TRAFFIC ENGINEERING COURSE(PWE 8322)

CAPACITY FOR TWSC

Instructor: Usama Elrawy Shahdah, PhDLecture # 04

Gap Acceptance Theory

Gap acceptance theory attempts to provide an analytical framework formodelling the process that drivers implement when they are engaged inmaking a manoeuvre from a stop controlled minor street across, or onto, anuncontrolled major street.

Vehicle “A” is on the minor street is attempting to cross the major street. The degree of difficulty that the driver of vehicle “A” faces in completing

this manoeuvre successfully depends on 3 factors:1) The size of the time headway required to complete the manoeuvre,2) The number and size of time headways that are available in the traffic

stream, and3) The relative priority of the movement the driver is attempting to make

relative to all other traffic movements.

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Required Gap Size

Drivers attempting to cross or enter the major traffic stream must use time gaps that are sufficiently large as to allow the manoeuvre to be completed and to allow for a safety buffer between the lead vehicle and the vehicle making the manoeuvre and between the vehicle making the manoeuvreand its following vehicle.

Vehicle “C” is waiting on the minor street approach to make a left-hand turn onto the major street. At time T, the driver of vehicle C decides that

the time headway between vehicle A and B is sufficient to permit the left turn movement and initiates the manoeuvre.

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Required Gap Size (cont.)The size of the time gap required to complete a manoeuvre is a function of many parameters, including:

1) The physical length of vehicle C. Consider the same time headway between vehicles A and B, but in this case

vehicle C is a tractor-trailer.

Even though the truck accelerates at the same rate as vehicle C in previous slide, the length of the truck occupies a greater portion of the roadway and

if vehicle B does not decelerate, there will be a collision.

2) Acceleration capabilities of vehicle C. A vehicle that accelerates quickly is able to clear the roadway in less time and

therefore can safely use a smaller gap than a vehicle that accelerates slowly.

3) Characteristics of the driver of vehicle C. A cautious driver generally requires a longer gap than a more aggressive driver.

Some drivers are not capable of accurately estimating the gap size, and therefore compensate by requiring longer gaps.

4) Width of the major roadway. The number of lanes that need to be crossed by vehicle C impacts the size of gap

required,

as the gap must be long enough to permit the vehicle to safely clear the conflict zone.

5) The type of manoeuvre being executed. left turn movement from a minor street requires a longer time gap than does a

through movement.

For the left turn movement, vehicle C is merging with a traffic stream and must not "cut off" the following vehicle (vehicle B ) such that a rear-end collision results.

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Usefulness of gaps

The usefulness of gaps depends on: whether or not drivers from the minor street will make

use of an available gap, and if they do, how many drivers will complete their

manoeuvre in the same gap. Need to be able to answer two questions:

1) Will drivers be able to use a gap of a specific duration (e.g. 10 seconds)?

2) If drivers are able to use the gap, How many drivers will depart in this gap?

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Capacity

The maximum number of vehicles that can complete one or more specific manoeuvres (e.g. left turn) from the minor street approach during 1 hour

Note: Estimates of delay and queue length can be made on the basis of capacity and the number of vehicles attempting to make the manoeuvre.

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Critical gap and follow-up time

Critical gap (tc) (Critical time headway) : The minimum gap size in the major steert that drivers in the minor street will accept.

Follow-up time (tf): if the gap is long enough, then more than one vehicle will complete

their intended manoeuvre in the gap. The time required for each non-first vehicle to initiate its manoeuvre

and clear the stop line is referred to as the “follow-up time” Minimum headway between two consecutive vehicles in the minor stream

Critical gap and follow-up time: varies from driver to driver and varies for a single driver from situation to situation. there exist distributions of values for tc and tf

Using distributions complicates the determination of capacity of the minor street approach.

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Critical gap and follow-up time

Assumptions to remove these distributions:1) Drivers are consistent in that a driver behaves the same way each time

she is faced with the same traffic conditions (i.e. drivers do not change behaviour with time).

2) The driver population is homogeneous in that all drivers behave in the same way and have the same gap size requirements (i.e. there is no distribution of driver types).

The errors incurred by these unrealistic assumptions counteract each other and that the resulting total error is small

Assume that tc and tf are constants instead of distributions. Different constant values of tc and tf are used for :

different minor street movements (e.g. left turn versus a through movement).

different vehicle types (e.g. trucks versus passenger cars)

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Estimating tc and tf (using filed data)

field data reflects a range of time headways used by 1, 2, 3, …, n vehicles.

Estimate the mean time gap used by 1 vehicle, the mean gap used by 2 vehicles, etc.

use linear regression to find the single value for tc and tfthat best fit the observed data.

Linear Relationship: n = at + b a: the slope of the line b: the intercept.

4 6 8 10 12

Gap Size (Sec)

Accepted gap by only One Vehicle

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The relationship between the number of vehicles using a gap and the gap size is a step function.

tf is the additional time required for another vehicle to use the same gap: tf = 1/a

The critical time headway is the time required for the first vehicle to use a gap.

Step function B reflects the assumption that the regression accounts for variance in driver behaviour and therefore point C represents the critical gap.

For this assumption: tc(B) = t0 + tf

t0 = -b/a (by setting t = 0 and n=0 in n = at + b)

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Critical Gap (tc)

Step function A reflects an assumption that: the regression line is a reflection of average driver behaviour but there is variance in driver behaviour (i.e. some drivers accept gaps less than the

gap size associated with point C) and therefore the critical gap size is assumed to be halfway between points t0 and C. For this assumption: tc(A) = t0 + 0.5tf

Step function B is more conservative than function A (larger value for tc).

HCM based on the assumptions associated with step function A.

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Example:

A gap use study has collected data for vehicles making a left-hand turn from a stop controlled minor street onto a one-way 1-lane major street.

Time headways were recorded between each pair of consecutive vehicles on the major street and the associated number of minor street vehicles using the gap to make a left turn.

Data were collected for 1,460 vehicles over a period of approximately 1 hour and 45 minutes.

The average gap size associated with each number of vehicles using a gap is shown in the next table

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Solution

regression coefficients: a = 0.30 and b = -1.2. Note: Do not use the data for 0 vehicles using a

gap tf = 1/a = 1/0.30 = 3.3 seconds

t0 = -b/a = -(-1.2)/0.3 = 4 seconds

tc = t0 + 0.5tf = 4 + 0.5*3.3 = 5.65 seconds

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Important Note

using the data for 0 vehicles using a gap a = 0.23 and b = -0.64 tf = 4.3 seconds and tc = 5.0.

These results are incorrect as they are biased by the large number of gaps in which no minor street vehicles were able to make their manoeuvre.

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Distribution of Gap Sizes

Please review lecture # 02

Vehicle Arrival Patterns (Poisson Distribution) Exponential Distribution of Gaps Shifted Exponential Distribution of Gaps

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Capacity Analysis for TWSC intersections

Capacity of a minor street movement is governed by: 1) The number of gaps in the opposing stream. 2) Size of gaps. 3) Characteristics of drivers use of gaps (i.e. tc and tf) 4) Priority to use suitable gaps.

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Interaction of Two Traffic Streams

Cx = capacity of minor street movement x (vph)

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Example:

Consider the intersection of two one-way streets. Find the capacity of the minor street in vph. λmajor = 900 vph = 0.25 veh/second; (average headway) = 4 seconds per vehicle tc = 8 seconds; tf = 4 seconds

𝑡𝑡

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Solution20

Solution

A Number of vehicles using the associated gap B Minimum gap size required for n vehicles to complete

manoeuvreC Calculation of exponent term for the exponential

distribution D Probability that headway is larger than the minimum

from column B E Probability that headway permits exactly n vehicles to

complete their manoeuvreF Number of gaps per hour that permit exactly n vehicles

to complete their manoeuvresG Number of minor street vehicles per hour that

complete their manoeuvres using the associated gap size

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Interaction of More Than Two streams

If streams are independent of each other, then

the probability of finding a suitable gap in both streams simultaneously is equal to

the product of the probability of finding a suitable gap in each stream individually.

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Priority of traffic streams

When calculating the capacity of movements for an intersection with more than two traffic streams, one must consider the rank/priority of that movement

For North America the HCM is the standard commonly used for TWSC analysis

Exhibit 17-3 illustrates the ranks of the different possible intersection movements

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Example

tc = 8 seconds; tf = 4 seconds for both minor street movements.

λ 2 = 900 vph; λ9 = 100 vph; λ10 = 60 vph. Determine: (a) the capacity of movement 9 and (b) the capacity of movement 10.

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Solution: part (a)

Movement 9 must yield right of way to movement 2 only capacity is a function of λ2 and not λ10.

The opposing volume of 900 vph and the critical gap and follow up times are the same as used in the previous example, capacity of movement 9 = 192 vph.

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Solution: part (b)

The capacity of movement 10 ≠ 192 vph because movement 9 uses some of the available gaps making them

unavailable for movement 10. We call this effect “impedance”.

The magnitude of this effect depends on the arrival rate and capacity of higher ranked movements.

The probability that a vehicle attempting to make movement 9 will be waiting for a gap can be estimated as the ratio of the demand for gaps by vehicles making movement

9 (i.e. λ9) and the capacity associated with movement 9 (cm,9). Therefore, the probability that there is no vehicle waiting to

make movement 9 (P0,9 ) when a gap becomes available for a vehicle making movement 10 is:

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P0,9 = probability that there is no queue waiting to make movement 9

λ9 = arrival rate for movement 9 (vph) cm,9 = capacity of movement 9 (vph)

The potential capacity (cp) of movement 10 = 192 vph, since this is the capacity if none of the available gaps were used by vehicles of movement 9.

The actual movement capacity must include the impedance effect and is computed as

cm,10 = 192 vph x 0.48 = 92 vph.

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HCM equation for determining capacity 30

Conflicting volumes (HCM 2000) 31

Conflicting volumes (HCM 2000) 32

Conflicting volumes (HCM 2000) 33

Conflicting volumes (HCM 2000) [1] If there is a right-turn lane on the major street, v3 or v6 should not be

considered. [2] If there is more than one lane on the major street, the flow rates in the

right lane are assumed to be v2/N or v5/N, where N is the number of through lanes. The user can specify a different lane distribution if field data are available.

[3] If right-turning traffic from the major road is separated by a triangular island and has to comply with a yield or stop sign, v6 and v3 need not be considered.

[4] If right-turning traffic from the minor road is separated by a triangular island and has to comply with a yield or stop sign, v9 and v12 need not be considered.

[5] Omit v9 and v12 for multilane sites, or use one-half their values if the minor approach is flared.

[6] Omit the farthest right-turn v3 for subject movement 10 or v6 for subject movement 7 if the major street is multilane.

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Conflicting volumes (HCM 2000) 35

HCM critical gap equation36

HCM follow-up time equation37

Adjustments on tc and tf

Adjustment ValuestcHV 1.0, Two-lane major street

2.0, Four-lane major street

tcG 0.1, Movements 9 and 120.2, Movements 7, 8, 10 and 111.0, Otherwise

tcT 1.0, With two stage process0.0, With single stage process

t3LT 0.7, Minor-street LT at T-intersection0.0, Otherwise

tfHV 0.9, Two-lane major street1.0, Four-lane major street

LTcTcGHVcHVcbc ttGtPttt 3−−++=

HVfHVfbf Pttt +=

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Considering Impedance

Rank 2 movements

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Considering Impedance

Then the capacity for the Rank 3 movement is computed as

Rank 3 movements

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Considering Impedance

Rank 4 movements41

Shared Lanes42

Control Delay43

Length of Queues44

Example

Consider the three-leg intersection illustrated below. Determine the average control delay and the resulting level of service for the minor street approach.

Assume: no heavy vehicles, no pedestrians, and all approaches are level (i.e. 0% grade). Assume an evaluation time period of 15 minutes and that the major street can be considered to be a two-lane roadway.

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Solution

We begin the analysis by denoted the rank of each movement as follows:

Rank Movement Rank 1: 2, 3, 5 Rank 2: 4, 9 Rank 3: 7

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Step 1: Rank 2 movements

RT from Minor Street (9) Note: Separate right turn lane on major street therefore don’t

include v3. vc,9 = 250 vph tc,9 = 6.2 seconds tf,9 = 3.3 seconds cp,9 = 794 vph cm,9 = 794 vph

LT from Major Street (4) vc,4 = v3 +v2 = 40 + 250 = 290 vph tc,4 = 4.1 seconds tf,4 = 2.2 seconds cp,4 = 1283 vph cm,4 = 1283 vph t t t t

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Step 2: Rank 3 movements

LT from Minor Street (7) Note that a separate turning lane is provided for movement 3, therefore

v3 need not be included when calculating vc,7. vc,7 = v2 +v5 + 2v4= 250 + 300 + 2(150) = 850 vph tc,4 = 7.1 – 0.7 = 6.4 seconds (Adjustment for T intersection) tf,4 = 3.5 seconds cp,4 = 334 vph

Must consider impedance created by Major Street LT vehicles. Normally, the minor street LT movement (7) is rank 4 and must

consider the impedance impacts from all higher priority streams including 1, 4, and 11.

However, for a T-intersection, movement 7 is rank 3 and is impeded only by movement 4 (LT from major street).

f7 = 1 – v4/cm,4 = 1 – 150/1283 = 0.88 cm,7 = cm,7 × f7 = 334 × 0.88 = 294 vph

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Step 3: Determine Shared Lane Capacity

,

40 120 54040 120294 794

yy

shy

y m y

VC

VC

+= = =

+

∑

∑

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Step 4: Determine Delay

Movement 4 (LT Major) cm,4 = 1283 vph v4 = 150 vph T = 0.25 hour

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Step 4: Determine Delay

Movement 7 & 9 (LT & RT Minor) csh = 540 vph vsh = v7 + v9 = 40 + 120 = 160 vph T = 0.25 hour d = 14.0 seconds

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Step 5: Determine Level of Service

Movement 4 LOS = A

Movements 7 & 9 LOS = B

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Step 6: Determine Average Queue Length

Movement 4 (LT Major) Q4 = 8.2 × 150/3600 = 0.34 vehicles

Movement 7 & 9 (LT & RT Minor) Qsh = 14.0 × 160/3600 = 0.62 vehicles

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Home Reading

Chapter # 17 in HCM 2000

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Thanks for your time55