op van maximaal 1000 iteraties
3) Offset de polyline en maak een nieuwe Brep van alle curven
4) Bepaal de Area-centroid van de Brep
5) Bepaal het verschil tussen de huidige centroid en de gezochte centroid
6) Als dit verschil minder is dan 1e-12 breken we af
7) Vermenigvuldig het verschil met vier en pas de polyline aan
8) herhaal (3 - 7)
Is dit min of meer wat je wilde? Het lijkt dat er ~50 iteraties nodig zijn voor een antwoord dat accuraat is binnen de 1e-12 eenheden.
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David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 5:26am on August 24, 2010
: C = 100-200
Animation start with A=1, B=50 and C goes from 100-200 : then A=1, B=51 and C = 100-200 and once B gets to 600, A turns to 2... and so forth
-Khizer…
I link a plugin for grasshopper i was programming ... it approximates the minimum boundingbox. I hope this helps. You have to use 25 to 50 cycles to get a good result.
image sample values when using the circle. So say you have originally 50 points. You plugged those 50 into image sample to get 50 values. Now you dispatched the 50 points into lets say 25 points each. Yet you are giving the original 50 to the new dispatched list of 25.
Dispatch your image sample data as well then use it for the circle component matching values a with a and b with b, or don't dispatch your points. (I am not sure why you dispatched in the first place)
Edit: this is in response to your first image. …
t and L,
so (L+(L/(3*π)))/2
and so on
I guess I could make theoretically make this value an input parameter.
With all the different features and controls people are asking for though, I'm concerned that if I keeping on adding them to one component it makes it an unwieldy 50 input monster.
I think it will be better to make several components each geared towards particular usage. For this it would be helpful to hear from all of you about what you are using or want to use it for.…
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…
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