are on their own paths, but the first branch contains 3 curves and the second one 2 curves. If you want the same result for all pairs of curves you'd need to split up the first and second branches, so that all curves are on their own branch.…
Added by Lars Renklint at 4:33am on September 6, 2009
ep is to understan the logics of what you want to do, in your case, build 4 point surfaces (u also need to know the right direction to build the surfaces). Then you can write an hipotetic list (by hand in a paper) of what you want. In your case the list was (0, 1, 3, 2) (2, 3, 5, 4) (4, 5, 7, 6), etc... if you can imagine building 2 lists, each one with the sequences (0, 2, 4, 6, etcc) and (1, 3, 5, 7, etc..) then you can manage with shift and graft to finally have four lists. A( 0 1 2 3 ...) B (1 3 5 etc..) C(3 5 7 etc..) D (2 4 6 etc..). And to achieve the 2 first lists, you need to get the odd and the pair numbers. The cull pattern does that amazingy well. With a pattern True-False you get de pair numbers, and with the False-True pattern you get de odd numbers.
Hope it was clear enough…
Added by Pep Tornabell at 5:32am on November 19, 2009
points 0, X-1, (2*x)-1, (3*X)-1, (4*X)-1, (5*X)-1 and then
1, X, (2*x), (3*X), (4*X), (5*X)
2, X+1, (2*x)+1, (3*X)+1, (4*X)+1, (5*X)+1
and so on till
5, X+4, (2*x)+4, (3*X)+4, (4*X)+4, (5*X)+4
How can I do this best?
Thanks,
Niels…
erations, is not it?
This is what I finally want and how I plan to do it:
In the starting tree, points are listed accordingly to the (11) 4-side-panels they belong to. I need to do a tree where each of the 11 lists contains not the points composing the panels but the points code of the points composing the panels. the point code is the number in the flattened list that refers to the points (like, panel 1 is made of points n 0,1,4,18, and so on). To do this, I suppose that I will use my tree of 11 lists of 33 true-false values, and apply it to cull 11 times a series of numbers from 0 to 32. I ll post it if it works!…
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