jaja, I tried a brute force method using Mathematica...lol I give it the list of values and it gave out instantly:
(1/9) * (57 + 48 * n + 15 * Cos((2 * n * Pi)/3) + Sqrt(3) * Sin((2 * n * Pi)/3) )
Added by Jesus Galvez at 7:42am on November 27, 2012
Hi,
Is there a way of setting the step size of a slider?
For instance if the limits are from 0 to 15 every 3 units, so i get 0,3,6,9,12,15, instead of 0, 1, 2 ...
Thanks
to -1 and return -1)
-4th iteration
for Z = 14
return -1 (because 0<Z<15 A is returned and has been set to -1 at the previous iteration)
-5th iteration
for Z=13
return -1
-and so on till
Z=0
return 1 (and set A to 1)
-next iteration
Z=1
return 1
-and so on till we reach 15 and go back the other way
Thanks a lot
…
cture, Rhino treats them as a single flat list. For example a surface can have 10 rows and 6 columns of control-points, resulting in a list of 60 points.
But 10 times 6 isn't the only way to get to 60. If you want to make a surface out of a list of 60 points, you'll also have to tell Rhino how those 60 points should be interpreted in terms of a grid. It could be 2*30, 3*20, 4*15, 5*12, 6*10, and all of the aforementioned products the other way around.
Sometimes there's only one way for a number of points to fit into a rectangular grid. For example if you provide 49 points, then 7*7 is the only way to make it work, but these cases are rare so we always demand you give us all the information required to actually make a rectangular grid of control-points from a linear collection.
As for "Why is it, sometimes we need to attach additional value into it?", this is usually because when you divide a domain or a curve into N segments, you end up with N+1 points. For example take the domain {0 to 5}, and divide it into 5 equal subdomains. You end up with {0 to 1}, {1 to 2}, {2 to 3}, {3 to 4} and {4 to 5}. However there are six numbers that mark the transitions between these domains 0, 1, 2, 3, 4 and 5. This is why you often have to add 1 to the UCount, because the number that controls the UCount often results in N+1 actual points.…
Added by David Rutten at 8:30am on December 25, 2014
t 2^15 are being listed. Later I will use this info to cull;
all instances where important nodes are not present,
geometric unstable trusses using Grubler's criterion,
the remainder will go to FEM for further analysis,
I can use this method but i require to separate each binary boolean
so they are stored separately like so;
{0} (0) 0 (1) 0 (2) 0 (3) 0 (4) 0 . . . . . . . . (15) 0
{1} (0) 1 (1) 0 (2) 0 (3) 0 (4) 0 . . . . . . . . (15) 0
{2} (0) 0 (1) 1 (2) 0 (3) 0 (4) 0 . . . . . . . . (15) 0 (I know there must be an easier scripted way how to do it) or else is there a way to decompose them directly from the method you sent me using standard Grasshopper components?. Any method would work.
Best,
Kane…
o now is select and group together list items by their index. {0:0} and {2:0} also {1:0} and {3:0}. They way I'm doing this at the moment is quite complicated and I'm sure there's an easier way to achieve the same result.
maybe someone in here knows a solution?
thanks a lot…
;1},{0;2},{0;3}, (note that the first item is NOT {0,0})
{1;1}{1;2},{1;3},
{2;1},{2;2},{2;3},
{3;1}{3;2},{3;3}...
Well I'll just upload an extract of the definition. The data path should be the same, but with the items in the last path shifted to the first path, and the items in the first path shifted to the second path and so on.
I'll keep trying. …