V-cubes > 11 are impossible or structurally unstable?

[BigSams]

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.

 

[Victor]

I think they are all possible but the cubes will become more rounded... Plus there are other Styles besides v cube try this
http://twistypuzzles.com/forum/viewtopic.php?f=9&t=20633&hilit=Nxnxn+designs

 

[Mike Hughey]

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.