My goal is to express the transformation between the black frame F1 and the other one F0 (Green, Red, Violet):

enter image description here

All what I know is the position X,Y Z of four points A, B, C, D wrt F0. My question is how can I know the transformation (3d rotation+ 3d translation) between the two frames?

enter image description here

  • $\begingroup$ Are the coordinates of $A,B,C,D$ $(0,0,0),(1,0,0),(0,1,0),(0,0,1)$ respectively? $\endgroup$ – Parcly Taxel Oct 18 '18 at 14:53
  • $\begingroup$ No, there are expressed with respect to the black frame, if you zoom in the first image you can read approximately their coordinates $\endgroup$ – Ja_cpp Oct 18 '18 at 14:55
  • $\begingroup$ I meant, those coordinates in the black frame. $\endgroup$ – Parcly Taxel Oct 18 '18 at 14:55
  • $\begingroup$ Oh yes that's correct, that's how I defined them wrt F1 $\endgroup$ – Ja_cpp Oct 18 '18 at 14:59

Translation + rotation can be expressed like this


Where $T$ first is transform that takes us to coordinate system where we rotate. and $R$ is rotation matrix, for example

$$R=\left[\begin{array}{ccc}\cos(\alpha)&\sin(\alpha)&0\\-\sin(\alpha)&\cos(\alpha)&0\\0&0&1\end{array}\right]$$ $$T=\left[\begin{array}{ccc}v_1&v_2&v_P\end{array}\right]$$ Where $v_P$ is vector along axis or rotation and $\{v_1,v_2\}$ is a basis for plane of rotation.

So what you have is some equations $Mw_1 = w_2$ where vectors in $w_2$ are coordinates for frame 2 and $w_1$ are same points in first frame. $w$ should be filled like this $w=[c_x,c_y,c_z,1]^T$ coordinate $x,y,z$ and don't forget the 1 at the end of vector.

The sin:s and cos:s make this slightly nonlinear, but there still exist quite okay methods to solve it. I think a few iterations of refining linear solver with some kind of correction should work.

  • $\begingroup$ Sorry I didn't get the meaning of the vector vP? and where should I use the coordinates of A, B, C, D because that's basically what I only know! $\endgroup$ – Ja_cpp Oct 18 '18 at 15:32
  • $\begingroup$ @Ja_cpp $v_P$ is the vector perpendicular to the plane of rotation. Are A,B,C,D four different points you know? Then you put them in $w_1,w_2$. $\endgroup$ – mathreadler Oct 18 '18 at 16:17

Suppose you know that

  • $(0,0,0)$ in the original $F_0$ frame is mapped to $(x_0,y_0,z_0)$ in the transformed $F_1$ frame
  • $(1,0,0)$ in $F_0$ maps to $(x_1,y_1,z_1)+(x_0,y_0,z_0)$ (the first vector can be found from the raw coordinates of $(1,0,0)$'s image by subtraction)
  • $(0,1,0)$ maps to $(x_2,y_2,z_2)+(x_0,y_0,z_0)$, $(0,0,1)$ maps to $(x_3,y_3,z_3)+(x_0,y_0,z_0)$

Then the transformation that takes a point expressed in $F_1$'s coordinates and returns it in $F_0$'s, or transforms $F_0$ to $F_1$, is $$\begin{bmatrix}x_1&x_2&x_2&x_0\\ y_1&y_2&y_3&y_0\\ z_1&z_2&z_3&z_0\\ 0&0&0&1\end{bmatrix}$$ using the homogenous coordinates trick of adding a fourth coordinate of 1 to each point to be transformed. To get the reverse transformation, simply invert this matrix.

  • $\begingroup$ So what you're saying is I just have to put the coordinates of the four points expressed in F0 frame like so [A B C D;0 0 0 1] and I got my transformation? $\endgroup$ – Ja_cpp Oct 18 '18 at 15:28
  • $\begingroup$ @Ja_cpp More like $[B-A,C-A,D-A,A;0,0,0,1]$ (assuming column vectors, that the mapping is as in the first comment I made on the question itself, and that the coordinates in this comment are those in the coloured $F_0$ frame). This is a generalisation of a neat trick I found for two-dimensional transformations. $\endgroup$ – Parcly Taxel Oct 18 '18 at 15:31
  • $\begingroup$ Okay got it and what kind of transformation is it? because I want to extract the rotation vector from it as I stated in the title of the post $\endgroup$ – Ja_cpp Oct 18 '18 at 15:56
  • $\begingroup$ Basically the translation will be the position of the origin of F1 in F0 (which is the point A) right? $\endgroup$ – Ja_cpp Oct 18 '18 at 16:06
  • $\begingroup$ @Ja_cpp Yes, and the rotation matrix (which is applied before the translation) is the upper-left 3×3 block of the 4×4 matrix I gave you. This may be converted to axis/angle using well-known formulas; see here (section Rotation matrix $\leftrightarrow$ Euler axis/angle). $\endgroup$ – Parcly Taxel Oct 18 '18 at 16:09

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