closest point to the very first would be removed from the list, so the initial list reduces from 100 to 98. From the 98 i pick one and search the remaining 97 for the closest. From the remaining 96 i pick again one and search in the 95,...
(The product I want to result is:
having a number of random lines in 3D space, produced by an even number of points as discribed, this shall be the initial springs for a ("selfadjusting") tensegrity. Each one of these lines (later springs in kangaroo) get divided in three areas - that means four points. These four points again are the "attractor points" of neighbor springs, so the strut "knows" where to set the next elastic connection,...the rest I´ll have to figure out)
angelos…
thankssssss
.74-.80 may is ok not big differents .. but i have to control each group of triangles :)
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sorry the version is
rhino 4.9
gr 0.8.0004 version
thanks :)
Added by architecture at 6:26am on December 20, 2011
rence not set to an instance of an object. (line: 80)
Both .dlls appear to have been successfully imported.
Thanks in advance,
Charles
RStatSystem rs = ri as RStatSystem;
List<Line> slines = new List<Line>();
foreach(RStatBeam b0 in rs.Beams) { <------ line 80
if (b0.StiffnessMultiplier < 0.3) continue;
slines.Add(new Line(b0.P0.ToPoint3d(), b0.P1.ToPoint3d())); }
A = slines;
…
ded a circle and been able to draw the two lines and cull out the correct distances but do not know how to pull out those two original lines - the one from the starting point to the circle division + the circle division to the end point.
To recap - I'm looking for a way to have 2 lines = 80" when the direct path between two points is 60". I do not want the 2 new lines to be equal in length but variable lengths.
thanks!
_patrick
…
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Added by MichaelD0112 at 12:25am on April 10, 2023