Sight Distance Problem

[ghim]

I'm a little bit struggling how to get the sight distance for right turning. See this problem below.

• 50mph posted speed limit
  
• only left and right turn movement will occur from the new driveway (i.e. there is no crossing traffic)
  
• Design Speed of 60mph
  
• Six-12 foot lanes and a 16 foot two way left turn lane(TWLTL)
  
• 0% grade
  
• Assume the majority of turning vehcles will be large combination trucks.
  

Thank you for your help in advance,

 

[mjoneswvu]

This appears to be an intersection sight distance problem. Refer to the AASHTO Green Book under intersection sight distance and refer to to a stop condition for a right turn. Hopefully this helps you.

 

[sac_engineer]

If I haven't misinterpreted the problem, the answer is determining the intersection sight distance (ISD) based on the time gap for large trucks to turn right and left from the driveway.

I'd use the following eq.

ISD = 1.47 x V x t

V = 60 mph

left turn:

t = 11.5 seconds (time for truck to turn left into 2-lane highway) + 0.7 x 5 (additional 4 lanes + left-turn lane) = 15 seconds

ISD (left turn) = 1.47 x 60 x 15 = 1323 feet

You can calc the ISD for right-turning trucks, but that would be less than those turning left, so the ISD for left-turning trucks would govern.

Please respond with the correct answer.

 

[ghim]

sac_engineer,

This question was to calculate for ISD using 'ISD=1.47 x V x t'

The solutoin says;

1. Left Turn:

t=11.5+0.7 x 3 (addtional lan crossed)= 13.6 sec----->> I thought additional lane crossed was 4 (3 lane + 1 left & right turn lane)

ISD=1.47 x 60 x 13.6 = 1199.52 ft

2. Right Turn:

t=10.5 (I'm not sure where this number came from)

ISD=1.47 x 60 x 10.5 = 930'

If you figure out anything, please let me know. I really want to know......

Thanks,

 

[Ambrug20]

I agree with sac_engineer, with some comments.

there are 6 lanes (assume 3 each direction) 2 lanes are dedicated for left turn and one for right (no traffic go straight). It change the calculations a bit, but the theory is right (I think)

 

[Dean]

ISD is designed for the longest time in this case left

6 lanes faciliuty is 3 add lanes = for each additional two lanes is 0.7 x 2= 1.4 footnote page 660

and 16 ' lane( page 661 last paragraph ) = 16/12=1.33 x 0.7 /2 = 0.455 time to cross the median

Total tg= 11.5 + 0.455 + 1.4 =13.355

ISD = 60 x 13.355 x 1.47 = 1177.9 feet

have to read the foot notes

 

[Breezluv32]

The solutoin says;

1. Left Turn:

t=11.5+0.7 x 3 (addtional lan crossed)= 13.6 sec----->> I thought additional lane crossed was 4 (3 lane + 1 left & right turn lane)

ISD=1.47 x 60 x 13.6 = 1199.52 ft

2. Right Turn:

t=10.5 (I'm not sure where this number came from)

ISD=1.47 x 60 x 10.5 = 930'

If you figure out anything, please let me know. I really want to know......

---------------------------------------------------------------------------------------------------------------

Ghim,

The way I see how they got the solution is as follows:

ISD = 1.47 * V * t

For left turns: t is given in Exhibit 9-54 as 11.5 sec. for Combination truck. Directly underneath that table, it says to add 0.7 seconds for trucks for each ADDITIONAL lane. So aside from the 1st lane, you would also cross 2 lanes plus the TWLTL for a total of 3 additional lanes crossed. From this... you have 11.5 + 0.7 x 3.

For right turns: t is given in Exibit 9-57 as 10.5 sec. for Combination truck. You would not adjust this number because the right turn will most likely turn into the 1st lane.

Hope this helps. Good Luck Friday!

 

[miloc]

The solutoin says;

1. Left Turn:

t=11.5+0.7 x 3 (addtional lan crossed)= 13.6 sec----->> I thought additional lane crossed was 4 (3 lane + 1 left & right turn lane)

ISD=1.47 x 60 x 13.6 = 1199.52 ft

2. Right Turn:

t=10.5 (I'm not sure where this number came from)

ISD=1.47 x 60 x 10.5 = 930'

If you figure out anything, please let me know. I really want to know......

---------------------------------------------------------------------------------------------------------------

Ghim,

The way I see how they got the solution is as follows:

ISD = 1.47 * V * t

For left turns: t is given in Exhibit 9-54 as 11.5 sec. for Combination truck. Directly underneath that table, it says to add 0.7 seconds for trucks for each ADDITIONAL lane. So aside from the 1st lane, you would also cross 2 lanes plus the TWLTL for a total of 3 additional lanes crossed. From this... you have 11.5 + 0.7 x 3.

For right turns: t is given in Exibit 9-57 as 10.5 sec. for Combination truck. You would not adjust this number because the right turn will most likely turn into the 1st lane.

Hope this helps. Good Luck Friday!

THANKS! I didn't have a practice problem like this one.

 

[simpatique]

using posted speed limit 50 mph , I got 1225 feet...... close enough

 

[simpatique]

posted and not designed speed limit unless there is more than 10 mph discrepency between

 

[jco0518]

i think in general, you'd have to use the greater one, so your design is conservative