ant to find all paths where exactly two items are the same like in branch {1}. How can I solve this in VB? Is there an easy "search in list" class or do I have to iterate over all items by myself ??
Regards Alex…
number of divisions on that curve as in the defintion (i.e. by 4). The offset in the def is slightly different and should cull two or three more curves as in the lists that show my aim below.
Basically I want to look into each branch of the groups of points from each closed curve . Marking in a list whether it contains a one or a zero (0= outside 1 = coincidents).
{0;0}0. 21. 22. 23. 2 {0;1} 0. 01. 22. 03. 2 {0;2}0. 01. 02. 03. 0 {0;3}0. 21. 22. 23. 2 {0;4}0. 21. 22. 23. 2 {0;5}0. 21. 22. 23. 2 {0;6}0. 01. 22. 23. 1 {0;7}0. 21. 22. 03. 0 {0;8}0. 21. 22. 23. 2 {0;9}0. 21. 22. 23. 2 {0;10}0. 21. 22. 23. 2 {0;11}0. 21. 22. 23. 2 {0;12}0. 21. 22. 23. 2 {0;13}0. 01. 22. 23. 0 {0;14}0. 21. 22. 23. 2
I want to create a list from these points. That marks each curve that pokes out, in a cull pattern as such:
20022210222202
Using a 1 where there are co-incidents in the curve points and the boundary. A 2 for true (outside points) and a 0 for containment. So I might be able to use the 1 in future developments - however if a true false list is easiest I can live with that.
So could I use F(x) function? - to look for 0 or 1's in each bunch of points and thus list as such for a cull pattern? or will Path mapper help me here? Or can I rely on simply grafting and splitting??
I am usure of the neatest solution and would love to learn. Hope you can direct me.rgrds
J.…
ements between mid axes of elements ( being perpendicular to both). On the attached image, the links are the small elements connectiong nodes 4 to 5, 6 to 7 and 8 to 9. All nodes (1 to 9 including 1', 2', 3') are defined in Karamba as fixed supports but nodes 1,1', 2, 2', 3, and 3' have hinges added with the beam joint component. The freed rotations are shown on the figure.
I wondered if that was the correct way of defining such a structure in Karamba bearing in mind that nodes 1', 2' and3' are free nodes in the reality.
Thanks again for your help !
Yousef…
Added some optimizations and runtime is down to ~35% of what it used to be. It actually seems to be 1.5 to 2 times faster than the old component.
--
David Rutten
david@mcneel.com
Poprad, Slovakia
Added by David Rutten at 8:34am on December 3, 2012
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!…
{0;1;0}N=6
{0;1;1}N=6
{0;1;2}N=5
{0;2;0}N=7
{0;2;1}N=8
{0;2;2}N=9
Can you shift and wrap any of the paths A B or C?
Say if I wanted to shift and wrap B by 1 to get the following...
{0;0;0}N=7
{0;0;1}N=8
{0;0;2}N=9
{0;1;0}N=3
{0;1;1}N=2
{0;1;2}N=5
{0;2;0}N=6
{0;2;1}N=6
{0;2;2}N=5…