Tiling a triangular grid using a specific pattern

Question

I have an array of Triangle objects that are equilateral triangles and have for x and y coordinates the center of the triangle :

The distance between this point and a, b or c is t_r. The distance between this point and A, B or C is t_R. The side of a triangle is t_size

I want to create a function that place each triangle on a grid following this order :

So this function gets the index of the triangle in the array and updates its x and y position. Do you have any idea how I could proceed ? You should note that when the triangles are upside down, their center point is still as if they were straight but I just draw them inverted. For example, the translation from the first triangle to the second isn't up and right but only right.

I really don't know how I could do this. I tried to find a pattern in each "level" (color) of triangles but I didn't find any, neither in the side where the next triangle will be that change for every triangle...

But center of triangle 2 is higher by t_R/2 (one third of the height) — MBo
– MBo, Commented Oct 11, 2018 at 11:36

tucuxi · Accepted Answer · 2018-10-11 11:48:36Z

This seems relatively straightforward. There is a limited number of jumps needed to go from one step to the next, and they are quite predictable too.

Let us enumerate the different jumps:

1 to 2 (also found in 8 to 9, or reversed in 4 to 5): (t_size * cos(30º), t_size * sin(30º))
2 to 3 (or any other vertical-up): (0, t_size)
3 to 4 (a reflection of 1 to 2): ( - t_size * cos(30º), t_size * sin(30º) )
4 to 5 (left as an exercise)
5 to 6 (another exercise)
6 to 7 (as in moving from 4 to 2; therefore = - (d12 + d23 + d34), where dXY is the vector used to move from X to Y)
6 to 1 = -d34

Now, let us continue the sequence using these displacements:

first ring: d12, d23, d34, d45, d56,
jump to next ring: d67
second ring: d61, d12, d23, d12, d23, d34, d23, d34, d45, d34, d45, d56, d45, d56, d61, d56, d61
jump to next: d67

In the next ring, you would extend the bolded segments (the "do a zigzag in this direction) to repeat them thrice, and in the n-th ring, you would extend them n times.

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