another example could be:
index 3 value 6
index 4 value 6
index 5 value 6
flipped and branched:
branch 6 index 0 value 3
branch 6 index 1 value 4
branch 6 index 2 value 5
Added by Ante Ljubas at 12:50pm on October 22, 2010
ards to the number before the start number...
i.e. 9, 0, 1, 2, 3, 4, 5, 6, 7, 8
then it will need to repeat this pattern (continuing to count upwards) and the repeat number is based on a slider (for example 3 in the case illustrated below):
9, 0, 1, 2, 3, 4, 5, 6, 7, 8
19, 10, 11, 12, 13, 14, 15, 16, 17, 18,
29, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29,
if anyone has any ideas on how to create this series it would be great
M.…
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.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
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?
I have this :
list 3 : 0 1 2 3 4 5 6
list 2 : 0 1 2 3 4 5 6
list 1 : 0 1 2 3 4 5 6
list 0 : 0 1 2 3 4 5 6
and I want to group the points of index 0 in a branch, the points of index 1 in another branch and so on.
I attached a file in which I generated the points.
Thank you in advance for your help !
Regards
Red…