2 Look PLL- problem!

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CubeX

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Hey guys,

I am just wondering if you don't know all the PLLs and that you had to learn 2 Look PLL as well, does it stuff your times when you have a 2 Look PLL?

When i don't have 2 Look PLL; just normal PLL i get about 20 sec, but when i suddenly need to do 2 Look (a PLL that i haven't learn't yet), i get like 26-27 sec.
 
I mean... how do i explain it.

Say for example you were doing an average of 5 3x3x3

and you haven't learnt all the PLLs and i had to use some 2 Look PLLs; since i haven't learnt the G perms.

I have solves that had normal PLL and i get 23 sec for them, but when i have a 2 Look PLL in my solves, i get a 26-27 and this stuffs up my average...
 
My question is, does that happen to everyone? Or is my 2 Look very different to them?
 
But what about Erik Akkersdijk's 2 Look OLL and PLL solves, he averages about 20 seconds with them? Is it his finger tricks...
 
But what about Erik Akkersdijk's 2 Look OLL and PLL solves, he averages about 20 seconds with them? Is it his finger tricks...

But he normally averages sub 13 at most! He's considerably slower using 2-look than 1 look.

It's his superior f2l lookahead, and just well practised algorithms.
 
You don't have to look for another PLLs, just use the one you remember you use..
 
Hey guys,

I am just wondering if you don't know all the PLLs and that you had to learn 2 Look PLL as well, does it stuff your times when you have a 2 Look PLL?

When i don't have 2 Look PLL; just normal PLL i get about 20 sec, but when i suddenly need to do 2 Look (a PLL that i haven't learn't yet), i get like 26-27 sec.
Yes.
 
Are you wondering why 2 algs is slower than 1? That's how I'm reading this. :s

Yeah, that's how I'm reading this, too...

Obviously, if the algs are about the same length -which they approximately are in PLL- then performing two is going to take longer (lets say, about twice the time :eek:) than one.

To this day, I still groan when I get a G perm (which seems like every solve!), but they aren't so bad to learn: Just get G(a) and G(c) done and then you can mirror for G(b) and G(d) or vice versa.

EDIT: I've just realised that I lied to you! What I meant to say was get G(a) and G(b) done and then mirror for G(c) and G(d) and vice versa, my bad. :cool:
 
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