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.…
, a1200)}) Is there any way I can make this list into {a1, a2, a3, a4, -a5, -a6, -a7, -a8, a9, a10, a11, a12, -a13, ..... , a1200} ? ( 4 positive signs then 4 negative signs and so on) - alternating every nth in general.
or
2. Is there any way (workaround) to get negative angle value from 2 vectors? I know 'ANGLE (of 2 vectors)' component by itself doesn't work and I know why too. I have feeling that the reflex angle output might be useful but again, matter of list manipulation.
Any help would be greatly appreciated. Thanks in advance.
Hyo
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а рабочее зеркало и на любые возникшие вопросы ,всегда ответит круглосуточная поддержка.Самые крутые матчи и ставки с высокими коэффициентами.Не теряйте время!…
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 ()…
gone with the wind topic: since this is utterly Academic the main issue here is to oversimplify LBS (in real life: a collection of columns/beams/slabs/X members + tube frame rigid members (shafts/elevators/cats/dogs)). Reason is that if we use the real "solids" (turned into meshes) as the "node" pool for the hinges required ... only HAL 9000 could solve it in "real-time" (for instance an E5 Xeon 1630 v3 takes ... several minutes). And this is ... er ... challenging I must say. This is a typical case where "simplifying" means "stupidity" almost instantly.
Spam on:
where's my collection of "bend-a-truss-that-looks-like-a-tower" K1 demo defs? Is in this workstation or in another? (blame Alzheimer).
Spam off.
More soon.…
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.
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