Hi guys:

anybody know how to find the 'a' distance for Case 'A' with no control. And, is there any errant on case 'B' regarding distance is either the 'Sum of totally length" or using the diagram just 'one way + half of the lane?'

thanks

I believe that you are referring to Intersection Sight Triangles. Refer to AASHTO Green Book (2004 Edition) Page 654 - Intersection Control.

Yes I am familiar with the section. They do not specific on how to determine distance 'a' on Case 'A' . If you read my message carefully, I am quoting from the green book. So, if anybody here know what is about please let me know.

tsli

Are you sure about this?

Please read the section carefully and if required start from Page 651 - Sight Triangles. The values in the Exhibit 9-51 (Table) and Exhibit 9-52 (Graph) represents the length of the legs of sight triangle for a given speed. In other words, these values are dimension "a" and "b" from Exhibit 9-50A.



I would like to give your suggestion back to you, read the section "carefully".

See I got an issue to the book. If the speed on highway 'A' was not provided or highway 'B' was not provided. how would you go about the leg then. There were no indication in how to compute the distance 'A' like case 'B' does beside looking up the chat or graph. Unless there is something that I had miss.

tsli

Again, read the discussion in the Green Book. It clearly states that the length of the leg should equal to the stoppping sight distance. However, since the drivers tend to slow down as they approach the intersection, the distance required is less than the SSD. This can be seen from the graph in Exhibit 9-52. All this is discussed in the book.

What you are asking is not possible. You can not calculate the stopping distance, or the length of the leg in this case, unless you are given the speed.