could accomplish this by adding something like '5.0 + Slider * 3.0' into the expression field... (and having an integer slider with domain 0 to 6).
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 10:19am on February 24, 2011
ve a Vertex [V] connected to four other Vertexs [N1-N4].
Each of the has a Value:
V ... 1
N1 ... 5
N2 ... 3
N3 ... 8
N4 ... 11
The Average Filter would set the Value of [V] to
(1+5+3+8+11)/5 = 5,6
The Median Filter would Sort Values and pick the middle one
1,3, [5], 8, 11
Hope that helped...…
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…
1+2+3+4+5+6 = 21
1+2+3+4+5+6+7 = 28
1+2+3+4+5+6+7+8 = 36
1+2+3+4+5+6+7+8+9 = 45
Is there a tool, that can do me that job?
How do I get this List {1,3,6,10,15,21,28,36,45}?…
Added by Ahmed Hossam at 2:19pm on September 22, 2013
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!…