BFC 32302: TRAFFIC ENGINEERING AND SAFETY

TRAFFIC SIGNAL TIMING: DESIGN PRINCIPLES

 Determination of Saturation Flow, S

Saturation flow is the maximum flow that can cross the stop line of the approach
where there is a continuous green signal indication and a continuous queue of vehicles
on the approach.

Saturation flow is expressed in passenger car units per hour.

Where there is no on-street parking,

i) where effective approach width, W > 5.5 m

S = 525 W

ii) where W < 5.5 m, refer to Table 1.1

Table 1.1: Relationship between effective lane width and saturation flow
W (m) 3.0 3.25 3.5 3.75 4.0 4.25 4.5 4.75 5.0 5.25
S 1845 1860 1885 1915 1965 2075 2210 2375 2560 2760
(pcu/hr)

Where there is on-street parking, W is reduced by LW where

LW = 1.7 – 0.9(Z – 7.6)

k
where Z = clear distance of the nearest parked car from the stop line
(> 7.6 m)
k = green time (secs)

If LW is negative, take LW as zero. For parked lorry or wide van, LW should be

increased by 50%.

S has to be corrected for effects of gradient, turning radius and the proportion of
turning traffic. (Refer Tables 1.2, 1.3 and 1.4 for correction factors)

Table 1.2: Correction factor for the effect of gradient, Fg

Correction Factor, Description
Fg
0.85 for upward slope of 5%
0.88 for upward slope of 4%
0.91 for upward slope of 3%
0.94 for upward slope of 2%
0.97 for upward slope of 1%
1.00 for level grade
1.03 for downward slope of 1%

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1.06 for downward slope of 2%

1.09 for downward slope of 3%
1.12 for downward slope of 4%
1.15 for downward slope of 5%

Table 1.3: Correction factor for the effect of turning radius, Ft

Correction Factor, Description
Ft
0.85 for turning radius R < 10 m
0.90 for turning radius where 10 m < R < 15 m
0.96 for turning radius where 15 m < R < 30 m

Table 1.4: Correction factors for turning traffic

% Turning Factor for right-turn, Factor for left-turn, Fl
Traffic Fr
5 0.96 1.00
10 0.93 1.00
15 0.90 0.99
20 0.87 0.98
25 0.84 0.97
30 0.82 0.95
35 0.79 0.94
40 0.77 0.93
45 0.75 0.92
50 0.73 0.91
55 0.71 0.90
60 0.69 0.89

 Determination of Y Value

y=q/S where y = ratio of flow to saturation flow

q = actual flow on a traffic-signal approach in pcu/hr
(refer to Table 1.5 for conversion to pcu)
S = saturation flow for the approach in pcu/hr

The y value for a phase is the highest y value from the approaches within that phase.

For the whole junction,

Y   yi where n = number of phases
i 1
yi = highest y value from the approach within phase i

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The Y value is a measure for the occupancy of the intersection.

Preferably, Y < 0.85.

If Y > 0.85, it is recommended that the geometrics of the intersection be upgraded to

increase capacity.

Table 1.5: Conversion factors to pcu

Vehicle type Equivalent pcu
value
Passenger cars 1.00
Motorcycles 0.33
Light Vans 1.75
Medium lorries 1.75
Heavy lorries 2.25
Buses 2.25

 Determination of Total Lost Time Per Cycle, L

Total lost time per cycle, L is given by:

n n
L   I  a    l
i 1 i 1

where I = intergreen time between phases

= R + a, where R = all red interval
a = amber time (assumed 3 seconds)
l = drivers reaction time at the beginning of green per phase.
(In practice, it is set at 2 seconds but 0 – 7 seconds can also be used)

To check for the adequacy of amber time, a, the following equation is used:

V W L
a 
2A V
where V = approach speed (km/h)
A = acceleration (taken as 4.58 m/s2)
W = width of intersection crossed (m)
L = length of vehicle (suggested 5.5 m)

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 Determination of Optimum Cycle Time, Co

Optimum cycle time, Co is the minimum average delay for the intersection, but this
delay is not greatly increased if the cycle time varies within the range of 0.75 to 1.50
of the calculated Co.

Co = 1.5 L + 5 (in seconds)

1–Y

For practical purposes, 45 s < Co < 120 s, although an absolute minimum of 25 s can
be used.

 Determination of Signal Settings

Effective green time is the green time plus the change interval minus the lost time for
a designated phase.

The total effective green time = cycle time – total lost time

g1 + g2 + … + gn = Co – L

Where n denotes the number of phases and gn is the effective green time for phase n.

For optimum conditions,

g1 = y1 (for a 2 phase cycle)

g2 y2

With the above ratio, the following formulas apply to each individual phase:

gn = Yn (Co – L) (in seconds)

Y

Where, gn = effective green time of the nth signal phase

Yn = calculated Y-value of the same signal phase

The actual green time, G = g + l + R

Controller setting time, K = G – a – R

=g+l–a

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 BFC 32302: TRAFFIC ENGINEERING AND SAFETY

Traffic Signal Timing: Design Example

Question

Design a 3 phase fixed-time traffic signal including pedestrian phase for the given
intersection.

The geometric and traffic flow values of an intersection are given as follows:

ALL DIMENSIONS IN METER

TURNING RADIUS = 10 m

GRADIENT N 3%
S

E 0% W

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 BFC 32302: TRAFFIC ENGINEERING AND SAFETY

Peak Hour Flows obtained from 16-hour classified traffic counts:

N
Morning peak:

100 35 55
150
150
175
195
200
165
40 75
65
Evening peak:

95 35 40
120
140
160
165
185
195
40 55 80

Solution:

Average value of the morning and evening peak:

From Left-turn traffic Straight traffic Right-turn traffic Total

Approach (pcu) (pcu) (pcu) (pcu)
NORTH (55+40)/2 = 48 (150+120)/2 = 135 (35+35)/2 = 35 218

SOUTH (65+40)/2 = 53 (165+195)/2 = 180 (40+55)/2 = 48 281

EAST (75+80)/2 = 78 (200+185)/2 = 193 (175+160)/2 = 168 439

WEST (100+95)/2 = 98 (150+140)/2 = 145 (195+165)/2 = 180 423

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 BFC 32302: TRAFFIC ENGINEERING AND SAFETY

 Determination of Saturation Flow, S

NORTH:

S = 1965 pcu/hr for W = 4.0 m (Table 1.1)

Fg = 1.09 for –3% grade (Table 1.2)

Fl = 0.98 for % left-turn = (48/218)100% = 22% (Table 1.4)
Fr = 0.90 for % right-turn = (35/218)100% = 16% (Table 1.4)

Adjusted SN = 1965  1.09  0.98  0.90 = 1889 pcu/hr

SOUTH:
S = 1965 pcu/hr for W = 4.0 m (Table 1.1)

Fg = 0.91 for +3% grade (Table 1.2)

Fl = 0.98 for % left-turn = (53/281)100% = 19% (Table 1.4)
Fr = 0.89 for % right-turn = (48/281)100% = 17% (Table 1.4)

Adjusted SS = 1965  0.91  0.98  0.89 = 1560 pcu/hr

EAST:

a) Left lane: S = 1915 pcu/hr for W = 3.75 m (Table 1.1)

Fg = 1.00 for 0% grade (Table 1.2)

Fl = 0.96 for % left-turn = (78/271)100% = 29% (Table 1.4)

Adjusted SEL = 1915  1.00  0.96 = 1834 pcu/hr

b) Right lane: S = 1915 pcu/hr for W = 3.75 m (Table 1.1)

Fg = 1.00 for 0% grade (Table 1.2)

Ft = 0.85 for R < 10 m (Table 1.3)

Adjusted SER = 1915  1.00  0.85 = 1628 pcu/hr

WEST:

a) Left lane: S = 1915 pcu/hr for W = 3.75 m (Table 1.1)

Fg = 1.00 for 0% grade (Table 1.2)

Fl = 0.93 for % left-turn = (89/243)100% = 40% (Table 1.4)

Adjusted SWL = 1915  1.00  0.93 = 1781 pcu/hr

b) Right lane: S = 1915 pcu/hr for W = 3.75 m (Table 1.1)

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Fg = 1.00 for 0% grade (Table 1.2)

Ft = 0.85 for R < 10 m (Table 1.3)

Adjusted SWR = 1915  1.00  0.85 = 1628 pcu/hr

Consider the following 3-phase traffic signal:

Phase 1: Phase 2: Phase 3:

 Determination of Y Value

PHASE 1 2 3
MOVEMENT WL EL WR ER N S
IDENTIFICATION
q 243 271 180 168 218 281
s 1781 1834 1628 1628 1889 1560
y = q/s 0.136 0.148 0.110 0.103 0.115 0.190
Y 0.148 0.110 0.190

Y = 0.148 + 0.110 + 0.190 = 0.448

Since Y < 0.85, we can proceed with the timing calculations.

 Determination of Total Lost Time Per Cycle, L

The amber time, a for the North and South approaches are the same since
vehicles from that approach has to travel 12.25m to clear the intersection.

Assume approach speed = 50 km/h = 13.9 m/s

V W L 13.9 12.25  5.5

a N/S =  =  = 2.79 sec  3 sec
2A V 2(4.58) 13.9

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The amber time, a for the East and West approaches are the same since
vehicles from that approach has to travel 9.00 m to clear the intersection.

13.9 9.00  5.5

a E/W =  = 2.56 sec  3 sec
2(4.58) 13.9

Therefore, take a = 3 sec

Assume all red interval, R = 2 seconds and driver reaction time, l = 2 seconds

Therefore intergreen time, I = a + R

=3+2
= 5 seconds

Total lost time, L =  (I – a) +  l

= 3(5 – 3) + 3(2)
= 12 seconds

 Determination of Optimum Cycle Time, Co

Optimum cycle time, Co = 1.5L + 5

1–Y
= 1.5(12) + 5
1 – 0.448
= 42 seconds

Design Co can be between 75% to 150% of the calculated Co.

For simplicity, take design Co = 60 seconds.

 Determination of Signal Settings

Total effective green time = Co – L

= 60 – 12
= 48 seconds

Effective green time, g = y (Co – L)

Y

g1 = 0.148(48) = 16 seconds

0.448

g2 = 0.110(48) = 12 seconds

0.448

g3 = 0.190(48) = 20 seconds

0.448

g1 + g2 + g3 = Co – L = 48 seconds  O.K.!

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Actual green time, G = g + l + R

G1 = 16 + 2 + 2 = 20 seconds

G2 = 12 + 2 + 2 = 16 seconds

G3 = 20 + 2 + 2 = 24 seconds

Controller setting time, K = G – a – R

K1 = 20 – 3 – 2 = 15 seconds

K2 = 16 – 3 – 2 = 11 seconds

K3 = 24 – 3 – 2 = 19 seconds

Check For Pedestrian Requirement For Green Time

(Note: Check only the critical pedestrian crossings where W is the widest)

Gped = 5 + (W/1.22) - I

= 5 + (9/1.22) - 5

= 7.38 seconds  8 seconds

In the calculation, green time available for critical pedestrian crossing is in

phase 1 where actual green time, G = 20 seconds.

Since G > Gped, therefore it is adequate.

TIMING SCHEDULE

Phase 1: 0 15 18 60
G A R

Phase 2: 0 20 31 34 60

R G A R

Phase 3:
0 R 36 G 55 58 60
R G AR
G 10 R A
R G
A R
 BFC 32302: TRAFFIC ENGINEERING AND SAFETY

DELAYS

- Average delay per vehicle on a particular arm (approach) is given by:

9  C(1  )2 x2 
d   
10  2(1  x ) 2qs (1  x ) 

where
C = cycle time (sec)
 = proportion of the cycle that is effectively green for the phase = g/C
qs = flow (pcu/sec)
x = degree of saturation = q/S

LEVEL OF SERVICE

- The level of service for signalised intersections are based on the delay, as given
in Table 1.6.

Table 1.6: Level of Service for Signalised Intersections (JKR)

LEVEL OF SERVICE DELAY PER VEHICLE

A < 5.0 seconds
B 5.1 to 15.0 seconds
C 15.1 to 25.0 seconds
D 25.1 to 40.0 seconds
E 40.1 to 60.0 seconds
F > 60.0 seconds

Example:

Using the results of the traffic signal timing design in the earlier example, determine
the following:

(a) Calculate the Delays

(b) Determine the LOS

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 BFC 32302: TRAFFIC ENGINEERING AND SAFETY

Solution:

(a) To calculate the delays and (b) determine the LOS, the designed Co = 60 sec
and designed green times are used.

The calculation of delays and LOS are tabulated as below:

Movement N S WL WR EL ER
C (sec) 60 60 60 60 60 60
g (sec) 20 20 16 12 16 12
 = g/C 0.333 0.333 0.267 0.200 0.267 0.200
S (pcu/hr) 1889 1560 1781 1628 1834 1628
q (pcu/hr) 218 281 243 180 271 168
qs(pcu/sec) 0.0606 0.0781 0.0675 0.0500 0.0753 0.0467
x = q/S 0.346 0.541 0.511 0.553 0.553 0.513
d (sec) 16.7 18.9 20.9 26.4 22.0 25.2
LOS C C C D C D

Sample Calculation:

For North:
9  C(1  )2 x2 
d   
10  2(1  x ) 2qs (1  x )  =

9  60(1  0.333) 2 0.346 2 


10  2(1  0.333 * 0.346) 2 * 0.0636(1  0.346) 

= 16.7 sec

From Table, the LOS is C

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