Uniform Subdivision of Equilateral Triangle

@Vikram_Subbaiah I replied in:

As the discussion to fix the icosahedron in my judgement is a seperate topic from Surface Subdivisions.

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Thanks @Vikram_Subbaiah. Unfortunately when running in your code today I found a problem inputting the surface which I am attempting to resolve in another thread. As soon as I get the surfaces oriented correctly I will continue chipping away at implementing the corrections to my script. I will add though; this is absolutely fascinating stuff!

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@PerfectArchCo You don’t try my code for equilateral subdivision? :cry:

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Awe. Sorry @Ricardo_Freitas . I did try your code – twice, same problem - but see now the context isn’t as clear as I thought it was. Please take no offense. VS’s is just whose I had up when I took the screen grab. The bigger picture is that your contribution to the thread is greatly appreciated, and even if it didn’t directly help me in this case it’s a great resource and for that I truly thank you. If we met in real life I would definitely raise a glass in graditude for your effort.

Displays yes @PerfectArchCo. It shows how to make the sub-division with a mathematical method from the coordinates of the vertices of the main triangle. It was exactly what I applied at Dynamo and it worked. I posted the DYN file in my first comment with the paper.

Below follows the transcript of the passage to which I refer

"**

7 Triangle Coordinates

**
If you know the 3D coordinates of the three vertices of a triangle in space and you wish to subdivide that triangle into n² smaller triangles and to determine the coordinates of the points of subdivision, the process is fairly simple.

Let’s use the example where n = 6 so the triangle is subdivided into 6² = 36 smaller triangles as in the following illustration:

If the 3D coordinates of the points at the top, bottom left and bottom right are A, B and C, respectively, then the coordinates of the interior points are given by the labels in the figure above, divided by 6. Thus the point labeled 2A + 3B + 1C will have coordinates A/3 + B/2 + C/6, et cetera. The coordinates are easily generated. Start from any vertex (say A, in which case the coordinates will be 6A + 0B + 0C – in other words, all A) and step along the lines to the desired vertex. Each time you step along a line in the direction from vertex Q toward vertex R, subtract 1 from the Q values and add 1 to the R value."

This is it.

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