behave like this:{a1;b2;c3;d1;e2;f3;g1;h2}
The ping pong matching would behave like this:
{a1;b2;c3;d2;e1;f2;g3;h2}
with a boolean option it could repeat the first and last like this:
{a1;b2;c3;d3;e2;f1;g1;h2}
If this already exist please let me know
Thanks
bye…
Added by Frane Zilic at 10:15pm on August 19, 2012
for waves, that is done with a 'phase shift', add 2*pi/4 radians (for a 90 deg shift) to your sin curve, could also be done using cos instead of sin for an inherent 90 deg shift
Added by mark zirinsky at 7:37pm on November 9, 2016
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
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 ()…