Slids of intellegent assignment

242 views

Published on

Published in: Business, Technology
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
242
On SlideShare
0
From Embeds
0
Number of Embeds
18
Actions
Shares
0
Downloads
7
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

Slids of intellegent assignment

  1. 1. Intelligent Urban Traffic Control System(KKKA 6424)Assignment no .1Traffic Light SettingSupervisorProf. Dr. Riza Atiq Abdullah OK RehmatPrepared by: Rasha salah ahmed P64799Sarah hazim P65407
  2. 2. Fig (1) (The Study area)Location of intersections
  3. 3. First IntersectionPhase Traffic volume Actualflow(pcu/hr)saturationflowperlane(pcu/hr)saturationflow(pcu/hr)Y=flow/saturationflowGreentime splitY / βˆ‘y(Y / βˆ‘y)*Geleft straight right1 111 88 199 1800 3600 0.1 0.33 82 95 201 296 1800 3600 0.1 0.33 83 85 100 185 1800 3600 0.1 0.33 8βˆ‘y=0.3 βˆ‘=24
  4. 4. 1. L(losttime)=3Γ—(3+2)=15sec2. Co=(1.5 𝐿 +51βˆ’βˆ‘π‘Œ=1.5 15 +51βˆ’0.3=39.28 takeCo=40sec3. Effectivegreentime(Ge)=Co-L=40-15=25sec(Multiplythisvalueingreentimesplit to getthefinalcolumn).4. Totalgreentime=24sec5. Cycletime=greentime+losttime=24+15=39sec
  5. 5. SecondintersectionPhase Trafficvolume Actualflow(pcu/hr)saturationflowperlane(pcu/hr)saturationflow(pcu/hr)Y=flow/saturationflowGreentimesplitY/βˆ‘y(Y/βˆ‘y)*Geleft straight right1 80 328 184 592 1800 3600 0.16 0.31 162 84 140 98 322 1800 3600 0.1 0.19 103 148 252 160 560 1800 3600 0.16 0.31 164 64 92 120 276 1800 3600 0.1 0.19 10βˆ‘y=0.52 βˆ‘=52
  6. 6. 1. L (lost time) =4Γ— (3+2) =20 sec2. Co=(1.5 𝐿 +51βˆ’βˆ‘π‘Œ=1.5 20 +51βˆ’0.52=72.9 take Co=73 sec3. Effective green time (Ge) =Co-L =73-20=53 sec(Multiply this value in green time split to get the final column).4. Total green time=52 sec5. Cycle time =green time + lost time=52+20=72 sec
  7. 7. ThirdintersectionPhase Trafficvolume Actualflow(pcu/hr)saturationflowperlane(pcu/hr)saturationflow(pcu/hr)Y=flow/saturationflowGreentimesplitY/βˆ‘y(Y/βˆ‘y)*Geleft straight right1 151 215 99 465 1800 3600 0.13 0.217 152 90 198 75 363 1800 3600 0.1 0.167 113 88 119 90 297 1800 3600 0.1 0.167 114 356 501 113 970 1800 3600 0.27 0.45 31βˆ‘y=0.6 βˆ‘=68
  8. 8. 1. L (lost time) =4Γ— (3+2) =20 sec2. Co=(1.5 𝐿 +51βˆ’βˆ‘π‘Œ=1.5 20 +51βˆ’0.6=87.5 take Co=88 sec3. Effective green time (Ge) =Co-L =88-20=68 sec(Multiply this value in green time split to get the final column).4. Total green time=68 sec5. Cycle time =green time + lost time=68+20=88 sec. To determine (offset time) T ideal from inter1 to 2:T ideal=πΏπ‘†βˆ’ 𝑄 Γ— β„Ž + π‘™π‘œπ‘ π‘  π‘‘π‘–π‘šπ‘’ Where:L=450mS=10m/secQ=14 carsh=2secLoss time=2 sec. T ideal=15 sec. To determine (offset time) T ideal from inter 2 to 3:L=650mS=10m/secQ=12 carsh=2 secLoss time=2 sec T ideal=39 sec
  9. 9. We choosethe maximum value ofCβ‚€ = 88 and make recalculation for the greentime, andthe new results inthe table below:Co=88 Ge=Co-L=88-20=68 secphase Intersection 1 Intersection 2 Intersection 3GreenTime = (Y/βˆ‘Y)*GeGreenTime = (Y/βˆ‘Y)*GeGreenTime = (Y/βˆ‘Y)*GeQ1 23 21 15Q2 23 13 11Q3 22 21 11Q4 13 31βˆ‘=68 βˆ‘=68 βˆ‘=68So the new diagramwill be like the following:
  10. 10. Pictures from the study area
  11. 11. THANK YOU

Γ—