I am currently writing up a program that needs to know the difference in degrees for each axis between two quaternions.

Now, suppose I have a starting quaternion Qs and I need to compute at each step the difference between my current orientation represented by the quaternion Qc. I do that: Q = Qc^-1 * Qs

Now I would like to get values from the quaternion Q that would represent for each axis the number of degrees that differ from Qc to Qs. Is that possible?

Edit: If that's not clear here are other explanations I know I can get the angle value from a quaternion by doing 2 * cos-1(q.w). Now I would like to get an idea on how to project this angle onto each axis. The expected output is something like:

From Qs to Qc there is a rotation of 30 degrees on the x axis, 20 degrees on the y axis and 0 degrees on the z axis

  • $\begingroup$ @Bacon not really. As far as I understood he wanted to do a specific rotation. That's not what I'm after. I want to go from a computer quaternion (which is the difference of two others) to angle values for each axis $\endgroup$ – LBes Aug 31 '16 at 14:08
  • $\begingroup$ Quaternions are still used? I'm in high school (which explains why I don't know about this), but I read that quaternions were developed to act as vectors but were not useful. $\endgroup$ – Ian Limarta Aug 31 '16 at 19:25
  • $\begingroup$ @IanLimarta well yes they are and they are really useful except for this very specific case. But most programming languages dealing with 3D use quaternions $\endgroup$ – LBes Aug 31 '16 at 19:27
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    $\begingroup$ @IanLimarta sounds like you have been reading one of those infamous backwards high school texts with lots of wrong stuff, or perhaps one written over 30 years ago. Quaternions are now more useful than ever, especially in graphics and computer vision. They have always stuck around in physics, although probably not as practically important as in modern graphics computing. There's this online tool called google that will help you find webpages that reflect these claims :) $\endgroup$ – rschwieb Sep 5 '16 at 22:08

It sounds like you are just looking for a way to convert your quaternion $Q$ to Euler angles, and there are numerous webpages on the internet that cover this very well already. Just google "quaternions to Euler angles" and you get things like these:




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  • $\begingroup$ Is it really necessary to extract the Euler angle representation? Personally I never saw the appeal, but apparently they are handy... $\endgroup$ – rschwieb Sep 6 '16 at 17:05
  • $\begingroup$ Did solve my problem with some coding workaround afterwards $\endgroup$ – LBes Sep 6 '16 at 23:00

So if I interpret your question right you already have the axis in ${\bf i,j,k}$ components (if you disregard the factor $\sin(\theta/2)$), so just scrap the real part (pretend it's zero). First let $\bf p_1,p_2$ be the quaternions after they have had a real part stripped away (and the rest renormalized so that square sum = 1).

We then have (from Wikipedia):

$${\bf p_1' = qp_1q^{-1} \Leftrightarrow p_1'q=qp_1 \hspace{1cm}} (1)$$

where $\bf q$ is one of the quaternions

$${\bf q} = \underset{a}{\underbrace{\cos(\alpha/2)}} + \underset{b}{\underbrace{\sin(\alpha/2)}} (b_1{\bf i}+ b_2 {\bf j}+ b_3{\bf k})$$

for the set of binary vector $${\bf b} = (b_1,b_2,b_3)^T \text{ such that } \cases {b_k \in \{0,1\}\\ \sum_{\forall i} b_i = 1}$$ or just simply $${\bf b} \in \{(1,0,0)^T,(0,1,0)^T,(0,0,1)^T\}$$

We now want to find the $\alpha$s for these so that the resulting rotation is as close as possible

$$\min_{\bf q} \left\|\bf qp_1q^{-1}-p_2 \right\| \Leftrightarrow \min_{\bf q} \left\|\bf qp_1-p_2q \right\|\hspace{1cm} (2)$$

is linear in the coefficients. Let us call them $a = \sin(\alpha), b = \cos(\alpha)$. However, we will at some point need to add some functionality to ensure $a^2+b^2 = 1$ (the trig 1, right?) and then it becomes non-linear.

A reasonably good approach would probably be to disregard the constraint, solve the linear (least squares) system then impose the constraint for the solution found by doing a renormalization update $$\cases{a = \frac{a}{a^2+b^2}\\b=\frac{b}{a^2+b^2}}$$ and then if needed iterate from the new position to refine the solution found.

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