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I've got a set of $N$ points $p_1,\dots,p_N$ that all belong to a real object. Consequently, there are $N-1$ vectors $\vec{v}_i$ when $\vec{v}_i$ points from $p_1$ to $p_i$.

Now, the object is rotated in some unknown way. $p_1$ stays in the same place (= no translation, just rotation), but all other points are now at their new location $p'_i$ - which means that the vectors also changed to $\vec{v}'_i$ (same length, but different directions).

I know all $p, p', \vec{v}$ and $\vec{v}'$ - using these values, how can I express the rotation via a rotation matrix?


I've tried to use cross-product to get the rotation axis and the scalar-product to get the rotation angle for a single vector, which enables me to compute a rotation matrix - but if I use different vectors I get different results!?

This is the way I do this:

$$\vec{a} = \frac{ \vec{v_2}\times\vec{v}_2' }{ |\vec{v_2}\times\vec{v}_2'| }$$ $$c = \frac{ \vec{v_2} * \vec{v}_2' }{ |\vec{v_2}| \cdot |\vec{v}_2'| }$$ $$s = sin(cos^{-1}(c))$$ $$t = 1 - c$$

With these values, the rotation matrix is (according to this website):

$$ R = \left( \begin{matrix} t*x*x + c & t*x*y - z*s & t*x*z + y*s\\ t*x*y + z*s & t*y*y + c & t*y*z - x*s\\ t*x*z - y*s & t*y*z + x*s & t*z*z + c \end{matrix} \right) $$

(with $\vec{a} = (x,y,z)^T$)

Thank you for any thoughts on this!

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By definition if you have rigid body motion then the axis and angle and common for all of them. Please show your calculation steps in case you missed something. –  ja72 Aug 29 '12 at 16:02
    
@ja72 That's what I thought, too. I've added my computations to the question. –  Niko Aug 29 '12 at 16:13
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1 Answer 1

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I am guessing this is in 3D. Then if you are given at least $3$ vectors that are linearly independent, say $\vec v_2, \vec v_3$ and $\vec v_4$, then your rotation matrix $R$ satisfies

$$ R\begin{pmatrix} \vec v_2 & \vec v_3 & \vec v_4 \end{pmatrix} = \begin{pmatrix} \vec v'_2 & \vec v'_3 & \vec v'_4 \end{pmatrix}. $$

You can then invert the matrix $\begin{pmatrix} \vec v_2 & \vec v_3 & \vec v_4 \end{pmatrix}$ to get $R = \begin{pmatrix} \vec v'_2 & \vec v'_3 & \vec v'_4 \end{pmatrix}\begin{pmatrix} \vec v_2 & \vec v_3 & \vec v_4 \end{pmatrix}^{-1}$.

If your transformation is consistent, any triple of $\vec v_i$ should give you the same $R$.

If you have fewer than $3$ linearly independent vectors, then your rotation matrix is not unique.

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