ee 3)
{5}
0 15
{6}
0 16
And I want to place points at every possible combination of these coordinates, treating Tree 1 as X coordinates, Tree 2 as Y coordinates, and Tree 3 as Z coordinates. Also, I would like the list of points to be a tree with paths corresponding to the coordinates. Wouldn't it be nice if I could plug these trees into a Point XYZ, with a new "branch cross reference" method, and get the following result?
{0:3:5}
0 {10.0, 13.0, 15.0}
{0:3:6}
0 {10.0, 13.0, 16.0}
{0:4:5}
0 {10.0, 14.0, 15.0}
{0:4:6}
0 {10.0, 14.0, 16.0}
{1:3:5}
0 {11.0, 13.0, 15.0}
{1:3:6}
0 {11.0, 13.0, 16.0}
{1:4:5}
0 {11.0, 14.0, 15.0}
{1:4:6}
0 {11.0, 14.0, 16.0}
{2:3:5}
0 {12.0, 13.0, 15.0}
{2:3:6}
0 {12.0, 13.0, 16.0}
{2:4:5}
0 {12.0, 14.0, 15.0}
{2:4:6}
0 {12.0, 14.0, 16.0}
In this form of cross referencing, every combination of individual branches from the different lists is used as separate input, and the output for each combination is put onto a branch in the result whose path is the concatenation of the input branch paths used.…
Added by Andy Edwards at 7:03pm on November 3, 2009
etting when I merge the three trees, but what I would like to get is:
essentially a tree with 27 branches, each with a single list of either 11 or 21 points.
{0} (N=11)
{1} (N=11)
...
{10} (N=21)
{11} (N=21)
...
{17} (N=11)
{18) (N=11)
{27} (N=11)
Any help would be greatly appreciated.
All the best,
Matt
…
Added by Matt Schmid at 3:06pm on December 4, 2010
this, you'll have no horizontal force at the roller, but you will have it at the pinned support. If you wouldn't, then the structure will be displaced.
Usually, in 2 dimensional structures, if you want to know if an articulated structure is isostatic (as opposed to hyperstatic, which is what you have right now) is to use the following formula:
b+c-2·n=0;
b being the number of bars, c the number of constraints you have and n the number of nodes. In your case: b=19, c=3 (displacements constrained in X, Z at your pinned support and only constrained in Z at your roller support) and n=11, so: 19+3-2·11=0.
I recommend you to download the app SW Truss, as it's very useful to check your results instantly.…
a specific domain, for example:
0.) 0 to 1 -----> 11 random values from 0 to 1 (0.245,0.678,0.36,0.78,.28,0.18........)
1.) 1 to 2 -----> 11 random values from 1 to 2 (1.26,1.36,1.01,1.68,1.26,1.96.........)
3.) 2 to 3 -----> 11 random values from 2 to 3 (2.96,2.45,2.78,2.56,2.98,2.10..........)
4.) 3 to 4 and so on where I have a data set containing 11 paths with 11 values and the values fall within the specific domain.
Like my post above I have the correct path but I need to feed it the correct seed to get different values for each number. I tried grafting a series similar to the last post but it scrambles my data. Thanks so much for the help!
…
in the desired order.
0 = 0
1 = 1
2 = 6
3 = 7
4 = 8
5 = 9
6 = 12
7 = 13
8 = 2
9 = 3
10 = 4
11 = 5
12 = 10
13 = 11
Where the first number is the index and the second number is the actual sorting key. Then you sort these keys while sorting your curves in parallel using the A input of the Sort component.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…