ap value = True
Shift List = 1 --> (B,C,D,A)
Shift List = 2 --> (C,D,A,B)
You can also use negative values.
Shift List = -1 --> (A,B,C)
Shift List = -2 --> (A,B)
and with Wrap = True
Shift List = -1 --> (D,A,B,C)
Shift List = -2 --> (C,D,A,B)
The most useful Shift List action I use is to either get rid of the first or last item in a list and sometimes both.
Shift list = -1 --> (A,B,C) Shift list = 1 --> (B,C)
In the example posted above you are creating a shift list value equal to its location along the curve. The first section = 0 doesn't get shifted, the second section gets a shift = 1, third = 2, forth = 3 and because the wrap value is set to true the fifth section gets back to 0, sixth = 1 etc etc. creating the twisting effect.
The "one more stupid question" answer is Mass Addition. You will find the component on the Math tab or you can type it into the Keyword search feature (by double clicking the canvas). This component has two outputs a total amount for each list and a partial set of results giving:
List (3,6,9,12)
{0} = 3
{1} = 3+6 = 9
{2} = 3+6+9 = 18
{3} = 3+6+9+12 = 30…
ep is to understan the logics of what you want to do, in your case, build 4 point surfaces (u also need to know the right direction to build the surfaces). Then you can write an hipotetic list (by hand in a paper) of what you want. In your case the list was (0, 1, 3, 2) (2, 3, 5, 4) (4, 5, 7, 6), etc... if you can imagine building 2 lists, each one with the sequences (0, 2, 4, 6, etcc) and (1, 3, 5, 7, etc..) then you can manage with shift and graft to finally have four lists. A( 0 1 2 3 ...) B (1 3 5 etc..) C(3 5 7 etc..) D (2 4 6 etc..). And to achieve the 2 first lists, you need to get the odd and the pair numbers. The cull pattern does that amazingy well. With a pattern True-False you get de pair numbers, and with the False-True pattern you get de odd numbers.
Hope it was clear enough…
Added by Pep Tornabell at 5:32am on November 19, 2009
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
If you right click on the 'Random' component, you will find an option for "Integer Numbers".
You might need to generate random integers of 3, 4, 5 or 6 and multiply that by ten?
{4}-0;3
{5}-6;7
{6}-5;7
{7}-5;6
Here it can be shown that there are two subgraphs containing 0,1,2,3,4 and 5,6,7. How can I use spiderweb (either using scripting or the components) to give me this result when I have many more vertices??
Thanks,
Sam…
lve around the tube. These diagonals then have to 'skip' for example 3 or 4 vertical ring sections before reaching the next corner of the polygon. In the current model a new diagonal starts from every division point..…