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
0, 5, 10, 15, 20
1, 6, 11, 16, 21
2, 7, 12, 17, 22
3, 8, 13, 18, 23
4, 9, 14, 19, 24
and if i'm here is because i'm not able... :)
can you help me?
thank you
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nput parameter and then set the named values on the second?
protected override void BeforeSolveInstance() { Param_Integer param0 = Params.Input[0] as Param_Integer; Param_Integer param1 = Params.Input[1] as Param_Integer; param1.ClearNamedValues();
GH_Structure<GH_Integer> data = param0.VolatileData as GH_Structure<GH_Integer>; if (data.IsEmpty) return; foreach (GH_Integer value in data.AllData(true)) { switch (value.Value) { case 1: param1.AddNamedValue("First option for 1", 11); param1.AddNamedValue("Second option for 1", 12); param1.AddNamedValue("Third option for 1", 13); break;
case 2: param1.AddNamedValue("First option for 2", 21); param1.AddNamedValue("Second option for 2", 22); param1.AddNamedValue("Third option for 2", 23); break;
case 3: param1.AddNamedValue("First option for 3", 31); param1.AddNamedValue("Second option for 3", 32); param1.AddNamedValue("Third option for 3", 33); break; } return; } }
--
David Rutten
david@mcneel.com…
Added by David Rutten at 1:56am on December 18, 2013
ied 12 times in X (always with the same gap) and for each time has a rotation of 30°. Then, I'd like these 12 items are copied n times in Y, shifting 1 module per time. At the end, the last 12 items have to be copied again n times in Y, shifting 1 module per time, but in the opposite way (like a Z).
See attached an example.
Thank you for your help!
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7, 9, 12 and 13 to be able to rotate freely around the y axis at nodes 2, 3, 6, 7, 10 and 11 respectively. The last 2 conditions, for elements 12 and 13, doesn't give any problems, but the first 4 does.
Any help?
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