ok, thanks, Isaw on that page before and from the begining I had a problem, because I tough these points which are like this one 3√3 / 7, the number 7 divide all the equation.
Thaks.
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
13 5 15 6 17 7 ... …
But it seems I have not been clear:
the input values (3,5,7,9,11,13) should be presented in the way of "number slider" parameter. So by increasing the value of the slider from minimum 3, to 5, 7, 9, 13, 15, 17... I should get an output numbers of 0,1,3,4,5,6,7...
So I if choose value 3 on the input "number slider", on the other end, I want 0 as an output.
Or if I choose value 5 on the input "number slider", I want 1 as an output, and so on.
Is this possible?…
shift. I realize I can use 'replace branch' but I do not have an available mask to utilize. I have simplified the problem to its simplest form so my question is understandable, however, the tree I am trying perform this operation on is a much larger 3 digit path address.
{1;3;2}
{2;3;4}
{3;5;4}
{4;3;7}
Change the above list to the list below.
{0;3;2}
{1;3;4}
{2;5:4}
{3;3;7}
I wish for a more robust arsenal of branch manipulation components. Most of the things I need to do are possible with the existing components, however, many operations take several components to perform even simple manipulations. Since branch/path manipulation is so integral to using GH successfully, it seems the GH community would be well served by enhancing the available path manipulation components.
Thanks,
Stan
…
le data is not ( data is give on the hour: 7:00, 8:00, 9:00 ...) and DAYSIM seems that it dose not calculate external horizontal diffuse illuminance.
Thanks
…
{4}-0;3
{5}-6;7
{6}-5;7
{7}-5;6
Here it can be shown that there are two subgraphs containing 0,1,2,3,4 and 5,6,7. How can I use spiderweb (either using scripting or the components) to give me this result when I have many more vertices??
Thanks,
Sam…