see a fairly comprehensive list of initialization code enabled keywords in the 0.8.0010 features.
BTW slider=50 gives you integers and slider=50.00 gives you real numbers to 2dp
If so, are you looking for further shorthanded keywords?…
Added by Danny Boyes at 4:08pm on September 1, 2011
y need to buy 2 Mega boards, or some combination of Megas and Unos. Again, power will likely be the most important consideration as each of those servos (if they're all going to be driven at the same time) will draw some amount of current, so you need to add up your loads to determine your power requirements.
HTH,
Andy…
do i get them to be in one list...
Example:
List 1:
1
2
3
4
5
etc....
List 2:
10.1
11.1
34.5
40
50
etc...
Merged List
1, 10.1
2,11,1
3,34.5
4,40
5,50
etc... …
Added by Eric Galipo at 3:10pm on January 13, 2016
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…
ll to one end of the field.
Here's another, slightly more involved way to approach it:
1. create an even, random field of points
2. sort them in the direction of the gradient (decompose the points into xyz and sort by z value, for instance)
3. plug the sorted points into the jitter component and use a slider to control the amount of jitter
4. use "split list" to extract one half of the list (use list length/2 as the index)
5. adjust the jitter amount - at 0 you'll get a solid block of the top half of the random points, at 1 you'll get a random set of 50% of the points, but somewhere in between you'll get the appearance of a gradient.
Let me know if any of this is unclear... hope this gets you started.
Andrew…
the Options. For example, if we look at the default settings in this order:
Population: Number of iterations / generation 50 - Galapagos tries 50 slider positions each generation. When it finishes 50, it looks at the results and takes from the best results based on your fitness.
Initial Boost: Factor for the first generation 2. You want to ensure Galapagos sees as much of the solution space as possible in order to not miss any potential solutions. The first generation is multiplied by this factor. If Population is 50, the first generation will be 50x2 = 100 slider positions.
Maintain and Inbreeding deal with what you keep between Generations.
Max Stagnant: Number of generations to try AFTER finding a better solution 50. If Galapagos finds a great solution in Generation 2 (Gen 0 = 100 tries, Gen 1 = 50 tries, Gen 2 = 50 tries) it will go another 50 Generations (50x50 tries) before it stops to ensure it did not miss anything.
Your solution space consists of 11 options, which is much less than any of the other parameters are suggesting. Galapagos flails wildly in your case because you told it to. You told it to try 50x50(+50 for initial boost) number of times to find the best value.
Hence why I do not think this is the best option. You said it, this is not an optimization problem. If it is not an optimization problem, why use a genetic algorithm solver which is predominantly used for optimizing parameters?
I wouldn't necessarily want to see the definition, I'm more curious about the data. For example, can you send the data for 10 structural members and some load cases? (again, I could be entirely oversimplifying it).
In any case, I changed Max. Stagnant to 5, Population to 11. So Galapagos will stop (5x11)+11 tries AFTER the best solution is found. It found the solution pretty quickly.…
Added by Luis Fraguada at 6:07am on September 7, 2016
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...
--
David Rutten
david@mcneel.com
Poprad, Slovakia…