he sunPath component works. For example if you want to simulate the hours from 8 to 16 it means you want 8 hours from 8 to 9, from 9 to 10,.... from 15 to 16 (8 hours duration period) so you get from the sunPath component (using default timeStep 1) the 9 sun position/vectors 8 9 10 11 12 13 14 15 16 (in the image the yellow suns). The things is that if you ask for a smaller timeStep for example 3 = 20 mins then the additional sun position (in the image the orange suns) are added also after the time limit of h16 so probably when you don't want/need. I understand that when you input a time period there is the ambiguity if the hours are the just 9 (the 9 inputs) or the 8 hours included between pairs of hours, but I would make in a way that it is possible to chose if the extra timeStep after the last hour are added or not. Thank you for your comments.
…
etc.
Group 2 - 1, 6, 11, 16, 21 etc.
Group 3 - 2, 7, 12, 17, 22 etc.
Group 4 - 3, 8, 13, 18, 23 etc.
Group 5 - 4, 9, 14, 19, 24 etc. "
except in data, the branches start at 0, so 'group 1' is branch 0
as for the order of your points, that depends on the input prior sorting...
yrs …
un similar models much faster. I'm just trying to understand if Pachyderm should provide comparable performance to commercial products to find out if I'm doing something wrong.…
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
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!…
ee 3)
{5}
0 15
{6}
0 16
And I want to place points at every possible combination of these coordinates, treating Tree 1 as X coordinates, Tree 2 as Y coordinates, and Tree 3 as Z coordinates. Also, I would like the list of points to be a tree with paths corresponding to the coordinates. Wouldn't it be nice if I could plug these trees into a Point XYZ, with a new "branch cross reference" method, and get the following result?
{0:3:5}
0 {10.0, 13.0, 15.0}
{0:3:6}
0 {10.0, 13.0, 16.0}
{0:4:5}
0 {10.0, 14.0, 15.0}
{0:4:6}
0 {10.0, 14.0, 16.0}
{1:3:5}
0 {11.0, 13.0, 15.0}
{1:3:6}
0 {11.0, 13.0, 16.0}
{1:4:5}
0 {11.0, 14.0, 15.0}
{1:4:6}
0 {11.0, 14.0, 16.0}
{2:3:5}
0 {12.0, 13.0, 15.0}
{2:3:6}
0 {12.0, 13.0, 16.0}
{2:4:5}
0 {12.0, 14.0, 15.0}
{2:4:6}
0 {12.0, 14.0, 16.0}
In this form of cross referencing, every combination of individual branches from the different lists is used as separate input, and the output for each combination is put onto a branch in the result whose path is the concatenation of the input branch paths used.…
Added by Andy Edwards at 7:03pm on November 3, 2009