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
circles that can be populated (for each radius size) is set as an integer (or slider)
(ie. radius 1.5 = 10 , radius 3= 6, radius 6 = 6, radius 9=4)
Conditions are:
1) Each of the circle has a radius of influence,
Radius of influence = double the radius of the circle)
(3, 6, 12, 18)
2) Any overlapping circles in either: Radius of influence or the Circles are removed so that
No circles overlap.
3) There must also be 4 circles set at the corner points of the grid - These must be circles with a radius of 3 or 6
If you can do that I will be amazed as i've been trying for weeks! :(
Ive attached a sketch of what im looking for…
it,
[3] the upper limit, [4] the slider position.
What do you think?
Matt
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Added by Matt Trimble at 7:54pm on November 17, 2009
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…
What I need to do, is to cycle through all the branched items of the X lists with every single item of the corresponding list of Y.
So:
{0,0}(N=1) with 1st item of Y{0},
{0,1}(N=1) with 1st item of Y{0}
...
{0,15}(N=1) with 1st item of Y{0} (and this is the first cycle).
Then (cycle 2)
{0,0}(N=1) with 2nd item of Y{0},
{0,1}(N=1) with 2nd item of Y{0}
...
{0,15}(N=1) with 2nd item of Y{0}
Then do the same thing with the X list {1,0} .. {1,15} with the corresponding Y{1}
As a result I need to have the following list structure:
{0,0}N=6, {0,1}N=6, {0,2}N=6 ... {0,15}N=6,
{1,0}N=15, {1,1}N=15, {1,2}N=15 ... {0,15}N=15.
If I use the component "shift path" (as suggested in another discussion) it will match the items per item-order instead of cycling through all the items of the first sublist...
Hope this makes sense.
Any suggestion?…
9 8 7 6
5 4 3 2 1 0
I am triangulating this surface. I want to select just the red vertices. As you can note, I just need the inner vertices of this surface. I could do it mannually, but if I want to change the mesh density later, I will have to pick all of them manually again later.
Can someone help me?
Tks
…
onsecutive points at the same height then your 'Break at discontinuities' component eliminates the middle point completely and then the 'Interpolate Curve' component gives a much bigger bump in the wrong direction. This was enough to get curves to meet from opposite sides.
I fixed this by changing the heights to 1.1 or 2.9, rather than 1.0 and 3.0, but it took a little while to work it out! Sigh.
I attach a new version. But I actually preferred it as it was before. See what you think!
Bob
p.s. in the first list, elements 11, 12, 23 and 24 go from 1 to 3; elements 17 and 18 go from 3 to 1. In the second list, elements 6, 17, 18 and 29 go from 1 to 3; elements 12 and 23 go from 3 to 1. Given the above fix, these can be easily seen.…
Added by Bob Mackay at 10:40pm on November 24, 2015