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I have an object with a position matrix and a rotation matrix (derived from a quaternion, but I digress). I'm able to translate this object along world-relative vectors, but I'm trying to figure out how to translate it along local-relative vectors. So if the object is tilted 45 degrees around its Z-axis the vector (1, 0, 0) would make it move to the upper right.

For world-space translations I simply turn the movement vector into a matrix and multiply it by the position matrix: position_mat = translation_mat * position_mat. For local-space translations I'd think I'd have to use the rotation matrix into that formula, but I see the object spin around instead when I apply a translation over time no matter where I multiply the rotation matrix.

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Just take the desired movement vector, multiply it by the object's rotation matrix, then use the result as before, i.e. convert it to a translation matrix and multiply it into the position_mat.

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    \$\begingroup\$ As far as I understand you, I want to do position_mat = (translation_mat * rotation_mat) * position_mat. I've tried this already but, as I said before, the translation doesn't go in a straight line. So if my object is tilted to the left and I try to translate it to go towards its local left (upper-left in world space) over a period of time, the object spins around instead. \$\endgroup\$
    – Aaron
    Jul 5, 2012 at 23:08
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    \$\begingroup\$ @Aaron, multiplying the matrices like that is not the same as what I said. You need to apply the rotation matrix to the translation vector, before converting it into a translation matrix. What you posted does something else: it also incorporates the rotation matrix into position_mat, applying the rotation to the whole object instead of just the translation vector. And since you do this every frame, of course the object spins around over time. \$\endgroup\$
    – Nathan Reed
    Jul 5, 2012 at 23:11
  • \$\begingroup\$ Thanks. I didn't realize the difference between the translation vector and the translation matrix. \$\endgroup\$
    – Aaron
    Jul 6, 2012 at 3:27

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