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.
lues. What I want to do is combine them so that the structure would be something like:
{4;0}
{4;1}
{4;2}
{4;3}
{5;0}
{5;1}
{5;2}
{5;3}
I tried the method here, but it didn't give me what I wanted, it was just tacking the new values onto the end, and not maintaining their paths. Any help would be appreciated. Thanks!…
Added by Dennis Goff at 8:13am on February 10, 2016
That is correct. In reality there are 3 major versions of .NET 1, 2, and 4 (different versions of mscorlib.dll)
Rhino 4 uses .NET 2 which includes 3 and 3.5.
Rhino 5 uses .NET 4
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
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…
ches it with the first branch in Tree B (and then the first branch in Tree C if more than two trees are involved).
I'm planning to add better branch matching logic, but I'm not going to touch it until I have a good idea about what's needed and how it can be accomplished without breaking existing files.
So, the branch "address" is only used to sort the branches in a single tree. Thus, a tree with the following branches is always sorted in the exact same way:
{0;0}
{0;1}
{0;2}
{0;3;0}
{0;3;1}
{1;6}
If you have another tree with different branches:
{0}
{1}
{2}
{3}
{4}
{5}
Then the matching will be:
{0;0} -> {0}
{0;1} -> {1}
{0;2} -> {2}
{0;3;0} -> {3}
{0;3;1} -> {4}
{1;6} -> {5}
As long as people adhere to your advice: "it is best for the addresses of each tree branch to be in the same format", there will be no problem. But it is at the moment extremely difficult to perform complex matchings.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 9:25am on August 11, 2010
3} N=2
....
{527} N=4
What I want to do is figure out the same the item in group A and retrieve their branch index so that I can get the corresponding item in group A. Is there any way to make it?…
a seed, and instead creating a pattern where each color has a seed/control slider for each row? For example, row 1: brown 2, tan 6, yellow 7, purple 3, repeat. row 2: brown 6, tan 1, yellow 4, purple 10, repeat. row 3: yellow 5, purple 1, brown 3, tan 10, repeat. row 4: purple 2, brown 7, tan 3, yellow 4, repeat. Then repeat that sequence up the wall? For each color, the number in the sequence should be adjustable.
Thank you again for your help!…