omponent that increases in the x-axis (example below).
A1 A2 A3 A4 A5 etc...B1 B2 B3 B4 B5 etc...C1 C2 C3 C4 C5 etc...D1 D2 D3 D4 D5 etc...
This is as far as I've gotten:
I have collected my points on the grid into a "List Length" component and input that into a "Series" which input into a "Function" with the expression Format("A{0}",x). The result labeling resembles the example below.
A1 A2 A3 A4 A5
A6 A7 A8 A9 A10
A11 A12 A13 A14 A15 etc...
Any help is appreciated.
Thank you in advance.…
structures. I think you can get นิสัย and learn more about the business. This architectural dance blurs the line between art and engineering. Because complex algorithms form a space that fosters creativity and collaboration. which is an outstanding platform for future business ideas.…
Added by MichaelD0112 at 1:34am on August 14, 2023
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.…
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…
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 ()…