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I've understand that quaternions do not have handness but rotation matricies derived from unit quaternions does.

The following formula is given by wikipedia for quaternion to rotation matrix conversion :

Given the unit quaternion $q = w + xi + yj + zk$ , the equivalent left-handed (Post-Multiplied) 3×3 rotation matrix is $$ Q = \begin{bmatrix} 1 - 2 y^2 - 2 z^2 & 2 x y - 2 z w & 2 x z + 2 y w \\ 2 x y + 2 z w & 1 - 2 x^2 - 2 z^2 & 2 y z - 2 x w \\ 2 x z - 2 y w & 2 y z + 2 x w & 1 - 2 x^2 - 2 y^2 \end{bmatrix} . $$

As mentioned, this formula is relative to a left-handed coordinate frame. What's the right-handed counterpart ?

Best : ) ,

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  • $\begingroup$ What have you tried? The difference between conventions is just a matter of quite straight forward calculations. $\endgroup$ – skyking Jan 29 '16 at 10:41
  • $\begingroup$ I've a bug in my software and i want to search for possible candidates : i want to be 100% sure that i'm not missing something with those handness conventions. $\endgroup$ – jcolafrancesco Jan 29 '16 at 10:52
  • $\begingroup$ Then it would maybe be a point to post what conventions you're using (fx your coordinate system, the way you apply rotation matrixes and the way you apply quaternions) and what formulas you're using. If nothing else it would show that you've done some own effort. $\endgroup$ – skyking Jan 29 '16 at 11:00
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Hint:

Given the unit quaternion $q = w + x \vec i + y\vec j + z \vec k=w+ \vec v$, the matrix $Q$ results from the representation of a rotation of a vector $\vec p=p_x \vec i+p_y \vec j+ p_z \vec k$ as: $$ Q(\vec p)= q \vec p q^{-1} $$ The resulting rotation is a rotation of angle $\theta$ around an axis oriented by a versor $\vec u$ such that:

$$ \cos \frac{\theta}{2}=\frac{w}{|q|} \qquad \sin \frac{\theta}{2}=\frac{|\vec v|}{|q|} $$ and $$ \vec u=\frac{\vec v}{|\vec v|} $$

This is a counter-clockwise rotation $R_{\vec u, \theta}$ around the axis $\vec u$.

You can find the clockwise rotation around the same axis changing the angle to $-\theta$, or inverting the orientation of the versor $\vec u$ (note that if you perform all the two transformation the rotation remain the same).

What does this means for the quaternion $q$ ?

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  • $\begingroup$ I'm i right if i say that right-handed version is just the transpose of the left-handed ? $\endgroup$ – jcolafrancesco Jan 29 '16 at 13:31
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    $\begingroup$ Change the sigh of $theta$ or the sigh of $\vec u$ is the same as take the conjugate of $q$, i.e. $\bar q=w-xi-yj-zk$ and this gives a matrix $Q'$ that is the transpose of $Q$... so : Yes! you are right. $\endgroup$ – Emilio Novati Jan 29 '16 at 13:39

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