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Suppose I have pairs of vector $\left\{(v_i,w_i)\right\}_{1 \leq i \leq n}$, and I want to find an angle $\theta$ that describes an optimal rotation that aligns all the pairs.

Now two possible cost functions came to my mind the first one is

$$ C_1(\theta) = \frac{1}{2n} \sum_i (\theta - \theta_i)^2 $$

where $\theta_i$ is the angle between $v_i$ and $w_i$. On the other end however since $\left\langle v_i,w_i \right\rangle = \left\lVert v_i \right\rVert \left\lVert w_i \right\rVert \cos \theta_i$

Clearly the two vectors are aligned if $\cos \theta_i = 1$ therefore I was thinking of minimizing

$$ C_2(\theta) = \frac{1}{2n} \sum_i \left(1 - \cos \left( \theta -\theta_i\right) \right)^2 $$

For both functions I'd assume all the angles, including the unknown are in the interval $[-\pi,\pi]$.

It is clear that if there's a perfect rigid transformations the solution would be the same for both cases, however in general which one is better?

Is there a known similar problem that uses $C_2$ instead of something like $C_1$?

Computationally speaking $C_1$ has a closed form solution

(Assume you have many many pairs $(v_i,w_i)$...)

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$C_1$ has a closed form if you know the angles $\theta_i$ ,your vectors are in $2d$and don't have issues with $2\pi$ type ambiguities. As for C_$2$ which is more standard, try this

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  • $\begingroup$ Interesting, how is my $C_2$ related to the wikipage? I don't see the same formula, is there a specific one you can point out just for sake of comparison? $\endgroup$ – user8469759 Dec 4 '18 at 9:39
  • $\begingroup$ By the way, in theory in my case $\theta_i$'s are known in my problem, I can easily retrieve them. $\endgroup$ – user8469759 Dec 4 '18 at 9:44
  • $\begingroup$ treat the vectors $v_i$ as the columns of $A$ and $u_i$ as the columns of$B$ ( as used in the wiki article). Applying a single rotation matrix $\Omega$ on the $v$'s now becomes $\Omega \{ v_1,\cdots v_n\}=\Omega A$ and the square of all the distances $\sum |\Omega v_i-u_i|^2$ is $|\Omega A-B|^2$ $\endgroup$ – user622715 Dec 4 '18 at 9:55
  • $\begingroup$ Though I understand the formula, I struggle a bit to see the resemblance with mine, am I supposed to be able to derive my $C_2$ from $\sum |\Omega v_i - u_i|^2$? $\endgroup$ – user8469759 Dec 4 '18 at 10:24
  • $\begingroup$ $v_i \cdot u_i= |v||u|\cos\theta_i$ so $\Omega v_i \cdot u_i=|v||u|\cos(\theta_i -\theta_\Omega)$ $\endgroup$ – user622715 Dec 4 '18 at 11:11

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