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
g definition but in diva for grasshopper in material it just appear dusty_med and not metal_railings and metal_treads. How I should write the correct definition?
void brightfunc dusty_med4 dirt dirt.cal -s 101 .25
dusty_med metal metal_railings005 .7 .7 .7 .3 .2
dusty_med metal metal_treads005 .5 .5 .5 .3 .2…
e), {1;2}(line), {1;3}(line)... and on the other side to have {0;0}(all lines except {0}(0)), {0;1} (all lines except {0}(1)), {0;2}(all lines except {0}(2)), {0;3}(all lines except {0}(3)), {1;0} (all lines except {1}(0)), {1;1} (all lines except {1}(1)), {1;2} (all lines except {1}(2)) ,{1;3} (all lines except {1}(3))...The first tree is easy to achieve, simply grafting a branch for each element, and the other, what I've done is to copy all lines of each tree ({0},{1},{2},{3}), to have them in all branches of each tree ({0;0}(elements of {0}), {0;1}(elements of {0}),,{1;0}(elements of {1}), {1;1}(elements of {1})..., and then remove in the first branch({0;1} the first element(0), in the second branch the second element, the third branch the third element...And so correctly you compare each line with all the other within each branched tree.Aaaaapufff XD…
ft , 4:3 (to use GH); it could be nice dock icon bars to the left or right working with laptop screen and put it on top working in GH with the 4:3 one.
This feature is cool, but not really usefull.
Best Regards.…
W, X, Y, Z}
----------------------------------------
and if I set this
SetA = {U, V, W, X, Y, Z}
SetB = {1, 2, 3}
Imap = {2, 2, 2}
I will get this?
result = {U, V, 3, 2, 1, W, X, Y, Z}
----------------------------------
And what if I set this?:
SetA = {U, V, W, X, Y, Z}
SetB = {1, 2, 3}
Imap = {2, 2}…
Added by Frane Zilic at 3:26pm on September 10, 2010
ctivity of vertices ordered sequentially, the order defining the direction of the normal, using 0 1 3 2 Causes an error this way. If it a quad face it seems odd to me that you would label the vertices in such an order, as an engineer, i have never seen it done as 0 1 3 2, it could be 3 2 1 0, 2 1 0 3 etc but going 0 1 3 2 is not acceptable, i will do a bit more reading on this.…
Added by Steve Lewis at 5:18pm on December 24, 2013