#Mapping Sphere to Cube

29 messages · Page 1 of 1 (latest)

hot python
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(Not in university, for a personal project)
I have a special equation for mapping from a cube to a sphere. It's particularly good for minimizing the distortion near the corners (attached).
However, I also need the inverse mapping. In other words, I need to solve the system of equations for x, y, z (position on a cube) given x', y', z' (position on a sphere). I've heard you're supposed to take 2 equations at a time and eliminate a variable, but I don't know how to do that with these square roots. Any help would be greatly appreciated.
Equation from: https://catlikecoding.com/unity/tutorials/cube-sphere/

24012831_0f071f41bc_o.gif 24013697_73e20d567a_o.jpg 03-gizmo-new-mapping.png
delicate martenBOT
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rancid ridge
hot python
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Oy vey. I mean I can approximate the solution by computing a look-up-table but... I'd really rather not do that if computing the exact solution is possible and isn't too complex computationally

rancid ridge
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Hmm... Perhaps it would be easier if we switched to spherical coordinates? Not sure, though.

hot python
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It might be simpler if I try to solve for just the top of the cube, so z = 1

rancid ridge
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Oh, that's true!

hot python
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At least for that I get:

x' = x * sqrt(0.5 - y^2/2 + y^2/3)
y' = y * sqrt(0.5 - x^2^2 + x^2/3)
z' = sqrt(1 - x^2/2 - y^2/2 + x^2*y^2/3)

maybe this helps?
1 = x'^2 + y'^2 + z'^2
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But I keep getting in the situation where I can't figure out how to further eliminate a variable

rancid ridge
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Then we have:
X = x√(1/2 - y^2/6)
Y = y√(1/2 - x^2/6)
Squaring and multiplying by 6 produces:
6X^2 = x^2 (3 - y^2)
6Y^2 = y^2 (3 - x^2)
Let's replace the equations by their sum and difference.
6(X^2 + Y^2) = 3x^2 - 2x^2 y^2 + 3y^2
2(X^2 - Y^2) = x^2 - y^2
From the second equation:
x^2 = 2(X^2 - Y^2) + y^2
Substituting into the first, we get:
6(X^2 + Y^2) = (3 - 2y^2)(2(X^2 - Y^2) + y^2) + 3y^2
And this is a quadratic equation in y^2, so the rest shouldn't be a problem.

hot python
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I multiplied that out to:
0 = 6y^2 - 2y^4 - 12y'^2 - 4y^2x'^2 + 4y^2y'^2 but that doesn't seem like a quadratic 💀

hot python
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When I try another way I get:
x'^2 = x^2 * (0.5 - 6y'^4 / (x^4 - 6x^2 + 9))
but fml that's got x^4

hard herald
hard herald
hot python
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I've spent 5 hours on this, please kill me

hard herald
hot python
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Solve for x, y, & z

hard herald
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i think

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y+mx+c

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x=my+c

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z = idk

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that's the best ik

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im in middle school literally

hot python
rancid ridge
hard herald
hot python
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+close