ital; aproveche esta oportunidad y vea la demostración en vivo: Lugar: Escuela Digital - Mexico D.F. Hora: 7:00 - 9:00 pm Con el apoyo de: ESCUELA DIGITAL - Metzli Valle metzli@escueladigital.com.mx…
nch, xno items in one list)2 divide the list lenght value by the numer of items per branch needed3A generate a list with the series component: the step equal to the target numer of items per branch; the no of items equals the number of target branches
3B generate a list with the series component: the first number of the series equals to the number of items needed (-1 to account for the 0 index); the step size again equal to the target number of itmes per branch as 3A4 feed 3A & 3B to a domain component thus identifying the start -3A- and end -3B- of the domains by which the list will be subdivided5 use a subset component with the domains above thus creating 19 branches with lists having 5 items eachfor lists which are subdivided into branches when the target number of branches is not a multiple of the number of items contained in the list:6 identify if the target number of branches is a multiple of the list by using the modulus component fed by the list lenght -1- and the target number of branches7 identify last index in the 3B series with the item component (reversed to take the last value fed)8 add 6+7 above which dill define the start of the domain that will pick up the remanent items not accommodated in 59 add (+1) to 7 above to define the end of the domain that will pick up the the remanent items not accommodated in 510 feed 8 & 9 to a domain component11 include 10 as part of the subset in 5I'm now trying to understand the components mentioned by Michael...
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Loop'. The fun part of the slower version is that you can see what it's doing while it's running. 'Fast Loop' gives no indication that it's working, so you want to test it with small numbers and be sure it's coded properly before bumping the iteration count up.
The GH profiler running the slow version showed between 1 and 1.5 seconds per loop, but the reality was more like ~10 seconds per loop toward the end of an 11 X 11 grid, or ~20 minutes total. It's easier to be patient because you know it's working.
The 'Fast Loop' finished the same grid in 1.6 minutes! An impressive improvement. I've been running it on a 30 X 30 grid (900 points) for ~23 minutes so far and see nothing yet. Not the ~12 minutes I had hoped for... Now 36 minutes on this loop for 900 points... hope it's not stuck. Not fast! Later - DONE!! Profiler says 59 minutes for 900 points but it was more like an hour and twenty minutes total. It succeeded, I have a single 'Closed Brep' from 900 extruded rings, baked to Rhino.
Another strategy to explore would be doing 'SUnion' on a smaller grid using the Anemone loop, then replicate it by moving it as needed to form a larger grid; then run the copies through another 'SUnion' loop. I went ahead and implemented that while waiting. It works and is fast! Started with 3 X 3 and ran the result again as 5 X 5 (9 X 25 = 225 total) in barely ~70 seconds!? Trying 36 X 36 now... 1,296 points appears to have succeeded in less than ten minutes! Though it seems to take quite awhile after the loop ends before control is restored to GH/Rhino. I'll let you do your own experiments and benchmarks.
I encapsulated the loop in a cluster called 'suLoop' (blue groups).
Internal of 'suLoop' cluster:
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Added by Joseph Oster at 11:14pm on March 22, 2017
e range of materials, including heavy-duty fabrics. They offer a range of embroidery machines that cater to different needs and budgets.best embroidery machine for custom designs…
nique name like in your example {0;0},{0;1},{0;2},{0;3} where each path contains 11 elements(curves,points,values,etc)
so if you want to restructure the structure first right click on path mapper and take create null mapping which reads your current path structure into the component
after that you can additional write on the left side to the existing path (i) as variable for the elements and then on the right side type in the structure you want to achieve
maybe {A;i}(B) so after that you have 11 paths where each have 3 elements
hope that helps…
each circle's border, let us say 1.0
3) So, the curve will end up with 5 points, in each point will have a circle, each circle will have a different Radius, but the distance in between the borders of each circle is always the same = 1.0 in this case.
4) The end result list here would be like this to evaluate a curve with these values and find the points on the curve:
List = 1, 5, 11, 19 etc If I use these values to eval a line, I will get the perfect points where I can draw the circles.
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e now contains (40x3x11=)1320 points with a branch structure of {0-39,0-2} i.e
{0;0}
{0;1}
{0;2}
{1;0}
...
{39;2}
with each branch containing 11 points.
I now want to create lines from the points on line {x;0} to {x;1}, {x;1} to {x;2}, and {x;2} to {x;0}, which will give me a triangular grid on each triangle. How do i do this? I think I need to split the tree 40 times to give me 40 single layered trees but I imagine there's a cleverer way of doing this!
This feel like it should be simple, but I;m having trouble working with the double-layered branch structure. Any help would be greatly appreciated!
Thanks,
Matt…
is shorthand for [0 to 8].
> 10 Any number larger than X. This notation is shorthand for [11 to infinity].
>= 5 Any number larger than or equal to X. This notation is shorthand for [5 to infinity].
--
David Rutten
david@mcneel.com
Seattle, WA…
Added by David Rutten at 9:27pm on November 3, 2013
e
7. True
8. True <-- this one
9. True
10. False
11. True
12. False
13. True
14. True <-- this one
15. True
16. False
17. True
18. False
19. True
20. True <-- this one
21. True
22. False
23. True
24. False
25. True
26. True <-- this one
27. True
28. False
29. True
30. False
31. True
32. True <-- this one
33. True
Any idea how I can solve this?
Thanks!…