CENG 6302 - CH 5-1 - Basic Principles of Intersection Signalization PDF
CENG 6302 - CH 5-1 - Basic Principles of Intersection Signalization PDF
CENG 6302 - CH 5-1 - Basic Principles of Intersection Signalization PDF
Chapter 5.1
Basic Principles of
Intersection Signalization
Interval
s = 3600/h
where s = saturation flow rate, vehicles per hour of green
per lane (veh/hg/ln)
h = saturation headway, seconds/vehicle (s/veh)
Start-Up Lost Time
8
l1 = ∑ ∆ i
i
where l1 = start-up lost time, s/phase
∆i = incremental headway (above “h” seconds)
for vehicle i, s
Start-Up Lost Time
9
gi = Gi + Yi – tLi
where gi = effective green time for movement(s) i, s
Gi = actual green time for movement(s) i, s
Yi = sum of yellow and all-red intervals for
movement(s) i, s (Yi = yi + ari)
yi = yellow interval for movement(s) i, s
ari = all-red interval for movement(s) i, s
tLi = total lost time for movement(s) i, s
This model results in an effective green time that may be fully
utilized by vehicles at the saturation flow rate (i.e. at an
average headway of “h” s/veh)
Capacity of an Intersection Lane or
Lane Group
13
Sample Problem
Critical-Lane and Time-Budget
14
Assuming the total lost time per phase (tL) to be constant for
all phases, the total lost time per signal cycle is:
Where L = lost time per cycle, s/cycle
L = N ⋅ tL tL = total lost time per phase (l1 + l2), s/phase
N = number of phases in the cycle
The total lost time in an hour depends upon the number of
cycles occurring in the hour:
(
LH = L ⋅ 3600
C
) Where LH = lost time per hour, s/hr
L = lost time per cycle, s/cycle
C = cycle length, s
Maximum Sum of Critical-Lane Volumes
19
1 3600
Vc = 3600 − Nt L
h C
For the intersection on slide 15:
Two-phase signal, cycle length of 60-sec, total lost times
of 4s/phase, and a saturation headway of 2.5s/veh
1 3600
Vc = 3600 − 2 * 4 * = 1,248 veh / hr
2.5 60
Maximum Sum of Critical-Lane Volumes
21
PHF (Peak Hour Factor) – to account for the flow rate in the
worst 15-minute period of the hour
v/c (volume to capacity ratio) – to provide some excess capacity
to avoid failure of individual cycles or peak periods on a
specific day
Finding an appropriate cycle length
23
The negative cycle length signifies that there is not enough time
within the hour to accommodate the demand with the required
green time plus the 12 s of lost time per cycle.
Sample Problem
The Concept of Left-Turn Equivalency
24
“In the same amount of time, the left lane discharges five
through vehicles and two left-turning vehicles, while the right
lane discharges eleven through vehicles”
s prev = sideal * f LT
s prev (3600 / h prev ) hideal
f LT = = =
sideal (3600 / hideal ) h prev
EXAMPLE:
An approach to a signalized intersection has two
lanes, permitted left-turn phasing, 10% left-turning
vehicles, and a left-turn equivalent of 5.0. The
saturation headway for through vehicles is 2.0
s/veh. Determine the equivalent saturation flow rate
and headway for all vehicles on this approach.
Delay as a Measure of Effectiveness
30
Overall Stable
Stable flow
(some phases fail)
throughout
Overall failure
(v/c > 1.0)
Components of Delay
37
ν ( R + tc ) = stc
s
R + tc = tc
v
s
R = tc − 1
v
R
tc =
s
− 1
v
Webster’s Uniform Delay Model
42
Substituting for tc
R vs
V = v( R + tc ) = v R + = R
s s − v
− 1
v
And for R
g vs
V = C 1 −
C s − v
Aggregate delay (in veh-secs for one signal cycle)
2
1 1 2 g vs
UDa = RV = C 1 −
2 2 C s − v
Webster’s Uniform Delay Model
43
2
g g
2
1 −
C 0.50 C 1 −
1 C
UD = C =
2 1 − g v
C c ( ) 1 −
g
C
X