square units. Then you have an integral number of fragments on each side. This means that if all fragments need to have the same surface area, you can only have the following possibilities for side A:
1 fragment = 100 square units
2 fragments = 50 square units each
3 fragments = 33⅓ square units each
4 fragments = 25 square units each
5 fragments = 20 " "
6 fragments = 16⅔ " "
etc.
For side B, the numbers are mostly different
1 fragment = 300 unit²
2 fragments = 150 unit²
3 fragments = 100 unit²
4 fragments = 75 unit²
For side C they are different still. Unless you join fragments across on both sides of the edges of the box, I very much doubt you'll be able to pull this off.
The solution I attached will create fragments as identical as possible, but it's a very boring outcome...
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
1+2+3+4+5+6 = 21
1+2+3+4+5+6+7 = 28
1+2+3+4+5+6+7+8 = 36
1+2+3+4+5+6+7+8+9 = 45
Is there a tool, that can do me that job?
How do I get this List {1,3,6,10,15,21,28,36,45}?…
Added by Ahmed Hossam at 2:19pm on September 22, 2013
g to keep the {15} {16} branch structure because its keeping the organization by floor level.
Any insight into how the Curve Plane Intersection is matching the branches for operation would be helpful as well.
Thanks!
…
} (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!…
ole size)
The problem I am seeing from what I hope to achieve is that there are too many holes. It seems the script continues until I can't fit in any more circles. Can we limit the overal number as a percentage or a maximum number of circles that can be generated = say 50 or 100. Can that be dynamically controlled?…
13;2} ... 20.{13;12}
21. {21;0}22. {21;1}23. {21;2} ... 41. {21;20}
42. {34;0}43. {34;1}44. {34;2} ... 75. {34;33}
76. {55;0}77. {55;1} ... ....
I want to grab the first 8 [0-7], the next 13[8-20], the next 21[21-42] etc
so i have the (known fibonacci seq) list of numbers on the left here:
C S
8 0
13 8
21 21
34 42
55 76
89 131
144 220
233 364
and i need the list on the right, so that i can select items using a Series (N=1 and S and C from the list above) and a List Item component.
the simple question is:
is there a component that can take a list and accumulate it in this way that I need?
if not, is there anyone that can point me to a simple relevant VB example so i could easily adapt it?
many thanks,
gotjosh…
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