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I subdivide a unit sphere, centered at origin, onto 20 spherical triangles. For the sake of argument let's take one such triangle $Ts$, in $\mathbb{R}^3$, that has vertices $Normalize(-1,0,g), Normalize(0,g,1)$, and $Normalize(1,0,g)$, where $g = \frac{1.0+\sqrt{5}}{2}$.

In the same time there is another equilateral planar triangle $Tp$, in $\mathbb{R}^2$, that has the following vertices $(0,0), (1,0),(0.5, \frac{\sqrt{3}}{2})$.

I'm looking for a projection (exact formula) that would create bijection of surface points of $Ts$ to surface points of $Tp$. It's crucial that such mapping maximize uniformity of distortions (not sure how to express this in strict math terms).

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  • $\begingroup$ Take any similitude carrying the vertices of $Ts$ to $Tp$, followed by the obvious "flattening" of $Ts$. $\endgroup$ – Aretino Feb 10 at 20:45
  • $\begingroup$ I'm good with transforming $Ts$ vertexes to $Tp$ ones. They match perfectly. However I'm clearly struggling with "flattening" part. With naive (inverse $Tp$, in $\mathbb{R}^2$ to $Ts$, in $\mathbb{R}^3$) mapping I'm getting points that are outside of the planar triangle. See the picture: i.snag.gy/GVYsAX.jpg Red marks represent roughly where the edge of my planar triangle is. $\endgroup$ – lhog Feb 10 at 22:09
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Suppose then you applied a suitable transformation, so that the vertices of the spherical triangle are the same as the vertices of the plane triangle, sitting then on plane $z=0$. Let $O=(x_0,y_0,z_0)$ be the center of the transformed sphere and $P=(x,y,z)$ any point on the spherical triangle. You can then:

  1. Apply a translation carrying $O$ at the origin: $P\to P-O=(x-x_0,y-y_0,z-z_0)$;

  2. Slide $P$ along $PO$ until its $z$ coordinate becomes $-z_0$: $$ P\to {-z_0\over z-z_0}P=\left(-{x-x_0\over z-z_0}z_0,-{y-y_0\over z-z_0}z_0,-z_0 \right); $$

  3. Translate back: $$ P\to P+(x_0,y_0,z_0)=\left({x_0z-xz_0\over z-z_0},{y_0z-yz_0\over z-z_0},0 \right). $$ Please ask if anything is not clear.

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  • $\begingroup$ I'm yet to try this. Let me ask you one question in advance though. $\endgroup$ – lhog Feb 11 at 23:57
  • $\begingroup$ Sincere "thank you" for your answer. I'm sure it's right, but unfortunately I've just managed to solve it myself. I guess I'll add my own answer for the sake of future reader's awareness. $\endgroup$ – lhog Feb 12 at 0:11
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I'm sure that the answer @Aretino provided works too, but I'm leaving my own solution for the sake of completeness.

So this is what I have done myself. As readers might have guessed my goal was to try to project surface points within icosahedron face onto respective spherical triangle. The issue I had was that with naive mapping spherical triangle would "take" planar surface points that are outside of planar triangle. See my lame picture, where I outlined the rough edge of planar triangle on of the sides: https://i.snag.gy/GVYsAX.jpg

Initially I did the mapping the following way:

  • get barycentric coordinates: $baryP = getTriangleBarycentric(pos, icoFace[0], icoFace[1], icoFace[2]$$);$

$pos$ is point on the sphere. $icoFace[0..3]$ are vertexes of an icosahedron and corresponding spherical triangle.

  • $planarPoint$ = $baryP.x * [0.0, 0.0] + baryP.y * [1.0, 0.0] + baryP.z * [0.5, \frac{\sqrt{3}}{2}]$

$planarPoint$ is two component vector that represents points of triangle I map from. $[0.0, 0.0] ; [1.0, 0.0] ; [0.5, \frac{\sqrt{3}}{2}]$ are points of triangle I do mapping from. Note: this triangle is similar (and equilateral) to original icosahedron planar triangle. It's just a matter of preference to work with "unit" triangle.

So that was the original approach and it didn't work.

What made it work is the following transformation of $pos$ before it's used to calculate barycentric coordinate corresponding to $pos$.

$triangleToPointDist = DistancePointToTriangle(pos, icoFace[0], icoFace[1], icoFace[2])$ $pos = pos * (1.0 - triangleToPointDist)$

These two lines essentially project $pos$ from being on spherical triangle surface to being on icosahedron's triangle.

The lack of projection from sphere to planar triangle of icodahedron was my only mistake.

P.S. As I do it for WebGL shader (a program that is executed on GPU), I just can't resist temptation to share the link to the excellent final result: https://shaderfrog.com/app/view/2360

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