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Consider a rotation around the Z axis, determined by angle $\theta$. If I change the polarity of the X or Y axis, the rotation becomes $- \theta$, and stays $\theta$ if both are inverted.

But what about if I change the polarity of the Z axis itself? Can someone help my fuzzy brain get some clarity on the matter?

Thanks

PaulSiML
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1 Answers1

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If I understand what you're asking correctly, you have a fixed point or line in $3$-$D$ space which you are rotating around the $Z$-axis. If so, then when you switch the polarity of the $Z$-axis, there's no change as such with the angle $\theta$. Instead, the difference is that there is a switch from rotating around the positive $z$ axis to now being around the negative $z$ axis, and vice versa.

However, if the point or line is represented by an equation involving the $X$, $Y$ and $Z$ co-ordinates, then there could be an effect if the $Z$ value changes to $-Z$ such that angle with the $Z$-axis is then different. In that case, then $\theta$ could change, possibly becoming $-\theta$ or even something completely different.

John Omielan
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  • Thanks for the answer! What about if I change the coordinate system from (X,Y,Z) to (X, Y, -Z)? If theta is the rotation in the 1st system, is it still theta in the 2nd? Or -theta? – PaulSiML Feb 18 '20 at 17:54
  • @PaulSiML I've expanded my answer to cover more situations, including I believe what you're asking about. If it doesn't handle what you're asking about, please let me know so I can explain further. – John Omielan Feb 18 '20 at 19:07
  • Thanks again, that is most helpful. Here are more details about my situation: I have rotation angles (around x, y and z) measured by a gyroscope, but I have to make use of them in a different coordinate system, where the polarity of all axes is inverted. In other words, I have Rx, Ry, Rz the rotation angles expressed in the (X,Y,Z) coordinate system, but I have to use them in a (-X, -Y, -Z) coordinate system. I am trying to both find what are the corresponding angles in (-X, -Y, -Z), as well as what happens to vectors. – PaulSiML Feb 18 '20 at 22:04
  • Actually, let me try to have a go at it: is it correct the whole transform is given by: first a homography H_invert_polarity to go from (-X, -Y, -Z) to (X, Y, Z), then the rotation homography H_rot, and finally a homography H_invert_polarity_inverse to go from (X, Y, Z) back to (-X, -Y, -Z). Correct? – PaulSiML Feb 18 '20 at 22:53
  • So then it may be hard to express just in terms of rotation angles, I'd better keep representing the whole rotation as a homography right? – PaulSiML Feb 18 '20 at 23:00
  • @PaulSiML Thank you for explaining your situation further. However, I've never previously learned or used homography (actually, I don't recall ever even hearing of the term before so I just quickly checked its definition in Wikipedia). As such, I'm hesitant to offer any feedback. It seems like your situation might be somewhat more complicated than what you were initially asking about in your question. To help ensure more qualified & knowledgeable people assist you, I suggest you ask a new question specifically about this, possibly referencing this question if you think it's appropriate. – John Omielan Feb 18 '20 at 23:09