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map points between two triangles in 3D space

Answer 1

A:

Simply add the vectors to each point. Point + Vector == new Point. This is basically the opposite of creating the Vector in the first place: V1 == (x1'-x1, y1'-y1, z1'-z1), so (x1', y1', z1') == (x1+V1x, y1+v1y, z1+V1z).

KeithS 2010-09-23 16:35:46

I don't think this answers the question. I think the OP means that (x, y, z) is an arbitrary point in the triangle.

Colin Fine 2010-09-23 16:52:29

Yep i feel too that this is not solution, because point should be affected by ALL three vectors V1,V2,V3 (because all vertices changes), not by just one V1.

0x69 2010-09-24 06:17:10

Answer 2

A:

Let

 x = αx1 + βx2 + γx3

Then x' = αx1' + βx2' + γx3' So

x' = α(x1+V1) + β(x2+V2) + γ(x3+V3)

Colin Fine 2010-09-23 16:36:59

This is a 3-D linear transformation. The problem requires a 3-D affine transformation.

andand 2010-09-24 01:58:26

I don't exactly understood. Did you mean: x' = α(x1+V1.x) + β(x2+V2.x) + γ(x3+V3.x) ?? Also How to calculate α,β,γ ?

0x69 2010-09-24 07:16:15

Answer 3

+1 A:

andand 2010-09-23 18:02:20

Thanks for deep analysis of the problem.

0x69 2010-09-26 19:27:33

Answer 4

+1 A:

You want a matrix transformation T such that T X = X', where X is the matrix whose columns are the co-ordinates of the vertexes of the first triangle and X' is the same for the second triangle. Multiplying each side by the inverse of X yields T = X' X^-1.

Steve Emmerson 2010-09-23 19:25:15

That's what I was thinking at first, but as I looked at it, it became clear that the problem as presented is underconstrained. I believe a unique solution is possible once the distance constraints are included, but finding that unique solution requires convex programming and can't be solved using the linear method you outlined.

andand 2010-09-23 21:31:56

@andand The matrix transformation takes all three vertexes to their counterparts. It is also a linear transformation; therefore, any point in the interior of the first triangle will be transformed into the interior of the second in a way that maintains proportional distance from the vertexes.

Steve Emmerson 2010-09-24 03:16:13

@Steve Emmerson: Suppose the origin is one of the points interior to the first triangle, but it is not a point interior to the second. This can't happen using linear transformations; the origin is always mapped to the origin. The problem as stated requires an affine (vice linear) transformation, so that the origin in the domain can be mapped to something other than the origin in the range.

andand 2010-09-24 03:28:51

@andand You're right. The X and X' in my answer should be constituted from the vectors from each triangle's center of mass (COM) to it's vertexes. Transforming a point in triangle 1 then comprises 1) subtracting the COM of the first triangle; 2) applying T; and 3) adding the displacement vector from triangle 1 COM to triangle 2 COM.

Steve Emmerson 2010-09-24 19:47:22

Answer 5

+2 A:

I think you might be looking for barycentric coordinates?

tfinniga 2010-09-23 19:39:22

+1 yes. Find the barycentric coordinates of a point on the source triangle and then locate those barycentric coordinates on the destination triangle.

phkahler 2010-09-24 16:20:25

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map points between two triangles in 3D space

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