# Defining a cost function to find a rotation

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 $$(v_i,w_i)$$...)

$$C_1$$ 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

• 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? – user8469759 Dec 4 '18 at 9:39
• By the way, in theory in my case $\theta_i$'s are known in my problem, I can easily retrieve them. – user8469759 Dec 4 '18 at 9:44
• 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 $|\Omega A-B|^2$ – user622715 Dec 4 '18 at 9:55
• 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 $\sum |\Omega v_i - u_i|^2$? – user8469759 Dec 4 '18 at 10:24
• $v_i \cdot u_i= |v||u|\cos\theta_i$ so $\Omega v_i \cdot u_i=|v||u|\cos(\theta_i -\theta_\Omega)$ – user622715 Dec 4 '18 at 11:11