ee 3)
{5}
0 15
{6}
0 16
And I want to place points at every possible combination of these coordinates, treating Tree 1 as X coordinates, Tree 2 as Y coordinates, and Tree 3 as Z coordinates. Also, I would like the list of points to be a tree with paths corresponding to the coordinates. Wouldn't it be nice if I could plug these trees into a Point XYZ, with a new "branch cross reference" method, and get the following result?
{0:3:5}
0 {10.0, 13.0, 15.0}
{0:3:6}
0 {10.0, 13.0, 16.0}
{0:4:5}
0 {10.0, 14.0, 15.0}
{0:4:6}
0 {10.0, 14.0, 16.0}
{1:3:5}
0 {11.0, 13.0, 15.0}
{1:3:6}
0 {11.0, 13.0, 16.0}
{1:4:5}
0 {11.0, 14.0, 15.0}
{1:4:6}
0 {11.0, 14.0, 16.0}
{2:3:5}
0 {12.0, 13.0, 15.0}
{2:3:6}
0 {12.0, 13.0, 16.0}
{2:4:5}
0 {12.0, 14.0, 15.0}
{2:4:6}
0 {12.0, 14.0, 16.0}
In this form of cross referencing, every combination of individual branches from the different lists is used as separate input, and the output for each combination is put onto a branch in the result whose path is the concatenation of the input branch paths used.…
Added by Andy Edwards at 7:03pm on November 3, 2009
identical knots on end ends of the curve, so the knot vector will look something like:
0,0,0,1,2,2,2
So one at each end and one in the middle.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 3:47am on January 7, 2013
nch, xno items in one list)2 divide the list lenght value by the numer of items per branch needed3A generate a list with the series component: the step equal to the target numer of items per branch; the no of items equals the number of target branches
3B generate a list with the series component: the first number of the series equals to the number of items needed (-1 to account for the 0 index); the step size again equal to the target number of itmes per branch as 3A4 feed 3A & 3B to a domain component thus identifying the start -3A- and end -3B- of the domains by which the list will be subdivided5 use a subset component with the domains above thus creating 19 branches with lists having 5 items eachfor lists which are subdivided into branches when the target number of branches is not a multiple of the number of items contained in the list:6 identify if the target number of branches is a multiple of the list by using the modulus component fed by the list lenght -1- and the target number of branches7 identify last index in the 3B series with the item component (reversed to take the last value fed)8 add 6+7 above which dill define the start of the domain that will pick up the remanent items not accommodated in 59 add (+1) to 7 above to define the end of the domain that will pick up the the remanent items not accommodated in 510 feed 8 & 9 to a domain component11 include 10 as part of the subset in 5I'm now trying to understand the components mentioned by Michael...
sn
…
a follow up question... how do I wrap a list onto itself at a certain frequency?
i.e. I want the list {1;2;3;4;5;6;7;8;9}
to become {1,4,7; 2,6,8; 3,6,9} wrapped every 3rd item
Added by Joshua Jordan at 5:30pm on November 17, 2012
Hi,
I want to divide curve with distance between points so it will be like this:
1--2---3----4-----5------6-------7-----, ...
with values in range 1 to 50, must be simple but im stuck..
tnx
ggle A
7. Toggle A
8. Toggle 2
9. Toggle 3
10. Toggle A
11. Toggle A
12. Toggle 3
I was thinking to use somehow slider and animate option....but without luck
Any idea would be appreciated…
segments (ie. polylines)
2 = conic section (ie. arcs, circles, ellipses, parabolas, hyperbolas)
3 = standard freeform curve
5 = smoother freeform curve
The higher the degree, the less effect a single control-point has on the curve, but the further that weak effect reaches. Degree=5 curves are smoother, but it's also harder to add local details to it without adding a lot of control points. Rhino supports curves up to degree=11, but you almost never need more than 5.…