ll these 12500 points.
Group 1 would represent the point located at 0, 5, 10, 15, 20 etc.
Group 2 - 1, 6, 11, 16, 21 etc.
Group 3 - 2, 7, 12, 17, 22 etc.
Group 4 - 3, 8, 13, 18, 23 etc.
Group 5 - 4, 9, 14, 19, 24 etc.
I can create the pattern but the selection of points are all the points in row 0 and then all the points in row 5 and so on.
I would like the selection of points to start at the bottom left, and sequentially continue to the right and then continue on the 2nd row (left to right & bottom to top). i am hoping the pattern i am trying to achieve is more understood with the quick screen capture I uploaded.
the end goal is to be able to select all the points in the grid that are in each pattern.
Thanks in advance for any guidance with this. …
Added by Alyne Rankin at 6:53am on October 11, 2017
the catenary shape
6) Get the mid point of all the heaxgons
7) Create surfaces to get the average normal of each hexagons
8) Project the hexagon on the normal plane
9) Move the plane using the normals
10) create mesh faces between first and second hexagons
11) create mesh faces to close the planar hexagon
12) Orient all the mesh faces on a flat grid to laser cut them
13) Join and weld the mesh
14) Thicken the mesh using Weaverbird to get a shell
I hope this helps. It is all one mesh at the end made of quads on the side and triangles on the top (yes 6 of them).
All the best,
Arthur
…
Create surface for wall (can be done manually)
2) Get vertical edge curve (can be done manually)
3) Divide curve randomly into 2" and 6" segments
4) Put a segment of predetermined length after each 2 or 6 inch segment (say 1/4") for reveals
5) get end points of segments
6) generate horizontal curves based on length of the surface's horizontal edge curve
7) split original surface with new curves
8) move reveals backward a certain distance opposite of original surfaces normal
9) extrude reveal curves backward the same distance so the 2" and 6" boards appear solid
I'm going for an effect similar to the boards in this image
…
dimension of matrices must be identical) and division is the same as multiplication (dimension must be in the order of A(mxn)*\/B(nxk) where n is the common dimension): to divide one element by another you just multiply it by 1/value (part or all of the elements can multiply while part or all of the elements divide):
so for example matrix addition of matrices A(2x2): {2,-1}{1,2} and B(2,2): {3,-5}{4,-2} will result in matrix C(2x2):{5,-6}{5,0}. subtraction of those matrices will result in D(2x2): {-1,4}{-3,4}
Division of matrices A(2x2): {2,0.5}{2,4} and B(2x1) :{2}{2} will result in matrix C(2x1): {1+0.25}{1+2}={1.25,3}. Multiplication of those matrices will result in D(2x1):{4+1}{4+8}={5,12}.…
glass panel).
2. This actually means that the parts on duty they don't differ that much. Meaning that we can use an "average" size (and "local" topology) acting as the Jack for all trades.
3. Meaning that we can effectively solve the abstract topology with an abstract app the likes of GH and then place in properly defined coordinate systems all the real-life bits and nuts ... closely "emulating" a pro solution (that could "adjust" the parts as well).
4. This means that one particular C# needs more lines of code since as it is it defines cable axis on a per nod to node basis ... but in fact these are defined as the min segment between curves (circles to be exact).
5. Additionally the end part of each strut differs depending on how many pairs of stabilizing cables are used (either 2 or 1). Meaning some lines of code more for defining the proper coordinate systems for the instance definitions.
6. This is the reason that I've postponed mailing to you the 4 horsemen (because PRIOR finishing the whole you MUST define what parts to use: the classic bottom-top design approach).
But in order to receive the Salvation (aka: Apocalypse) you MUST answer correctly to a simple puzzle:
Provided that money is no object, pick your car:
1. Ferrari 245 (Less is more)
2. Lancia Stratos (Lethal).
3. Cobra 427 (Men only)
4. Ford GT40 (Mama mia)
5. Ariel Atom (Mental)
6. Aston Zagato GTB4 (Sweet Jesus)
7. Fulvia HF Fanalone (THE racer)
8. Lambo Miura (Enough said)
9. Lotus Elise (Just add lightness)
10. Alfa Romeo 8C Competizione (In red)…
which will result in creating a check for each branch with one item to be 'matched' (in this case to see if it is included in the domain) with all three domains in the single list. so in the end you get a data tree with 8 branches (derived from the 8 grafted values) with 3 items in each branch, that are the check of each initial value with the 3 items domain list.
thank you for sharing the other option too!.
cheers
alex…
byte-accuracy red, green, blue channels) = 27 bytes. More likely 28 bytes as colours are probably stored as 32-bit integers, allowing for an unused alpha channel.
28 * 800,000 equals roughly 22 megabytes, which is way down from 9 gigabytes. That's a 400 fold memory overhead, which is pretty hefty.
Grasshopper stores points as instances of classes, so on 64-bit systems it actually takes 64+64+3*8 = 152 bytes per point*, which adds up to 122MB, still way less than 9GB. It would be interesting to know where all the memory goes...
* Grasshopper points also store reference data, in case they come from the Rhino document. This data will not exist, but even so it will require 64-bits of storage.…
Added by David Rutten at 4:13pm on December 11, 2014