of lines, etc) but I can't see a way to add the text I need where I need it. If I could get each line for the print run to generate automatically, I can put the rest in manually, so just need something like:
... ; I would do the previous to this manuallyG1 X10 Y5 Z3
G1 X5 Y5 E5
G1 X5 Y15 E10
... ; I would do the rest manually
for a 5 mm line from [10, 5, 3] to [5, 5, 3], followed by a 10mm line from there to [5, 15, 3]. Any pointers greatly appreciated.
Ewan…
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
est of the best)
Crucial DDR4 2133 ECC (what else?)
4* WD RE 500 in Raid combo (not shown)
Some stupid 2.5'' HD thingy (avoid 2.5'' disks)
No SSD thingy
Corsair CPU cooler (Tequila replaced the OEM liquid: it works)
…
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 ()…
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
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