Cartesian coordinates of a regular tetrahedron

[randomMesh]

I am trying to render a regular tetrahedron via drawVertexPrimitiveList().

The wikipedia article says the coordinates are

• (1, 1, 1)
  (−1, −1, 1)
  (−1, 1, −1)
  (1, −1, −1)

Code: Select all

vertices[0].Pos = core::vector3df(halfSize, halfSize, halfSize);
vertices[1].Pos = core::vector3df(-halfSize, -halfSize, halfSize);
vertices[2].Pos = core::vector3df(-halfSize, halfSize, -halfSize);
vertices[3].Pos = core::vector3df(halfSize, -halfSize, -halfSize);
const u16 indices[12] = {  0, 1, 3,  0, 3, 2,  0, 2, 1,  3, 1, 2 };

works well, the above code looks like this


Now i want to have the tetrahedron rotated, so that one corner points up the Y-axis. The centroid should be at (0.0f, 0.0f, 0.0f).

So i tried

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vertices[0].Pos = core::vector3df(0.0f, halfSize, 0.0f);
vertices[1].Pos = core::vector3df(-halfSize, -halfSize, -halfSize);
vertices[2].Pos = core::vector3df(halfSize, -halfSize, -halfSize);
vertices[3].Pos = core::vector3df(0.0f, -halfSize, halfSize);

This unfortunately results in a non-regular tetrahedron.
I know i could just rotate the node, but i really want to know the position of the vertices.

The formulas for regular tetrahedron in the article don't help, since my math skills are... uhm.. poor.

Can anyone help?

[DavidJE13]

I can't vouch for these numbers because I just worked them out back-of-the-envelope style, but they should be correct;

Code: Select all

vertices[0].Pos = core::vector3df(0.0f, r, 0.0f);
vertices[1].Pos = core::vector3df(r*sqrt(2.0)*2.0/3.0, -r/3.0, 0.0);
vertices[2].Pos = core::vector3df(-r*sqrt(2.0)/3.0, -r/3.0, r*sqrt(2)/sqrt(3));
vertices[3].Pos = core::vector3df(-r*sqrt(2.0)/3.0, -r/3.0, -r*sqrt(2)/sqrt(3));

I've left it in terms of sqrt(2) and sqrt(3) because there's probably some constants somewhere set to these values.
r = radius (distance from centre of mass to each vertex). It's equal to 4/3 of the total height, and the base of the pyramid is at y = -r/3.

[DavidJE13]

For anybody interested, the maths is quite simple - uses nothing more than trig and a little intuition.

Assume we're building a tetrahedron around a point, with a radius of 1 (to simplify the math a little).
Intuition: The centre of mass is the centre of mass in all directions. Therefore if we have a point at -1 y, and know that the other 3 points form a plane perpendicular to y (i.e. all have the same y value), they must all be at +1/3 y, else the centre of mass would be elsewhere.
Now build a triangle with height = 1/3, hypotenuse = 1. By Pythagoras the remaining dimension (width) is sqrt( 1^2 - (1/3)^2 ) = sqrt( 8/9 ) = sqrt(2) * 2 / 3
Here define the second angle of rotation by setting z to 0, and this has given us the second point. Now it's just an equilateral triangle base with radius sqrt(2) * 2 / 3. Using the centre of mass argument a second time, or using the cos(60)=1/2 identity, we find that the x of the remaining 2 points is -sqrt(2) / 3
Finally the sin(60)=sqrt(3)/2 identity tells us that the z values of the last 2 points are +/- sqrt(2) * sqrt(3) / 3, which simplifies to +/- sqrt(2/3)

[randomMesh]

Thank you very much, works like a charm.

One last little question:
Say the length of each edge should be 1000.0f. What's the math to compute radius r?

I tried f32 r = 1000.0f*sqrt(3)/sqrt(2)/2; which results in edge length = 1000.000008, but
i don't know if there's a more elegant and faster solution.

Again, thank you for your help, your post cleared some things up.

[DavidJE13]

Your formula is perfectly correct and elegant. Nothing will be more accurate short of using 64-bit precision or replacing all r's with that formula and simplifying them (and that may not be any better)

for a little more programming elegance you can use M_SQRT2 (=sqrt(2)) and M_SQRT1_2 (=1/sqrt(2)) in the place of some sqrt calls for speed. I'd also recommend making your own SQRT3 constant (and maybe a SQRT1_3 constant too) to save some calculations.