I came a cross the problem (Test Master #27 transportation): see attached file.Why do we have to find percentage of clear zone and why we use 14’/20’
and 14’ (not 16’)/22’.
Why don’t we use data from the table? What is that data any way in this case?
Did I puzzle enough? :bananapowerslide:
Tanya,
While your question was a little confusing, I think I know what you are asking. It looks like your problem #27 (attached .pdf) is referencing the 2002 AASHTO RDG and I am using the 2006, but I don't think it matters. This was a fun exercise, thanks.
I know what you are thinking - why can't you simply go to Table 3.1 and get the Foreslope and Backslope Clear-zone distances directly from the table? The reason is because there is a V ditch in the clear-zone. This now directs you to Figure 3.6. Plot your Foreslope slope and your Backslope slope on the figure - if your plotted point falls outside of the shaded area (the Preferred channel cross slope), you need to solve for an adjusted clear-zone (you can't simply use Table 3.1 directly).
Please see Example G, page 3-26 (2006 RDG). It's a perfect example, very similar to your problem #27.
Step 1.
14'/20' = 70% (14' = edge of traveled way to ditch FL and 20' = lower range of foreslope Clear-zone from Table 3.1)
14'/22' = 63.6% (22' = upper range of foreslope Clear-zone from Table 3.1)
Step 2.
(100%-70.0%) x 14' = 4.2' (14' = lower range of backslope CZ from 3.1)
(100%-63.6%) x 16' = 5.8' (16' = upper range of backslope CZ from 3.1)
The adjusted blackslope CZ is 4.2' - 5.8', or 4' - 6' (Answer B ).
I hope I helped.