Finding 3D co-ordinates of UV(1,1)?
Author
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.
Form the matrices
and
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]
[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]
Quote:Original post by Pragma
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).
What are u1,v1, u2,v2, u3,v3?
What are x1,y1,z1, x2,y2,z2, x3,y3,z3?
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.
(u1, v1) = texture coordinate of first vertex
(x1, y1, z1) = position of first vertex
(u2, v2) = texture coordinate of second vertex
... etc.
Quote:Original post by Pragma
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.
The xyz part is a matrix of a triangle? If you took the inverse of such a matrix, what exactly would that represent?
Author
Quote:Original post by samster581
The 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...
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:
Now this starts to look like a pair of matrix equations. We can write them as
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
and plug this answer in to the second equation to get
[Edited by - Pragma on May 3, 2010 5:37:59 AM]
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 z3Now 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]
Author
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)
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)
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:
And here's the output:
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
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+02This topic is closed to new replies.
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