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Triangular Francisco Method (for 3x3)

qqwref

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I have invented a new method. The name comes from
[19:23:42] <Kirjava> You're now chatting with a random stranger. Say hi!
[19:23:42] <Kirjava> You: can you solve a rubiks cube?
[19:23:42] <Kirjava> Stranger: yes
[19:23:42] <Kirjava> You: what method do you use?
[19:23:42] <Kirjava> Stranger: triangular francisco method
Let's begin! Yes, this was 17 minutes ago.

Step 1: The Triangle
Solve a 1x2x2 block on D, and then add two adjacent corners. It's probably possible to do this all in inspection. It should make a triangle, like so:
imagecube.php


Step 2: Middle Layer
Solve the edges of the middle layer. Hold the triangle on D so that the free tcorner is at DRF; use u moves to move the middle layer around. Now, for an edge on UF, you can insert it into the FR slot using r U r' or R U' R' depending on its orientation. For an edge at DF you can insert it with r U' r' (or if it's in the wrong orientation r U2 r' R U' R'). For an edge at DR, just do a y turn and then you can do either of those algs with L/l/U.

Step 3: DR/DF + EO
You're going to be using <M,U> moves here. Solve one edge just by inserting it blindly, and then do the second one with an algorithm so that the top layer edges get oriented as well. If you don't want to learn algorithms you can just insert both and then do an OLL.

Step 4: CLS (DRF + CO)
Out of the bottom two layers, only the DRF corner is left. Use a CLS algorithm to place the corner while orienting the top layer corners (or, if you don't want to learn extra algorithms, you can just insert it with an F2L alg and then do an OLL).

PLL
Easy stuff. And you're done! :D
 
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FOr the first step, I find it simpler for the middle layer edges to just put one corner in for now. Then, once you get there, make an F2L pair.
 
Interesting.. I'd like to see the algs for inserting the DR and DF edges while simultaneously orienting the LL edges. CLS I can find on Lucas' site.
 
Well althuogh interesting, its look impractical, considering you need to learn 139 algs total if your going all that way, and just the the algorithm portion(3b-5) is about 30 moves average and 2 and 3a at minimum is 15 probably averaging at 18 moves so not including step one your already at 48 moves. It jsut seems impractical, but correct me if I'm wrong here.
 
Well althuogh interesting, its look impractical, considering you need to learn 139 algs total if your going all that way, and just the the algorithm portion(3b-5) is about 30 moves average and 2 and 3a at minimum is 15 probably averaging at 18 moves so not including step one your already at 48 moves. It jsut seems impractical, but correct me if I'm wrong here.

Step 1 is something like 6-8 moves on average if you build it all at once. And... yeah, it might be slightly more moves than most speedsolving methods, but it's still a lot of fun. I'm getting ~23 seconds average.
 
But I can see the fact that its even more braindead than frid, but if only there was a way to split up the CLS to lower the alg count.
 
Well that kind of defeats the purpose of the method then. The 2nd Insertion + EO is wasted and your putting in an edge earlier jsut to take in out. Be sides the whole B2 idea, it looks alot liek a BM.
 
Even if this method has been created before, I give credit for the great name. ;U
 
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