Chapter 20-4e
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Transcript of Chapter 20-4e
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Chapter 20 2
Four critical aspects of signalized intersectionoperation discussed in this chapter
1. Discharge headways, saturation flow
rates, and lost times
2. Allocation of time and the critical lane
concept
3. The concept of left-turn equivalency
4. Delay as a measure of service quality
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Chapter 20 3
20.1.1 Components of a Signal Cycle
Cycle lengthPhase
Interval
Change interval
All-read interval(clearance interval)
Controller
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Chapter 20 4
Signal timing with a pedestrian signal: Example
Interval Pine St. Oak St. %
Veh. Ped. Veh. Ped.
1 G-26 W-20 R-31 DW-31 36.4
2 FDW-6 10.9
3 Y-3.5 DW-29 6.4
4 R-25.5 AR 2.7
5 G-19 W-8 14.5
6 FDW-11 20.0
7 Y-3 DW-5 5.5
8 R-2 AR 3.6
Cycle length = 55 seconds
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Chapter 20 5
20.1.2 Signal operation modes and left-turn
treatments & 20.1.3 Left-turn treatments
Operation modes:
Pretimed (fixed) operation
Semi-actuated operation
Full-actuated operation
Master controller,
computer control, adaptive
traffic control systems
Left-turn treatments:
Permitted left turns
Protected left turns
Protected/permitted
(compound) or
permitted/protected left turns
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Chapter 20 6
Factors affecting the permitted LT
movement
LT flow rate
Opposing flow rate
Number of opposing
lanes
Whether LTs flow
from an exclusive LTlane or from a shared
lane
Details of the signal
timing
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Chapter 20 7
CFI (Continuous Flow Intersection
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Chapter 20 8
DDI (Diverging Diamond Interchange)
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Chapter 20 9
Four basic mechanisms for building an
analytic model or description of a signalizedintersection
Discharge headways at a signalized intersection
The critical lane and time budget concepts
The effects of LT vehicles
Delay and other MOEs (like queue size and the
number of stops)
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Chapter 20 10
20.2 Discharge headways, saturation flow,
lost times, and capacity
1 2 3 4 5 6 7
h
Vehicles in queue
(i) Start-up lost time
nhlT
il
hs
!
(!
!
1
1 )(
3600Saturation flow rate
C
gsc
earyl
llt
aryY
tYGg
iii
L
iii
Liii
!
!
!
!
!
2
21
Capacity
Cycle length(Sample problem in p.467)
Effective
green
Startup lost time
Clearance lost time
Total lost time
Extension
of green
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20.2.6 Saturation flow rates froma nationwide survey
Chapter 20 11
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Chapter 20 13
20.3.2 Finding an Appropriate Cycle Length
)/3600)(/(1
/36001
min
hcvP F
NtC
h
NtC
c
Ldes
c
L
!
!
Desirable cycle length, incorporating
PHF and the desired level of v/c
ii
i
i
svratioflowY
Y
LC
)/(_
1
55.1
1
0
!
!
!
J
The benefit of longer cyclelength tapers around 90 to 100
seconds. This is one reason why
shorter cycle lengths are better.
N = # of phases. Larger N, more
lost time, lower Vc.
(Review the sample problem on page 473.)
Doesnt this look like the Webster model?
Eq. 20-13
Eq. 20-14
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Chapter 20 14
Websters optimal cycle length model
!
! J
1
0
1
55.1
i
isv
CC0 = optimalcyclelengthforminimum delay,sec
= Totallosttimepercycle,sec
Sum (v/s)i = Sumofv/sratiosforcriticallanes
Delay is not so sensitive
for a certain range of
cycle length This is the
reason why we can round
up the cycle length to,
say, a multiple of 5seconds.
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Chapter 20 15
20.3.2 Finding an Appropriate Cycle Length
(Review the sample
problem on page 473)
Marginal gain in
Vc decreases as
the cycle length
increases.
Desirable cycle length, Cdes
Cycle
length
100%
increase
Vc 8% increase
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Chapter 20 16
20.4 The Concept of Left-Turn (and Right-
Turn) Equivalency
In the same amount of time, the left
lane discharges 5 through vehiclesand 2 left-turning vehicles, while the
right lane discharges 11 through
vehicles.0.3
2
511
:
1125
!
!
!
LT
LT
E
and
E
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Chapter 20 17
Left-turn vehicles are affected by
opposing vehicles and number of
opposing lanes.
The LT equivalent increases as the opposing flow increases.
For any given opposing flow, however, the equivalent
decreases as the number of opposing lanes is increased.
5
1000 1500 1900
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Chapter 20 18
Left-turn consideration: 2 methods
Given conditions: 2-lane approach
Permitted LT
10% LT, TVE=5
h = 2 sec for through
Solution 1: Each LTconsumes 5 times more
effective green time.
vphgplh
s
hh
ave
avg
128680.2
36003600
sec/80.2)00.2)(9.0()00.10)(1.0(
!!!
!!
Solution 2: Calibrate a factor that would multiply the saturation flow
rate for through vehicles to produce the actual saturation flow rate.
? A ? A
714.0)15(10.01
1
)1(1
1
)0.1)(1(
18002
3600
!!!
!!
!!
LTLT
LTLTLTavg
LT
EP
hPhEP
h
h
hf
vphgpls vphgpls 1286)714.0(1800 !!
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Chapter 20 19
20.5 Delay as an MOE
Common MOEs:
Delay
Queuing
No. of stops (orpercent stops)
Stopped time delay: The time avehicle is stopped while waiting to
pass through the intersection
Approach delay: Includes stopped
time, time lost for acceleration and
deceleration from/to a stop
Travel time delay: the difference
between the drivers desired total time
to traverse the intersection and the
actual time required to traverse it.
Time-in-queue delay: the total time
from a vehicle joining an intersection
queue to its discharge across the stop-
line or curb-line.
Control delay: time-in-queue delay +
acceleration/deceleration delay)
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Chapter 20 20
20.5.2 Basic theoretical models of delay
At saturation flow rate, s
Uniform arrival
rate assumed, v Here we assumequeued vehicles
are completely
released during
the green.
Note that W(i) is
approach delayin this model.
The area of the
triangle is the
aggregate delay.
Figure 20.10
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Chapter 20 21
Three delay scenarios
This is great.This is acceptable.
You have to do something
for this signal.
A(t) = arrival
function
D(t) = discharge
function
UD = uniform
delay
OD = overflow delay due to
randomness (random delay).
Overall v/c < 1.0
OD = overflow delay
due to prolonged
demand > supply
(Overall v/c > 1.0)
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Chapter 20 22
Arrival patterns compared
HCM uses the Arrival Type factor to adjust the delay computed as
an isolated intersection to reflect the platoon effect on delay.
Isolated intersections
Signalized arterials
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Chapter 20 23
Websters uniform delay model, p480
-
-
!!
-
!!
!!!
-
!
vs
vs
C
gCst
C
g
Cvs
v
vs
vR
t
stvtvRtRv
C
gCR
c
c
ccc
1
1
1
The area of the triangle is the aggregated
delay, Uniform Delay (UD).
-
-
!! vs
vs
C
g
CVheightRbaseUDa
2
2
12
1
):)(:(2
1
UDa
Total approach delay
To get average approach
delay/vehicle, divide this by vC
? A ? Asv
CgCUD
!1
1
2
2
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Chapter 20 24
Modeling for random delay, p.481
UD = uniform
delay
OD = overflow delay due to
randomness (in reality random
delay). Overall v/c < 1.0
? A ? A
? A
? ACgcvvccvv
cv
sv
CgCD
!
2312
22
65.0
/121
1
2
Adjustment term for
overestimation
(between 5% and 15%)
Analytical model
for random delay
D = 0.90[UD + RD]
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Random delay
derivation
Chapter 20 25
Chapter 20.
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Chapter 20 26
Modeling overflow delay
? A
? A
? A
? A 2
)(1//1
/1
21
1
2
22
CgCcvCg
CgC
sv
CgCUDo
!
!
!
because c = s (g/C), dividebothsides
byv andyouget(g/C)(v/c) = (v/s).
And v/c = 1.0.
cvT
cTvTTODa !!22
1 2
The aggregate overflow delay is:
Because the total vehicledischarged during Tis cT,
12
! cvT
OD
See the right column of p.482 for the
characteristics of this model.
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Average overflow delay between
T1 and T2
Chapter 20 27
12
21
! cvTT
OD
Average delay/vehicle =
(Area of trapezoid)/(No.
vehicles within T2-T1).
Derive it by yourself.
Hint: the denominator is
c(T2-T1).
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Chapter 20 28
20.5.3 Inconsistencies in random and
overflow delay
? A ? A ? A ? ACgcvvc
cvvcv
svCgCD
!
2312
22
65.0
/1211
2 1
2! cvTOD
The stochastic modelsoverflow delay is
asymptotic to v/c = 1.0
and the overflow
models delay is 0 at
v/c =0. The real
overflow delay is
somewhere between
these two models.
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Chapter 20 29
Comparison of various overflow delay model
20.5.4 Delay model in the HCM 2000
The 4th edition dropped the HCM 2000 model (I dont know why). It
looks like Akceliks model (eq. 20-26).
These models try to address delays for 0.85
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Chapter 20 30
20.5.5 Sample delay computations
We will walk through sample problems(pages 495-496).
Start reading Synchro 6.0 User Manual
and SimTraffic 6.0 User Manual. We
will use these software programsstarting Wednesday, October 21, 2009.