Page 123 - Lecture Notes DCC3113

Given : Assume :
Car = 1.00 pcu Amber time, a = 3 sec
Bus = 2.25 pcu Lost Time, ℓ = 2 sec
Heavy Vehicle = 1.75 pcu All red interval, R = 2 sec
Motorbike = 0.33 pcu

From the data given, determine :
i. Optimum cycle time both phases
ii. The actual green time for each phase
iii. The time diagram for each phase

Solution:

Unit Lane Group North/Utara South/Selatan East/Timur West/Barat
Car 259 264 580 557
Motorbike 150 118 158 124
Q, Flow (pcu/hr)
Bus 47 46 43 35
Heavy Vehicle 58 62 40 45
259 x 1 + 150 x 0.33
Flow, q (pcu/hr) + 47 x 2.25 + 58 x 514.94 798.89 755.42
1.75 = 515.75
Width (mm) 6.0 6.0 7.0 7.0
Fg 1.00 1.00 0.88 1.12
Fr 0.95 0.99 0.95 0.98
Fl 0.98 0.99 1 1
S = 525x Width, (pcu/hr) 525 x 6 = 3150 525 x 6 = 3150 525 x 7= 3675 525 x 7= 3675
3150 x 1 x 0.95 x 3150 x 1 x 0.99 x 3675 x 0.88 x 0.95 3675 x 1.12 x 0.98
S’ actual
0.98 = 2933 0.99 = 3087 x 1 = 3072 x 1 = 4033
y = q / S’ 0.18 0.17 0.26 0.19
y max 0.18 0.26

Y = y1 + y2
= 0.18 + 0.26
= 0.44 ≤ 0.85 (OK)

Time between green, I = R + a
= 2 + 3
= 5 sec

Total lost time, L = ( I - a ) + ( I - a )2 + ℓ 1 + ℓ 2
= (5 - 3) + (5 - 3) + 2 + 2
= 8 sec

i. Optimum Cycle, Co = (1.5L + 5)
1 - Y

= (1.5(8) + 5)
1 - 0.51

= 30.4 ≈ 35 sec ≤ 120 sec (OK)

Effective green , gn = yn (Co – L)
Y

g for each phase, :

Phase 1 N/S : g1 = 0.18 (35 – 8) = 11.04 sec ≈ 11 sec
0.44

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