ceros. Parametrización, panelización y análisis en Grasshopper, así como el proceso de manufactura digital para maquinaria de corte Láser y CNC.
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byte-accuracy red, green, blue channels) = 27 bytes. More likely 28 bytes as colours are probably stored as 32-bit integers, allowing for an unused alpha channel.
28 * 800,000 equals roughly 22 megabytes, which is way down from 9 gigabytes. That's a 400 fold memory overhead, which is pretty hefty.
Grasshopper stores points as instances of classes, so on 64-bit systems it actually takes 64+64+3*8 = 152 bytes per point*, which adds up to 122MB, still way less than 9GB. It would be interesting to know where all the memory goes...
* Grasshopper points also store reference data, in case they come from the Rhino document. This data will not exist, but even so it will require 64-bits of storage.…
Added by David Rutten at 4:13pm on December 11, 2014
eone would know how to group each data position from each branch ... meaning, for instance, that I have a list of data with 3 branches, each branch containing 28 data instances (see attached image). Now, how would I get the (0;0)+(1;0)+(2;0) to be grouped in a new branch, as well as (0;1)+(1;1)+(2;1) and so on?
Also, I need for the setup to able to take in variables, so that the 3 branches might change to 6, and the 28 data instances might to change to anything between 10 and 100.
Thanks in advance,
/Claus
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pen Brep"; I didn't know it worked on flat surfaces. And I think it's only fair to include in your benchmark the considerable time 'SUnion' takes in this example: 21.9 seconds for 121 rings and likely much more with 400 or 1,000+ rings.
Then I noticed the pattern doesn't match. Checked the circles and they are the same. The distance between them, however, is different: 7 instead of 6. When I change that value to 6, the Python fails badly. All the holes and gaps are gone, which destroys the pattern:
I can't do the "two phase" approach on an 11 X 11 grid, but I can do 6 X 6 and 2 X 2 to get a 12 X 12 grid (40 'SUnion' operations) in 28 seconds total. That beats your benchmark of ~37 seconds for an 11 X 11 grid, if you include the 'SUnion' in your code.
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t item (see the image), is it possible to do this in another way (quickly) ?
Is it possible to divide that curve into 2 separate curves using a point that i've used for the division?
Thanks…
Added by luca.pavarin at 4:08pm on January 7, 2010
es, and these sum up to ~7100. I shall see if I can post a screenshot of everything.
I have no idea how to script that, sorry. But maybe could be a more optimized workflow - just feed one object B at a time, maybe that makes the math behind it more relaxed.
I should emphasize that it is not about 'fault' in the operation, and rather a very slow calculation - the difference does eventually get calculated with no errors whatsoever, the only problem is the unbelievably inefficient, or unproductive time it takes to do so. I think the problem could be the proportion of the objects, one very large, and one very small (28 / 6 / .1cm vs. .05cm), maybe that does something funky to the bounding box calls, I have no idea. And one other thing I suspect is the number of faces in the object, as I progress to let's say 700 cuts, I have then created 700*4 new faces in the object. I don't know if this is indeed something of concern. …
Added by DumDaDaDum at 7:03am on September 29, 2011
iders that control the total amount ( U&V) and the branches are formed by picking i.e the U and the points will then be the number on the V slider...
I realize that this question may have been asked before, but I cannot solve it with the path mapper and it seems so easy
M…
Mmm really nice post. Usually light electronic Trip Hop, or spanish Hip-Hop.
Portishead
Radiohead
Massive Attack
Propellerhead
SFDK
Juaninacka
Tote King
Best Regards ;)