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# Finding 3D co-ordinates of UV(1,1)?

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Given 3 vertices in 3D space, each with a texture co-ordinate, how do you find the 3D co-ordinates of UV(1,1)(0,0)(1,0)and (0,1). Assuming the 3 texture co-ordinates are all different, < 1.0 and > 0.0. Sounds really simple until you try to work it out. :( Any help greatly appreciated.

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Form the matrices
 [u1 u2 u3] [v1 v2 v3] = U [ 1  1  1]

and
 [x1 x2 x3] [y1 y2 y3] = X [z1 z2 z3]

Then the point corresponding to a particular UV is given by X * inverse(U) * vec3(u,v,1). For example X * inverse(U) * vec3(1,1,1) is the point corresponding to a UV of (1,1).

[Edited by - Pragma on May 1, 2010 8:52:19 PM]

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Quote:
 Original post by PragmaForm the matrices [u1 u2 u3] [v1 v2 v3] = U [ 1 1 1]and  [x1 x2 x3] [y1 y2 y3] = X [z1 z2 z3]Then the point corresponding to a particular UV is given by X * inverse(U) * vec3(u,v,1). For example X * inverse(U) * vec3(1,1,1) is the point corresponding to a UV of (1,1).

What are u1,v1, u2,v2, u3,v3?
What are x1,y1,z1, x2,y2,z2, x3,y3,z3?

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He said he had 3 vertices in 3d space, each with a texture coordinate. So
(u1, v1) = texture coordinate of first vertex
(x1, y1, z1) = position of first vertex
(u2, v2) = texture coordinate of second vertex
... etc.

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Quote:
 Original post by PragmaHe said he had 3 vertices in 3d space, each with a texture coordinate. So(u1, v1) = texture coordinate of first vertex(x1, y1, z1) = position of first vertex(u2, v2) = texture coordinate of second vertex... etc.

The xyz part is a matrix of a triangle? If you took the inverse of such a matrix, what exactly would that represent?

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Quote:
 Original post by samster581The xyz part is a matrix of a triangle? If you took the inverse of such a matrix, what exactly would that represent?

Yes, it's for a triangle, specifically to find the flat(or planar) projection of a UV textured triangle.

@Pragma 10 years since I did any matrix maths so taking a while to work out the equations, simplify them and test it with real numbers, thanks for the inverse of a matrix btw OMG :-D
Seriously, if this works, your a mega genius, how the hell did you figure out that solution? Guess it's knowing the power of matrices...

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Maybe it's easier just to think about it as a system of linear equations. Every point in the plane of the triangle can be written as a linear combination of the vertices with coefficients a,b,c that sum to one. Suppose this point has UV of (u,v) and coordinates (x,y,z). Then the equations it satisfies are:
u = a u1 + b u2 + c u3v = a v1 + b v2 + c v31 = a + b + cx = a x1 + b x2 + c x3y = a y1 + b y2 + c y3v = a z1 + b z2 + c z3

Now this starts to look like a pair of matrix equations. We can write them as
 [u1 u2 u3] [a]     [a]    [v1 v2 v3]  = U  = [v] [ 1  1  1] [c]     [c]   [1] [x1 x2 x3] [a]     [a]   [x] [y1 y2 y3]  = X  = [y] [z1 z2 z3] [c]     [c]   [z]

This is where the matrices U and X came from. Now we can use the first matrix equation to solve for a, b and c as
 [a]         = U^-1 [v][c]        [1]

and plug this answer in to the second equation to get
[x]     [a]            [y] = X  = X * U^-1 [v][z]     [c]            [1]

[Edited by - Pragma on May 3, 2010 5:37:59 AM]

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Can't get it to work :(
Either i'm doing something fundamentally wrong, screwed up the inverse or it just doesn't work. Using a column vector, that's ok isn't it?

Some real data if you want to try it.
x,y,z -41.3953 -7.9605 -97.3297 uv(0.12500, 0.12500)
x,y,z 34.4961 -45.9345 27.6143 uv(0.81250 0.43750)
x,y,z -20.6976 39.8266 -4.0867 uv(0.31250 0.62500)

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Yes, everything should be a column vector. Maybe check your matrix inversion?

The method works fine for me. Here's some octave code I used to test it:
X = [ -41.3953,  34.4961, -20.6976;       -7.9605, -45.9345,  39.8266;      -97.3297,  27.6143,  -4.0867];U = [ 0.12500, 0.81250, 0.31250,      0.12500, 0.43750, 0.62500,      1.00000, 1.00000, 1.00000];M = X * inverse(U);M * [0.12500; 0.12500; 1]M * [0.81250; 0.43750; 1]M * [0.31250; 0.62500; 1]

And here's the output:
ans =  -41.3953   -7.9605  -97.3297ans =   34.496  -45.934   27.614ans =  -20.6976   39.8266   -4.0867

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Also if you want to compare to your code, the matrix I get is
M =   1.1039e+02   1.0959e-04  -5.5194e+01  -1.1895e+02   1.4018e+02  -1.0614e+01   1.1690e+02   1.4265e+02  -1.2977e+02

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