• Welcome to the Speedsolving.com, home of the web's largest puzzle community!
    You are currently viewing our forum as a guest which gives you limited access to join discussions and access our other features.

    Registration is fast, simple and absolutely free so please, join our community of 40,000+ people from around the world today!

    If you are already a member, simply login to hide this message and begin participating in the community!

V-cubes > 11 are impossible or structurally unstable?

BigSams

Member
Joined
Jan 3, 2009
Messages
215
For reasons I can't mention until some later date, I need an answer to the following question (no guesses please):

Verdes says that V2 to V11 are possible to construct, but is Vx for \( 12\le x\in\mathbb{N} \) possible to construct using his designs?

Could have sworn it used to say somewhere on his site that his designs allow any size to be constructed but structurally unstable for greater than V11, but my memory might be wrong.
 

Mike Hughey

Administrator
Staff member
Joined
Jun 7, 2007
Messages
11,305
Location
Indianapolis
WCA
2007HUGH01
SS Competition Results
YouTube
Visit Channel
I do remember a discussion somewhere (not on the V-Cubes site, but somewhere else) that he had additional modifications that he knew of to get past 11, but they weren't as carefully fleshed out and so were not patented.

The patent is here. If you look at the second page of the description, it has a little bit about the issue:
The practical reasons why the present invention finds application up to the cube N=11 are the following:

a) A cube with more layers than N=11 would be hard to rotate due to its size and the large number of its separate pieces.

b) When N>10, the visible surfaces of the separate pieces that form the acmes of the cube lose their square shape and become rectangular. That's why the invention stops at the value N=l 1 for which the ratio of the sides b/a of the intermediate on the acmes rectangular is 1, 5.
 
Top