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
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!…
{1;1;4}{1;1;5}{1;2;0}{1;2;1}{1;2;2}{1;2;3}{1;2;4}{1;2;5}{1;3;0}{1;3;1}{1;3;2}{1;3;3}{1;3;4}{1;3;5}
etc...
and I want to format as a text it so it replaces the innermost branch with a letter so {0;0;1} would read A-0-1. I am able to replace all the symbols using replace text but am no sure if there's a way to convert a number to a letter.…
Added by Ryan Whitby at 12:40pm on February 3, 2015
{0;0;0;0 -1, 2... 19 ) N=5
i know is basic but im still trying to get my head around lists/trees and how to manipulate them properly
thanks for your time
sn…
each circle's border, let us say 1.0
3) So, the curve will end up with 5 points, in each point will have a circle, each circle will have a different Radius, but the distance in between the borders of each circle is always the same = 1.0 in this case.
4) The end result list here would be like this to evaluate a curve with these values and find the points on the curve:
List = 1, 5, 11, 19 etc If I use these values to eval a line, I will get the perfect points where I can draw the circles.
…