Rock Outlet Protection Source Equation
Rock Outlet Protection Source Equation
(OP)
I'm looking for the equation(s) used to establish the Rock Outlet Protection nomographs used by so many jurisdictions. You know, the ones based on Tw < or > .5 dia., enter the figure with a Q and Pipe dia. to get a d50 then further to get length of level section. I think the source is SCS, perhaps a Maryland office. In this day of spreadsheets, why use figures !?





RE: Rock Outlet Protection Source Equation
RE: Rock Outlet Protection Source Equation
TW<0.5 Do La=(1.8*(q/(Do^.5))) + (7*Do)
W=(3*Wo)+La
TW>0.5 Do La=3*(q/(Do^.5))
W=(3*Wo)+(0.4*La)
D50= (0.016/Tw)*(q^1.33)
Where Tw can not be computed use Tw=0.2*Do
Do=maximum inside culvert height in feet (or diameter of round pipe)
Wo=maximum inside culvert width in feet (or diameter of round pipe)
q=unit discharge = Q/Wo (CFS per foot for conduit design storm or 25 yr storm, whichever is greater)
La= length of apron
W=width of apron (Width should be at least 3 times the culvert width)
Hope this helps!!
Kate