PM Depth Problem - ISD - Chapter 9 of the Green Book

[blondebedhead]

This is a question from my School of PE Class -

[SIZE=12pt]A manufacturer plans to build a new plant on SR 39, which is a windy, four‐lane, undivided highway in the hills west of town. The plant will generate a significant amount of combination truck traffic. SR 39 has a design speed of 50 mph. In evaluating possible sites for the new plant, you consider the required intersection sight distance. What is the required intersection sight distance (ft) for left turns from the plant onto SR 39?[/SIZE]

[SIZE=12pt](A) 501[/SIZE]

[SIZE=12pt](B) 897[/SIZE]

[SIZE=12pt]© 846[/SIZE]

[SIZE=12pt](D) 740[/SIZE]

I have the answer as B but I disagree with 0.7 seconds being added to tg so let me know if anyone could help me out understand the explanation better for adding tg.

Reference Table 9‐5, p. 9‐37
ISD = 1.47Vtg
V = 50 mph;

tg = 11.5 sec + 0.7 sec = 12.2 sec (NOTE: 0.7 sec is for crossing the second lane for combination truck)
ISD = 1.47 * 50 * 12.2 = 896.7 ft

​

 

[John QPE]

897 feet is correct.

Add 0.7s for each additional lane crossed for the left turn movement. You will be crossing 2 lanes because this is a 4-lane road.

 

[Yakfish PE PS]

The note under Table 9-5 states: ...or 0.7 s for trucks for each additional lane, from the left, in excess of one, to be crossed by the turning vehicle.

You have to read the whole statement. Yes you are crossing two lanes, but you only count one lane in your calculation. 0.7 s is correct.