ización de estructuras, panelización de superficies, gestión y conexión con tablas de datos, automatización de dibujo, programación visual … Adjuntamos el temario del cuso. El contenido del curso ha sido revisado y ampliado, gracias a la experiencia de nuestros anteriores. Está orientado a profesionales y estudiantes de arquitectura y diseño en general.
Será impartido por dos Authorized Rhino Trainers en Madrid, en la calle Bailén. Tiene un formato intensivo de 18 horas; el horario es: viernes, de 17 a 21; sábado, de 10 a 14 y de 16 a 20; y domingo, de 11 a 14 y de 16 a 19. El número de asistentes está limitado a un máximo de 8 personas.…
Added by Miguel Vidal at 11:11am on December 17, 2009
arted to fluctuate in a pattern. Like a standing wave with big amplitude :(
See figure:
Before going to bed I start a new optimization where:
Population: 20
Initial boost: 5
Maintain: 10% (thought maintaining more of the good solution would help the solver in the right "direction")
Inbreeding: +90% (cause the best solutions will be near each other)
After 8 hours the fitness iteration look like this figure:
…
Added by Pil Lauridsen at 1:02am on February 10, 2014
t "AddMesh" command.
On a related note, does anyone know how to modify an existing Plankton Mesh to get a valid new one in Python? For example, I fed a Plankton mesh called "PMesh" in to a Python script component with the following code:
PMesh.Vertices.Add(0, 0, 0)
PMesh.Vertices.Add(20, 0, 0)
PMesh.Vertices.Add(20, 20, 0)
PMesh.Vertices.Add(0, 20, 0)
PMesh.Faces.AddFace(0, 1, 2, 3)
The resulting Plankton mesh has the new vertices, but the face doesn't show up. Do I have to add something to the half-edge lists as well?
Thanks,
Dave…
ld go
Slider at 1 -> 1°, 2°, 3°.
Slider at 2 -> 2°, 4°, 6°
Slider at 3, etc.
And when the angle was too much that it surpassed 90 before connecting the 20 points, that line was not created. Maybe using a series instead of a range...mmm
Everything helps, thanks
…
ceros.
Public concerné /
Architectes et designers, utilisateurs de Rhino souhaitant paramétrer Rhinocéros à l’aide de Grasshopper, programme
associant des composants et une structure de graphe interagissants avec le modèle Rhino.
Une bonne connaissance de Rhinocéros est nécessaire. La langue de la formation est le français.
Structure et Objectif de la formation /
La formation se déroule sur 3 jours : les 2 premières journées sont consacrées aux « fondamentaux » de Grasshopper
avec en préambule une introduction au design et à l’architecture paramétrique et leurs impacts dans la conception, la
création et la construction.
La troisième journée sous forme d’atelier est dédiée à l’étude de cas concrets proposés par les stagiaires, qui, quelques
jours avant la formation, pourront envoyer leurs projets par mail à info AT rhinoforyou DOT com
Les stagiaires, après la formation, pourront rester en contact avec les formateurs de HDA par le biais du blog
complexitys.com et le twitter @HDA_Paris. La durée de cette formation permettra d’atteindre une autonomie et une
bonne compréhension basée sur des exemples concrets.
3 Formules possibles /
3 jours ( Initiation+Atelier ) : du lundi 20 septembre au mercredi 22 septembre
2 jours ( Initiation ) : lundi 20 et mardi 21 septembre
1 jour ( Atelier ) : mercredi 22 septembre
Programme ind icatif des notions traitéES pendan t la formation /
Introduction à la conception Paramétrique . Rhinoscript, Grasshopper: différences et similarités . Interface
graphique de Grasshopper . Objets, Données, Listes . Opérateurs scalaires : La mathématique de
Grasshopper . Gestions des données : la logique de Grasshopper . Vecteurs, Points, Lignes, Surfaces : La
géométrie de Grasshopper . Listes, Arbres, Branches . Le dessin paramétrique: exercices divers et exemples
. Références, Bibliographie, Support de cours . Ateliers d’architecture et design paramétrique (3ème jour) .
Moda lité de la formation /
Venir avec un PC portable équipé de Rhinocéros version 4.0 SR 7 et de la dernière version du plug-in
Grasshopper (téléchargeable sur www.grasshopper3d.com).
Le coût du stage est de 350 € HT/jour par personne.
Réserver votre place dès que possible car les places sont limitées à 10 participants maximum.
Inscriptions et renseignements: Jacques Hababou, info AT rhinoforyou DOT com
Pour en savoir plus sur l’architecture paramétrique: www.complexitys.com…
ut at this point in the quarter, I don't have time to learn to do this properly. I'll definitely have to come back to it though, as I'd like to become more proficient. Thank you for the help, though. I'll have plenty to dig through when I get back to it.…
ber of mesh vertices is defined as (precision_+1)^2.So if you would like to have its beam, diffuse and ground-reflected components as well, that means 3 * 8760 values per single point.Example: if you set your precision_ input to 20, the number of values would be a couple of millions:
(20+1)^2 * 8760 * 3 = 11 589 480 hourly values
Check the attached definition below. The outputs that you need are: "Ebeam", "Ediffuse", "Eground".They contain annual hourly values for each tilt and azimuth combination (that's what upper mesh vertices represent) in a data tree.…
ay how many valid permutations exist.
But allow me to guesstimate a number for 20 components (no more, no less). Here are my starting assumptions:
Let's say the average input and output parameter count of any component is 2. So we have 20 components, each with 2 inputs and 2 outputs.
There are roughly 35 types of parameter, so the odds of connecting two parameters at random that have the same type are roughly 3%. However there are many conversions defined and often you want a parameter of type A to seed a parameter of type B. So let's say that 10% of random connections are in fact valid. (This assumption ignores the obvious fact that certain parameters (number, point, vector) are far more common than others, so the odds of connecting identical types are actually much higher than 3%)
Now even when data can be shared between two parameters, that doesn't mean that hooking them up will result in a valid operation (let's ignore for the time being that the far majority of combinations that are valid are also bullshit). So let's say that even when we manage to pick two parameters that can communicate, the odds of us ending up with a valid component combo are still only 1 in 2.
We will limit ourselves to only single connections between parameters. At no point will a single parameter seed more than one recipient and at no point will any parameter have more than one source. We do allow for parameters which do not share or receive data.
So let's start by creating the total number of permutations that are possible simply by positioning all 20 components from left to right. This is important because we're not allowed to make wires go from right to left. The left most component can be any one of 20. So we have 20 possible permutations for the first one. Then for each of those we have 19 options to fill the second-left-most slot. 20×19×18×17×...×3×2×1 = 20! ~2.5×1018.
We can now start drawing wires from the output of component #1 to the inputs of any of the other components. We can choose to share no outputs, output #1, output #2 or both with any of the downstream components (19 of them, with two inputs each). That's 2×(19×2) + (19×2)×(19×2-1) ~ 1500 possible connections we can make for the outputs of the first component. The second component is very similar, but it only has 18 possible targets and some of the inputs will already have been used. So now we have 2×(18×2-1) + (18×2-1)×(18×2-1) ~1300. If we very roughly (not to mention very incorrectly, but I'm too tired to do the math properly) extrapolate to the other 18 components where the number of possible connections decreases in a similar fashion thoughout, we end up with a total number of 1500×1300×1140×1007×891×789×697×...×83×51×24×1 which is roughly 6.5×1050. However note that only 10% of these wires connect compatible parameters and only 50% of those will connect compatible components. So the number of valid connections we can make is roughly 3×1049.
All we have to do now is multiply the total number of valid connection per permutation with the total number of possible permutations; 20! × 3×1049 which comes to 7×1067 or 72 unvigintillion as Wolfram|Alpha tells me.
Impressive as these numbers sound, remember that by far the most of these permutations result in utter nonsense. Nonsense that produces a result, but not a meaningful one.
EDIT: This computation is way off, see this response for an improved estimate.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 12:06pm on March 15, 2013
point @ 20%, curve 3, point @ 40% and so on. If I change X, the percentage of each curve changes. This way, I will end up with a wave-like pattern in my array of curves.
Upon your request, here is the code :)…