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I found in this article here that the distance between two quaternions can be calculated in many ways, one of them is using "Modified" Euclidean distance (See the link under "Norm of the difference of quaternions", or you can see below equation)

$$d(1,2)=2\times\min(||q_1-q_2||,||q_1+q_2||)$$

where $q_1$ and $q_2$ are normalized quaternion.

The question is, in that article it mentioned that the range values is $(0,2{\sqrt 2})$. Thus the maximum value is $2{\sqrt 2}$. But if we calculate the distance manually, assuming that $q_1$ is $(0,0,0,0)$ and $q_2$ is either $(1,1,1,1)$ or $(-1,-1,-1,-1)$, then the distance will be $4$.

Can someone explain why the maximum value is $2{\sqrt 2}$ instead of $4$? I guess I'm not quite understand quaternion, that's why I'm confuse. Please lead me to some links or explain it to me, thank you

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The values you're substituting aren't normalised. Without loss of generality, let $q_2=(1,0,0,0)$. Then $d(q_1,q_2)$ is maximised if $q_1$ is at maximal distance from both $(1,0,0,0)$ and $(-1,0,0,0)$, e.g. for $q_1=(0,1,0,0)$. Then $\|q_1-q_2\|=\|q_1+q_2\|=\sqrt2$, and the result follows.

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  • $\begingroup$ I'm sorry, I realize that my normalize method is wrong, thank you for pointing it out for me $\endgroup$ – Richard Aug 10 '16 at 8:28

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