where each branch contains all the points generated by dividing each curve, so if you divide into 10 segments, you'll get:
{0;0}(N = 11)
{0;1}(N = 11)
{0;2}(N = 11)
{0;3}(N = 11)
{0;4}(N = 11)
Where the second integer in the curly brackets refers back to the index of the curve in the original list.
Another way to look at this data is to see it as a table. It's got 5 rows (one for each original curve) and 11 columns, where every column contains a specific division point.
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David Rutten
david@mcneel.com
Poprad, Slovakia…
branches in each A's list of B's, or remove its ends etcso that if I want to remove the last B in every A{0;1},{0;2},{0;3},{0;4},{0;5},{0;6}{1;1},{1;2},{1;3},{1;4}{2;1},{2;2},{2;3},{2;4},{2;5}would become{0;1},{0;2},{0;3},{0;4},{0;5}
{1;1},{1;2},{1;3}
{2;1},{2;2},{2;3},{2;4}I guess the question is do I need to figure out the cull pattern- each B may have different lengths...…
1. Duplicate the first list.(24-->48) and Graft(48 branchs of 1 item each)
2. Graft the second list.(48 branchs of 1 item each)
3. merge two list
4. Join Curves
5. Fillet Curve
{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…
Maybe with numbers it would be easier to understand.
If there is 1 insert to the list 5
If there is a 5 insert to the list 6 or 4
So if i start from 1 one possible list would be: {1, 5, 6}