xtract A1, A2, B1, B2 as one set, A2, A3, B2, B3 as the second set, A3, A4, B3, B4...etc. as the third set and so on. How can I get about doing this?
Any help would be much appreciated!
Thanks,
Ben…
سلام آقای جان بیلی کاوه اشکوه هستم اپراتور سی ان سی از ایران وطراح دکوراسیون داخلی اگر ممکن هستش فایل اصلی میزcosinosa baram mail konid adressmail man hast
(kavehoshkooh2290@gmail.com)بسیار متشکرم
Added by kavehoshkooh at 3:25pm on September 11, 2015
some i7 if ECC memory sounds an oddity to you). If the model is big you'll need a decent Kepler Quadro as well ... say a K4200 (I hate game cards including Titan).
http://www.tsplines.com/
Alternatively use a top dog subdivision app (Modo eats them all for breakfast) but that works with DX and that brings us back to game cards.…
nt B2[i] so B1[i]<=0 means no new connections allowed for point i ,so point i is deleted from B1, B2 updated accordingly.
Initialization:
B1: max number of connections x number of points
B2: all the points
B3: nothing (well null or something, need to create the branch)
Algo:
Get first point in B2, get his allowed number of connections N in B1, find N closest points in B2, create lines in B3, update B2 accordingly. Erase points with max connections (including the first point)
Next
Stop when no points available
At end of loop, B3 stores the created lines.
…
ng long in the x axis and three in the y and they don't all intersect each other... I wrote a script to Boolean difference them but its not working like i want it to . I included a rhino result that id like to achieve in the file. THX -ethan
heres the script:
import rhinoscriptsyntax as rs
b1 = []for i in range(b1L): b1.append (x)print b1bb= len(b1)print bbb2 = []for j in range(b2L): b2.append (y)print b2bc = len(b2)print bc
def bool ():....for i in range (bb):........for j in range(bc):............a = rs.BooleanDifference( b1,b2, False).....return (a) a = bool ()…