owing:
{0}0. d1. e2. f
3. g4. h5. i
{1}0. a1. b2. c
3. g4. h5. i
{2}0. a1. b2. c
3. e
4. f
5. g
Thought maybe I could use relative Item but I cant figure out how to do an offset that includes multiples.
…
Good to hear it helped.
As for the slider you can just change the limits to Min=1 and Max=4, so you'll get the values you want (.2, .4, .6, .8). This works for me fine.
Is it like this:
If a beam is connected from nod 0 to 1 and from 1 to 4. Another from 2 to 3 and from 3 to 5.
Node 1 and 3 have the same coordinates, but are they rigidly connected or not?
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
53 → 53 → 63 → 74 → 74 → 84 → 9
As you can see from the above list the connection sequence comes in waves of three, where each group of similar indices on the left is associated with a group of three incrementing indices on the right.
Some combination of Series components will probably generate this list, but it'll only work for the first ring, the second one will need a different connection pattern. It is perhaps better to just encode the integer pairs by hand. But then you cannot change your mind about the number of sides later.…
Added by David Rutten at 10:39am on October 21, 2015