Old-Pochmann - single flipped edge - confused!

[MichaelErskine]

Sorry if this has been asked before (not really sure what to search for!)

I'm a beginner using Old-Pochmann and I have a single flipped edge cycle such as the following situation (scrambed with White on U, Green on F) :-

R U R D B F' L' R' F2 D' B F D L2 F' D2 U B' F2 D' R F L' R' F'

OK, I solve with Yellow on top, Orange on front. My buffer position is Upper-Right and my very first piece is Blue-Yellow already in the Yellow-Blue position! If I swap this with UL (White-Red, using a T-Perm) and then swap it back but upside down then the White-Red will become Red-White and all subsequent edges will be flipped too. Is this what people do?

I imagine I'm missing a trick and there's a more sensible approach.

EDIT: OK, I found a similar thread and I'm slowly getting the idea - it really messes with my memorisation though. I suppose I need to remember that the first piece was initially kicked out but in what orientation? This might become clearer to me as I try this solve again.

 

[byu]

Here's what you've got to do. First, solve all the cycles, just as you normally would. I believe there is parity in this case, but that's not exactly important right now. Next, you need to shoot to one side of BD, and then the other side. So your memorization would be:

BD DB

That would happen at the very end or the very beginning. Try it, shoot to BD, with l Jb-Permutation, then shoot to DB with l2 Ja-Permutation.

Then after that, fix parity with the R-Permutation.

 

[Stefan]

OK, I solve with Yellow on top, Orange on front. My buffer position is Upper-Right and my very first piece is Blue-Yellow already in the Yellow-Blue position!

Sounds like you're solving the buffer piece. Don't. Ignore that one. Just solve all the other pieces and your buffer piece will automatically be solved, too.

then the White-Red will become Red-White and all subsequent edges will be flipped too.

I understand why white/red is flipped, but it shouldn't affect any subsequent pieces.