0
$\begingroup$

I am not really sure if I understand the phenomenon of gimbal lock correctly.

Say I have a vector $\begin{pmatrix} x\\ y\\ z \end{pmatrix}$.

And I want to keep the vector's length fixed but move it in a given direction with respect to the $x, y$ or $z$ axis - i.e. rotate it in that direction.

So, for instance, if I want to rotate it $30$ degrees about the $z$-axis, I would multiply by the matrix $$\begin{pmatrix} \cos(30°) & -\sin(30°) & 0\\ \sin(30°) & \cos(30°) & 0\\ 0 & 0 & 1\end{pmatrix}_.$$

And likewise for the other two axes. Will some sequence of these rotations eventually cause "gimbal lock?" Or will no problem arise using this method?

$\endgroup$
0
$\begingroup$

Gimbal lock occurs when one of the rotation matrices reduces to the identity. Then you effectively reduce one degree of freedom.

Let $R_x(\alpha)$ denote a rotation matrix around $x$ by $\alpha$.

Then, a general rotation can be written as $R = R_x(\alpha) R_y (\beta) R_z(\gamma)$. Suppose that $R_x(\alpha)$ becomes the identity map. Then $R = R_y(\beta) R_z(\gamma)$ in the new coordinate frame, and hence there is no longer any notion about "rotation around $x$."

$\endgroup$
  • $\begingroup$ Right, but I'm doing rotations in sequence, and only about x, y, or z. I.e. I do not have "general rotations" - my rotations are only $R_x(\alpha), R_y(\beta), R_z(\gamma)$, but no combination of these. My coordinate frame is not changing. $\endgroup$ – user2258552 May 30 '14 at 23:52
  • $\begingroup$ When you apply a rotation matrix, the next rotation applies along the new coordinate axes. $\endgroup$ – Emily May 31 '14 at 0:04
  • $\begingroup$ But my axes remain x, y, and z - I'm not changing the coordinate axes. $\endgroup$ – user2258552 May 31 '14 at 4:53
  • $\begingroup$ You're missing the key. $ R_1$ rotates around your first set of axes. $ R_2$ rotates around the set of axes determined by $ R_1$, whether you like it or not. Play with some examples and see. $\endgroup$ – Emily May 31 '14 at 14:47
  • $\begingroup$ Hmm. Well, it indeed locked, so I guess I was wrong. I should review the math again, because this is too simple for me to misunderstand like this. $\endgroup$ – user2258552 May 31 '14 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.