Search Results - 湖北福彩30选5玩法『9TBH·COM』158计划网时时彩20192023年3月19日6时0分33秒.H5c2a3.jeusyfvdq

Comment on: Topic 'Can I sort curves in branches by the X value of the start point?': Mmmm, for example {5;19} and {5;20} are not sorted How can I change the script so that the it starts with {0;0} instead of {1;0}?; Added by Bas Goris at 7:22am on February 17, 2017

Event: Grasshopper Cloud Event - NY - FULL: and Ronnie of StudioMode and David Fano of DesignReform will also be attending. RSVP has been closed on this event. Space is limited to 50 people. Please attend if you do RSVP. Agenda - 12:00-1:00 Arrival, informal discussion 1:00 - 1:15 Introductions 1:15 - 2:00 Project presentation 1 (30 minutes + 15 min QA) - David Lee - Clemson - 3D pattern environments using volumetric proxies. 2:00 - 2:45 Project Presentation 2 (30 minutes + 15 min QA) - P. Casey Mahon - Organic Abstractions (30 minutes + 15 min QA) 2:45 - 3:45 David Rutten - New work in GH (30 min QA) 3:45 - 4:30 Sameer Kumar AIA - KPF - Project presentation 3 (30 minutes + 15 min QA) 4:30 - 5:15 Chris Wilkins - Clemson - Urban Renewal and parametric urban development studies in Grasshopper. 5:15 - 6:00 David Rutten - Scripting in GH (15 min QA) After 6:00 conversations may move down the street for more discussion. If you would like to present your project at the Cloud please email: scottd@mcneel.com…; Added by Scott Davidson at 4:24pm on September 22, 2009

Comment on: Topic 'Combinations of length k from set m': indexes later and the k value is just 5. The goal is to output all combinations of length 5 from the list of integers 0 to 19. I placed the list in brackets in order to make it iterable (I don't really know what this means.) Somehow I need to get the combinations out of the itertools.combination output.…; Added by Jackson Bewley at 12:22pm on April 4, 2013

Comment on: Topic 'how to rotate every nth item a gradual specific amount of degrees?': , 75, 90.. etc}, then use a Duplicate component to turn it into {0, 0, 0, 0, 15, 15, 15, 15, 30, 30, 30, 30, 45, 45, 45, .... etc}. -- David Rutten david@mcneel.com Poprad, Slovakia…; Added by David Rutten at 9:41pm on March 7, 2010

Comment on: Topic 'recursion in c#': n; n = (int) (Math.Log10((double) nfaces / 20.0) / Math.Log10(4.0)); double t = (1.0 + Math.Sqrt(5.0)) / 2.0; double c = Math.Sqrt(1 + (1.0 + Math.Sqrt(5.0)) * (1.0 + Math.Sqrt(5.0)) / 4.0); //Icosaedron Rhino.Geometry.Mesh mesh = new Rhino.Geometry.Mesh(); mesh.Vertices.Add(-1 / c, t / c, 0); mesh.Vertices.Add(1 / c, t / c, 0); mesh.Vertices.Add(-1 / c, -t / c, 0); mesh.Vertices.Add(1 / c, -t / c, 0); mesh.Vertices.Add(0, -1 / c, t / c); mesh.Vertices.Add(0, 1 / c, t / c); mesh.Vertices.Add(0, -1 / c, -t / c); mesh.Vertices.Add(0, 1 / c, -t / c); mesh.Vertices.Add(t / c, 0, -1 / c); mesh.Vertices.Add(t / c, 0, 1 / c); mesh.Vertices.Add(-t / c, 0, -1 / c); mesh.Vertices.Add(-t / c, 0, 1 / c); mesh.Faces.AddFace(0, 11, 5); mesh.Faces.AddFace(0, 5, 1); mesh.Faces.AddFace(0, 1, 7); mesh.Faces.AddFace(0, 7, 10); mesh.Faces.AddFace(0, 10, 11); mesh.Faces.AddFace(1, 5, 9); mesh.Faces.AddFace(5, 11, 4); mesh.Faces.AddFace(11, 10, 2); mesh.Faces.AddFace(10, 7, 6); mesh.Faces.AddFace(7, 1, 8); // 5 faces around point 3 mesh.Faces.AddFace(3, 9, 4); mesh.Faces.AddFace(3, 4, 2); mesh.Faces.AddFace(3, 2, 6); mesh.Faces.AddFace(3, 6, 8); mesh.Faces.AddFace(3, 8, 9); // 5 adjacent faces mesh.Faces.AddFace(4, 9, 5); mesh.Faces.AddFace(2, 4, 11); mesh.Faces.AddFace(6, 2, 10); mesh.Faces.AddFace(8, 6, 7); mesh.Faces.AddFace(9, 8, 1); for (int i = 0; i < n; i++) { Rhino.Geometry.Mesh mesh_rec = new Rhino.Geometry.Mesh(); recursiveSubdivision(mesh, ref mesh_rec); mesh = mesh_rec; } mesh.Vertices.CombineIdentical(true, true); mesh.Normals.ComputeNormals(); mesh.Compact(); A = mesh; } // <Custom additional code> public void recursiveSubdivision(Mesh mesh, ref Mesh mesh_rec) { int ind = 0; for (int i = 0; i < mesh.Faces.Count; i++) { Point3f A = new Point3f(mesh.Vertices[mesh.Faces[i].A].X, mesh.Vertices[mesh.Faces[i].A].Y, mesh.Vertices[mesh.Faces[i].A].Z); Point3f B = new Point3f(mesh.Vertices[mesh.Faces[i].B].X, mesh.Vertices[mesh.Faces[i].B].Y, mesh.Vertices[mesh.Faces[i].B].Z); Point3f C = new Point3f(mesh.Vertices[mesh.Faces[i].C].X, mesh.Vertices[mesh.Faces[i].C].Y, mesh.Vertices[mesh.Faces[i].C].Z); ind = mesh_rec.Vertices.Count; mesh_rec.Vertices.Add(A);//0 mesh_rec.Vertices.Add(middle(A, B)); mesh_rec.Vertices.Add(B); mesh_rec.Vertices.Add(middle(B, C)); mesh_rec.Vertices.Add(C); mesh_rec.Vertices.Add(middle(C, A)); mesh_rec.Faces.AddFace(ind + 0, ind + 1, ind + 5); mesh_rec.Faces.AddFace(ind + 1, ind + 3, ind + 5); mesh_rec.Faces.AddFace(ind + 1, ind + 2, ind + 3); mesh_rec.Faces.AddFace(ind + 5, ind + 3, ind + 4); } } public Point3f middle(Point3f A, Point3f B) { Point3f AB = new Point3f(); float length; AB.X = (A.X + B.X) / 2; AB.Y = (A.Y + B.Y) / 2; AB.Z = (A.Z + B.Z) / 2; length = (float) Math.Sqrt(AB.X * AB.X + AB.Y * AB.Y + AB.Z * AB.Z); AB.X = AB.X / length; AB.Y = AB.Y / length; AB.Z = AB.Z / length; return AB; }…; Added by Laurent DELRIEU at 12:58pm on August 6, 2016

Comment on: Topic 'Partition Data': ould say partition branches makes more sense than partition data as a name. …; Added by Michael Pryor at 9:48pm on November 4, 2013

Topic: Tree Branch Replacement: 0;5} n8 I can easily retrieve a specific tree branch, then modify the data, lets say to each have 6 items in their list instead of 8. the end resut i want is: {0;0} n6 {0;1} n8 {0;2} n8 {0;3} n8 {0;4} n8 {0;5} n6 But i cant seem to get the "replace branch" componet to work In reality i have a tree that consists of 108 branches, so exploding the tree and entwining (which is my normal work around) is impractical any help is appreciated!…; Added by Ryan Grande at 2:09pm on April 18, 2013

Topic: Search in lists: ant to find all paths where exactly two items are the same like in branch {1}. How can I solve this in VB? Is there an easy "search in list" class or do I have to iterate over all items by myself ?? Regards Alex…; Added by Alexander Greil at 10:01am on April 21, 2013

Comment on: Topic 'Specific results(outputs) for specific inputs?': 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?…; Added by djordje at 6:37pm on August 5, 2011

Topic: how can I use, "MeshVertexColorList.SetColor Method (MeshFace, Color)" ?: try.Mesh mesh = new Rhino.Geometry.Mesh(); mesh.Vertices.Add(0.0, 0.0, 1.0); //0 mesh.Vertices.Add(1.0, 0.0, 1.0); //1 mesh.Vertices.Add(2.0, 0.0, 1.0); //2 mesh.Vertices.Add(3.0, 0.0, 0.0); //3 mesh.Vertices.Add(0.0, 1.0, 1.0); //4 mesh.Vertices.Add(1.0, 1.0, 2.0); //5 mesh.Vertices.Add(2.0, 1.0, 1.0); //6 mesh.Vertices.Add(3.0, 1.0, 0.0); //7 mesh.Vertices.Add(0.0, 2.0, 1.0); //8 mesh.Vertices.Add(1.0, 2.0, 1.0); //9 mesh.Vertices.Add(2.0, 2.0, 1.0); //10 mesh.Vertices.Add(3.0, 2.0, 1.0); //11 mesh.Faces.AddFace(0, 1, 5, 4); mesh.Faces.AddFace(1, 2, 6, 5); mesh.Faces.AddFace(2, 3, 7, 6); mesh.Faces.AddFace(4, 5, 9, 8); mesh.Faces.AddFace(5, 6, 10, 9); mesh.Faces.AddFace(6, 7, 11, 10); mesh.Compact(); mesh.VertexColors.SetColor(mesh.Faces[0], Color.Black); A = mesh; } ------------------------------------------------------------------- but it seems, it doesn't work. I can't see any change of color. what did I wrong? Thanks!…; Added by y.KWON at 3:01pm on September 18, 2012