Intersection Sight Distance

[Calixico]

I am trying to understad this problem:

Given a vehicle stopped on a minor road (south east corner) at a major road intersection which is located to the north of the minor road. The design speed is 60 mph for the major road. The minor road approach grade is 3.4%. The minor road is a single lane. The vehicle will be making a left turn unto the major road and has to cross three (east bound) travel lanes (36 feet) in order to enter into the inside westbound lane which is 24 feet (2 lanes). The westbound lane on this major road is. The vehicle will be turning into the inside lane to proceed westbound on the major road. There is also a 5-feet median existing between the east bound and west bound lanes on the major road. What is the design intersection sight distance for passenger car?

The solution in the book states that this is a case B1 (left turn from a minor road from stop), AASHTO, page 660, exhibit 9-54.

and the followings solution steps:

ISD = 1.4V * tg

with, V=60mph, the equation becomes ISD=(1.47*60MPH)+tg = 88.2*tg.....

where tg= time gap at design speed.

so for passenger car: 88.2 [7.5+(0.5*2.42) +0.2*3.4)] = 828 feet

This is where I get confused.

I understand every steps except the 0.5* 2.42.

I know that the 7.5 is for the time gap in second (AASHTO exhibit 9-54) and 0.5 seconds is for the left turn unto the two-way highways, with more than two lanes.

I know that the 0.2 is added b/c greater that 3 % grade, you add 0.2 for each percent grade > 3%. But I don't know where 2.42 is coming from. I know it has to do with the width of the lanes, but how?

My calculation for the lanes; I am coming up with 36 feet for the east bound travel lanes (on major road) plus the 5 foot median and the inside (west bound) travel lane on the major road (where the vehicle will be turning into) is 12 feet, totaling (36+5+12)= 53 feet

AASTHO say to convert median width to equivalent lanes, but the median is only 5-feet.

Could someone explain where this is coming from please.

 

[Calixico]

Checked the 2004 Green Book when I got to work today. Pg. 660 (I think), first half of the page. It's pretty explicit.
7.5 seconds for passenger car turning left onto a two lane road.

Add an extra 0.5 for each additional lane. Three East Bound Lanes and a 5' median, means 7.5 + 2(0.5) + (5/12)*(0.5) so, that is exactly where the 2.52 is coming from.

Thanks major highway. I appreciate your helping with this matter. I actually read that today also, but was still a little confused on the "adding of the 0.5% per lane in addition to the 7.5%.

Since the Greenbook states for the user to add this amount for each lane in excess of two lanes, wouldn't it be just an addition of 1* 0.2 since technically you only cross one additional lane (not counting the last lane you arrive in to go westbound?), since two is already accounted for in the three eastbound lanes, thus leaving one to be added to the equation. Or we should include the arriving lane (westbound) lane to the computation----as shown in the calculation.

I just want to know so for sure so i don't make this silly mistake in the exam----even though it's self explantory that we account for the arriving lane.

Thnaks again!

 

[Calixico]

A couple things:
1. It is 7.5 seconds time gap (not percent). This is actually pretty important to note, because it makes sense of the equation:

1.47 (converts MPH to FPS) * 60 MPH = 88.2 FPS

Since Sight Distance is measured in feet, you need to eliminate the per second part of the 88.2 fps. I know you probably understand this, and I'm not trying to belittle you by any means, but my point to this is that you can solve some problems on the exam simply by solving for the units... or at least that will narrow it down usually.

2. The 7.5 per AASHTO is the time gap to do the following:

- Accelerate from a stop

- Cross one lane of traffic

- Enter the lane of traffic the design vehicle is turning into

3. The additional 0.5 seconds per lane are for any additional lanes over the criteria in number 2 above. In this case, that number is two lanes plus a 5 foot median which we treat as a partial 12' lane (5/12).

I don't have the green book with me at home, but that was what I remember from looking it up this morning from the office.

Major:

Got it...Appreciate all your help..

Cheers!!!