was stopped and the solver returned 3 options with the highest fitness score.
(Details are here: http://dimak1999.blogspot.com/2011/02/i-have-not-had-chance-to-use-galapagos.html)…
This must be a bug because its true for dividing by all odd numbers:
(i-1)/3
(i-2)/5
(i-3)/7
(i-4)/9
(i-5)/11
....
(i-n)/2n+1
And you can't make it work for even numbers
Added by Danny Boyes at 5:06pm on January 13, 2010
ggle A
7. Toggle A
8. Toggle 2
9. Toggle 3
10. Toggle A
11. Toggle A
12. Toggle 3
I was thinking to use somehow slider and animate option....but without luck
Any idea would be appreciated…
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…
Thank you Marios,
but I want to put boxes on all the plane: if I divide U domain in 4 parts and V domain in 3 parts I'll have 12 boxes with 12 different heights in W
How can be done?
regards
Maurizio
thank you very much for your reply pieter,I'm wondering if there were more than 3 expressions how will it be?I mean if it was X, X+12, X+24, X+36 and more...and that depended on a slider.
ve a Vertex [V] connected to four other Vertexs [N1-N4].
Each of the has a Value:
V ... 1
N1 ... 5
N2 ... 3
N3 ... 8
N4 ... 11
The Average Filter would set the Value of [V] to
(1+5+3+8+11)/5 = 5,6
The Median Filter would Sort Values and pick the middle one
1,3, [5], 8, 11
Hope that helped...…
this, you'll have no horizontal force at the roller, but you will have it at the pinned support. If you wouldn't, then the structure will be displaced.
Usually, in 2 dimensional structures, if you want to know if an articulated structure is isostatic (as opposed to hyperstatic, which is what you have right now) is to use the following formula:
b+c-2·n=0;
b being the number of bars, c the number of constraints you have and n the number of nodes. In your case: b=19, c=3 (displacements constrained in X, Z at your pinned support and only constrained in Z at your roller support) and n=11, so: 19+3-2·11=0.
I recommend you to download the app SW Truss, as it's very useful to check your results instantly.…