e
7. True
8. True <-- this one
9. True
10. False
11. True
12. False
13. True
14. True <-- this one
15. True
16. False
17. True
18. False
19. True
20. True <-- this one
21. True
22. False
23. True
24. False
25. True
26. True <-- this one
27. True
28. False
29. True
30. False
31. True
32. True <-- this one
33. True
Any idea how I can solve this?
Thanks!…
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…
the next challenge i am confronted with is that all 5 circles of the above-mentioned radi are placed in each point. My aim is to randomly distribute ONE of the created circles in each point, instead of 5 in each point.…
om frame -5 till frame -10 a frame spacing of 100mm is used,
etc.
Frame 0 is located on X=0 mm
Frame -5 will be on X=-500 mm
Frame -6 will be on X=(X of frame -5) -25 = -525 mm
Frame -11 wil be on X= ((X of frame -10) -10 = ?? mm
etc.
Cheers,
Bas…
element connectivities in text, when matched with the corresponding coordinates of the nodes from dupPt, it makes all the polygons again (Green lines in the picture below), exactly as the original.
So Im thinking its like reconstructing everything again from the original shape with new relation between coordinates and element connectivities.
So in this case, eg: 0 1 2 3 4 5 makes the hexagon, 21 20 0 5 to makes the trapz.
Well here it misses the '0' in the end to close the hexagon here, the same with trapz. However the 'string join' and cset somehow close it to 0 1 2 3 4 5 0 to makes the hexagon, and 21 20 0 5 21 for trapz.
Im very glad now.
…