and Ronnie of StudioMode and David Fano of DesignReform will also be attending.
RSVP has been closed on this event. Space is limited to 50 people. Please attend if you do RSVP.
Agenda -
12:00-1:00 Arrival, informal discussion
1:00 - 1:15 Introductions
1:15 - 2:00 Project presentation 1 (30 minutes + 15 min QA) - David Lee - Clemson - 3D pattern environments using volumetric proxies.
2:00 - 2:45 Project Presentation 2 (30 minutes + 15 min QA) - P. Casey Mahon - Organic Abstractions (30 minutes + 15 min QA)
2:45 - 3:45 David Rutten - New work in GH (30 min QA)
3:45 - 4:30 Sameer Kumar AIA - KPF - Project presentation 3 (30 minutes + 15 min QA)
4:30 - 5:15 Chris Wilkins - Clemson - Urban Renewal and parametric urban development studies in Grasshopper.
5:15 - 6:00 David Rutten - Scripting in GH (15 min QA)
After 6:00 conversations may move down the street for more discussion.
If you would like to present your project at the Cloud please email: scottd@mcneel.com…
nts me this:
[[0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7]]
this is what I wanted but how to convert this to tree in grasshopper?
In grasshopper I just get:
8x IronPython.Runtime.List…
ep is to understan the logics of what you want to do, in your case, build 4 point surfaces (u also need to know the right direction to build the surfaces). Then you can write an hipotetic list (by hand in a paper) of what you want. In your case the list was (0, 1, 3, 2) (2, 3, 5, 4) (4, 5, 7, 6), etc... if you can imagine building 2 lists, each one with the sequences (0, 2, 4, 6, etcc) and (1, 3, 5, 7, etc..) then you can manage with shift and graft to finally have four lists. A( 0 1 2 3 ...) B (1 3 5 etc..) C(3 5 7 etc..) D (2 4 6 etc..). And to achieve the 2 first lists, you need to get the odd and the pair numbers. The cull pattern does that amazingy well. With a pattern True-False you get de pair numbers, and with the False-True pattern you get de odd numbers.
Hope it was clear enough…
Added by Pep Tornabell at 5:32am on November 19, 2009
branches in each A's list of B's, or remove its ends etcso that if I want to remove the last B in every A{0;1},{0;2},{0;3},{0;4},{0;5},{0;6}{1;1},{1;2},{1;3},{1;4}{2;1},{2;2},{2;3},{2;4},{2;5}would become{0;1},{0;2},{0;3},{0;4},{0;5}
{1;1},{1;2},{1;3}
{2;1},{2;2},{2;3},{2;4}I guess the question is do I need to figure out the cull pattern- each B may have different lengths...…
ards to the number before the start number...
i.e. 9, 0, 1, 2, 3, 4, 5, 6, 7, 8
then it will need to repeat this pattern (continuing to count upwards) and the repeat number is based on a slider (for example 3 in the case illustrated below):
9, 0, 1, 2, 3, 4, 5, 6, 7, 8
19, 10, 11, 12, 13, 14, 15, 16, 17, 18,
29, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29,
if anyone has any ideas on how to create this series it would be great
M.…