output will show a tree with 3 branches of 4 integers each that I can pass on to other components. What is the best way to do it?
I have tried creating a tree and using a for loop to do so, but it didn't work.
Thank you for your help.
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cture, Rhino treats them as a single flat list. For example a surface can have 10 rows and 6 columns of control-points, resulting in a list of 60 points.
But 10 times 6 isn't the only way to get to 60. If you want to make a surface out of a list of 60 points, you'll also have to tell Rhino how those 60 points should be interpreted in terms of a grid. It could be 2*30, 3*20, 4*15, 5*12, 6*10, and all of the aforementioned products the other way around.
Sometimes there's only one way for a number of points to fit into a rectangular grid. For example if you provide 49 points, then 7*7 is the only way to make it work, but these cases are rare so we always demand you give us all the information required to actually make a rectangular grid of control-points from a linear collection.
As for "Why is it, sometimes we need to attach additional value into it?", this is usually because when you divide a domain or a curve into N segments, you end up with N+1 points. For example take the domain {0 to 5}, and divide it into 5 equal subdomains. You end up with {0 to 1}, {1 to 2}, {2 to 3}, {3 to 4} and {4 to 5}. However there are six numbers that mark the transitions between these domains 0, 1, 2, 3, 4 and 5. This is why you often have to add 1 to the UCount, because the number that controls the UCount often results in N+1 actual points.…
Added by David Rutten at 8:30am on December 25, 2014
a seed, and instead creating a pattern where each color has a seed/control slider for each row? For example, row 1: brown 2, tan 6, yellow 7, purple 3, repeat. row 2: brown 6, tan 1, yellow 4, purple 10, repeat. row 3: yellow 5, purple 1, brown 3, tan 10, repeat. row 4: purple 2, brown 7, tan 3, yellow 4, repeat. Then repeat that sequence up the wall? For each color, the number in the sequence should be adjustable.
Thank you again for your help!…