First use a series component with start=1, step=42, count=3
Use the output to create a new series component with start=existingseries, step=1, count=11
vector * number
8. number * point
9. point * number
10. complex * complex
11. colour * colour
12. colour * number
13. number * colour
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David Rutten
david@mcneel.com
Seattle, WA…
Added by David Rutten at 10:39pm on November 12, 2010
en 3 of them, and one poolyline between two of them.
It would also be very nice if i could control it so that only the successive ones can be connected
so if {0:0:0} has 8 points and {0:0:1} has 8, as do {0:0:5} and {0:0:6} i would like to have this as two polylines, not one continoous that would in this case jump three branches (or curves that are shorter).
Does this make any sense?…
Added by Dusan Bosnjak at 2:08pm on September 28, 2009
} (N=11) {0;1} (N=11) {0;2}(N = 11) {0;3}(N = 11) {0;4}(N = 11)
2. I run the Points that are coming out from the Divide Curve Components through the Path Mapper components with this definition:
{A;B} (i) > {A} (i)
3. I run data coming out from Path Mapper component through:
a) Parameter Viewer component and the result is:
{0} N=11 (data with 1 branches)
b) Point > Panel and the result is:
collection of 11 point (N=11) which is the exactly the same as the collection of point belonging to {0;4} (N = 11).
So, here is the question:
why the collection of points coming out from the Path Mapper {A;B} (i) > {A} (i) component is the same as the collection of points belonging to the curve {0;4}(N = 11) ?
Anyway ... It 's the first time I ask a question here... so I would like to thank you for what you do with your work! Thank you! You are really great!…
would like to group the paths based on their item count (n) values resulting in a tree which should look something like this:
{0;0} (3)
{0;1} (2)
{0;2} (2)
{0;3} (1)
in other words, all paths with 2 items are under one path, all with 6 items in another, and so on.
I feel that the pathmapper should be able to do this very easily but cannot figure out what the expression should be... I have tried searching the forum but have not had much luck!
Any ideas? Thanks a ton!…
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
ll these 12500 points.
Group 1 would represent the point located at 0, 5, 10, 15, 20 etc.
Group 2 - 1, 6, 11, 16, 21 etc.
Group 3 - 2, 7, 12, 17, 22 etc.
Group 4 - 3, 8, 13, 18, 23 etc.
Group 5 - 4, 9, 14, 19, 24 etc.
I can create the pattern but the selection of points are all the points in row 0 and then all the points in row 5 and so on.
I would like the selection of points to start at the bottom left, and sequentially continue to the right and then continue on the 2nd row (left to right & bottom to top). i am hoping the pattern i am trying to achieve is more understood with the quick screen capture I uploaded.
the end goal is to be able to select all the points in the grid that are in each pattern.
Thanks in advance for any guidance with this. …
Added by Alyne Rankin at 6:53am on October 11, 2017
r itself is compiled against .NET 3·5.
I recommend you use Rhino4, .NET 3·5 and VS 2008 as a debugging platform.
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David Rutten
david@mcneel.com
Poprad, Slovakia…