, CC... etc
once i find the right component to do this, wrapping the list back to (AA, BB) shouldn't be overly difficult, but i just can't seem to find the right component to start with.
Thanks,
Alex…
.01)
A = bb
The problem is that it doesn't matter which way I flip the input surface normal directions, the trim always returns the larger surface with a circular hole. Actually what I want is just the circular hole. The documentation implies that the orientation of the cylinder should affect which piece is returned.
Any suggestions?
Steven…
g these times itself). If it works on selection alone, it would probably implement faster.
Theoretically, does this mean the total solving time of the definition is the 'chain of components' that takes the longest time? In the picture above, it would be the chain consisting 'point-curve-divideDistance'?
Because that still adds up only to 97%, I am assuming the Point and Slider component start solving in parallel, and the two Divide components also start solving in parallel?…
nologically label them (there are currently 65 points and this is labelled as in the file i've attached). However, what i'm actually after is to reformat these points into an x and y style grid.(a1, a2, a3, a4, a5)(b1, b2, b3, b4, b5)(c1, c2, c3, c4, c5)(d1, d2, d3, d4, d5) etc.Any ideas/help how this can be made possible would be great.Thanks in advance…
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 ()…