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 ()…
CA, DA, DC)Two of those diagonal lengths are obviously redundant but they allow you to simply shift the array to get at different rotational permutations. This makes the search for the nearest mean a bit more straightforward since, in the context of panel clustering, you'd need to consider all rotational permutations of each one.…
Added by David Reeves at 5:26am on November 9, 2014
tecture. Hochbau | University of Innsbruck . A simple random, but at the same time organised growth routine. 5 iterations for this image.
Link to the course here:
http://www.exparch.at/index.php?option=com_content&task=view&id=1054&Itemid=87
View full size...as per Pieter's suggestion…
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…