ersect (2, 3, 4, 5, 6) with the line and the ones which do not intersect (0, 1, 7). Intersect is done! But how to get the non intersecting vectors (0, 1, 7)?
So I e. g. could deselect vectors 2, 3, 4, 5, 6 so I would display/use only vectors 0, 1, 7 and the bounced ones.
Appreciate your help!
Rudi…
Like you've done. Use List Item and input an integer which represents the curve you want to select. Your slider will need to be able to handle integers 0, 1, 2, 3, 4, and 5 since you have 6 curves.
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
maybe not.
Ex: if i have 3 rectangles sent in at {0;0} and i make a component to return the 2nd edge and 4th edge of each of them, i need to have a new branch level so that i will have 2 edges in each of three new branches
{0;0;0}(N=2)
{0;0;1}(N=2)
{0;0;2}(N=2)
if i simplified back into {0;0} then i would have N=6 but the data wouldn't be divided as logically into meaningful sets.
(its getting late - i hope this makes sense)... good luck and good night...…
Tree:
{0;0;0} N = 2
{0;0;1} N = 1
{0;0;2} N = 3
{0;1;0} N = 5
{0;1;1} N = 8
{0;1;2} N = 10
If we apply the aforementioned mapping to this tree, we'll end up with the following result:
{0;0} N = 6
{0;1} N = 23
Basically {0;0;0}, {0;0;1} and {0;0;2} are combined into a single path {0;0} as we disregard the third index because "C" is no longer present in the target mapping.
Because we only use the Mapper to modify paths, we do not lose any data items, though we might lose some of the paths.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 1:03pm on August 25, 2010
points 0, X-1, (2*x)-1, (3*X)-1, (4*X)-1, (5*X)-1 and then
1, X, (2*x), (3*X), (4*X), (5*X)
2, X+1, (2*x)+1, (3*X)+1, (4*X)+1, (5*X)+1
and so on till
5, X+4, (2*x)+4, (3*X)+4, (4*X)+4, (5*X)+4
How can I do this best?
Thanks,
Niels…
Could someone please tell me if it is possible to input a non repetitive relative item pattern as shown below. I am interested in something like this
0 {0;0}
1 {1;0}
2 {1;6+i}
3 {0;6+i}
Thanks, Dan
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.
…