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
} (N=11) {0;1} (N=11) {0;2}(N = 11) {0;3}(N = 11) {0;4}(N = 11)
2. I run the Points that are coming out from the Divide Curve Components through the Path Mapper components with this definition:
{A;B} (i) > {A} (i)
3. I run data coming out from Path Mapper component through:
a) Parameter Viewer component and the result is:
{0} N=11 (data with 1 branches)
b) Point > Panel and the result is:
collection of 11 point (N=11) which is the exactly the same as the collection of point belonging to {0;4} (N = 11).
So, here is the question:
why the collection of points coming out from the Path Mapper {A;B} (i) > {A} (i) component is the same as the collection of points belonging to the curve {0;4}(N = 11) ?
Anyway ... It 's the first time I ask a question here... so I would like to thank you for what you do with your work! Thank you! You are really great!…
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.…
vector * number
8. number * point
9. point * number
10. complex * complex
11. colour * colour
12. colour * number
13. number * colour
--
David Rutten
david@mcneel.com
Seattle, WA…
Added by David Rutten at 10:39pm on November 12, 2010
1:
{0} N=5
{1} N=6
and a tree 2:
{0;0} N=1
{0;1} N=1
...
{0;18} N=1
{1;0} N=1
{1;1} N=1
...
{1;19} N=1
I want to basically modify Tree 1 in order to get
{0;0} N=5
{0;1} N=5
...
{0;18} N=5
{1;0} N=6
{1;1} N=6
...
{1;19} N=6
which is the Tree 1 with the structure of Tree 2 and the items of Tree 1 {0} branch populated at each {0;*} branch of Tree 2 and the items of Tree 1 {1} branch populated at each {1;*} branch of Tree 2.
…
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!
…
i mean, i want a slider that can do 3 sides, 4 sides, 5, 6, 7, 8, 9 and 10. for the grids because I dont want to use a fixed grid shape such as square grid (4 sides only).