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…
ectural context.
We will work with 5 projects that meet the conditions to be re-thinked nowadays in parametrical terms. { FOA : World Trade Center / MAD OFFICE : 800M Tower / TOYO ITO : Relaxation Park / FOSTER & PARTNERS : Gherkin Tower / FREE : Soumaya Museum }
We will produce dynamic objects defined by parameters and modified in realtime.
We will learn to discretize, facet and panel in a non homogeneous way.
We will work with restrictions to determine our shapes, obtaining unpredictable results.
And we will take the control of what we produce : by measuring , visualizing and optimizing our parametric forms.
2 Groups / 30 students max :
Thursdays : 25 Apr ,02 ,09 ,16 & 23 May
Saturdays : 27 Apr ,04 ,11, 18 & 25 May
Fee : 73 € / student…
Added by Carlos Bañón at 10:26am on April 18, 2013
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.…
ee 3)
{5}
0 15
{6}
0 16
And I want to place points at every possible combination of these coordinates, treating Tree 1 as X coordinates, Tree 2 as Y coordinates, and Tree 3 as Z coordinates. Also, I would like the list of points to be a tree with paths corresponding to the coordinates. Wouldn't it be nice if I could plug these trees into a Point XYZ, with a new "branch cross reference" method, and get the following result?
{0:3:5}
0 {10.0, 13.0, 15.0}
{0:3:6}
0 {10.0, 13.0, 16.0}
{0:4:5}
0 {10.0, 14.0, 15.0}
{0:4:6}
0 {10.0, 14.0, 16.0}
{1:3:5}
0 {11.0, 13.0, 15.0}
{1:3:6}
0 {11.0, 13.0, 16.0}
{1:4:5}
0 {11.0, 14.0, 15.0}
{1:4:6}
0 {11.0, 14.0, 16.0}
{2:3:5}
0 {12.0, 13.0, 15.0}
{2:3:6}
0 {12.0, 13.0, 16.0}
{2:4:5}
0 {12.0, 14.0, 15.0}
{2:4:6}
0 {12.0, 14.0, 16.0}
In this form of cross referencing, every combination of individual branches from the different lists is used as separate input, and the output for each combination is put onto a branch in the result whose path is the concatenation of the input branch paths used.…
Added by Andy Edwards at 7:03pm on November 3, 2009
I used "panel = rg.NurbsSurface.CreateFromCorners(cpts[l],cpts[l+1],cpts[l+3],cpts[l+2])"to generate planar surface. But when I want to join this surfaces, it went wrong.
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!…
First use a series component with start=1, step=42, count=3
Use the output to create a new series component with start=existingseries, step=1, count=11