ards to the number before the start number...
i.e. 9, 0, 1, 2, 3, 4, 5, 6, 7, 8
then it will need to repeat this pattern (continuing to count upwards) and the repeat number is based on a slider (for example 3 in the case illustrated below):
9, 0, 1, 2, 3, 4, 5, 6, 7, 8
19, 10, 11, 12, 13, 14, 15, 16, 17, 18,
29, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29,
if anyone has any ideas on how to create this series it would be great
M.…
What if for every set {0}, {1}, etc. i have 3 items that are always sorted in a different way. F.ex. {0} i need 1, in {1} i need 2 and in {3} i need 0?
hope anyone can help
Added by Stefano Rossi at 6:26am on November 21, 2013
number of divisions on that curve as in the defintion (i.e. by 4). The offset in the def is slightly different and should cull two or three more curves as in the lists that show my aim below.
Basically I want to look into each branch of the groups of points from each closed curve . Marking in a list whether it contains a one or a zero (0= outside 1 = coincidents).
{0;0}0. 21. 22. 23. 2 {0;1} 0. 01. 22. 03. 2 {0;2}0. 01. 02. 03. 0 {0;3}0. 21. 22. 23. 2 {0;4}0. 21. 22. 23. 2 {0;5}0. 21. 22. 23. 2 {0;6}0. 01. 22. 23. 1 {0;7}0. 21. 22. 03. 0 {0;8}0. 21. 22. 23. 2 {0;9}0. 21. 22. 23. 2 {0;10}0. 21. 22. 23. 2 {0;11}0. 21. 22. 23. 2 {0;12}0. 21. 22. 23. 2 {0;13}0. 01. 22. 23. 0 {0;14}0. 21. 22. 23. 2
I want to create a list from these points. That marks each curve that pokes out, in a cull pattern as such:
20022210222202
Using a 1 where there are co-incidents in the curve points and the boundary. A 2 for true (outside points) and a 0 for containment. So I might be able to use the 1 in future developments - however if a true false list is easiest I can live with that.
So could I use F(x) function? - to look for 0 or 1's in each bunch of points and thus list as such for a cull pattern? or will Path mapper help me here? Or can I rely on simply grafting and splitting??
I am usure of the neatest solution and would love to learn. Hope you can direct me.rgrds
J.…
Any chance of a % notation that allows for the selection of a modulo pattern of branchs/items i.e. {0;1}[%3] gets b,f,j,m,q,t,x
And {0;1}[%!3] gets you the flip side
Added by Danny Boyes at 12:42am on November 4, 2013
s points"
(the set didnt change from the picture above)
so param viewers show me (with 3 columns points)
1 list with 3 branches : {0;0}/{0;1}/{0;2}
an other with 4 branches : {0;0;0}/{0;0;1}/{0;0;2}/{0;0;3}
I tried to merge those 2 and then input the list in a path mapper but could'nt manage to make it work to get :
{0;0;0}/{0;0}/{0;0;1}/{0;1}/{0;0;2}/{0;2}/{0;0;3}
Am I using the right component?…
rsection part.
I plan to set up a unit plane and then use a separate function to reset it to the correct location etc.
Any ideas to why the code below might cause the crash?
Dim CornerPts(3) As On3dPoint
CornerPts(0) = New On3dPoint(0, 0, 0)
CornerPts(1) = New On3dPoint(1, 0, 0)
CornerPts(2) = New On3dPoint(1, 1, 0)
CornerPts(3) = New On3dPoint(0, 1, 0)
Dim CutPlane As New OnSurface = RhUtil.RhinoCreateSurfaceFromCorners(CornerPts(0), CornerPts(1), CornerPts(2), CornerPts(3))
print(cutplane.IsValid)…
3)
0:1 (N=3)
Is there a way to do this with the current tree utilities or is there a planned component such as the path mapper that has been previously discussed?
Best regards
Danny…
Added by Danny Boyes at 3:59am on November 13, 2009