qqwref has apparently done this, too. I'll remind him sometime and let him explain.
oh, sorry, i didn't know about that.
any way, here's part 1:
(I'll continue writing my thoughts down when i find the time, probably on friday)
as i have said,
I have been thinking about Stefan's method for speedsolving the Rubik's Clock (to be found here:
http://www.stefan-pochmann.info/spocc/speedsolving/clock/) and this is the result:
Somewhere in his tutorial, he mentioned, one should memorize the first move on the backside before getting started.
Actually, this works perfect for the first three moves.
Firstly, you allign the centre clock with the upper edge clock. You can see, what you need to do right away.
Secondly, since you alligned the center clock and the upper edge clock using the LD wheel with only the LD pin sticking up, the distance between center clock and lefthand edge clock hasn't changed, so you can see your second move right away aswell.
Thirdly, since you moved both lefthand clock and the bottom clock in your first move, but not on your second move, when you used the UR wheel with only the UR pin sticking up, the distance between them didn't change. That's what your third move will be.(-> distance between left edge clock and bottom edge clock)
The fourth move, e.g. alligning the center clock with the right edge clock is the first time, you have to do some simple maths. You have changed the starting position of the right edge Clock once: when you alligned the center clock with the upper edge clock (first step). Also, you have changed the distance between the center clock and the right edge clock once: when you alligned the center clock with the bottom edge clock. (third step)
Hence, the distance between the center clock and the right edge clock is the starting distance between center and right edge minus your first move minus your third move.
ok, that's what i've written down so far.
I'll leave, giving you all of the formulas for the calculations you need to make and come back later and explain them in detail.
I'll also try and make an example later on.
(Please note that i'm well asure of the fact, that one could abreviate the formulas, but i have decided to keep them this way, in order for better remembrance)
Notaion first:
(Before I forget: Unless I write sth different, i am always talking about
distances between certain clocks)
F: Front
B: Back
L: Left
R: Right
U: Up
D: Down
C: Center
R12: distance between right edge clock and 12 O'clock
Solve the Clock in this order:
1. Front edges
FCU
FCL
FLD
FCR: (CR-FCU-FLD)
F12: (FR12-FCL)
Flip the clock.
2. Back edges (same as on frontside)
BCU
BCL
BLD
BCR: (CR-BCU-BLD)
B12: (BR12-BCL)
3. Corners
BDR(h) (the 'h' means the hour, where the clock is in the beginning)
-FCU
-FCR
-F12
=BDR
BDL(h)
+BCU
+BCR
+B12
=BDL
BUR(h)
+BR12
-FLD
-FCR
-F12
=BUR
BUL(h)
-FR12
+BLD
+BCR
+B12
=BUL
12: - BUL (or whichever corner was last unequal to 0)
NOTE the similarity between BDR & BDL clocks and the similarity between BUR and BUL clocks.
Ok, hope this made sense so far.
