square units. Then you have an integral number of fragments on each side. This means that if all fragments need to have the same surface area, you can only have the following possibilities for side A:
1 fragment = 100 square units
2 fragments = 50 square units each
3 fragments = 33⅓ square units each
4 fragments = 25 square units each
5 fragments = 20 " "
6 fragments = 16⅔ " "
etc.
For side B, the numbers are mostly different
1 fragment = 300 unit²
2 fragments = 150 unit²
3 fragments = 100 unit²
4 fragments = 75 unit²
For side C they are different still. Unless you join fragments across on both sides of the edges of the box, I very much doubt you'll be able to pull this off.
The solution I attached will create fragments as identical as possible, but it's a very boring outcome...
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David Rutten
david@mcneel.com
Poprad, Slovakia…
roperties drop box menu (on the right) choose "Material".
3) Choose "Basic" from the radio buttons below
4) Check "Texture"'s checkbox
5) Define your texture "Map file". In your case, use "Carrots small.jpg" file.
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which I understand analyses only 2 octave bands (500 Hz and 2 kHz) instead of the 8 bands (for which the STI component requires background noise.)
Or have I misunderstood this metric and the three values mean something else?
For reference / a minimal definition I have the open office example (gha and 3dm attached) which includes 20 receivers and the sti results show three values for each (see the image).
Thanks in advance Roly…
ause if i increase the maximus or minimun radio it turn into diagonal lines. this lines are created between nearest 4 points of each one.
Maybe im doing something wrong.
Sorry about my bad english.
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hing not acceptable. To build up the final results I tried to summarize some steps and script them but I got stuck.
1) Voronoi subdivision
2) Voronoi vertices connected to the center
3) Tween curves between center and boundary
4) Control of U and V parameters
Point 4 is really important because it will allow me to control the strength of the springs in the two directions in kangaroo. Voronoi curve is going to be an anchor and the point is going to pull the mesh.
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had):
You can see that the 4 rings have different paths {0;0;0},...,{0;0;3}. That's why [Dif] component will not treat them as a "set of Breps"(leave your mouse pointer over the component to see what inputs it expects) but separately. So it will do 4 subtractions (the output will be 4 Breps).
But if you flatten this data tree into a single list you get this:
Now all 4 rings are in a single list and [Dif] component can treat them as a "set of Breps"
Notice also how the wires change (single line for one object, double line for one list, double dashed line for a data tree). Now the output of [Dif] is one Brep(all four rings subtracted from the rod).
Hope this is more clear now... Still, check the links for a better understanding of trees :)…
f honeybee in series. Once one ends, the next begins. But we are limited to 8 parallel processors for each unique iteration. So running 500 iterations is tough. I would love to take 100 computers and run 5 iterations each, preferably using 8 parallel processors per iteration. Like a render farm, but set up to compute unique gh produced iterations. Has anyone explored this?
Thanks for your help,
-Leland…
Tetrahedron: 24 Symmetries
Pyramid: 8 Symmetries
Design space = 24 X 8 = 192 permutations
So I decided to write a simple orientation script to iterate over all permutations. And this is the result. Below are some technical notes.
I used the vertices of the shapes for creating a 3 point plane, and used it for orientation.
I used compound transform to combine multiple steps of transformation.
The cross reference component is very handy, generating all the possible combinations without worrying too much about data tree.
The spatial relationship and the basic grammar A -> A + B and B -> A + B
The basic grammar and possible marker positions.
All results in 6 iteration steps
All results in 6 iteration steps (Top View)…