Relative rotation between quaternions
$begingroup$
Say I have a quaternion q which describes how to get from frame 0 to frame 1, and a quaternion r which describes how to get from frame 0 to frame 2. To get the "quaternion difference" between q and r, I do
$$
q_{d} = q^{-1} r
$$
This is however in frame 1. How do I get the quaternion difference in frame 0?
quaternions
asked Dec 22 '13 at 21:31
user113514user113514
13
$endgroup$
add a comment |
$begingroup$
Say I have a quaternion q which describes how to get from frame 0 to frame 1, and a quaternion r which describes how to get from frame 0 to frame 2. To get the "quaternion difference" between q and r, I do
$$
q_{d} = q^{-1} r
$$
This is however in frame 1. How do I get the quaternion difference in frame 0?
quaternions
asked Dec 22 '13 at 21:31
user113514user113514
13
$endgroup$
$begingroup$
What does 'frame' mean here in this context?
$endgroup$
– Berci
Dec 22 '13 at 21:57
$begingroup$
say we have a world frame of reference, in this frame q represents a rotation around x of 90 degrees. So the quaternion rotation is frame specific. I guess this is what I meant to say.
$endgroup$
– user113514
Dec 22 '13 at 22:22
$begingroup$
How do you mean '$q^{-1}r$ is in frame $1$ and not in frame 0'? Aren't all quaternions present in each frame? (I still don't clearly get what frames really want to be..)
$endgroup$
– Berci
Dec 22 '13 at 22:52
$begingroup$
Yes, but the same rotation looks different in different frames. So the quaternions corresponding to a rotation about world x would look like (w,x,y,z) [cos(t), sin(t), 0, 0]. But in an arbitrary frame the y and z components need not be 0.
$endgroup$
– user113514
Dec 23 '13 at 2:39
$begingroup$
By frame, I meant coordinate frame [a set of three orthogonal axes]. I guess my question is given q and r defined in a coordinate frame, what coordinate frame is $q^{-1} r$ defined in? Is that clear?
$endgroup$
– user113514
Dec 25 '13 at 2:57
add a comment |
$begingroup$
Say I have a quaternion q which describes how to get from frame 0 to frame 1, and a quaternion r which describes how to get from frame 0 to frame 2. To get the "quaternion difference" between q and r, I do
$$
q_{d} = q^{-1} r
$$
This is however in frame 1. How do I get the quaternion difference in frame 0?
quaternions
asked Dec 22 '13 at 21:31
user113514user113514
13
$endgroup$
Say I have a quaternion q which describes how to get from frame 0 to frame 1, and a quaternion r which describes how to get from frame 0 to frame 2. To get the "quaternion difference" between q and r, I do
$$
q_{d} = q^{-1} r
$$
This is however in frame 1. How do I get the quaternion difference in frame 0?
quaternions
quaternions
asked Dec 22 '13 at 21:31
user113514user113514
13
asked Dec 22 '13 at 21:31
user113514user113514
13
asked Dec 22 '13 at 21:31
user113514user113514
13
asked Dec 22 '13 at 21:31
user113514user113514
13
asked Dec 22 '13 at 21:31
user113514user113514
13
13
$begingroup$
What does 'frame' mean here in this context?
$endgroup$
– Berci
Dec 22 '13 at 21:57
$begingroup$
say we have a world frame of reference, in this frame q represents a rotation around x of 90 degrees. So the quaternion rotation is frame specific. I guess this is what I meant to say.
$endgroup$
– user113514
Dec 22 '13 at 22:22
$begingroup$
How do you mean '$q^{-1}r$ is in frame $1$ and not in frame 0'? Aren't all quaternions present in each frame? (I still don't clearly get what frames really want to be..)
$endgroup$
– Berci
Dec 22 '13 at 22:52
$begingroup$
Yes, but the same rotation looks different in different frames. So the quaternions corresponding to a rotation about world x would look like (w,x,y,z) [cos(t), sin(t), 0, 0]. But in an arbitrary frame the y and z components need not be 0.
$endgroup$
– user113514
Dec 23 '13 at 2:39
$begingroup$
By frame, I meant coordinate frame [a set of three orthogonal axes]. I guess my question is given q and r defined in a coordinate frame, what coordinate frame is $q^{-1} r$ defined in? Is that clear?
$endgroup$
– user113514
Dec 25 '13 at 2:57
add a comment |
$begingroup$
What does 'frame' mean here in this context?
$endgroup$
– Berci
Dec 22 '13 at 21:57
$begingroup$
say we have a world frame of reference, in this frame q represents a rotation around x of 90 degrees. So the quaternion rotation is frame specific. I guess this is what I meant to say.
$endgroup$
– user113514
Dec 22 '13 at 22:22
$begingroup$
How do you mean '$q^{-1}r$ is in frame $1$ and not in frame 0'? Aren't all quaternions present in each frame? (I still don't clearly get what frames really want to be..)
$endgroup$
– Berci
Dec 22 '13 at 22:52
$begingroup$
Yes, but the same rotation looks different in different frames. So the quaternions corresponding to a rotation about world x would look like (w,x,y,z) [cos(t), sin(t), 0, 0]. But in an arbitrary frame the y and z components need not be 0.
$endgroup$
– user113514
Dec 23 '13 at 2:39
$begingroup$
By frame, I meant coordinate frame [a set of three orthogonal axes]. I guess my question is given q and r defined in a coordinate frame, what coordinate frame is $q^{-1} r$ defined in? Is that clear?
$endgroup$
– user113514
Dec 25 '13 at 2:57
$begingroup$
What does 'frame' mean here in this context?
$endgroup$
– Berci
Dec 22 '13 at 21:57
$begingroup$
What does 'frame' mean here in this context?
$endgroup$
– Berci
Dec 22 '13 at 21:57
$begingroup$
say we have a world frame of reference, in this frame q represents a rotation around x of 90 degrees. So the quaternion rotation is frame specific. I guess this is what I meant to say.
$endgroup$
– user113514
Dec 22 '13 at 22:22
$begingroup$
say we have a world frame of reference, in this frame q represents a rotation around x of 90 degrees. So the quaternion rotation is frame specific. I guess this is what I meant to say.
$endgroup$
– user113514
Dec 22 '13 at 22:22
$begingroup$
How do you mean '$q^{-1}r$ is in frame $1$ and not in frame 0'? Aren't all quaternions present in each frame? (I still don't clearly get what frames really want to be..)
$endgroup$
– Berci
Dec 22 '13 at 22:52
$begingroup$
How do you mean '$q^{-1}r$ is in frame $1$ and not in frame 0'? Aren't all quaternions present in each frame? (I still don't clearly get what frames really want to be..)
$endgroup$
– Berci
Dec 22 '13 at 22:52
$begingroup$
Yes, but the same rotation looks different in different frames. So the quaternions corresponding to a rotation about world x would look like (w,x,y,z) [cos(t), sin(t), 0, 0]. But in an arbitrary frame the y and z components need not be 0.
$endgroup$
– user113514
Dec 23 '13 at 2:39
$begingroup$
Yes, but the same rotation looks different in different frames. So the quaternions corresponding to a rotation about world x would look like (w,x,y,z) [cos(t), sin(t), 0, 0]. But in an arbitrary frame the y and z components need not be 0.
$endgroup$
– user113514
Dec 23 '13 at 2:39
$begingroup$
By frame, I meant coordinate frame [a set of three orthogonal axes]. I guess my question is given q and r defined in a coordinate frame, what coordinate frame is $q^{-1} r$ defined in? Is that clear?
$endgroup$
– user113514
Dec 25 '13 at 2:57
$begingroup$
By frame, I meant coordinate frame [a set of three orthogonal axes]. I guess my question is given q and r defined in a coordinate frame, what coordinate frame is $q^{-1} r$ defined in? Is that clear?
$endgroup$
– user113514
Dec 25 '13 at 2:57
add a comment |
1 Answer
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$begingroup$
You are confusing quaternions with vectors in physical space. As the comments say, there is no such thing as having a quaternion "in" a certain frame anymore than there is such a thing as having a rotation matrix in a certain frame--both of these transform between frames and their components depend only on the relative rotation of the starting and ending frames. If $q:0rightarrow 1$ and $r:0rightarrow 2$ and my guess is you are looking for some third quaternion $p:1rightarrow 2$ then since $r = pq$ you have $p = q^{-1}r = q^{*}r$ (since all rotations are unit quaternions). Note that for given $q$ and $r$ this is totally unambiguous since $q^{-1}r neq rq^{-1}$ or any other weird permutation of $q$, $r$ and the inverse map: $q^{-1}r$ and only $q^{-1}r$ delivers the quaternion you are looking for.
Finally, to clarify some terminology, this is not the quaternion difference, which is perfectly well-defined using the Hamilton algebra (though useless for rotational analysis), but rather the relative rotation between frames 1 and 2. The same idea would hold were you to represent the rotations as matrices.
answered Jun 4 '16 at 1:27
SZNSZN
2,708720
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
You are confusing quaternions with vectors in physical space. As the comments say, there is no such thing as having a quaternion "in" a certain frame anymore than there is such a thing as having a rotation matrix in a certain frame--both of these transform between frames and their components depend only on the relative rotation of the starting and ending frames. If $q:0rightarrow 1$ and $r:0rightarrow 2$ and my guess is you are looking for some third quaternion $p:1rightarrow 2$ then since $r = pq$ you have $p = q^{-1}r = q^{*}r$ (since all rotations are unit quaternions). Note that for given $q$ and $r$ this is totally unambiguous since $q^{-1}r neq rq^{-1}$ or any other weird permutation of $q$, $r$ and the inverse map: $q^{-1}r$ and only $q^{-1}r$ delivers the quaternion you are looking for.
Finally, to clarify some terminology, this is not the quaternion difference, which is perfectly well-defined using the Hamilton algebra (though useless for rotational analysis), but rather the relative rotation between frames 1 and 2. The same idea would hold were you to represent the rotations as matrices.
answered Jun 4 '16 at 1:27
SZNSZN
2,708720
$endgroup$
add a comment |
$begingroup$
You are confusing quaternions with vectors in physical space. As the comments say, there is no such thing as having a quaternion "in" a certain frame anymore than there is such a thing as having a rotation matrix in a certain frame--both of these transform between frames and their components depend only on the relative rotation of the starting and ending frames. If $q:0rightarrow 1$ and $r:0rightarrow 2$ and my guess is you are looking for some third quaternion $p:1rightarrow 2$ then since $r = pq$ you have $p = q^{-1}r = q^{*}r$ (since all rotations are unit quaternions). Note that for given $q$ and $r$ this is totally unambiguous since $q^{-1}r neq rq^{-1}$ or any other weird permutation of $q$, $r$ and the inverse map: $q^{-1}r$ and only $q^{-1}r$ delivers the quaternion you are looking for.
Finally, to clarify some terminology, this is not the quaternion difference, which is perfectly well-defined using the Hamilton algebra (though useless for rotational analysis), but rather the relative rotation between frames 1 and 2. The same idea would hold were you to represent the rotations as matrices.
answered Jun 4 '16 at 1:27
SZNSZN
2,708720
$endgroup$
add a comment |
$begingroup$
You are confusing quaternions with vectors in physical space. As the comments say, there is no such thing as having a quaternion "in" a certain frame anymore than there is such a thing as having a rotation matrix in a certain frame--both of these transform between frames and their components depend only on the relative rotation of the starting and ending frames. If $q:0rightarrow 1$ and $r:0rightarrow 2$ and my guess is you are looking for some third quaternion $p:1rightarrow 2$ then since $r = pq$ you have $p = q^{-1}r = q^{*}r$ (since all rotations are unit quaternions). Note that for given $q$ and $r$ this is totally unambiguous since $q^{-1}r neq rq^{-1}$ or any other weird permutation of $q$, $r$ and the inverse map: $q^{-1}r$ and only $q^{-1}r$ delivers the quaternion you are looking for.
Finally, to clarify some terminology, this is not the quaternion difference, which is perfectly well-defined using the Hamilton algebra (though useless for rotational analysis), but rather the relative rotation between frames 1 and 2. The same idea would hold were you to represent the rotations as matrices.
answered Jun 4 '16 at 1:27
SZNSZN
2,708720
$endgroup$
You are confusing quaternions with vectors in physical space. As the comments say, there is no such thing as having a quaternion "in" a certain frame anymore than there is such a thing as having a rotation matrix in a certain frame--both of these transform between frames and their components depend only on the relative rotation of the starting and ending frames. If $q:0rightarrow 1$ and $r:0rightarrow 2$ and my guess is you are looking for some third quaternion $p:1rightarrow 2$ then since $r = pq$ you have $p = q^{-1}r = q^{*}r$ (since all rotations are unit quaternions). Note that for given $q$ and $r$ this is totally unambiguous since $q^{-1}r neq rq^{-1}$ or any other weird permutation of $q$, $r$ and the inverse map: $q^{-1}r$ and only $q^{-1}r$ delivers the quaternion you are looking for.
Finally, to clarify some terminology, this is not the quaternion difference, which is perfectly well-defined using the Hamilton algebra (though useless for rotational analysis), but rather the relative rotation between frames 1 and 2. The same idea would hold were you to represent the rotations as matrices.
answered Jun 4 '16 at 1:27
SZNSZN
2,708720
answered Jun 4 '16 at 1:27
SZNSZN
2,708720
answered Jun 4 '16 at 1:27
SZNSZN
2,708720
answered Jun 4 '16 at 1:27
SZNSZN
2,708720
2,708720
add a comment |
add a comment |
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$begingroup$
What does 'frame' mean here in this context?
$endgroup$
– Berci
Dec 22 '13 at 21:57
$begingroup$
say we have a world frame of reference, in this frame q represents a rotation around x of 90 degrees. So the quaternion rotation is frame specific. I guess this is what I meant to say.
$endgroup$
– user113514
Dec 22 '13 at 22:22
$begingroup$
How do you mean '$q^{-1}r$ is in frame $1$ and not in frame 0'? Aren't all quaternions present in each frame? (I still don't clearly get what frames really want to be..)
$endgroup$
– Berci
Dec 22 '13 at 22:52
$begingroup$
Yes, but the same rotation looks different in different frames. So the quaternions corresponding to a rotation about world x would look like (w,x,y,z) [cos(t), sin(t), 0, 0]. But in an arbitrary frame the y and z components need not be 0.
$endgroup$
– user113514
Dec 23 '13 at 2:39
$begingroup$
By frame, I meant coordinate frame [a set of three orthogonal axes]. I guess my question is given q and r defined in a coordinate frame, what coordinate frame is $q^{-1} r$ defined in? Is that clear?
$endgroup$
– user113514
Dec 25 '13 at 2:57