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.
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David Rutten
david@mcneel.com
Poprad, Slovakia…
dont get you, i am saying sleect numbers in range 1 to 10, starting from 1 with a step of 2.
1 to 10 by 3 = 1 4 7 10
1 to 10 by 5 = 1 6
1 to 10 by 1 = 1 to 10 = 1 2 3 4 5 6 7 8 9 10
Added by Steve Lewis at 3:15pm on November 11, 2013
This must be a bug because its true for dividing by all odd numbers:
(i-1)/3
(i-2)/5
(i-3)/7
(i-4)/9
(i-5)/11
....
(i-n)/2n+1
And you can't make it work for even numbers
Added by Danny Boyes at 5:06pm on January 13, 2010
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).
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David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 10:19am on February 24, 2011
nts me this:
[[0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7]]
this is what I wanted but how to convert this to tree in grasshopper?
In grasshopper I just get:
8x IronPython.Runtime.List…