Homework 3 Solutions

0lt 943 957 and 9 phf eb t wb t nb lt sb lt 554 502

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Unformatted text preview: 54 502 115 108 616 558 384 ( ) ( Part (b) and (c), Three Phase: Direction EB LT WB LT Original 283 287 Volumes Modified Volumes (3.0 LT 331 335 SB T 528/2 Lanes 360 276 294 NB T 495/2 Lanes SB T 528/2 Lanes ) )) ( EB T WB T NB LT SB LT 554 502 115 108 616 Perm., 1.05 LT protect, .9 PHF) 558 384 360 276 294 NB T 495/2 Lanes SB T 528/2 Lanes 276 294 ( ) ( ) ( Part (b) and (c), Four Phase: Direction EB LT WB LT Original 283 287 Volumes Modified Volumes (3.0 LT 331 335 ( ) )) ( EB T NB LT SB LT 502 115 108 616 ) WB T 554 Perm., 1.05 LT protect, .9 PHF) ( ( NB T 495/2 Lanes 558 134 126 ( ) ( ) ( ) ( )) ( Part (d): The most suitable phasing plan is ________________ because ________________ Give a good reason! - The most-suitable phasing plan is the three phase signal design because it has the lowest v/s ratio (0.703 compared to 0.706 and 0.726). This indicates that it should be operating better than the other two options. Also, it doesn’t exceed Webster’s max cycle length of 85 seconds for a three phase signal. Part (e): Yellow Time = 3 sec...
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