4, 3, 7, 7, 7, 8
would need to be
5, 6, 7, 8, 4, 3, 7, 8
ideas?
current workaround (kangaroo dupPt component) removes all duplicates from list, tho does maintain order.
thx…
triangulate my mesh faces thus i was getting 7 points per polygon. 7 outputs needed. It works with 7 outputs even if 3 of them are not being used.
Problem solved!
Thanks so much for your help!
…
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
{4}-0;3
{5}-6;7
{6}-5;7
{7}-5;6
Here it can be shown that there are two subgraphs containing 0,1,2,3,4 and 5,6,7. How can I use spiderweb (either using scripting or the components) to give me this result when I have many more vertices??
Thanks,
Sam…
ersect (2, 3, 4, 5, 6) with the line and the ones which do not intersect (0, 1, 7). Intersect is done! But how to get the non intersecting vectors (0, 1, 7)?
So I e. g. could deselect vectors 2, 3, 4, 5, 6 so I would display/use only vectors 0, 1, 7 and the bounced ones.
Appreciate your help!
Rudi…
ea being the further up the column, the more dramatic the effect. For example:at 1", the effect is at 1at 3", the effect is at 3at 7", the affect is at 7, ect.I figure it must be some form of formula, but I am uncertain how to call upon other number sliders, or how to set it up.Thanks!…
This must be a bug because its true for dividing by all odd numbers:
(i-1)/3
(i-2)/5
(i-3)/7
(i-4)/9
(i-5)/11
....
(i-n)/2n+1
And you can't make it work for even numbers
Added by Danny Boyes at 5:06pm on January 13, 2010