y you have a mesh with 100 vertices (points). The first one is at index 0, the second one at index 1, then 2, 3 etc. all the way to 99. A face might connect vertices 0, 1, 22 and 23.
You typically don't use this kind of low level method to create a mesh, though of course there's nothing stopping you. Most meshes are either the result of some operation on existing meshes, the approximate mesh of a surface/brep or based on one of the mesh primitives such as Plane, Box or Sphere.
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David Rutten
david@mcneel.com
Poprad, Slovakia…
For example.
If you have two lists of points.
List A List B
{0;0;0}(0) {0;0}(0)
{0;0;1}(0) {0;1}(0)
{0;2}(0)
{0;3}(0)
{0;4}(0)
And you want to merge the two lists so that the two points in list A are the end points.
Merge Lists Results:
{0;0}(0)
{0;0;0}(0)
{0;0;1}(0)
{0;1}(0)
{0;2}(0)
{0;3}(0)
{0;4}(0)
Because of their path structures the order is wrong from a simple merge so Flattening now is out of the question.
Path Mapper
{A;B} --> {A;B+1}
{A;B;C} --> {A;C*6}
---------------------
Results:
{0;0} --> {0;0+1} = {0;1}
{0;1} --> {0;1+1} = {0;2}
{0;2} --> {0;2+1} = {0;3}
{0;3} --> {0;3+1} = {0;4}
{0;4} --> {0;4+1} = {0;5}
{0;0;0} --> {0;0*6} = {0;0}
{0;0;1} --> {0;1*6} = {0;6}
Now with the Path Structures similar when they are re-ordered the results will have the two points of list A as the end points.
Question 2
why did the curve-line intersection lose the path structure? Both trees had 38 branches.
Both trees had 38 Paths but Tree A had more Items, 147 compared to 38 in Tree B.
So you get this happening:
{0;0;0;0;0;0}(0) compared to {0;0;0;0}(0) results: Null {0;0;0;0;0;0}(0)
Base result paths on longest
{0;0;1;0;0;0}(0) compared to {0;0;0;1}(0) results: Null {0;0;1;0;0;0}(0)
{0;0;2;0;0;0}(0) compared to {0;0;0;2}(0) results: Yes {0;0;2;0;0;0;0}(0)
Add a branch to contain result
{0;0;3;0;0;0}(0) compared to {0;0;0;3}(0) results: Yes {0;0;3;0;0;0;0}(0)
{0;0;3;0;0;0}(1) compared to {0;0;0;3}(0) results: No {0;0;3;0;0;0;1}(0)
{0;0;4;0;0;0}(0) compared to {0;0;0;4}(0) results: Yes {0;0;4;0;0;0;0}(0)
{0;0;4;0;0;0}(1) compared to {0;0;0;4}(0) results: Yes {0;0;4;0;0;0;1}(0)
{0;0;5;0;0;0}(0) compared to {0;0;0;5}(0) results: Yes {0;0;5;0;0;0;0}(0)
{0;0;5;0;0;0}(1) compared to {0;0;0;5}(0) results: Yes {0;0;5;0;0;0;1}(0)
{0;0;5;0;0;0}(2) compared to {0;0;0;5}(0) results: Yes {0;0;5;0;0;0;2}(0)
...... etc
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the map? For example in one list I want curves 5, 20, 21, 22, 23, 60. In another I want curves 1, 37, 40. In another maybe 70-80. And in the last, all curves that aren't specified in those three lists. Is there a way to partition the lists as such?…
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…
Tree:
{0;0;0} N = 2
{0;0;1} N = 1
{0;0;2} N = 3
{0;1;0} N = 5
{0;1;1} N = 8
{0;1;2} N = 10
If we apply the aforementioned mapping to this tree, we'll end up with the following result:
{0;0} N = 6
{0;1} N = 23
Basically {0;0;0}, {0;0;1} and {0;0;2} are combined into a single path {0;0} as we disregard the third index because "C" is no longer present in the target mapping.
Because we only use the Mapper to modify paths, we do not lose any data items, though we might lose some of the paths.
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David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 1:03pm on August 25, 2010
I'm trying to retrieve data from a tag heuer timing system through rs 232. Firefly ports available says com ports 1 and 3 are available. Is one of those two the rs 232 9 pin? Thanks for your help.