s 8, 4, 2, 10, 1, 3, 8, 4, 2, 0. But then for the end result to maintain all numbers above 5 but replace all numbers below with a defined number..Let's say zero. So then the list would read...8, 0, 0, 10, 0, 0, 8, 0, 0.…
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. "
except in data, the branches start at 0, so 'group 1' is branch 0
as for the order of your points, that depends on the input prior sorting...
yrs …
and says "repeat every Nth element", meaning say if N was 3, I would get A B C C D E F F G H. I would also like to be able to then insert a list of 3 elements in between each of the repeats. For example, I wanted to insert: 1. 1
2. 2
3. 3 into the list, I would get A B C 1 2 3 C D E F 1 2 3 F G H. It seems like an easy task, but I cannot for the life of me figure out how to do it, I must be making a very basic mistake. I am new to grasshopper, so any and all help I could get would be GREATLY appreciated!…
or each branch goes to col 1, element 2 to col 2 and so on. The issue I am having is how to deal with branches that have less than 5 elements. Notice how in the attachment the number 8 is at row 2 where I want it to be in row 3.…
Added by jon kontuly at 1:03pm on September 30, 2014
0;3} - 2 curves
{1;1} - 2 curves
{1;2} - 2 curves
{1;3} - 2 curves
{1;1} - 2 curves
{2;2} - 2 curves
{2;3}- 2 curves
And what I want.
{0} - 6 curves
{1} - 8 curves
{2} - 4 curves
I have tried some different stuff whit the path mapper tool, but I am not to skilled in using it. I imagine it can do the work for me?.
If anyone can help me out, I would be glad.
…
53 → 53 → 63 → 74 → 74 → 84 → 9
As you can see from the above list the connection sequence comes in waves of three, where each group of similar indices on the left is associated with a group of three incrementing indices on the right.
Some combination of Series components will probably generate this list, but it'll only work for the first ring, the second one will need a different connection pattern. It is perhaps better to just encode the integer pairs by hand. But then you cannot change your mind about the number of sides later.…
Added by David Rutten at 10:39am on October 21, 2015
ements between mid axes of elements ( being perpendicular to both). On the attached image, the links are the small elements connectiong nodes 4 to 5, 6 to 7 and 8 to 9. All nodes (1 to 9 including 1', 2', 3') are defined in Karamba as fixed supports but nodes 1,1', 2, 2', 3, and 3' have hinges added with the beam joint component. The freed rotations are shown on the figure.
I wondered if that was the correct way of defining such a structure in Karamba bearing in mind that nodes 1', 2' and3' are free nodes in the reality.
Thanks again for your help !
Yousef…