Difference in rotation matrix
$begingroup$
I have two objects A and B, with a start transformation matrices MA1 and MB1 (they include translation, rotation and scale). End matrix of the first object is MA2. How do I apply the same rotation as a rotation of object A from MA1 to MA2 ( difference) to object B and get MB2?
matrices transformation rotations
asked Dec 19 '18 at 11:31
pazduhapazduha
61
$endgroup$
add a comment |
$begingroup$
I have two objects A and B, with a start transformation matrices MA1 and MB1 (they include translation, rotation and scale). End matrix of the first object is MA2. How do I apply the same rotation as a rotation of object A from MA1 to MA2 ( difference) to object B and get MB2?
matrices transformation rotations
asked Dec 19 '18 at 11:31
pazduhapazduha
61
$endgroup$
add a comment |
$begingroup$
I have two objects A and B, with a start transformation matrices MA1 and MB1 (they include translation, rotation and scale). End matrix of the first object is MA2. How do I apply the same rotation as a rotation of object A from MA1 to MA2 ( difference) to object B and get MB2?
matrices transformation rotations
asked Dec 19 '18 at 11:31
pazduhapazduha
61
$endgroup$
I have two objects A and B, with a start transformation matrices MA1 and MB1 (they include translation, rotation and scale). End matrix of the first object is MA2. How do I apply the same rotation as a rotation of object A from MA1 to MA2 ( difference) to object B and get MB2?
matrices transformation rotations
matrices transformation rotations
asked Dec 19 '18 at 11:31
pazduhapazduha
61
asked Dec 19 '18 at 11:31
pazduhapazduha
61
asked Dec 19 '18 at 11:31
pazduhapazduha
61
asked Dec 19 '18 at 11:31
pazduhapazduha
61
asked Dec 19 '18 at 11:31
pazduhapazduha
61
61
add a comment |
add a comment |
1 Answer
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oldest
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First you extract rotation matrix from transformation matrix. You take top-left 3×3 submatrix $T$ and use polar decomposition to find rotation matrix $U$. If you know that scale is the same on all axes, then you can just divide $T$ by cubic root of determinant: $U=T/sqrt[3]{det T}$.
Finally, $U_{1to2} = U_{A2} U_{A1}^{-1}$
answered Dec 19 '18 at 11:44
Vasily MitchVasily Mitch
2,6791312
$endgroup$
$begingroup$
I have the problem in the final step. I useB1 *(U1→2)^-1for objectBand this looks like it applies the rotationU1→2to objectBaround the object'sBcoordinate system. It works as needed if MA1 is identity.
$endgroup$
– pazduha
Dec 20 '18 at 13:35
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First you extract rotation matrix from transformation matrix. You take top-left 3×3 submatrix $T$ and use polar decomposition to find rotation matrix $U$. If you know that scale is the same on all axes, then you can just divide $T$ by cubic root of determinant: $U=T/sqrt[3]{det T}$.
Finally, $U_{1to2} = U_{A2} U_{A1}^{-1}$
answered Dec 19 '18 at 11:44
Vasily MitchVasily Mitch
2,6791312
$endgroup$
$begingroup$
I have the problem in the final step. I useB1 *(U1→2)^-1for objectBand this looks like it applies the rotationU1→2to objectBaround the object'sBcoordinate system. It works as needed if MA1 is identity.
$endgroup$
– pazduha
Dec 20 '18 at 13:35
add a comment |
$begingroup$
First you extract rotation matrix from transformation matrix. You take top-left 3×3 submatrix $T$ and use polar decomposition to find rotation matrix $U$. If you know that scale is the same on all axes, then you can just divide $T$ by cubic root of determinant: $U=T/sqrt[3]{det T}$.
Finally, $U_{1to2} = U_{A2} U_{A1}^{-1}$
answered Dec 19 '18 at 11:44
Vasily MitchVasily Mitch
2,6791312
$endgroup$
$begingroup$
I have the problem in the final step. I useB1 *(U1→2)^-1for objectBand this looks like it applies the rotationU1→2to objectBaround the object'sBcoordinate system. It works as needed if MA1 is identity.
$endgroup$
– pazduha
Dec 20 '18 at 13:35
add a comment |
$begingroup$
First you extract rotation matrix from transformation matrix. You take top-left 3×3 submatrix $T$ and use polar decomposition to find rotation matrix $U$. If you know that scale is the same on all axes, then you can just divide $T$ by cubic root of determinant: $U=T/sqrt[3]{det T}$.
Finally, $U_{1to2} = U_{A2} U_{A1}^{-1}$
answered Dec 19 '18 at 11:44
Vasily MitchVasily Mitch
2,6791312
$endgroup$
First you extract rotation matrix from transformation matrix. You take top-left 3×3 submatrix $T$ and use polar decomposition to find rotation matrix $U$. If you know that scale is the same on all axes, then you can just divide $T$ by cubic root of determinant: $U=T/sqrt[3]{det T}$.
Finally, $U_{1to2} = U_{A2} U_{A1}^{-1}$
answered Dec 19 '18 at 11:44
Vasily MitchVasily Mitch
2,6791312
answered Dec 19 '18 at 11:44
Vasily MitchVasily Mitch
2,6791312
answered Dec 19 '18 at 11:44
Vasily MitchVasily Mitch
2,6791312
answered Dec 19 '18 at 11:44
Vasily MitchVasily Mitch
2,6791312
2,6791312
$begingroup$
I have the problem in the final step. I useB1 *(U1→2)^-1for objectBand this looks like it applies the rotationU1→2to objectBaround the object'sBcoordinate system. It works as needed if MA1 is identity.
$endgroup$
– pazduha
Dec 20 '18 at 13:35
add a comment |
$begingroup$
I have the problem in the final step. I useB1 *(U1→2)^-1for objectBand this looks like it applies the rotationU1→2to objectBaround the object'sBcoordinate system. It works as needed if MA1 is identity.
$endgroup$
– pazduha
Dec 20 '18 at 13:35
$begingroup$
I have the problem in the final step. I use
B1 *(U1→2)^-1 for object B and this looks like it applies the rotation U1→2 to object B around the object's B coordinate system. It works as needed if MA1 is identity.$endgroup$
– pazduha
Dec 20 '18 at 13:35
$begingroup$
I have the problem in the final step. I use
B1 *(U1→2)^-1 for object B and this looks like it applies the rotation U1→2 to object B around the object's B coordinate system. It works as needed if MA1 is identity.$endgroup$
– pazduha
Dec 20 '18 at 13:35
add a comment |
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