--limited to in order to keep some consistancy with typical LSE style-- --more optimal solutions may be possable if the L and F faces are utilized-- --by slightly modifying the algorithim, any edge of the F2B can be fliped-- 3flip on top= M2 U' R U' M' U2 M2 U' r' 9stm 3flip on top + 2flip on D= rM'U M U2 M2 U r'M U' M 11stm 4flip on top + DF= R U M2 U2 M U r'M U' M' 10stm 4flip on top + DB= R U M U2 M2 U r'M U' M' 10stm adj2flip on top + DF= r U M U2 M2 U R' 7stm adj2flip on top + DB= r U' M2 U2 M' U' r'M U' M' 10stm opp2flip on top + DF= R U M U2 M2 U r' 7stm opp2flip on top + DB= R U' M' U2 M2 U' r' 7stm 1flip= M' U R U' M' U2 M2 U' R' 9stm 1flip on top + 2flip on D= rM' U' M' U2 M2 U r' 8stm = 8.8 = 9avg stm --Since the average stm is greater than that of the worst standard 4a case, it would not be wise to intentionally place a misoriented edge during a solve unless doing so would save you ~3 moves. However, the UL and UR edges are often easily manipulated during the standard 4a due to it's highly intuitive nature. Knowing this, I would place the number of moves you would have to save at 4 to justify solving a block with a misoriented edge. It is(seems) highly unlikely for such a large payoff to occure during a speedsolve, except in some fluke cases of great luck. Indeed, these algorithms should only be used in order to eficiently fix mistakes and potentially save you from too poor a result. So learn them well-- Happy Cubing, ~Ross :)