ft):

$L = \frac{A S ^{2}}{100 ( 2 h _{1} + 2 h _{2} ) ^{2}}$

Under these standard values, this reduces to:

$L = \frac{A S ^{2}}{2158}$

Sag vertical curve (based on standard headlight criteria):

$L = \frac{A S ^{2}}{400 + 3.5 S}$

Case 2: When $S > L$

Crest vertical curve:

$L = 2 S - \frac{200 ( h _{1} + h _{2} ) ^{2}}{A}$

Under standard values:

$L = 2 S - \frac{2158}{A}$

Sag vertical curve:

$L = 2 S - \frac{400 + 3.5 S}{A}$

Where:

$A$ = absolute value of the difference in grades ( $%$ )
$S$ = sight distance
$h_{1}$ , $h_{2}$ = eye and object heights

Example: Crest vertical curve length

A crest vertical curve connects a $+ 2%$ grade to a $- 3%$ grade. The required stopping sight distance is $S = 500$ ft, using standard eye and object heights. Find the minimum curve length.

$A = ∣ g_{2} - g_{1} ∣ \times 100 = ∣ - 0.03 - 0.02∣ \times 100 = 5$

Assume Case 1 ( $S \leq L$ ): $L = \frac{A S ^{2}}{2158} = \frac{5 \times 50 0 ^{2}}{2158} = 579.2$ ft

Check: $L = 579.2$ ft $> S = 500$ ft, so $S \leq L$ holds and Case 1 was the right choice.

Answer: $L \approx 579.2$ ft

Horizontal curves

A horizontal curve is defined by its radius, central angle, and related geometric elements (tangent length, chord length, external distance, and middle ordinate).

Equations:

$R = \frac{5729.58}{D}$

$R = \frac{L C}{2 sin ( I /2 )}$

$T = R tan (I /2) = \frac{L C}{2 cos ( I /2 )}$

$L = R (\frac{I}{180} π)$

$M = R [1 - cos (I /2)]$

$\frac{R}{E + R} = cos (I /2)$

$\frac{R - M}{R} = cos (I /2)$

$c = 2 R sin (d /2)$

$l = R d (\frac{π}{180})$

$E = R (\frac{1}{cos ( I /2 )} - 1)$

Where:

$c$ = length of sub-chord
$d$ = angle of sub-chord (the sub-chord formulas for $c$ and $l$ use $d$ ; every whole-curve formula uses $I$ )
$D$ = degree of curve (arc definition)
$e$ = superelevation ( $%$ )
$E$ = external distance
$f$ = side friction factor
$I$ = intersection angle (also $Δ$ )
$L$ = length of curve, from PC to PT
$L C$ = length of long chord
$M$ = length of middle ordinate
$PC$ = point of curve
$P I$ = point of intersection
$PT$ = point of tangent
$R$ = radius
$S$ = sight distance ( $f t$ )
$T$ = tangent distance
$V$ = design speed ( $m p h$ )

Example: Horizontal curve station

A horizontal curve has radius $R = 1, 000$ ft and central angle $Δ = 40°$ . The PC is at station $20 + 00$ . Find the station of the PT.

Curve length: $L = R (\frac{Δ}{180} π) = 1, 000 \times (\frac{40}{180}) π = 698.1$ ft

Station of PT $=$ station of PC $+ L = (20 + 00) + 698.1$ ft

Answer: Station $26 + 98.1$

Additional horizontal curve equations

Side friction factor (based on superelevation):

$0.01 e + f = \frac{V ^{2}}{15 R}$

Spiral transition length:

$L = \frac{3.15 V ^{3}}{RC}$

Where:

$C$ = rate of increase of lateral acceleration

Sight distance (to see around obstruction):

$H SO = R [1 - cos (\frac{28.65 S}{R})]$

Where:

$H SO$ = Horizontal sight line offset, measured from the center of the inside lane ( $f t$ )
$R$ = radius to the center of the inside lane ( $f t$ )
$S$ = stopping sight distance ( $f t$ ); the angle $28.65 S / R$ is in degrees

Traffic signal timing

These equations are used to estimate key signal timing intervals (yellow, red clearance, and pedestrian minimum green).

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

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

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

Where:

$t$ = driver reaction time ( $sec$ )
$v$ = vehicle approach speed ( $f t / sec$ ; convert mph using 1 mph = 1.47 ft/sec)
$W$ = width of intersection, curb-to-curb ( $f t$ )
$l$ = length of vehicle ( $f t$ )
$y$ = length of yellow interval to nearest 0.1 sec ( $sec$ )
$r$ = length of red clearance interval to nearest 0.1 sec ( $sec$ )
$G_{p}$ = minimum green time for pedestrians ( $sec$ )
$L$ = crosswalk length ( $f t$ )
$S_{p}$ = pedestrian speed ( $f t / sec$ ), default 3.5 $f t / sec$
$N_{p e d}$ = number of pedestrian in interval
$a$ = deceleration ( $f t / se c^{2}$ )
$\pm G$ = percent grade divided by 100 (uphill grade “+”)

Example: Yellow interval timing

An intersection approach has a design speed of $v = 44$ ft/sec (30 mph), driver reaction time $t = 1.0$ sec, and deceleration $a = 10$ ft/sec $^{2}$ on level ground ( $G = 0$ ). Find the yellow interval $y$ .

$y = t + \frac{v}{2 a + 64.4 G} = 1.0 + \frac{44}{2 ( 10 ) + 64.4 ( 0 )} = 1.0 + \frac{44}{20} = 1.0 + 2.2$

Answer: $y \approx 3.2$ sec

Stopping sight distance

Stopping sight distance combines perception-reaction distance and braking distance. Intersection sight distance is based on the time gap needed to enter or cross the major road.

$SS D = 1.47 V t + \frac{V ^{2}}{30 ( \frac{a}{32.2} \pm G )}$

$I S D = 1.47 V_{maj or} t_{g}$

Where:

$a$ = deceleration ( $f t / se c^{2}$ )
$\pm G$ = percent grade divided by 100 (uphill grade “+”)
$SS D$ = stopping sight distance ( $f t$ )
$I S D$ = intersection sight distance ( $f t$ )
$t$ = driver reaction time ( $sec$ )
$t_{g}$ = time gap for vehicle entering roadway ( $sec$ )
$V$ = design speed ( $m p h$ )
$V_{maj or}$ = design speed of major road ( $m p h$ )

Peak hour factor

Peak hour factor (PHF) compares the hourly volume to the peak 15-minute flow rate within that hour.

$P H F = \frac{Hourly volume}{4 \times V _{15}}$

Where:

$P H F$ = peak hour factor
$V$ = hourly volume (veh/hr)
$V_{15}$ = peak 15-min. volume (veh/15 min)

Basic freeway segment highway capacity

Equations

For $v_{p} \leq B P_{a d j}$

$S = FF S_{a d j}$

For $B P_{a d j} < v_{p} \leq c_{a d j}$

$S = FF S_{a d j} - [\frac{FF S _{a d j} - \frac{c _{a d j}}{D _{c}}}{( c _{a d j} - B P _{a d j} ) ^{a}}] (v_{p} - B P_{a d j})^{a}$

Free-flow speed (FFS) equation

This equation predicts $FFS$ from a base value and subtracts adjustments for geometric and operational conditions.

$FFS = BFFS - f_{L W} - f_{R L C} - 3.22 \cdot TR D^{0.84}$

Where

$FFS$ = free flow speed of basic freeway segment ( $m p h$ )
$BFFS$ = base free flow speed of basic freeway segment (default: 75.4 mph)
$f_{L W}$ = adjustment for lane width ( $m p h$ )
$f_{R L C}$ = adjustment for right-side lateral clearance ( $m p h$ )
$TR D$ = total ramp density ( $r am p s / mi$ )

Demand flow rate equation

Demand flow rate converts the observed demand volume into an equivalent passenger-car flow rate under base conditions.

$v_{p} = \frac{V}{P H F \times N \times f _{H V}}$

Where:

$v_{p}$ = demand flow rate under equivalent base conditions ( $p c / h / l n$ )
$V$ = demand volume under prevailing conditions ( $v e h / h$ )
$P H F$ = peak-hour factor
$N$ = number of lanes in analysis direction
$f_{H V}$ = adjustment factor for presence of heavy vehicles in traffic stream

Heavy vehicle adjustment factor

$f_{H V} = \frac{1}{1 + P _{T} ( E _{T} - 1 )}$

Where:

$P_{T}$ = proportion of single-unit trucks and tractor-trailers in the traffic stream
$E_{T}$ = passenger-car equivalent (PCE) of single unit truck or tractor-trailer in traffic stream, typically $2.0$ on level terrain and $3.0$ on rolling terrain

Density equation

Density is flow per lane divided by mean speed.

$D = \frac{v _{p}}{S}$

Where:

$D$ = density ( $p c / mi / l n$ )
$v_{p}$ = demand flow rate ( $p c / h / l n$ )
$S$ = mean speed of traffic stream under base conditions ( $m p h$ )

Example: PHF, demand flow rate, and speed

A freeway segment carries hourly volume $V = 3, 600$ veh/hr with peak 15-min volume $V_{15} = 1, 000$ veh, over $N = 3$ lanes. Trucks are $P_{T} = 10%$ of the stream on level terrain ( $E_{T} = 2.0$ ). $FF S_{a d j} = 70$ mph, $c_{a d j} = 2, 300$ pc/h/ln, $B P_{a d j} = 1, 300$ pc/h/ln, $D_{c} = 45$ pc/mi/ln, $a = 2.0$ . Find $P H F$ , $v_{p}$ , and $S$ .

$P H F = \frac{V}{4 \times V _{15}} = \frac{3 , 600}{4 \times 1 , 000} = 0.90$

$f_{H V} = \frac{1}{1 + 0.10 ( 2.0 - 1 )} = 0.909$

$v_{p} = \frac{V}{P H F \times N \times f _{H V}} = \frac{3 , 600}{0.90 \times 3 \times 0.909} = 1, 467$ pc/h/ln

Since $B P_{a d j} = 1, 300 < v_{p} = 1, 467 \leq c_{a d j} = 2, 300$ : $S = 70 - [\frac{70 - \frac{2 , 300}{45}}{( 2 , 300 - 1 , 300 ) ^{2}}] (1, 467 - 1, 300)^{2} = 70 - 18.89 (0.167)^{2} \approx 69.5$ mph

Answer: $P H F = 0.90$ , $v_{p} = 1, 467$ pc/h/ln, $S \approx 69.5$ mph

Traffic flow relationships

Greenshields model

The Greenshields model assumes a linear relationship between speed and density, which leads to a parabolic flow-density relationship.

$S = S_{y} (1 - \frac{D}{D _{J}})$

$V = S \cdot D = S_{y} D (1 - \frac{D}{D _{J}})$

$V_{ma x} = \frac{S _{y} D _{J}}{4}$

$D_{o} = \frac{D _{J}}{2}$

Where:

$D$ = density ( $v e h / mi / l n$ )
$S$ = speed ( $m p h$ )
$V$ = flow ( $v e h / h r / l n$ )
$V_{ma x}$ = maximum flow ( $v e h / h r / l n$ )
$D_{o}$ = optimum density ( $v e h / mi / l n$ )
$D_{J}$ = jam density ( $v e h / mi / l n$ )
$S_{y}$ = theoretical speed ( $m p h$ )

Gravity model

The gravity model estimates trips between zones using productions, attractions, and impedance (via friction factors), with optional socioeconomic adjustments.

$T_{ij} = \frac{P _{i} A _{j} F _{ij} K _{ij}}{\sum A _{j} F _{ij} K _{ij}}$

Where:

$T_{ij}$ = number of trips from Zone i to Zone j
$P_{i}$ = trips produced in Zone i
$A_{j}$ = trips attracted to Zone j
$F_{ij}$ = friction factor (inverse of travel time between i and j)
$K_{ij}$ = socioeconomic adjustment factor

Logit models

Logit models use a utility value for each alternative and convert those utilities into probabilities.

Utility function:

$U_{i} = \sum β_{k} x_{k}^{i}$

Probability (2 modes):

$P (A) = \frac{e ^{U_{A}}}{e ^{U_{A}} + e ^{U_{T}}}$

Probability (n modes):

$P (x) = \frac{e ^{U_{x}}}{\sum e ^{U_{i}}}$

Traffic safety equations

Crash rates at intersections

$RME V = \frac{A}{A D T \times 365} \times 1, 000, 000$

Where:

$RME V$ = crash rate per million entering vehicles
$A$ = average number of crashes per year
$A D T$ = average daily traffic entering the intersection

Crash rates for roadway segments

$R V M V = \frac{A}{A D T \times T} \times 1, 000, 000$

Where:

$R V M V$ = crash rate per million vehicle miles
$A$ = number of crashes during the study period
$A D T$ = average daily traffic
$T$ = time (days in study period) × length (miles)

Crashes prevented

$Crashes prevented = N \times CR (\frac{A D T _{after}}{A D T _{before}})$

Composite reduction factor

$CR = C R_{1} + (1 - C R_{1}) C R_{2} + (1 - C R_{1}) (1 - C R_{2}) C R_{3} + \dots + (1 - C R_{1}) (1 - C R_{2}) \dots (1 - C R_{m - 1}) C R_{m}$

Highway pavement design

AASHTO structural number equation

The structural number $SN$ is a weighted sum of layer thicknesses, adjusted by drainage coefficients for unbound layers.

$SN = a_{1} D_{1} + a_{2} D_{2} m_{2} + a_{3} D_{3} m_{3} + \dots + a_{m} D_{m} m_{m}$

Where:

$SN$ = structural number for pavement
$a$ = layer coefficient
$D$ = thickness of layer (inches)
$m$ = drainage coefficient

Example: Pavement structural number

A flexible pavement has an asphalt surface ( $a_{1} = 0.44$ , $D_{1} = 4$ in), a base course ( $a_{2} = 0.14$ , $D_{2} = 8$ in, $m_{2} = 1.0$ ), and a subbase ( $a_{3} = 0.11$ , $D_{3} = 6$ in, $m_{3} = 0.9$ ). Find $SN$ .

$SN = a_{1} D_{1} + a_{2} D_{2} m_{2} + a_{3} D_{3} m_{3} = (0.44 \times 4) + (0.14 \times 8 \times 1.0) + (0.11 \times 6 \times 0.9)$

$SN = 1.76 + 1.12 + 0.594$

Answer: $SN \approx 3.47$

Load equivalency factor (LEF)

Example: Applying a load equivalency factor

A pavement section carries $200$ passes per day of a tandem axle with $L EF = 0.5$ relative to the standard 18,000-lb single axle. Find the number of equivalent single axle loads (ESALs) contributed per day.

$ES A L = passes \times L EF = 200 \times 0.5 = 100$

Answer: $100$ ESALs/day